[Grundlehren der mathematischen Wissenschaften] Finite Groups II Volume 242 ||

Grundlehren der mathematischen Wissenschaften 242

A Series of Comprehensive Studies in Mathematics

Editors

M. Artin S. S. Chern J. L. Doob A. Grothendieck E. Heinz F. Hirzebruch L. Hormander s. Mac Lane W. Magnus C. C. Moore J. K. Moser M. Nagata W. Schmidt D. s. Scott J. Tits B. L. van der Waerden

Managing Editors

B. Eckmann s. R. s. Varadhan

B.Huppert N.Blackbum

Finite Groups II

Springer-Verlag Berlin Heidelberg New York 1982

Bertram Huppert

Mathematisches Institut der Universitat SaarstraBe 21 D-6500 Mainz

N orman Blackburn

Department of Mathematics The University GB-Manchester M13 9 PL

Library of Congress Cataloging in Publication Data. Huppert, Bertram, 1927-. Finite groups II. (Grundlehren der mathematischen Wissenschaften; 242). Bibliography: p. Includes index. I. Finite groups. I. Blackburn, N. (Norman). II. Title. III. Series. QA 17l.B 577. 512'.22. 81-2287.

ISBN-13: 978-3-642-67996-4 DOl: 10.1007/978-3-642-67994-0

e-ISBN-13: 978-3-642-67994-0

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© Springer-Verlag Berlin Heidelberg 1982

Softcover reprint of the hardcover 1 st edition 1982

2141/3140-543210

In Memoriam

Reinhold Baer (1902-1979)

Richard Brauer (1901-1977)

Preface

17):~t? L It CIFDr- ! wei! unsre Weisheit Einfalt ist,

From "Lohengrin", Richard Wagner

At the time of the appearance of the first volume of this work in 1967, the tempestuous development of finite group theory had already made it virtually impossible to give a complete presentation of the subject in one treatise. The present volume and its successor have therefore the more modest aim of giving descriptions of the recent development of certain important parts of the subject, and even in these parts no attempt at completeness has been made.

Chapter VII deals with the representation theory of finite groups in arbitrary fields with particular attention to those of non-zero characteristic. That part of modular representation theory which is essentially the block theory of complex characters has not been included, as there are already monographs on this subject and others will shortly appear. Instead, we have restricted ourselves to such results as can be obtained by purely module-theoretical means.

In Chapter VIII, the linear (and bilinear) methods which have proved useful in questions involving nilpotent groups are discussed. A major part of this is devoted to the classification of Suzuki 2-groups (see §7); while a complete classification is not obtained, the result proved is strong enough for an application to the determination of the Zassenhaus groups in Chapter XI. The standard procedure involves the use of Lie rings, and rather than attempting a theory of the connection between nilpotent groups and Lie rings, we give a number of applications to such topics as the length of the conjugacy classes of p-groups (§9), fixed point free automorphisms of nilpotent groups (§ 1 0), the restricted Burnside problem (§12) and automorphisms of p-groups (§13). In many of these considerations, the finiteness of the group is a relatively unimportant condition, and the last two of these applications depend on the Magnus-Witt theory of the lower central series of free groups, which is described in § 11.

The ground-breaking investigations of P. Hall and G. Higman on the theory of p-soluble groups form the basis of Chapter IX. These arose from the restricted Burnside problem and led first to a solution for

VIII Preface

exponent 6 (see 1.15). Then however there followed far-reaching theorems for composite exponents (4.10, 4.13, 4.17). Besides various estimates of the p-length of a p-soluble group in terms of the structure of its Sylow p-subgroups (§5), we deal with some theorems about fixed point free automorphisms of soluble groups (§6). Finally we discuss the derived notion of p-stability, which will be of considerable use in Chapter X.

The three chapters in this volume are thus all concerned with relations between finite groups and linear algebra, but otherwise they are rather independent of one another, apart from occasional technical references, of course.

The authors must apologize for the length of time which readers have had to wait for this volume. They promise that Volume III will be available within a matter of months.

It is a great pleasure to thank the many colleagues who have helped us in the preparation of this volume and its successor. In this respect the second author must give pride of place to Philip Hall, who first stimulated his interest in the subject more than 25 years ago by combining patient encouragement of a naturally pessimistic student with lectures of a beauty which seems to be lost to subsequent generations. With the writing of the book the greatest help was given by W. Gaschiitz and his associates in Kiel, where each year since 1967 our sketches were read and exhaustively studied. The participants in these discussions in the course of the years were H. Bender, D. Blessenohl, W. Gaschiitz, F. Gross, K. Johnsen, O.-V. Kramer, H. Laue, K.-V. Schaller and R. Schmidt. We are most grateful for the hospitality of the Mathematics Department in Kiel, without which this kind of work would not have been possible. Also we are indebted for financial assistance, enabling the two of us to meet reasonably often, to the National Science Foundation, the Alexander von Humboldt-Stiftung and the University of Manchester.

In the laborious proof-reading B. Hartley (Manchester), O. Manz, 1. Pense and W. Willems (Mainz) all spent a great deal of time helping us, and we offer them our most sincere thanks. Also we thank the Manchester secretaries Kendal Anderson, Rosemary Horton and Patricia McMunn for the enormous amount of help they have given us with the typing and preparation of the manuscript.

Finally our thanks are due to Springer-Verlag and to the typesetters and printers for their patience with us and for the excellent quality of the production of this book.

July, 1981 Bertram Huppert, Mainz Norman Blackburn, Manchester

Contents

Chapter VII. Elements of General Representation Theory . ....... .

§ 1. Extension of the Ground-Field. . . . . . . . . . . . . . . . . . . . . . . . . . 4 § 2. Splitting Fields ....................................... 27 § 3. The Number of Irreducible Modular Representations ...... 32 § 4. Induced Modules ..................................... 44 § 5. The Number ofIndecomposable K(f)-Modules. . . . . . . . . . . .. 63 § 6. Indecomposable and Absolutely Indecomposable Modules.. 71 § 7. Relative Projective and Relative Injective Modules. . . . . . . .. 81 § 8. The Dual Module. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 97 § 9. Representations of Normal Subgroups ................... 123 §1O. One-Sided Decompositions of the Group-Ring ............ 147 §11. Frobenius Algebras and Symmetric Algebras. . . . . . . . . . . . .. 165 §12. Two-Sided Decompositions of Algebras. . . . . . . . . . . . . . . . .. 174 §13. Blocks of p-Constrained Groups ........................ 184 §14. Kernels of Blocks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 189 §15. p-Chief Factors of p-Soluble Groups ..................... 203 §16. Green's Indecomposability Theorem ..................... 223 Notes on Chapter VII . ..................................... 237

Chapter VIII. Linear Methods in Nilpotent Groups . . . . . . . . . . . . .. 238

§ 1. Central Series with Elementary Abelian Factors . . . . . . . . . .. 239 § 2. Jennings'Theorem .................................... 252 § 3. Transitive Linear Groups .............................. 266 § 4. Some Number-Theoretical Lemmas ...................... 270 § 5. Lemmas on 2-Groups .................................. 275 § 6. Commutators and Bilinear Mappings .................... 286 § 7. Suzuki 2-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 299 § 8. Lie Algebras ......................................... 316 § 9. The Lie Ring Method and an Application ................ 326 §10. Regular Automorphisms ............................... 349 §11. The Lower Central Series of Free Groups ................. 366

x Contents

§12. Remarks on the Burnside Problem ....................... 385 §13. Automorphisms of p-Groups ........................... 396 Notes on Chapter VIII ..................................... 404

Chapter IX. Linear Methods and Soluble Groups. . . . . . . . . . . . . . .. 405

§l. Introduction .......................................... 407 §2. Hall and Higman's Theorem B ........................... 419 §3. The Exceptional Case ................................... 429 §4. Reduction Theorems for Burnside's Problem ............... 449 §5. Other Consequences of Theorem B ....................... 464 §6. Fixed Point Free Automorphism Groups. . . . . . . . . . . . . . . . .. 476 §7. p-Stability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 492 §8. Soluble Groups with One Class of Involutions: . . . . . . . . . . . .. 503 Notes on Chapter IX ...................................... 514

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 515

Index of Names ........................................... 525

Index .. .................................................. 527

Index of Symbols

Autl XII sp(m) XII Yn(m) XII, 239 0"(05),0"1, ... ,,,.(05),0"(05) XIII,

407 mm XII Am, zm, Km XII n' XIII H(V) 12 S(V) 12 V~ 15 Q(21) 33 T(21) 35 Vm 44 <»V 51 V* 97 R(6) 155 L(6) 155 Sr(m:) 155 Sz{m:) 155 An9l(S), AnvJ(21) 155

Pm (V) 190 ~(V) 193 m(V) 194 ~(.?4) 194 m(~) 194 K~ 203 A~ 204 An(m) 242 Kn(m) 248 A(n,8) 294 l(m:) 318 gn 320 g(n) 323 adx 338 Cix) 338 (m: on m) 409 ep(m) 449 'rn(m) 449 e;(m) 449 SA(2, p) 494

Terminology and Notation

In this volume, the same terminology and notation as in Volume 1 will be used, with the following exceptions.

1. The identity mapping on a set X will be denoted by 1x. 2. The identity element of the group G) will be denoted by 1m or 1. 3. Bya section of a group G) is meant a group of the form b/Sl, where

Sl <l b ~ G). 4. If X is any algebraic system, Aut X denotes the group of all

automorphisms of X. The group of inner automorphisms of the group G) is denoted by Inn G).

5. The set of Sylow p-subgroups of the finite group G) will be denoted by Sp(G)).

6. The lower central series (III, 2.2) of the group G) will be denoted by

here Yn(G)) = [Yn-l (G)), G)] for n > 1. 7. If G) is a group and m is a positive integer, G)m = (xmlx E G).

Thus G)mn ::; (G)m)". 8. Let A be a commutative ring with identity and let G) be a group.

The group-ring of G) over A (I, 16.6) will be denoted by AG). 9. Let G) be a finite group. The field K is called a splitting field of

G) if KG)/J(KG)) is the direct sum of complete matrix algebras over K, where J(KG)) is the Jacobson radical of KG). Thus by V, 11.2a), K is a splitting field of G) if and only if K is a splitting field of KG)/J(KG)).

This definition is not the same as that given in V, 11.2b), but the two definitions reduce to the same thing when IG)I is not divisible by char K.

10. A KG)-module M is called absolutely irreducible if (i) M is irreducible and (ii) Hom Km (M, M) = K.

This definition is equivalent to that given in V, 11.8; this is proved in VII,2.2.

11. The unit matrix will be denoted by I. 12. If 7r is a set of primes, the complementary set of primes is denoted

Terminology and Notation XIII

by n'; thus

n' = {pip is a prime, p ¢ n}.

13. If n is a set of primes, the product of all the normaln-subgroups of the finite group GJ is denoted by O,,(GJ). Thus O,,(GJ) is the maximal normaln-subgroup of GJ. Clearly O,,(GJ) is a characteristic subgroup of GJ and O,,(GJjO,,(GJ)) = 1.

More generally, suppose that n1 , n2 , ••• are sets of primes. We define a characteristic subgroup 0", ..... ",(GJ) of GJ by induction on i: for i > 1,

0"""" ",(GJ)jO"""" "'-1 (GJ) = O",(GJjO"""" "'-1 (GJ).

Examples. a) The Fitting subgroup of GJ is TIp Op(GJ). b) If p is a prime, the upper p-series of GJ (VI, 6.1) is

c) The maximal p-nilpotent normal subgroup of GJ is 0p',p(GJ). For if in is a normal p-nilpotent subgroup of GJ, the normal p-complement .R of in is a characteristic p'-subgroup of GJ. Hence .R <J GJ,.R :$ 0p,(GJ), in0p,(GJ)jOp.(GJ) is a normal p-subgroup and in :$ 0p',p(GJ).

14. If n is a set of primes, O"(GJ) is defined to be the intersection of all the normal subgroups in of GJ for which GJjin is a n-group. Thus GJjO"(GJ) is the maximaln-factor group of GJ, and O"(GJ) is a characteristic subgroup of GJ.

Example. If GJ is a p-nilpotent group, OP(GJ) is the normal p-complement of GJ.

15. If IJ is a free group, a group-basis of IJ is a subset X of IJ such that X generates IJ and any mapping of X into a group is the restriction of some homomorphism of IJ. Such a set always exists, by definition of a free group (I, 19.1).

Chapter VII

Elements of General Representation Theory

In Chapter V, classical representation theory was studied. This is the theory of the group-ring K(f) and the K(f)-modules, where K is an algebraically closed field of characteristic O. (Many theorems remain valid under the hypothesis that K is algebraically closed and that char K does not divide the order of (f)). In this case, K(f) is semisimple and all K(f)modules are completely reducible. For many purposes it is therefore sufficient to handle the irreducible representations.

In this chapter we shall study the group-ring K(f) and the K(f)-modules when K is an arbitrary field. Thus we are concerned above all with the case when char K = p and p is a prime divisor of the order of the group; for short we call this the modular case. In this case the Jacobson radical of K(f) is non-zero and not all K(f)-modules are completely reducible. The number of isomorphism types of irreducible K(f)-modules is the pi-class number of (f), as long as K is sufficiently large (§ 3). The irreducible modules are determined by K<D/J(K(f)), and the divergence of K(f) from semisimplicity is determined by J(K(f)). Unfortunately there is no general procedure known for the determination of dimKJ(K(f)). But by using the technique of lifting idempotents from the theory of algebras, certain facts about the direct decompositions of K(f) into right ideals and twosided ideals can be established (§ 10, 12). The decomposition of K(f) into two-sided ideals leads to the theory of blocks and is central for the further development of the theory; unfortunately no general method for finding the number of blocks is known. More detailed assertions are made by taking into account the fact that the group-ring possesses a certain self-duality, namely, it is a symmetric algebra (§ 11). Among the consequences of this self-duality is the fact that projective and injective K(f)-modules are the same. If

is a decomposition of K(f) into indecomposable right ideals Pi' then all types of indecomposable projective K(f)-modules occur among the Pi.

2 VII. Elements of General Representation Theory

Further, each Pi has just one maximal submodule, namely p;J(Kffi), and P; has just one minimal submodule Si' which is isomorphic to P;/p;J(Kffi). Also Pi is determined to within isomorphism by Si' Thus the top and bottom composition factors of P; are known, but the complete composition structure of Pi can only rarely be determined. We therefore restrict ourselves to the investigation of the multiplicities of the composition factors of p;. This yields the Cartan matrix C of Kffi. The calculation of the elementary divisors of C from the centralizers of the p' -elements of ffi is possible by deep theorems of Richard Brauer, which, however, will not be presented in this chapter.

In this way some information about the indecomposable projective Kffi-modules can be obtained, but the general indecomposable Kffimodule is almost unapproachable. If char K = p, there is only a finite number oftypes of indecomposable Kffi-modules if and only if the Sylow p-subgroups of ffi are cyclic (§ 5). On the one hand this fact leads in the further development of the theory to the deep results of Brauer and Dade on groups with cyclic Sylow p-subgroups, but on the other hand it presents difficulties for the development of the general theory which have not yet been overcome.

In spite of these difficulties, some useful general facts about Kffimodules have been proved. Among these are the theory of the induced module and the reciprocity theorems (§ 4), the theorems of Clifford type about the relations between Kffi-modules and Km-modules for a normal subgroup m of ffi (§ 9) and the duality theory of Kffi-modules (§ 8).

What are the aims of a general theory of group-rings? We mention here two lines of development.

(1) If ffi is a p-soluble group, then the p-chief factors of ffi yield irreducible Kffi-modules in a natural way, where K = GF(p). (It is not very important that K need not be a splitting field for 63, since the theory ofthe Schur index is trivial for finite fields.) On the one hand, one would like to know what place these representations, obtained so directly from the structure of 63, have in the general theory (§ 15). On the other hand, abundant knowledge of irreducible Kffi-modules of a given p-soluble group ffi is often necessary for the construction of more complicated p-soluble groups.

(2) Another application is much better developed; analogously to local number theory there is a local theory of the characters of ffi over C. This is developed in the following way.

Let L be a field of characteristic 0 with a non-Archimedean valuation, let 0 be the ring of integers in L, let p be the maximal ideal of 0 and let K = o/p. We choose L large enough to be a splitting field for 63. As in V, 12.5, each L63-module may be regarded as obtained from an 063-module by extending the domain of coefficients. If M is an o63-module,

VII. Elements of General Representation Theory 3

MjMp can be made into a K(fj-module in a natural way. Thus the theory of K(fj-modules appears as a first approximation to the theory of o(fjmodules. If 0 is supposed to be complete with respect to its valuation, then as in Hensel's lemma we can build up the o(fj-module from the K(fj-module by successive approximation. The result is a theory of characters in 0 and thus a local representation theory for the prime p. Amongst other results this yields refinements of the classical orthogonality relations which have been drawn upon for the proofs of deep assertions about the structure of finite groups. The "local to global" step from the local theory to a theory of D(fj-modules, where D is a Dedekind ring, has been only partially successful up to now. The results thus obtained have played no part in the structure theory of finite groups.

The results of this chapter and the consequent modular representation theory are above all the work of Richard Brauer. Since 1936 he has systematically built up this theory and made it into a more and more delicate instrument for the investigation of finite groups.

We shall assume that the reader is familiar with the following simple facts about projective and injective modules. The proofs may be found in MACLANE [1 J. Let 9l be an arbitrary ring with 1.

(1) An 9l-module P is called projective if any diagram

can be completed by adding y; more precisely, if V, Ware 9l-modules, II. E Hom\R(p, W), f3 E Hom\R(V, W) and f3 is an epimorphism, there exists Y E Hom\R(P, V) such that II. = yf3 (p. 20).

(2) An 9l-module is projective if and only if it is a direct summand of a free module. Any finitely generated projective 9l-module is a direct summand of a finitely generated free 9l-module (p. 21).

(3) If P is a projective 9l-module and

is an exact sequence of 9l-modules, then there exists an 9l-submodule pi of W isomorphic to P such that W = VII. EB pi (p. 24).

(4) If


is an exact sequence of 9l-modules and P is a projective 9l-module, then

is an exact sequence of Abelian groups. If 9l is a K-algebra, this is an exact sequence of vector spaces over K (p. 24).

(5) Direct summands of projective modules are projective. Direct sums of projective modules are projective.

(6) An 9l-module J is called injective if any diagram

can be completed by adding y; more precisely, if V, Ware 9l-modules, a E Hom\ll(V, W), f3 E Hom!JI(V, J) and a is a monomorphism, then there exists y E Hom!JI(W, J) such that f3 = ay (p. 92).

(7) An 9l-module J is injective if and only if any diagram

can be completed by adding y; here 6 is a right ideal of 9l (p. 92). (8) An injective submodule of an 9l-module is a direct summand

(p.92). (9) Direct summands of injective modules are injective. Direct sums

of a finite number of injective modules are injective.

§ 1. Extension of the Ground-Field

In this section we consider the behaviour of group-rings and modules under extension of the ground field.

§ 1. Extension of the Ground-Field 5

1.1 Definition. (V, 11.1) Suppose that the field L is an extension of the field K.

a) If m: is a K-algebra, then m: ®K L becomes an L-algebra, multiplication being given by

for all at, a2 in m: and At, A2 in L. We denote this algebra by m: L. If {at, ... , an} is a K-basis of m: and

n

aiaj = L Cijkak k=t

with Cijk E K, then {at ® 1, ... , an ® 1} is an L-basis of m:L and

n

(ai ® l)(aj @ 1) = L Cijk(ak ® 1). k=t

In particular dimL m:L = dimK m:. b) If V is an m:-module, the vector space V ®K L becomes an m:L-

module VL if we put

for v E V, a E m: and At, A2 E L. We have dim L V L = dim K V. c) If m: is a K-algebra and m is a K-subspace of m:, there is an L

homomorphism B of m ®K L into m: @K L in which (b @ A)B = b @ A (b E m, A E L). We write im B = mL. Note that B is a monomorphism, for if T is a K-basis of L, m @K L = EBtETm ® t and m: ®K L = EBtETm: ® t. If m is a subring of m:, mL is a subring of m:L; if m is an ideal of m:, mL is an ideal of m:L.

1.2 Lemma. Suppose that L is an extension of the field K. a) If m: t , ... , m:k are K-algebras, then

b) If m: is a K-algebra and (m:)m is the complete matrix ring of degree mover m:, then ((m:)m)L ~ (m:dm.

c) If m: is a K-algebra and 3 is a two-sided ideal of m:, then (m:/3)L ~ m:L/3L·


d) If'l1 is afinite-dimensional K-algebra, then J('l1)L ~ J('l1L). e) If V and Ware 'l1-modules and V ;;> W, then (V/W) ®K Land

VL/WL are isomorphic 'l1L -modules.

Proof. a) It is easily checked that there is an isomorphism ':J of ('l11 EB ... EB 'l1k)L onto ('l1 1)L EEl ... EB ('l1k)L such that

b) There is an isomorphism f3 for which

c) There is an L-algebra epimorphism y of 'l1L onto ('l1/3)L in which

(a ® A)Y = (a + 3) ® A (a E 'l1, A E L).

If T is a K-basis of L, 'l1L = EBI ET 'l1 ® t, so ker y = 3L. d) By V, 2.4a), J('l1) is nilpotent. Suppose that J('l1)" = O. Then

(J('l1)dn = O. Thus J('l1)L is a nilpotent ideal of 'l1L. Hence by V, 2.4b), J ('l1)L ~ J ('l1L).

e) The proof is similar to that of c). q.e.d.

1.3 Examples. Suppose that L is an extension of the field K. a) We have (Km)L ~ Lm.

By l.la), (Km)L has the L-basis {g ® llg E m} and

Hence the mapping ':J of (Km)L into Lm given by

is an L-algebra isomorphism of (Km)L onto Lm. b) By 1.2a) and b), for

k

'l1 ~ EB(K)n" i=l

we get immediately


k

ill:L ~ EB(L)n, i=l

In the following lemma, some elementary facts about fields are collected for later use.

1.4 Lemma. a) Suppose that 0 i= f E K[tJ and L is an extension field of K. Let f = TIi=l gr' be the decomposition off in L[tJ with pairwise nonassociated irreducible polynomials gi. Then

r

(K[tJ/.fK[tJ) L ~ L[tJ/fL[tJ ~ EB L[tJ/gi'L[tJ. i=l

b) Suppose K = GF(q) and Li = GF(qn,) (i = 1, 2). Let d be the greatest common divisor and k the least common multiple of nl and n2. Then

with d direct summands on the right. c) Let Ll be a separable extension of K and L2 any extension of K.

Then

where the fields Fi are separable extensions of L2 .

Proof a) We have

(K[tJ/fK[tJ)L ~ (L ®K K[tJ)!(L ®K fK[tJ)

~ L[t J/f L[t J,

The mapping IY. of L[tJ into EBi=l L[tJ/gim'L[tJ given by

(by 1.2c))

hlY. = (h + gT'L[tJ, ... , h + g;"L[tJ) (h E L[tJ)

is obviously an L-algebra homomorphism, and hE ker IY. if and only if gi' divides h for all i = 1, ... , r. Thus ker IY. = f L[t J. By the Chinese remainder theorem for the principal ideal ring L[t], the system of congruences


h == hi (mod g~i) (i = 1, ... , r)

has a solution h in L[t] for any given hi' ... , hr. Hence r:t. is an epimorphism. Thus

r

L[t]/JL[t] = L[t]/kerr:t. ~ EB L[t]/giiL[t]. i=1

b) We have Ll ~ K[t]/JK[t] for some irreducible and separable polynomial J in K[t] of degree nl' As J has no multiple roots in any extension field of K, we have a decomposition J = gl ... gr in L2 [t] with irreducible, pairwise non-associated polynomials gi' Thus by a),

r

Ll ®K L2 ~ EBL2[t]/gi L2[t]. i=1

If ri is the degree of gi' we have

Writing mi = n2ri we get

r

Ll ®K L2 ~ EBGF(qmi) i=l

with n2lmi' As this decomposition of Ll ®K L2 into simple algebras is unique, we have also nllmi' by symmetry. Hence klmi for all i = 1, ... , r.

In Li the identical relation a qn, = a holds. Hence aq• = a. For any C = Ljaj ® bj with aj ELl' bj E L2, we conclude, since Ll ®K L2 is commutative and char K = p, that

This identical relation in Ll ®K L2 is naturally inherited by all the epimorphic images GF(qm,) of Ll ®K L2. Thus the polynomial t q• - t has qm, roots in GF(qm,), and hence mi ~ k. Since klmi' this forces mi = k for all i. Comparing dimensions over K,

and thus r is the greatest common divisor of n1 and n2'

§ I. Extension of the Ground-Field 9

c) As LJ : K is separable, we have

LJ = K(a) ~ K[t]lfK[t]

for some irreducible, separable polynomial f in K[t]. Hence in L2 [t] we obtain

r

f = flgi i=J

for non-associated irreducible, separable polynomials gi' Thus 1.4 a) implies that

r

LJ ®K L2 ~ L2[t]lfL2[t] ~ EB L2 [t]lgi L2[t]. i=J

As gi is separable, L2[t]lgiL2[t] is a field Fi> which is separable over L2 • q.e.d.

Now we can describe the behaviour of the radical of a group-ring under extension of the ground-field. We remark that all results in this section remain true for K-algebras 2l which can be defined over a finite subfield Ko of K, that is, for which there exists a Ko-algebra 2lo such that 2l ~ 2lo ®Ko K.

1.5 Theorem. Suppose that L is an extension of the field K. a) We have

and

b) If {aJ' ... , as} is a K-basis of J(K(fj), it is also an L-basis of J(L(fj). In particular for all fields L of characteristic p we have

Proof a) If char K = 0, we have

J(K(fj) = ° = J(K(fj ®K L) (by l.3a) and V, 2.7).


Hence we assume char K = p. We write Ko = GF(p) and show first that

By Wedderburn's theorem,

r

KoGJ/J(KoGJ) ~ EJj(F;)n" ;=1

where the F; are finite division rings. By another well-known theorem of Wedderburn, the F; are fields. Hence F; = GF(pd,) for some d;. We conclude that

r

~ EJj(F; ®Ko K)n, (by 1.2a) and b)). ;=1

Iffi is an irreducible polynomial of degree d; in Ko[t], we have

As fi has no multiple roots in any extension field of Ko, we have a decomposition

s,

fi = TI g;j' j=l

with irreducible polynomials g;j from K[t], pairwise non-associated for fixed i. By 1.4a), it follows that

s, F; ®Ko K ~ EJj K[t]/g;jK[tJ.

j=l

We denote the field K[t]/gijK[t] by F;j. Thus

Hence

r Sj

(Ko<V/J(Ko<V)h ~ EBEB(Fij)n, ;=1 j=l

is semisimple. By 1.2d),

since also

is semisimple,

Now suppose that char K = p and L is any extension of K. By 1.3a) and the results proved already,

Also by 1.2d),

Equality of the dimensions shows that

As

J(K<V) @K L = J(K<V @K L).

(K<V/J(K<V))L ~ (K<V @K L)/(J(K<V) @K L) (see 1.2c))

= (K<V ®K L)/J(K<V ®K L)

is semisimple, we see that

J«K(f)/J(K(f)))d = O.


b) In a) we saw that dimKJ(K(f) = dimL J(L(f). Hence we have only to prove that J(K(f) ~ J(L(f). Let .3 be the L-subspace of L(f) spanned by J(K(f). Since J(K(f) is a nilpotent ideal of K(f), .3 is a nilpotent ideal of L(f). Thus J(L(f) 2 .3 2 J(K(f). q.e.d.

To show that complete reducibility is preserved under field extensions, we first prove a simple lemma, which will also be frequently used later on.

1.6 Lemma. Let '!I be an algebra of finite dimension over a field K and let V be an '!I-module .

. a) V jVJ('!I) is a completely reducible '!I-module. If W is a submodule of V sitch that VjW is completely reducible, then W 2 VJ('!I).

b) The set

AnvJ('!I) = {vlv E V, vJ('!I) = O}

is a completely reducible submodule of V. If W is a completely reducible submodule of V, then W ~ AnvJ('!I).

Proof a) As '!I = '!IjJ('!I) is a semisimple algebra of finite dimension over K and VjVJ('!I) can be regarded as an '!I-module, VjVJ('!I) is completely reducible as an '!I-module and also as an '!I-module. Conversely, ifVjW is completely reducible, then (VjW)J('!I) = ° and thus VJ('!I) ~ W.

b) By similar arguments the submodule W is completely reducible if and only if WJ('!I) = 0, that is, if W ~ AnvJ('!I). q.e.d.

1.7 Definition. Let '!I be an algebra of finite dimension over a field K and let V be an '!I-module.

a) We call the largest completely reducible factor module V jVJ('!I) of V the head of V and denote it by H(V).

b) We call the largest completely reducible submodule AnvJ('!I) of V the socle of V and denote it by S(V).

1.8 Theorem. Suppose that V is a K(f)-module and L is an extension of K. Then the L(f)-module VL is completely reducible if and only if V is completely reducible.

Proof By 1.5,


Hence VLJ(K(fj ®K L) = 0 if and only if VJ(K(fj) = O. But by 1.6, VJ(K(fj) = 0 and VLJ(K(fj ®K L) = 0 are equivalent to the complete reducibility of V and VL respectively. q.e.d.

1.9 Remark. In 1.5 a), we showed that (K(fjjJ(K(fj))L is semisimple for every extension L of K. This property is called the separability of K(fjjJ(K(fj). By a theorem of Wedderburn (see DEURING [2], p. 23), in any algebra m for which mjJ(m) is separable, there exists a semisimple subalgebra 6 (~mjJ(m)) such that m = 6 EEl J(m). (This is only a direct sum of K-vector spaces; in general 6 is not an ideal of m.) For any two such complements 6 1 , 6 2 of J(m), there existsj E J(m) such that

(see CURTIS and REINER [1], p. 491). The proof of this theorem is technically related to the proof of the Zassenhaus theorem about the existence and conjugacy of complements of normal Hall subgroups in finite groups. The existence of complements of J(K(fj) in K(fj, however, has not found important applications in representation theory.

1.10 Lemma. Let K be any field of prime characteristic p. Then

k

KffijJ(Kffi) = EB(Li)n" i=1

where the Li are finite, separable extensions of K.

Proof We put Ko = GF(p). Then we have

m

Ko(fjjJ(Ko(fj) = EB(Fi)n, i=1

for finite division rings Fi , which by Wedderburn's theorem are fields. This implies that

(by 1.2c)) m

~ EB (Fi ®Ko K)n, (by 1.2a), b)). i=1

As Fi is separable over Ko, it follows from l.4c) that

mi Fi @Ko K ~ EB Fij

j=l

for separable extensions Fij of K. Hence

m mj

K(fj/J(K(fj) ~ EBEB(Fi)n, i=l j=l

1.11 Theorem. Let K be a field of prime characteristic p.

q.e.d.

a) If <p is the character of a K(fj-module V, then <p(g) = <p(gp,) for any element g of (fj, where gp, is the p' -part of g.

b) Let X be the set of p'-elements of (fj. Let V1, ••• , Vk be all the irreducible K(fj-modules to within isomorphism, and let <Pi be the character of Vi' Then <P1' ... , <Pk are K-linearly independent on X.

Proof a) Let pam be the order of g, where (m, p) = 1. Then there exist integers i, j such that im + jpa = 1, and gp, = gjpa. Hence if ~ is an eigen-value of the linear transformation v --+ vg of V (in some extension of K), ~im+jp. = ~ and ~mp. = 1. Since char K = p, it follows that ~m = 1 ane! ~jp. = ~. But <p(g) is the sum of the various eigen-values ~ and <p(gp,) is the corresponding sum of the ~jp., so <p(gp,) = <p(g).

b) By Lemma 1.10, we have

k

K(fj/J(K(fj) ~ EB(Li)n i ,

i=l

where Li is separable over K. Let Di be the corresponding epimorphism of K(fj onto (Li)ni and let Yi be the faithful irreducible representation over K of (Li)n on the module of row vectors with coefficients in Li. Then D1Y1' ... : DkYk are the inequivalent irreducible representations of K(fj and every irreducible representation of K(fj is equivalent to one of D1 Y1' ... , DkYk' We suppose the numbering so chosen that DiYi is the representation of (fj on Vi' Thus if a E K(fj,

If C E (Li)ni' it is easy to see that

so

<Ma) = tr L,:K(tra6i) (i = 1, ... ,k).

Since Li is separable over K, there exists bi E Li such that tr L K bi = 1 (see BOURBAKI [1], p. 155). Let ai be an element of Kffi for which

(

bi 0 ... 0) ai6i = ~ ~ ... ~ ,

o 0 ... 0

Then (MaJ = tr L,: K bi = 1 and c/Jiai) = 0 (j # i). Hence c/Jl' ••• , c/Jk are linearly independent. It follows from a) that their restrictions to ); are also linearly independent. q.e.d.

In particular, it follows from 1.11 that if c/Ji(g) = c/Jj(g) for all 9 E ffi, then i = j. This is not true for arbitrary representations over a field of finite characteristic, although it is true for arbitrary representations over C.

1.12 Lemma. Let V, W be Kffi-modules and let L be an extension of K. Then

If V = W, the two sides of this are isomorphic L-algebras.

Proof The proof is identical with that of V, 11.9. q.e.d.

We have seen that complete reducibility is preserved under field extensions. We study now how irreducible modules behave under Galois extensions of the ground field.

1.13 Definition. Let L be a Galois extension of K with Galois group ~. Let V by any Lffi-module. For every '7 E ~, we define the Lffi-module V~ as follows.

Let V~ be an Abelian group for which there is an additive isomorphism l~ of V onto V~. We define the structure of a vector space over L on V~ by the formula


If {V 1, ... , vn} is an L-basis of V, then {V11~, ... , Vnl~} is an L-basis of Vw Hence dimL V = dimL V~.

We now define the structure of an L(fj-module on V~ by putting

(Vl~)g = (Vg)l~ (v E V, g E (fj).

It is easy to see that V~ indeed becomes an L(fj-module. We consider the matrix representation of (fj on V~.

Suppose that {Vi> ... , vn } is an L-basis of V and that

n

Vig = I aij(g)Vj (i = 1, ... , n), j=l

n

= I (aij(g)f/)(V}~). j=l

Hence the matrix representation of (fj on V~ is conjugate under f/ to the matrix representation of (fj on V. Thus if X is the character of (fj on V, the character tjJ of (f) on V~ is given by

n

tjJ(g) = I aii(g)f/ = x(g)f/. i=l

We write tjJ = x.,. If W is an L(fj-submodule of V, it is obvious that Wl~ is an L(fj

submodule of V~. Hence V~ is irreducible if and only if V is irreducible.

1.14 Lemma. Let L be a Galois extension of K with Galois group ~. Let V be a K(fj-module and let W be an irreducible L(fj-module. If W is isomorphic to a submodule of VL, then for all f/ E ~ the algebraic conjugate W~ is also isomorphic to a submodule ofVL •

Proof. By assumption, there exists an a such that

o =P a E Hom L6) (W, Yd.


We consider the mapping a' = l;la(1 (8) '1) of W~ into VL• We show first that rl is L-linear.

Suppose that W E W~, A ELand

WI;la = LVi ® Ai' i

where Vi E V and Ai E L. Then

(AW) a' = ((A'1-1)(wl;1))a(1 ® '1)

= ((A'1-1)(wl;la))(1 ® '1)

= ~)Vi (8) Ai(A'1- 1))(1 ® '1). i

= LVi ® (Ai'1)A i

= A((wl;la)(l (8) '1)) = A(wa').

Next we show that a' is an Lffi-homomorphism. For W E W~ and g E ffi, we have

(wg) a' = (wg)l;la(l ® '1)

= ((WI;l )g)a(l ® '1)

= ((wz;lex)g)(l ® '1)

= (~>ig ® Ai)(1 (8) '1)

= Lvig ® Ai'1 i

= (~Vi (8) A)(1 ® '1)g

= wl;la(1 ® '1)g = (wa')g.

(by definition of W~)

(as ex is an Lffi-homomorphism)

Since a #- 0, a' #- 0. As W~ is irreducible, it follows that it is isomorphic to a submodule of VL• q.e.d.

1.15 Lemma. Let K be a field of prime characteristic, L any extension of K and V an irreducible Kffi-module. Then


for non-isomorphic irreducible LGl-modules W1 , ..• , Wr •

Proof. The fields Li, occuring in 1.10, are by Wedderburn's theorem antiisomorphic to the endomorphism rings of the irreducible KGlmodules. Hence, in particular, F = HomK(!;(V, V) is a field. By 1.8, VL is completely reducible. Suppose that

where the Hi are the homogeneous components of VL; this means that

H. ~ w. EF> •.• EF> W. l t Sj ,

for irreducible LGl-modules Wi' and Wi 't Wj for i =I j. We obtain

r r

Hom L(!; (VL' Vd ~ EBHomL(!; (Hi' Hi) ~ EB(Fi)s" i=l i=l

where Fi = HomL(!; (Wi> Wi) is a division algebra. But by 1.12,

Hence HomL(!; (VL' Vd is commutative, and this forces Si = 1 for all i. q.e.d.

1.16 Theorem. Let L be a Galois extension of K with Galois group ~, and let V be an LGl-module. We denote V, regarded as a K(fj-module, by Yo.

a) Vo ®K L ~ EB~EV V~. b) If x, Xo are the characters of V, Vo respectively, then

Xo(g) = I x(g)'1 = trL:KX(g)· ~EV

c) If V is a completely reducible L(fj-module, Vo is a completely reducible K(fj-module.

d) If V is an irreducible L(fj-module, then

Vo ~ W EF> .•• EF> W

for some irreducible K(fj-module W.


e) Suppose that K is of prime characteristic and V is an irreducible L(fj-module. We define the subfield Kx of L by the formula

Kx = K(x(g)lg E (fj).

Let

and let {111' ... , 11m} be a transversal ofU in~. Then

Vo ~ WEB' ~ . EB W,

where W is irreducible, s = (L: K) and Wl ~ EBf=1 V~i' In particular, Vo is irreducible if and only if L = Kx'

Proof a) We again denote by l~ the mapping of V onto V~ as in 1.13. We define a mapping (} of Vo ®K L into EB~ES V~ by putting

for v E Vo, A E L. Obviously (} is well-defined. We show that (} is L-linear. For v E Vo and ,11' ,12 in L, we have

(A2(v ® A1»(} = (v ® A1 A2)(} = «A1A2)(v/~»~Es

= A2«A1(VI~»~Es) = A2((v ® A1)(})'

For v E Vo, A ELand g E (fj, we obtain

«v ® A)g)(} = (vg ® A)(} = (A«vg)/~»~Es

= (A«VI~)g»~ES = «A(VI~»g)~ES

= «A(VI~»~Es)g = «v ® A)(})g.

Hence (} is an L(fj-homomorphism. Now

diml(Vo ®K L) = dimK Vo = (L: K)diml V = diml EB V~. ~ES

Thus, to prove that (} is an isomorphism, it suffices to show that ker (} = O. Let {V1' ... , vn} be an L-basis of V and {a1' ... , am} a K-basis of

L. Then


{aiVjli = 1, ... , m;j = 1, ... , n}

is a K-basis of Yo' Hence every element v of Vo ®K L can be written in the form

v = I aiVj ® Aij i,j

If v E ker Q, we obtain

(Aij E L).

for every 1'/ E ~. As {VI' ... , vn} is an L-basis of V, this implies that

for all 1'/ E ~. But by the separability of Lover K,

(see BOURBAKI [1], p. 119). Hence Aij = 0 for all i, j and v = O. This shows that ker p = O.

b) This follows at once from a). c) If V is completely reducible, then

o = VJ(Ufj)

:2 VJ(K(fj)

= VoJ(K(fj).

(by 1.6)

(by 1.5b))

Hence Vo is a completely reducible K(fj-module, by 1.6. d) Let W be an irreducible K(fj-submodule of Yo' Then by 1.12 and

a),

o "# HomK6; (W, yo) ®K L ~ HomL6; (WL' (Vo)d

= HomL6;(WL, EB V,,). "ED

Hence HomL6;(Wl> V,,) "# 0 for some 1'/ E~. By 1.8, WL is completely reducible, so V" is isomorphic to a submodule of WL. Thus by 1.14, V


also is isomorphic to a submodule of Wv If W1 and W2 are irreducible submodules of Yo, then (W1)L and (W2)L have direct summands isomorphic to V. Thus

This forces W1 ;;:; W2 • Hence

Vo ;;:; WED··· ED W s

for some s. e) Now suppose that K is of prime characteristic. Let X be the char

acter of V. If X = X~, then V ;;:; V~ by 1.11. Thus the isomorphism type of V~, appears among the V~ exactly lUI = (L: Kx) times. Hence it follows from a) and d) that

m

WL ED .; . ED WL ;;:; Vo ®K L ;;:; EB V~ ;;:; EB(V~iED ... ED V~,). ~€s i=l (L:K,)

By 1.15, the multiplicity of any irreducible direct summand of WL is 1. Hence

m

W L;;:; EB V~i and s = (L: Kx)· q.e.d. i=l

1.16 has the important cohsequence that in the case of finite fields the Schur index is 1.

1.17 Theorem (R. BRAUER [4]). Let L be afinitefield and V an irreducible L(fj-module. Let K be a subfield of L such that X(g) E K for all g E (fj,

where X is the character of V. Then the representation of (fj on V can be realized over K; that is, there exists an irreducible K(fj-module W such that V;;:; WL•

Proof Certainly L is a Galois extension of K. Now all V~ have the same character and are therefore isomorphic, by 1.11. Thus by 1.16e), WL ;;:; V. q.e.d.

1.18 Theorem. Suppose that L is a Galois extension of K with Galois group ~.

a) Let V be an irreducible L(fj-module. Then there exists exactly one irreducible K(fj-module W (to within isomorphism) such that V is isomorphic to a direct summand of W L'

b) (cf V, 13.3) Suppose that W is an irreducible K(fj-module. Then

W L ~ Vi EB ... EB Vr

with irreducible L(fj-modules Vi' and all Vi are conjugate under automorphisms in i) in the sense of 1.13. Every isomorphism type appears among the Vi with the same multiplicity. Finally r divides (L: K).

Proof a) We have

K(fj/J(K(fj) ~ E8 Wi' i

where the Wi are irreducible K(fj-modules. Thus it follows that

(K(fj/J(K(fj))L ~ E8(Wi)u i

and by 1.8, the (Wi)L are completely reducible L(fj-modules. However,

(K(fj/J(K(fj))L ~ (K(fj ®K L)/(J(K(fj) ®K L) (by 1.2e))

= (K(fj ®K L)/J(K(fj ®K L) (by 1.Sa))

~ L(fj/J(Ufj) (by 1.3a)).

As the irreducible L(fj-module V is isomorphic to a direct summand of L(fj/J(L(fj) (considered as an L(fj-module), V is isomorphic to a direct summand of some (WJL'

Suppose that V is isomorphic to a direct summand of (Wi)L and (WA. Then by 1.12,

This shows that HomK<D (Wi' W) =1= O. Since ~ and Wj are irreducible, Wi ~ Wj'

b) Let V be isomorphic to an irreducible direct summand of WL. By 1.14 every conjugate V~ of V (11 E i») is also isomorphic to a direct summand of the completely reducible module WL.

We consider the K(fj-module Yo. By 1.16d),

Vo ~ U EB ... EB U

for some irreducible K&-module U. By 1.16a),

EB V~ ~ Vo @K L ~ UL 81 ... 81 UL· ~E f>

Thus UL and W L contain a submodule isomorphic to V. Hence U ~ W, by a). From

(1) EBV., ~ VO@K L ~ WL81 ···81 WL, ~Ef>

we conclude that

where each Vi is isomorphic to one of the conjugates V~ of V. Further, if s is the number of summands WLin (1), rs = I~I = (L: K), so r divides (L: K). As the '1 for which V~ ~ V form a subgroup of~, all isomorphism types of irreducible modules appear in EB~Ef> V~ with the same multiplicity. Hence by (1), this is also true for WL. q.e.d.

1.19 Corollary. Suppose that L is a Galois extension of K and W is an irreducible K&-module. If dimK W is coprime to (L: K), then WL is an irreducible L&-module.

Proof By 1.18,

where the Vi are irreducible l(fj-modules all of the same dimension, and r divides (L: K). But we also have

This forces r = 1, so W L is irreducible. q.e.d.

For the sake of completeness, we show that the analogue of Theorem 1.18b) is also true for indecomposable modules.

1.20 Theorem. Suppose that L is a Galois extension of K with Galois group i>. Suppose further that W is an indecomposable K<f>-module. Then

where the Vi are indecomposable l(fj-modules and all the Vi are conjugate under automorphisms in ~ in the sense of 1.13. Every isomorphism type appears among the Vi with the same multiplicity, which is a divisor of I~I = (L: K).

Proof (DRESS). Suppose that

t

WL ~ EB(Vi EEl';,' EEl Vi), i=1

where the Vi are indecomposable, pairwise non-isomorphic l(fj-modules. This implies that

t

(WL)o ~ EB (Vi EEl . " . EEl Vi)O' i=1

Since W is a K<f>-module, (WL)o is the direct sum of (L: K) copies of W, so by the Krull-Schmidt theorem,

Hence

WL EEl ... EEl WL ~ (V1)o ®K L ~ EB (Vl)~' ~E,6

by 1.16a). Let

and let 111' ... , 11 s be a transversal of U in ~. Then

s

WL EEl· .. EEl WL ~ EB (Vl)~ ~ EB«Vl)~i EEl ... EEl (Vl)~)' ~Eli i=1

where each direct sum of the (Vd~i contains lUI copies. By the KrullSchmidt theorem, then,

s

WL ~ EB«Vd~i EEl ... EEl (Vl)~)' i=1

where the number k of copies of (VI)~i is independent of i and divides I~I. q.e.d.

1.21 Theorem. Suppose that L isan extension of K and that V, Ware K<f>-modules such that W ®K L is isomorphic to a direct summand of V ®K L. Then W is isomorphic to a direct summand ofV.

Proof a) First suppose that W is indecomposable. It follows from the hypothesis that there exist mappings

such that (XP = 1. By 1.12, HomUfi(W ®K L, V ®K L) ~ HOmK(f)(W, V) ®K L, so it follows that

for certain (Xi E HomK(f)(W, V), Pj E HomK(f)(V, W). Thus (XiPj lies in the K-algebra ~ = HomK(f)(W, W). Suppose that (XiPj E J(~) for all i, j. Then (XiPj ® AiA} E J(~) ®K L ~ J(~l) by 1.2d), so

1 = I (XiPj ® AiA} E J(~d, i,j

a contradiction. Hence there exist i, j for which (XiPj If J(~). It follows from V, 2.4b) that the right ideal .3 of ~ generated by (XiPj is not a nil ideal. Thus .3 contains an element (XiPjy which is not nilpotent. Since W is indecomposable, it follows from Fitting's lemma (I, 10.7) that (XiPjy is an automorphism of W. Thus (XiPj is a non-singular linear transformation of W, and it follows easily that W(Xi ~ Wand

V = W(Xi E9 (ker PJ

b) To deal with the general case, write

where WI' ... , Wn are indecomposable K<f>-modules. We use induction on n. By a), V ~ WI E9 U for some K<f>-module U. But also Vl ~ Wl E9 X for some L<f>-module X, so

By the Krull-Schmidt theorem,

UL :;;;; (W2)L EEl ... EEl (Wn)L EEl X

:;;;; (W2 EEl ... EEl Wn)L EEl X.

By the inductive hypothesis, W2 EEl ... EEl Wn is a direct summand of U, so W is a direct summand of V. q.e.d.

1.22 Theorem (DEURING [1], E. NOETHER). Let V, W be K<f>-modules. If there exists an extension L of K such that VL, WL are isomorphic L<f>modules, then V, Ware isomorphic K<f>-modules.

Proof This follows at once from 1.21. q.e.d.

Exercises

1) Let L be a Galois extension of K with Galois group ~. Let p be an absolutely irreducible representation of <f> in L for which tr p(g) E K for all g E <f>. Prove the following.

a) Since p and pq (11 E ~) have the same trace, they are equivalent, and there exist matrices A(I1) with coefficients in L such that

for all 11 E ~ and all g E <f>. b) For all 111, 112 in ~, we have

for some c(111' 112) E LX. c) From A(111 (112"3» = A«"1 "2)"3)' derive

Hence c is a 2-cocycle on ~ with values in L x (in the sense of I, § 17). d) Suppose that c is a co boundary. Then

for some be,,) E LX. If we write A'(,,) = b(,,)A(,,), then A'("1 "2) = (A'("1)"2)A'("2). Hence A' is a cocycle on ~ with values in the group

§ 2. Splitting Fields 27

GL(n, L) (where n is the degree of p). An extension of Hilbert's Theorem 90 says that H1(f), GL(n, L)) = 1 (see SERRE [1], p. 159). This means that A'(17) = (B17)B- 1 for some B E GL(n, L). Then

for all 17 E f) and g E (f;.

e) Show that p can be realized in K if and only if c is a co boundary.

2) Now prove Theorem 1.17 by using the fact that H2(f), LX) = 1 if K is finite and L is a Galois extension of K with Galois group f).

3) Give an example of a reducible representation that cannot be realized in the smallest field containing all the traces of the matrices in the representation.

§ 2. Splitting Fields

We now turn to the discussion of splitting fields. Most of the results of this section are true for algebras m for which m/J(m) is separable (cf. 1.9), but for the sake of simplicity we shall often restrict ourselves to group-rings.

2.1 Definition. Let m be an algebra over K of finite dimension. a) An irreducible m-module V is called absolutely irreducible if

Homm(V, V) = K. (Observe that in general, neither Homm(V, V) = K nor the irreducibility of V implies the other).

b) K is called a splitting field of m if

k

m/J(m) ~ EJj(K)ni i=1

for some ni'

c) If {f; is a group, a splitting field of K{f; is called a splitting field for (f;.

2.2 Lemma. Let V be an irreducible K{f;-module. Then the following statements are equivalent.

a) V is absolutely irreducible. b) For any extension L of K, the L{f;-module VL is irreducible. c) If R is the algebraic closure of K, then VR is irreducible.


Proof a) => b): By 1.8, VL is completely reducible and by 1.12,

Hence ° and 1 are the only idempotents in Homur; (VL' Vd, and the completely reducible Lffi-module V L is therefore irreducible.

b) => c): This is trivial. c) => a): VR is irreducible by assumption, so by Schur's lemma and

1.12,

This shows that

Hence HomK(f;(V, V) = K and V is absolutely irreducible. q.e.d.

The two concepts introduced in 2.1 are closely related.

2.3 Theorem. The following assertions are equivalent. a) Every irreducible Kffi-module is absolutely irreducible. b) K is a splitting field for Kffi.

Proof Let V1, •.. ,Vk be all the irreducible Kffi-modules (to within isomorphism). As ViJ(Kffi) = 0, the Vi are all the irreducible KffijJ(Kffi)modules. By Wedderburn's theorem,

k

KffijJ(Kffi) ~ EB (\t;)n" ;=1

where the division algebra \t; is antiisomorphic to

By 2.1, the absolute irreducibility of Vi means that HomK(f;(V;, V;) = K. As the Wedderburn decomposition is unique, the assertion follows.

q.e.d.

2.4 Theorem. a) Let K be a splitting field for Kffi and V1 , ••• , Vk all the irreducible K ffi-modules (to within isomorphism). Then for any extension


L of K, the Uf)-modules (Vi)L (i = 1, ... , k) are all the irreducible Lffimodules (to within isomorphism).

b) Let L be an algebraically closed extension of K and V~, ... , V~ all the irreducible Lffi-modules (to within isomorphism). Suppose that for every i there exists a Kffi-module Vi (necessarily irreducible) such that V; ~ (V;)L' Then K is a splitting field for Kffi and V1, ... , Vk are all the irreducible K ffi-modules (to within isomorphism).

Proof a) As K is a splitting field for Kffi, we have

k

KffijJ(Kffi) ~ EB(K)n, i=l

for some ni' By 2.3, Vi is absolutely irreducible, hence (V;)L is irreducible by 2.2. For i =f:. j, by 1.12,

This shows that (V;)L it (VA for i =f:. j. We also have

LffijJ(Lffi) ~ (Kffi ®K L)jJ(Kffi ®K L)

= (Kffi ®K L)/(J(Kffi) ®K L) (by 1.5a))

(by 1.2c)) k k

~ EB((K)n, ®K L) ~ EB(L)n, (by 1.2a), b)). i=l i=l

By Wedderburn's theorem, LffijJ(Lffi) and Lffi have exactly k irreducible modules (to within isomorphism), and these must be (V1)L, ... , (Vk)L'

b) Let W be any irreducible Kffi-module. Then WL contains an irreducible submodule, and this is isomorphic to some V;. Hence by 1.12,

As Vi and Ware irreducible Kffi-modules, we deduce from HomK6i(Vi, W) =f:. 0 that W ~ Vi'

We show that all the Vi are absolutely irreducible. Since L is algebraically closed, L contains a subfield Lo isomorphic to the algebraic closure K of K. From

we see that Vi ®K Lo is irreducible. Hence Vi is absolutely irreducible by 2.2, and by 2.3, K is a splitting field for K(fj. q.e.d.

2.5 Lemma. Let K be any field, L an algebraically closed extension of K and (fj a finite group. Then there exists a splitting field F for (fj such that K c::; F c::; Land F : K is a Galois extension.

Proof Let Lo be the algebraic closure of the prime field P of L in L. As Lo is algebraically closed, Lo is a splitting field for (fj. Let V~, ... , V~ be all the irreducible Lo(fj-modules to within isomorphism. Let {Vi1' ... ,

vi,n.} be an Lo-basis for V; and ,

n,

vijg = I ajl(g)Vi/ (g E (fj, ajl(g) E Lo)· 1=1

Let M be the subfield of L generated over the prime field P of L by all the finitely many aJl(g) and their algebraic conjugates over P. As every ajz(g) is algebraic over P, we have (M : P) < 00.

L

K F

M p

If char K = 0, then M : P is trivially a separable extension. If char K > 0, then P and M are finite, so again M : P is separable. Thus M : P is a Galois extension. We put F = K(M). From Galois theory we know that F : K is also a Galois extension and K c::; F c::; l.

We still have to show that F is a splitting field for (fj. It suffices to show that M is a splitting field for (fj. We define an M(fj-module Vi by

and

n,

Vi = EB MVij j=1

n,

vug = I ajl(g)Vi/ (g E (fj). 1=1


Then (Vi)Lo ~ V;. Thus by 2.4b), M is a splitting field for (f). q.e.d.

The determination of splitting fields for (f) is a very delicate problem in characteristic o. In V, 19.11 we proved that if n is the exponent of (f), the field of n-th roots of unity over the rational field 4) is always a splitting field for (f). But also smaller fields may suffice, and in general there is no uniquely determined smallest splitting field.

We now show that for characteristic p the answer to this question is much simpler. This is due basically to the fact that every finite division ring is commutative.

2.6 Theorem (R. BRAUER). a) Let L be an algebraically closed field of characteristic p and ¢l' ... '¢k the characters of all the irreducible L(f)modules. Then

K = GF(P)(¢i(g)lg E (f), i = 1, ... , k)

is the unique smallest splitting field for (f) of characteristic p (to within isomorphism) .

b) Suppose that (fj is of exponent pbm, where (p, m) = 1. Let f be the smallest positive integer such that pI == 1 (m). Then GF(pI) is a splitting field for (fj and all subgroups of (fj.

Proof a) By 2.5, there exists a splitting field F for (fj such that F: GF(p) is a Galois extension and F s;;; L. Let Vi (i = 1, ... , k) be the types of irreducible F(fj-modules. Then by 2.4a), the (Vih are the types of irreducible L(fj-modules. Hence the character of Vi is also the character of (VJL' so by assumption has values in K. Thus by 1.17, there exist irreducible K(fj-modules V; such that Vi ~ (Wk Then

(VJL ~ (V; ®K F) ®F L ~ V; ®K L,

and so by 2.4b) K is a splitting field for (fj. It is obvious that K is the smallest splitting field, for the values ¢i(g)

of the irreducible characters must belong to every splitting field, by 2.4a).

b) Choose L as in a). The field GF(pI) contains all m-th roots of unity in L. Let Qi be a representation of (f) with character ¢i' and let Gi1(g), ... , Gin;(g) be the eigen-values of Qi(g) (in L). As

and char L = p,


Hence Gij(gr = 1, so Gij(g) E GF(pf) and (/J;(g) E GF(pf). Now we apply a). q.e.d.

§ 3. The Number of Irreducible Modular Representations

Suppose first that K is an algebraically closed field the characteristic of which does not divide the order of the group (f). The following statements about the irreducible representations of the group-ring K(f) hold.

(1) If h(f) is the class-number of (f), there are exactly h(f) irreducible representations of K(f) (to within isomorphism) (V, 5.1).

(2) If the degrees of the irreducible representations of K(f) are nj (i = 1, ... , h(f)), then

h(g)

1(f)1 = L nf· j=l

This relation follows from the assertion that the semisimple algebra K(f) is the direct sum of complete matrix algebras of degree nj over K (V, 5.1).

(3) In addition we have the assertion that the nj are divisors of 1(f)1 (V, 5.12).

What appears in place of these assertions when K is an algebraically closed field of arbitrary characteristic p? In place of(2) we obviously have

(2') dimK K(f)jJ(K(f) = Lnt, j

where the nj are the K-dimensions of the irreducible K(f)-modules and J(K(f) is the Jacobson radical of K(f). If p is a divisor of 1(f)1, then by V, 2.7, K(f) is not semisimple, but we know little about the dimension of J(K(f). Thus the formula (2') is of little use.

Assertion (3) still holds if (f) is p-soluble (see SWAN [1] and 9.21), but is false in general (see Example 3.10). It is conjectured that in all cases the highest power of p dividing nj is a divisor of I (f) I. 1

However a general form of (1) is known, which will be the subject of this section.

1 This conjecture is wrong. S. Norton showed recently that McLaughlin'S simple group of order 27 • 36 . 53 . 7 . 11 has in characteristic 2 an absolutely irreducible module of dimension 29 . 7.

§ 3. The Number of Irreducible Modular Representations 33

3.1 Definition. If 21 is an algebra over the field K, we denote by Q(21) the K-subspace of 21 spanned by all elements ab - ba with a, b E 21.

3.2 Lemma. a) If 21 is a complete matrix algebra over K, then

Q(21) = {AlA E 21, tr A = O},

and dim K 21/Q(21) = 1. b) If<f> is afinite group, Q(K<f» is spanned over K by all elements of the

form x - y, where x, yare conjugate elements of <f>. If {g l' ... ,gh} is a complete set of representatives of the conjugacy classes of <f>, then {gi + Q(K<f»li = 1, ... , h} is a K-basis of K<f>/Q(K<f».

c) If 21 1, ... , 21m are ideals of 21 and 21 = EBi";l 21i> then

m

Q(21) = EB Q(21J. i;l

Proof a) Suppose that 21 is the algebra of all n x n matrices with coefficients in K. We denote by X the K-subspace which consists of all such matrices of trace o. Since tr(AB - BA) = 0 for all A, B in 21, we have Q(21) s; X. Let {Euli,j = 1, ... , n} be the usual basis of 21 for which EijEkl = JjkEil . Then

for i '" j and

for i = 1, ... , n. Thus we obtain

and so Q(21) = X. Hence dimK 21/Q(21) = dimK 21/X = 1. b) If x, yare conjugate elements of <f>, x = g-lyg for some g E <f> and

x - y = g-l(yg) - (yg)g-l E Q(K21).

It is clear that Q(K<f» is spanned over K by all uv - vu with u, v in <f>; since however vu = u- 1(uv)u, it follows that Q(K<f» is spanned by all x - y, where x, yare conjugate elements of <f>. Now the gi and all g - gi


for which g is conjugate to but distinct from gi form a K-basis of K(f). Since Q(K(f)) is spanned by the g - gi for which g is conjugate to gi' it follows that

{gi + Q(K(f))/i = 1, ... , h}

is a K-basis of K(f)/Q(K(f)). c) This is obvious. q.e.d.

It should be observed that statement (1) on page 32 is an easy consequence of Lemma 3.2. In the modular case, however, we need further preparations.

3.3 Lemma. Suppose that char K = p and 2l is an algebra over K. a) If a E Q(2l), then aP E Q(2l). b) If a, b are elements of2l, then

for all natural numbers m.

Proof We first prove b) for m = 1. Clearly, (a + b)P - aP - bP is the sum of the 2P - 2 products C 1 ... cP' where Ci = a or Ci = b, but not all Ci are equal to a and not all Ci are equal to b. Now if 2 ~ i ~ p,

Thus

Hence (a + b)P - aP - bP lies in Q(2l). To prove a), we first observe that

But by b) for m = 1,

(ab - ba)P - (ab)P + (ba)p E Q(2l).

Hence (ab - ba)p E Q(2l). It now follows from b) for m = 1 that cP E Q(2l) for all C E Q(2l).

We prove b) by induction on m. The case m = 1 has already been settled. If m > 1, write

thus c E Q(m) by the inductive hypothesis. By the case m = 1,

But cP E Q(m) by a), so the assertion is clear. q.e.d.

3.4 Defmition. Suppose that char K = p and that m is an algebra over K. We define

T(m) = {ala E m, aP'" E Q(m) for all sufficiently large m}.

3.5 Lemma. T(m) is a K-subspace of m containing Q(m) and all the nilpotent elements of m.

Proof By 3.3a) and Definition 3.4, Q(m) s; T(m). If t E T(m) and c E K, then for all sufficiently large m,

since Q(m) is a K-subspace.lftl' t2 are in T(m), then it follows from 3.3b) that for all sufficiently large m,

Thus T(m) is a K-subspace of m. If a is a nilpotent element of m, then aP" = 0 E Q(m) for some m, so

a E T(m). q.e.d.

3.6 Lemma. a) Ifm is a complete matrix algebra over K, then

T(m) = Q(m) = {AlA E m, tr A = O},

b) Suppose that char K = p, <D is a finite group and that {g 1, ... , gk} is a complete set of representatives of those conjugacy classes of (£) which consist of elements of order prime to p (the pi-elements). Then {gi + T(K<D)li = 1, ... , k} is a K-basis of K<DjT(K<D).


c) If m: 1 , .•• , m:m are ideals in m: and m: = EB:'!:l m:i, then T(m:) = EBi=l T(m:i)'

Proof a) m: contains an element Ell for which Ei1 = Ell and tr Ell = 1. By 3.2a), Ell ~ Q(m:). For all m we have

Ef; = Ell ~ Q(m:),

so Ell ~ T(m:). Since T(m:) ;2 Q(m:) and dimK m:/Q(m:) = 1, it follows that T(m:) = Q(m:).

b) Suppose that pa is the highest power of p which divides l(fjl. For g E (fj, write g = gpgp' = gp,gp' where gp is a p-element and gp, a p'element. Since g and gp, commute, we see that for all n :?: 0,

Thus g - gp' E T(K(fj). Now gp' is conjugate to gi for some i, and gp' - gi E Q(K(fj) by 3.2b). Thus g - gi E T(K(fj). Hence the gi + T(K(fj) span K(fjjT(K(fj). To show that they are linearly independent, suppose that

k

r = L cigi E T(K(fj) i=l

with Ci E K. If l(fjl = paq, choose m so that pm == 1(q) and

By 3.3b),

_ _ p'"p'"_ pm ( k )pm k k ° = i~ Cigi = i~ Ci gi = i~ Ci gi mod Q(K(fj).

By 3.2b), Ci = ° (i = 1, ... , k). c) This follows at once from Definition 3.4 and 3.2c). q.e.d.

3.7 Lemma. Ifchar K = p and m: is an algebra offinite dimension over K, then

Proof By V, 2.4, J(m:) is nilpotent. Hence J(m:) s;;; T(m:) by 3.5. Trivially,

Q(2{jJ(2{» = (Q(2{) + J(2{))jJ(2{).

It follows at once that

T(2{)jJ(2{) S;; T(2{jJ(2{)).

Suppose that r + J(2{) E T(2{jJ(2{)). Then for all sufficiently large m,

rpm + J(2{) = (r + J(2{))pm E (Q(2{) + J(2{))jJ(2{).

Since Q(2{) S;; T(2{) and J(2{) S;; T(2{), it follows that rpm E T(2{). From the definition of T(2{), it then follows that r P' E Q(2{) for all sufficiently large n, and so r E T(2{). Hence T(2{jJ(2{)) = T(2{)jJ(2{). q.e.d.

3.8 Theorem (R. BRAUER [6]). Let K be afield of characteristic p and 2{ afinite dimensional K-algebra. Suppose that 2{jJ(2{) is a direct sum of complete matrix algebras over K. Then dimK 2{jT(2{) is the number of isomorphism types of irreducible 2{-modules.

Proof. Suppose that

where 2{1' ... , 21m are complete matrix algebras over K. Thus m is the number of isomorphism types of irreducible 2l-modules. By 3.7 and 3.6c),

Hence

dimK 2{jT(21) = dimK(2{jJ(2{))j(T(2{)jJ(2{))

m

= I dimK 2{ijT(2{J i=1

By 3.6a), dimK 2{;jT(21i ) = 1. Thus dimK 2{jT(2{) = m. q.e.d.

3.9 Theorem (R. BRAUER [2]). Let K be a splitting field of characteristic p for <V. Then the number of isomorphism types of irreducible K<V-modules is equal to the number of conjugacy classes of <V which consist of p'elements.

Proof. This follows at once from 3.8 and 3.6b). q.e.d.


3.10 Example (R. BRAUER, NESBITT [1]). We determine the irreducible representations of (fj = SL(2, p) over an algebraically closed field K of characteristic p.

a) First we prove that SL(2, p) has exactly p conjugacy classes which consist of p' -elements. Since conjugate elements have the same trace, this may be done by showing that p'-elements of SL(2, p) with the same trace are conjugate and that every element of GF(p) is the trace of some p'-element in SL(2, p).

Certainly there exist p'-elements in SL(2, p) with trace 2 and -2, namely I and -I. If s E GF(p) and s =1= ±2, put

A(s) = ( ° 1). -1 s

Thus A(s) E SL(2, p) and tr A(s) = s. We show that A(s) is a p'-element. Otherwise the p-component A(s)p of A(s) is conjugate to an element

for some a =1= 0, and A(s) is conjugate to an element in the centralizer of C. But the centralizer of C in SL(2, p) consists of elements of the form

and all these elements have trace ±2. Thus A(s) is a p'-element. Let A be a p' -element of SL(2, p) of trace s. Then

f(t) = t2 - st + 1

is the characteristic polynomial of A. If s = ±2,J(t) = (t ± 1)2 and A thus has the eigen-values 1, 1 or - 1, - 1. Hence A is conjugate in GL(2, p) to a matrix of the form ±C (C as above), and A2p = I. Since A is a p'-element, it follows easily that A = ±I.1f s =1= ±2, then A is not a scalar multiple of I, so there is a vector Vl such that Vl and V2 = v1A are linearly independent. By the Cayley-Hamilton theorem,

° = f(A) = A 2 - sA + I.

Thus


Hence A is conjugate in GL(2, p) to A(s). We have to show that A is conjugate even in SL(2, p) to A(s). Suppose that

X-I AX = A(s)

with X E GL(2, p). If we can find Y E CGL (2,p)(A(s)) such thatdet XY = 1, then

(Xy)-l A(XY) = y- 1 A(s) Y = A(s)

and XY E SL(2, p). It thus suffices to show that the determinants of the elements of CGL (2,p)(A(s)) take all values in GF(p) X • A simple calculation shows that

Thus we have to show that the quadratic form x 2 + sxy + y2 represents all values in GF(p). This is trivial for p = 2. If p is odd, then

If 1 - is2 is not a square, it is trivial that this form represents all values in GF(p). Otherwise, 1 - is2 is a non-zero square since s i= ±2; thus the form is equivalent to x 2 + y2 and represents all values, by II, 10.6.

b) We now construct irreducible KG>-modules Vm (0 ::; m ::; p - 1) for which dimK Vm = m + 1 (cf. V, 5.13). It then follows from a) and 3.9 that these Vm are all the irreducible KG>-modules to within isomorphism.

Let Vm be the K-vector space of homogeneous polynomials of degree m in the independent variables x and y. If A = (au) E SL(2, p), we put

( i m-i)A ( )i( )m-i x y = allx + a12Y a21x + a22Y .

Then Vm obviously becomes a KG>-module of K-dimension m + 1. We now show that Vm is irreducible for 0 ::; m ::; p - 1. Let W be a non-zero submodule of Vm and suppose that

n

o i= f = I ajxjym-j E W, j=O

where an i= 0 and n ~ m. For t E GF(p), we put

S(t) = G ~) and T(t) = C ~). Then

n n

IS(t) = L a)x + ty)jym- j = L jj(x, y)t j E W j=O j=O

for suitable polynomials fj, where 10 = I and In = anym. But

Since in GF(p)

we obtain

p-1

L t-i(IS(t)) E W. 1=1

p-1. {-I L t' = 1=1 0

if p - 1 divides i,

otherwise,

p-1 p-1 n n p-1

L t-i(jS(t)) = L c i L jjtj = L jj L tj-i = - L jj 1=1 1=1 j=O j=O 1=1 p-1ij-i

= {-/; for ~ ~ i ~ P - 2, 1 ~ i ~ n, - 10 - I p-1 for I = 0, P - 1 = n = m.

Since 10 = lEW, it follows that /; E W for all i = 1, ... ,n. In particular, since In = anym, we have ym E W. A further application of the same method shows that W also contains the element

1=1 1=1

for 0 ~ j ~ m < p - 1

and 0 < j < p - 1 = m,

for m = p - l,j = O.

As yP-1 E W has already been shown in the case m = p - 1, it follows

§ 3. The Number ofIrreducible Modular Representations 41

that xiym-i E W for allj such that ° ~ j ~ m. Thus W = Vm and Vm is irreducible for ° ~ m ~ p - 1.

c) Now (but not earlier), we can also determine dimKJ(KG». From what we have proved it follows that

dimKJ(KG» = dimK KG> - f i2 = p(P2 - 1) - ~(p + 1)(2p + 1) i=l

p(p + 1)(4p - 7) 6

We thus see that dimK J(KG» accounts for a quite substantial part of the dimension of KG>.

d) It is easy to find cases where the dimension of an irreducible module does not divide the order of the group. Namely SL(2, 7) has an irreducible module of dimension 5, but 5 does not divide the order of SL(2,7).

3.11 Theorem (BERMAN [1]). Let K be any field. Suppose that G> is a finite group and define 1 to be (i) the whole of G> if char K = 0, (ii) the set of pi -elements of G> if char K is a prime p. Let m be (i) I G> I if char K = 0, (iiJ the greatest pi -divisor of I G> I if char K = p. Let L = K(~) for some primitive m-th root of unity ~. Let A be the set of integers afor which there exists a K-automorphism IX of L such that ~IX = ~a. Then there is an equivalence relation '" on 1 in which x '" y if and only if y is conjugate to x a for some a E A, and the number of inequivalent irreducible representations of G> in K is the number of equivalence classes under "'.

Proof If a E A, then (a, m) = 1, so the mapping x -+ x a is a bijective mapping of 1 onto 1. It follows easily that '" is an equivalence relation. Let s denote the number of equivalence classes. Thus s is the dimension of the K-space M of mappings f of 1 into K having the property that f(x) = f(y) whenever y is conjugate to x a for some a E A. Now if <PI' ... , <Pr are the characters of all the inequivalent irreducible representations of G> in K, then <PI' ... , <Pr are K-linearly independent on 1; this follows from 1.11 if char K = p and from V, 5.8 if char K = 0. It is to be shown that r = s, and this will be accomplished if we show that

<PI' ... , <Pr span M. Let F be a splitting field for G> such that F : K is a Galois extension

(cf. 2.5) and F ;2 L. First, suppose that

<Pi = ljJI + ljJ2 + ...

is the decomposition of <Pi into absolutely irreducible constituents in F. Thus if X E X, each I/Ij(x) is a sum of powers of ~ and I/Ij(X a) is the sum of the corresponding powers of ~a. Hence if CI. is a K-automorphism of F and ~CI. = ~a,

Thus <p;(x)CI. = <Pi (X a). Since <p;(x) E K, it follows that <p;(x) = <Pi (X a). Hence <Pi E M.

Next suppose that X is the character of an absolutely irreducible representation of (fj in F. By 1.18, X is a component of some <Pi and

where X, X', ... are conjugate under K-automorphisms of F. Since <P1' ... , <Pr are linearly independent, <Pi ::j:. 0 and ri is not divisible by char K. Hence La XCI., where XCI. runs through all distinct algebraic conjugates of X, is a K-multiple of <Pi'

It is to be proved that any element (J of M is a K-linear combination of <Pl' ..• , <Pr· Then (J is a class-function on X into F. By 1.11 or Y, 5.8, the absolutely irreducible characters of (fj are F-linearly independent on X, and by 3.9, their number is the number of classes of (fj contained in X. Thus the space of class-functions on X is spanned by the absolutely irreducible characters, and we can write

where Ci E F and Xl' Xz,' .. are distinct absolutely irreducible characters of (fj.

Let CI. be an automorphism of F over K. Then ~CI. = ~a for some a E A. Hence for every x E X, we obtain

L C;(xiCl.)(X) = L CiXi(Xa) = (J(x a) = (J(x) = L CiXi(X), iii

So Ci = Cj whenever XiCl. = Xj' Hence (J is an F-linear combination of the La XiCl. (summed over the different algebraic conjugates of Xi), and from above it follows that (J is an F-linear combination of the <Pi' say

r

(J = L di<Pi i=l

with di E F. Since, however, (j and the tPi all have values in K, we obtain for every automorphism r:t. of F over K

r

(j = I (dir:t.)tPi· i=l

As the tPi (i = 1, ... , r) are linearly independent over K, there exist Xj EX (j = 1, ... , r) such thatthe matrix (tPi(x))(i,j = 1, ... , r) is nonsingular. Hence the di are uniquely determined by

r

(j(Xj) = I ditPi(x) (j = 1, ... , r). i=l

It follows that di = dir:t. E K. So (j is a K-linear combination of the tPi> as required. q.e.d.

3.12 Theorem. a) The number of isomorphism types of irreducible Qmmodules is equal to the number of conjugacy classes of cyclic subgroups ofm.

b) ffr is the number of real conjugacy classes of m and 2s is the number of non-real ones, then r + s is the number of isomorphism types of irreducible ~m-modules.

Proof a) In the notation of 3.11, A consists of all integers prime to I ml, so x '" y if and only if <x) is conjugate to <y).

b) In this case a E A if and only if a == ± 1( 1m D, so x '" y if and only if x is conjugate to y ± 1. The assertion follows since the class C is real if and only if C = C- 1. q.e.d.

Exercises

4) Let p be an odd prime, K an algebraically closed field of characteristic p and m = GL(2, p).

a) Show as in 3.10a) that m has exactly p(p - 1) conjugacy classes of p' -elements.

b) Let m operate on Vm = EBi'!:o Kxiym-i as in 3.10b) and let Pm denote the representation of m on Vm• Define Pm.n by


for 9 E (f). Show that the Pm,n with 0 :;:;: m :;:;: p - 1 and 0 :;:;: n :;:;: p - 2 are representatives of all the irreducible representations of (f) over K. (To show that the representations Pm,n are pairwise inequivalent, consider

g= ( 0 1) -t s '

where (t) = GF(pr, and show that tr Pm(g) =I 0 for some choice of s.)

§ 4. Induced Modules

The principal tool for the construction of modules over the group-ring K(f) is the operation of forming the induced module, and we shall now discuss the formal properties of this. We shall prove the reciprocity theorems of Nakayama, which are generalizations of the Frobenius reciprocity theorem. We shall obtain two dual assertions which both reduce to the Frobenius reciprocity theorem in the case of completely reducible modules. For the sake of logical clarity, we shall temporarily use not only the induced modules but also the coinduced modules. With applications to group-rings over rings of p-adic integers in mind, we consider group-rings A(f) of finite groups (f) over an arbitrary commutative ring A with identity.

4.1 Definition. Let U be a subgroup of m. a) If V is an AU-module, we put

By defining (v ® a)g = v (8) ag for all v E V, a E Am and gEm, \f!J becomes an A(f)-module, the module induced from V.

b) If V1, Vz are AU-modules and rx E HomAU (V1, Vz), we put rxo; = rx ® 1. Thus rxo; is a mapping of V~ into V:. In fact, rxo; is an A(f)-homomorphism of Vr into Vf, for if gEm,

«v ® a)g)rxO; = (v ® ag)(rx ® 1) = vrx (8) ag = (vrx ® a)g

= «v ® a)rxo;)g

for all v E V1, a E Am.

§ 4. Induced Modules 45

c) IfW is an Am-module, we denote by Wu the AU-module obtained from W by restricting the domain of operators to AU.

4.2 Theorem. Suppose that U ~ m. Then·o; is a covariant exact functor from the category of AU-modules into the category of Am-modules. This means the following.

a) If V1 , V2 , V3 are AU-modules and IX E HomAU(V1 , V2),

p E HomAU(V2 , V3), then (1Xp)0; = 1X00po;.

b) If

is an exact sequence of AU-modules, then

is an exact sequence of Am-modules. If IX is the inclusion mapping, this implies that

Proof. a) For V1 E V1 and a E Am, we have

(V1 ® a)lXo; po; = «V1 ® a)(1X ® l»(P ® 1) = V1IXP ® a

= (V1 ® a)(lXp)o;.

b) We have to show that (1) 1X0; is a monomorphism; (2) im 1X0; = ker po;; (3) po; is an epimorphism. Obviously (3) holds. To prove (1) and (2), we consider ker yO; and im lj for y E HomAU(W1 , W2), where W1 , W2 are any AU-modules. Let T be a transversal of U in m. Thus

Am = EB(AU)t lET

and

W;o; = EBW; ® t. lET

Now if

L Wr ® t E kerY<!\ rET

then

L WrY ® t = 0, rET

hence WrY = 0 and Wr E ker Y for all t E T. Thus

Similarly,

Now we conclude that

ker rJ.<D = ker rJ. ®AU A(fj = 0,

hence (1) holds; and

hence (2) holds. q.e.d.

The following theorem contains the most important formal properties of induced modules.

4.3 Theorem (D. G. HIGMAN [2]). Suppose that U :::; (fj and that T is a transversal of U in (fj (that is, lUx n TI = 1 for all x E (fj).

a) Suppose that V is an AU-module and 11 is the mapping of V into (V<D)u defined by V11 = v ® 1 (for v E V). Then 11 is an AU-monomorphism of V onto a direct summand of(V<D)u·

b) IfW is an A(fj-module and e is the mapping of(Wu)<D into W defined by

( L Wr ® t)e = L wrt (WrEW), rET rET


then e is an Am-epimorphism of(Wu)'" onto Wand

c) If W is an Am-module and /1 is the mapping of W into (Wu)fD defined by

W/1 = L wt-1 Q9 t (w E W), lET

then /1 is an Am-monomorphism of W into (Wu)fD; /1 is independent of the choice of the transversal T and (W/1)u is a direct summand of ((Wu)fD)U' Further

w/1e = 1m: Ulw

for all w E W.Iflm: UI has an inverse in the ring A, then

Proof. a) Obviously, 11 is an AU-monomorphism. We have

VfD = EB V ® t = (V ® 1) EB V', lET

where

V' = EB V ® t. lET. ~u

It remains to show that V' is an AU-module. But if t ~ U and u E U then tu = u't', where u' E U, t' E T and t' ~ U; thus

(v ® t)u = vu' Q9 t' E V'

for all v E V. b) Suppose that gEm. For each t E T, write tg = (tgt'-l)t', where

t' E T and tgt'-l E U. As t runs through T, so does t'. Hence

(L: WI ® t)g = L: wl(tgt'-l) ® t'. lET I' ET


Applying e, we obtain

((I Wt ® t)g)e = I wttgt,-lt' = I wttg = ((I Wt ® t)e)g. tET t' E T tET tET

It follows that e is an Affi-epimorphism of (Wut, onto W. Obviously

(Wu)rJ n ker e = O.

Since

I Wt ® t - (I wtt) ® 1 E ker e, tET tET

it follows that

c) Suppose that T' is another transversal of U in ffi. Then given t E T, we can write t = utt' with Ut E U and t' E T'. Then

I wt-1 ® t = I wt'-lU;-l ® utt' = I wt,-l ® t'. tET t'ET' t'ET'

Thus J.l is independent of the choice of the transversal. It follows that for all g E ffi,

(wg)J.l = I wgC1 ® t = I wgt-1 ® (tg-1)g tET tET

Hence J.l is an Affi-homomorphism. Since

(Wu)(f) = EB Wu ® t, tET

J.l is a monomorphism. IfW' = EBtET.t~UWU ® t, then as in a), W' is an AU-submodule. Now W' is a complement of (WJ.l)u. For on the one hand, if

wJ.l = L wC! ® t E WJ.l n W', tE T


then w = O. And on the other hand, ifT n U = {tl}, we have

I Wt ® t - I (wt.td t - I ® tEW'. tET tET

This shows that

Finally

wJ1e = I wt-It = 1m: ulw. tET

If 1m: UI is a unit in A, write e' = 1m: UI-Ie. Then J1e' = 1 and (y - ye' J1)e' = 0 for all y E (Wu)lD. Thus

We shall now characterize VlD by a universal property.

4.4 Theorem. Suppose that U ~ m. a) Suppose that V is an AU-module, 11 is the mapping in 4.3a) and W is

an Am-rtWdule. For each IX E HomAU(V, Wu), there exists exactly one r:x' E HomA(!i(VlD, W) such that 0( = 1]0('.

b) Suppose that V is an AU-module, V' is an Am-module and Q E HomAU(V, V~). Suppose that, given an Am-rtWdule Wand given IX E HomAU(V, Wu), there always exists exactly one IX' E HomAlD(V', W) such that QIX' = IX. Then V' and VlD are isomorphic Am-modules.

Proof a) There exists an A-homomorphism IX' of VlD = V ®AU Am into W such that (v ® a)IX' = (vlX)a for all v E V and a E Am, since IX is an AU-homomorphism. Since

((v ® a)g)cx' = (v ® ag)IX' = (vlX)(ag) = ((vlX)a)g = ((v ® a)IX')g


for all g E G>, rl is an AG>-homomorphism. Further

vl1a' = (v ® l)a' = (va)1 = va

for all v E V, so rJa' = a. To establish the uniqueness of a', it suffices to show that if fJ E HomA(l;(V(l;, W) and rJfJ= 0, then fJ = O. This is so because

o = (vrJfJ)g = ((v ® l)fJ)g = (v ® g)fJ

and the v ® g (with v E V, g E (fj) generate V(l; as an A-module. b) We have the following commutative diagrams.

where Q = I1Q' and 11 = (!I1'· It follows that rJ = rJ(Q'I1') and Q = Q(rJ' Q'). We also have the following commutative diagrams.

V'

On account of the uniqueness assertions, it follows from these that Q'I1' is the identity mapping on V(l; and 11' Q' is the identity mapping on V'. Thus (!' is an A(fj-isomorphism of V(l; onto V'. q.e.d.

A conceptual proof of the first reciprocity theorem of Nakayama follows easily from 4.4 (cf. V, 16.6).

4.5 Theorem (NAKAYAMA). Suppose that U :$ (fj, that V is an AU-module and that W is an A(fj-module. Then

(This is an isomorphism of A-modules, of course).


Proof Let /'f be the AU-monomorphism of V into (Vm)u defined in 4.3a). If f3 E HomAm(Vm, W), then /'ff3 E HomAU(V, Wu). We put f39 = /'ff3; thus 9 is an A-monomorphism of HomAm(Vm, W) into HomAU(V, Wu). Given rt E HomAU(V, Wu), by 4.4 there exists a unique rt' such that IX = /'fIX' = 1X'9. Thus 9 is an isomorphism. q.e.d.

Before we evaluate the assertion of 4.5, we prove a dual assertion. Our proofs will be most transparent if we temporarily introduce a functor m. dual to .m.

4.6 Definition. If m is a finite group and V is an A-module, H = HomA(Am, V) is, of course, just the direct sum of Iml copies of V. H becomes an Am-module if we define cjJg (for cjJ E H, gEm) by

a(cjJg) = (ga) cjJ (a E Am).

For

a((cjJg)g') = (g'aHcjJg) = (gg'a)cjJ = a(cjJ(gg')).

If now U ::; m and V is an AU-module, HomAu(Am, V) is an Amsubmodule of H, for if cjJ E HomAu(Am, V) and gEm, then

(au)(cjJg) = (gau)cjJ = ((ga)cjJ)u = (a(cjJg»u

for all a E Am, u E U. The Am-module HomAu(Am, V) is called the Am-module coinduced by V and will be denoted by mv.

If V1 , V2 are AU-modules and rt E HomAU(V1 , V2), then for any cjJ E mV1 = HomAu(Am, V1) we put cjJ(mrt) = cjJlX. Thus mrt is a mapping of mV1 into mv2 . In fact, mrt is an Am-homomorphism of mV1 into mV2 , since if cjJ E mV1 and gEm, then

a((cjJg)(mrt» = (a(cjJg))rt = ((ga)cjJ)rt = a((cjJ(mrt))g)

for all a E Am.

We now prove for mV the assertions dual to 4.3-4.5.

4.7 Theorem. Suppose that U ::; (f). Then m. is a covariant exact functor from the category of AU-modules into the category of Am-modules.

Proof Suppose that Vb V2 , V3 are AU-modules and that a E HomAU(V1, V2), 13 E HomAU(V2, V3). Then obviously lD(aI3) = (lDa)(!DI3). Thus !D. is a covariant functor.

Let T be a transversal of U in m; thus m = U lET Ut. For 4> E IDV1, 4> E ker !Da is equivalent to

o = a(4) IDa) = (a4»a for all a E Am.

This means that a4> E ker a. Hence

Obviously, im (\ja s;;: HomAu(Am, im a). Suppose that 13 E HomAu(A(\'), im a). Given t E T, t-1 f3 = vIa for some VI E VI. Now Am = EBIETt-1 AU is a free AU-mudule with basis {t-1It E T}. Thus there exists

such that t- l 4> = VI for all t E T. Then

t- 1(4)(!Da)) = t- l 4>a = vIa = t- 1f3.

This shows that 4>(!Da) = 13. Thus

It now follows easily that !D. is an exact functor. q.e.d.

4.8 Theorem. Suppose that U ::;; m and that T is a transversal of U in m. a) Suppose that V is an AU-module. The mapping (} defined by

is an AU-epimorphism of (!DV)u onto V, and ker (} is a direct summand of (ffiV)u·

b) IfW is an Am-module, the mapping a defined by

a(wa) = wa (w E W, a E Am)

is an Am-monomorphism ofW into ID(Wu), and (Wa)u is a direct summand of (<5(Wu))u·

c) If W is an Am-module, the mapping r defined by


4>r = L (t-l4»t (4) E ID(WU)) IE T

is an Am-epimorphism of ID(WU) onto W. r is independent of the choice of the transversal T, and (ker r)u is a direct summand of (ID(WU))U. Further

war = 1m: Ulw

for all w E W. If 1m: UI has an inverse in the ring A, then

Proof a) Since

(4)u){! = 1(4)u) = u4> = (14))u = (4){!)u

for 4> E IDV and u E U, (! is an AU-homomorphism. If v E V, define 4> E HomA(Am, V) by putting u4> = vu for u E U and g4> = 0 for gEm - U. Then 4> E IDV and 4>{! = v. Thus {! is an epimorphism. Now

ker{! = {4>I4>E IDV, 14> = O} = {4>I4>E IDV,u4> = 0 foralluEU}.

If

then V' is an AU-submodule and

b) If WE W, then wa E ID(WU) since

(au)(wa) = w(au) = (wa)u = (a(wa))u

for all a E Am, u E U. a is an Am-homomorphism since if wE Wand gEm,

a((wg)a) = (wg)a = (ga)(wa) = a((wa)g).

If wa = 0, then

o = l(wa) = wI = w;


thus a is an A(f)-monomorphism of W into ID(Wu). If

x = {4>/4> E 6i(Wu), (AU)4> = O},

X is an AU-submodule of (6i(Wu»u. If wa E Wa n X, then

o = l(wa) = wl = w,

so Wa n X = O. Further, for all 4> E 6i(Wu),

1(4) - (l4»a) = 14> - 14> = o.

As 4> and (l4»a are AU-homomorphisms, it follows that 4> - (l4»a E X and thus (ID(WU»U = (Wa)u EB X. Hence

c) Suppose that T is another transversal of U in (f). Then, given t E T, we can write t = utt' with Ut E U and t' E T. Then

tET t' E l' t'ET'

Thus t is independent of the choice of the transversal T. It follows that for all gEm,

(4)g)t = I (t-1(4>g»t = I ((gC 1)4»t tET tET

= I ((tg- 1t 14>)(tg-1)g = (4)t)g. lET

Thus t E HomAID((!;(Wu), W). Suppose that Tn U = {t 1 }. Since Am is a free AU-module with

basis {C 1 /t E T}, it follows that given w E W, there exists 4> E HomAu(A(f), Wu) such that t1'14> = wt1'1 and C 14> = 0 for t E T, t #- t 1 . Thus 4>t = w. Hence t is an Am-epimorphism of 6i(Wu) onto W. Put

Then Y is an AU-submodule of (6i(Wu»u. If 4> E ker t n Y, then

o = 4>t = I (t- 14»t = (t114»t1· IE T


This forces tl1e/> = 0, and since e/> E Y, e/> = O. Thus ker r n Y = O. If e/> E (!;(Wu), we define", E (!;(Wu) by

and

t11", = - L (t-le/»tt11. t, ",rET

Then'" E kerr and e/> - '" E Y. Thus

(!;(Wu) = ker r EB Y.

Finally, if w E W,

rET rET

Suppose that Im:ul has an inverse in A. If we write

then (Tr' = 1, so (e/>r'(T - e/»r' = 0 for alle/> E m(Wu). Thus

(!;(Wu) = im (T EB ker r' = W(T EB ker r. q.e.d.

We also characterize mV by a universal property.

4.9 Theorem. Suppose that U :s; m. a) Suppose that V is an AU-module, W is an Am-module and

(L E HomAJWu, V). Then there exists precisely one (L' E HomA(!;(W, (!;V) such that (L = (L' Q, where Q is the mapping defined in 4 .8a).

w

b) Suppose that V is an AU-module, V' is an Am-module and rr E

HomAU(V~, V). Suppose that, given an Am-module W and

rx E HomAU(WU, V), there always exists exactly one rx' E HomAo;(W, V') such that rx = rx' n. Then V' and o;v are isomorphic A<fJ-modules.

Proof a) Define the mapping wrx' (w E W) of A<fJ into V by

a(wrx') = (wa)rx (a E A<fJ).

Then wrx' E o;V, and rx' is an A-homomorphism such that

(wrx') Q = 1 (wrx') = wrx

for all w E W. Thus rx = rx' Q. Further, rx' is an A<fJ-homomorphism, since if a E A<fJ, w E Wand 9 E <fJ, then

a«wrx')g) = (ga)(wrx') = (wga)rx = a«wg)rx').

To verify the uniqueness of rx', it obviously suffices to show that if fJ E HomAo;(W, o;V) and fJQ = 0, then fJ = O. Thus we must show that g(wfJ) = 0 for all 9 E <fJ, w E W. But

g(wfJ) = l«wfJ)g) = l«wg)fJ) = (wg)(fJQ) = O.

b) The assertion of b) follows similarly to that of 4.4b). q.e.d.

The second reciprocity theorem of Nakayama follows immediately from 4.9.

4.10 Theorem (NAKAYAMA). Suppose that U :::; <fJ, that V is an AU-module and that W is an A(f)-module. Then

Proof The mapping fJ --. fJQ yields an A-homomorphism, and as in 4.5, this is an isomorphism of HomAo;(W, o;V) onto HomAU(WU, V) by 4.9a).

q.e.d.

So far we have only rarely used the finiteness of <fJ. (In 4.3c) and 4.8c) the finiteness of I (f): U I was used.) The essential point is that for finite 1(f):UI,·o; and 0;. are the same.

4.11 Theorem. Suppose that U :5 (!) and that I (!) ; UI is finite. For each AU-module V there is an A(f)-isomorphism Jiv ojo;V onto VO; such that if rx E HomAU(V1, V2), the diagram

ffJv1 ffJ:x ,ffJ v2

",j j'" V!6

1 rt.'"

, Vf

is always commutative. (Thus the I1v define a natural equivalence between the functors <l'i. and .<l'i " see MACLANE [1], p. 29.) In particular, V<l'i ~ <l'iV.

Proof Let T be a transversal of U in (!). For each AU-module V, we define a mapping I1v of <l'iV = HomAu(A(!), V) into V<l'i = V ®AU A(!) by putting

<Pl1v = L t-l<p @ t (<p E <l'iV). lET

Obviously I1v is an A-monomorphism. Since A(!) is a free AU-module with basis {ellt E T}, there exists <p E HomAu(A(!), V) such that the C l <p take preassigned values in V. Thus I1v is surjective, and so I1v is an A-isomorphism. It is an A(!)-isomorphism, since if g E (!) and tg = ult' with UI E U and t' E T, then

lET lET lET

I' ET I'ET

To verify that the I1v define a natural equivalence between'<l'i and <l'i., we have to establish the commutativity of the diagram displayed in the theorem. For <p E <l'iVl , we have

<P(<l'ir:t.)l1v2 = L el(<p(<l'ir:t.)) ® t = L el<pr:t. ® t lET lET

= L (t-l<p @ t)(r:t. ® 1) = <Pl1v1 (r:t.<l'i). q.e.d. lET

We shall now show that Theorems 4.5 and 4.10 can be used to prove a generalization of the Frobenius reciprocity theorem. We replace the ring A by a field K. We shall need a simple lemma.

4.12 Lemma. Let K be an arbitrary field. Let W be an irreducible K(!)module and let X be an arbitrary K(!)-module.


a) There exists a uniquely determined smallest submodule X' of X such that X/X' is a completely reducible Kffi-module and all composition factors of X/X' are isomorphic to W. If X/X' is the direct sum of k submodules isomorphic to W, then

In particular, if K is a splitting field for ffi, we have

b) There exists a uniquely determined largest submodule X" of X such that X" is a completely reducible Kffi-module and all composition factors of X" are isomorphic to W. If X" is the direct sum of I submodules isomorphic to W, then

I n particular, if K is a splitting field for ffi, then

Proof a) Suppose that X/Xl and X/X2 are completely reducible factor modules of X. By 1.6, X/(X1 n X2) is also completely reducible. If all the composition factors of X/Xl and X/X2 are isomorphic to W, so are those of X/(X1 n X2), since

Suppose that 0 # a E HomKm(X, W). As W is irreducible, X/ker a ~ W. It follows from the above that ker a ;2 X'. Thus if

where Wi ~ W, then

k

HomKm(X, W) ~ HomKm(X/X', W) ~ EB HomKm(Wi, W). i=l

It follows that


If K is a splitting field for (f;, then HOmKtli(W, W) = K, whence

b) The existence of X" is proved similarly, and so as in a),

dimKHomKtli(W, X) = dimKHomKtli(W, X") = IdimKHomKtli(W, W). q.e.d.

4.13 Theorem. Let K be an arbitrary field and suppose that U :::;; (f;.

Suppose further that V is an irreducible KU-module and that W is an irreducible K(f;-module.

a) Let k be the multiplicity ofW as a composition factor in the head of Vtli, and let I be the multiplicity of V as a composition factor in the socle of Wu. Then

In particular, ifK is a splitting field for U and (f;, then k = I. b) Let m be the multiplicity of W as a composition factor in the socle of

Vtli, and let n be the multiplicity of V as a composition factor in the head of Wu. Then

If K is a splitting field for U and ffi, then m = n.

Proof. a) We have

k dimK HomKtli(W, W) = dimK HomKtli(VC», W)

= dimK HomKU(V, Wu)

= IdimKHomKu(V, V)

b) Similarly

mdimKHomKtli(W, W) = dimKHomKtli(W, Vtli)

= dimKHomKtli(W, tliV)

= dimKHomKU(WU, V)

= ndimKHomKU(V, V)

(by 4.12a))

(by 4.5)

(by 4.12b)).

(by 4.12b))

(by 4.11)

(by 4.10)

(by 4. 12a)). q.e.d.

4.14 Example. Let K be a splitting field for m. We apply 4.13 with U = 1 and V = K. Then VI» = KG>. By 4. 13a), the multiplicity of an irreducible KG>-module W in the head KG>/J(KG» of KG> is equal to the multiplicity of K in the soc1e of Wu, and this is dim K W. This assertion is, of course, a consequence of the Wedderburn theorem long known to us. But 4.13b) yields a new result, namely, the multiplicity ofW in the soc1e of VI» = KG> is equal to the multiplicity of K in Wu/(Wu)J(KU), which is also dim K W. This result will also be a consequence of our considerations in § 11.

For later use, we add the following lemma.

4.15 Lemma. a) Suppose that U $ G>. Let V be a KU-module and let W be a KG>-module. Then

b) If'R <l G> and W is a KG>-module, then

where the KG>-module structure of K(G>/'R) is given by ('Rg 1)g2 = 'Rg 1g2

Proof a) Let T be a transversal of U in G>. We have

(V ®K Wur» = EB (V ® W) ® t tET

and

VI» ®K W = EB (V ® t) ® w. tET

We define a K-linear mapping rx from (V ®K Wu)1» into VI» ®K W by

((v ® w) ® t)rx = (v ® t) ® wt (v E V, W E W, t E T).

It is easily checked that rx is bijective. For g E G> and tg = uti with u E U and t' E T, we have

and

(((v ® w) ® t)g)rx = ((v ® w) ® tg)rx = ((v ® w) ® ut')rx

= ((vu ® wu) ® t')rx = (vu ® t') ® wut'

§ 4. Induced Modules

«(v ® w) ® t)r:x)g = «v ® t) ® wt)g = (v ® tg) ® wtg

= (vu ® t') ® wut'.

Hence r:x is a KG>-isomorphism. b) Let K be the trivial K91-module of dimension 1. By a),

61

The assertion follows at once since W'JI ®K K ~ W'JI and KOi ~ K(G>j91). q.e.d.

We mention finally that the first cohomology group can be defined by using coinduced modules.

4.16 Theorem. Let V be an AG>-module, H = HomA(AG>, V) and J = HomAOi(AG>, V), considered as AG>-modules as in 4.6.1f

then Hl(G>, V) ~ Zj(C + J).

Proof. If ¢ E H and the mapping f of G> into V is defined by

f(g) = g</J - 1</J (g E G»,

then ¢ E Z if and only if fE Zl(G>, V). For ¢ E Z is equivalent to ¢g 1 - </J E J for all 9 1 E G>, or

for all 9 1, 9 2 E G>. But this is equivalent to

or

Hence if ¢ E Z and we putf(g) = g¢ - 1¢ for all 9 E G>, the mapping </J --+ f + Bl(G>, V) is a homomorphism of Z into Hl(G>, V). Given f E Zl(G>, V), we define </J E H by


Since f(l) = 0, f(g) = g¢ - 1¢, so ¢ E Z. Hence the above mapping is an epimorphism. We must prove that the kernel M of it is C + J. Now

M = {¢I¢ E Z, there exists v E V such that g¢ - 1¢ = v(g - 1) for all g E (fj},

C = {¢I¢EH,¢g = ¢ forallgE(fj}

= {¢I¢ E H, g¢ = 1¢ for all g E (fj}

and

J = {¢I¢ E H, (xg)¢ = (x¢)g for all x, g E (fj}

= {¢I¢ E H, g¢ = (l¢)g for all g E (fj}.

Hence C + J £; M. Conversely, suppose ¢ E Z and g¢ - 1¢ = v(g - 1). If t/! E J is defined by gt/! = vg, then ¢ - t/! E C. Hence C + J = M.

q.e.d.

Exercises

5) Prove the following by applying the universal properties 4.4 and 4.9. a) If V is a projective AU-module, then VID is a projective A(fj-module. b) If V is an injective AU-module, then IDV is an injective A(fj-module.

6) Deduce from the universal properties that if U1 :$; U2 :$; (fj and V is a KU1-module, then

7) (WILLEMS [1]). Suppose that U :$; (fj and that K is a field. Let V be a KU-module. If Vffi is a completely reducible K(fj-module, then V is a completely reducible KU-module. (Hint: Let W be a KU-submodule of V. Take a projection 1t E HomKffi(Vffi, Vffi) with im 1t = WID. Write

(v ® t)n = L vnt,t' ® t', t'ET

§ 5. The Number ofIndecomposable KG>-Modules 63

where T is a transversal of U in ffi. Show that n!p!, is a KU-projection of V onto W, where {t 1 } = Tn U.)

§ 5. The Number of Indecomposable K(fj-Modules

If char K does not divide I (fjl, any K(fj-module is completely reducible, and knowledge of all irreducible K(fj-modules is to a large extent sufficient for the description of all K(fj-modules. If, however, K(fj is not semisimple, a description of all indecomposable Kffi-modules is required. Unfortunately, this can only be carried out in a few cases, because if char K = p, the number of indecomposable K(fj-modules is finite if and only if the Sylow p-subgroups of (fj are cyclic. This will be shown in this section.

It is easy to see that in case of an infinite field K of characteristic p there are infinitely many non-isomorphic indecomposable K(fj-modules of dimension 2 for the elementary Abelian group (fj = <g1) x <g2) of type (p, p). Namely, for any a E K, consider the K(fj-module with K-basis {V1' v2 } in which

But in Theorem 5.1 the emphasis is on the fact that indecomposable modules of arbitrarily large dimensions exist.

We begin by considering p-groups.

5.1 Theorem (D. G. HIGMAN [3]). Suppose that K is afield of characteristic p and that (fj is a non-cyclic p-group. Then there exist indecomposable K(fj-modules of K-dimension 1, 3,5, ....

Proof (CURTIS, REINER [1]). It follows from III, 7.1 that any non-cyclic p-group possesses an elementary Abelian factor group of type (p, p). Thus, to prove the theorem, we may assume that (fj is elementary Abelian of type (p, p). Suppose then, that (fj = <g1) x <g2) and gf = 1.

Let n be a natural number and let V be a vector space over K of dimension 2n + 1 with K-basis {vo, V1' .•• , Vn, W 1, ••. , wn}. Since char K = p, V becomes a K(fj-module if we put

(0 ~ i ~ n),

(1 ~ i ~ n).

We show that V is an indecomposable K(fj-module. Let n be the projection of V onto the K-subspace W = (W I' ... ,wn) with ker n = (vo, VI' ... , Vn)· Obviously gI - 1 induces a K-isomorphism of W onto (VI' ... , vn) and g2 - 1 induces a K-isomorphism of W onto (vo, VI' ... , Vn-I)'

Suppose that V = VI ffi V2 is a non-trivial decomposition of V as a K(fj-module. Let ni = dimK Vin (i = 1, 2). Then

so

Hence

Thus dimK Vi < 2ni + 1 for either i = 1 or i = 2. Hence for this i, ni > O. Since

and dimK Vin = ni' it follows that

But

So

Note that for all V E V, we have V - vn E (vo, ... , Vn), so

(V - vn)(g - 1) = 0

for all g E (fj, and v(g - 1) = vn(g - 1). Thus

§ 5. The Number of Indecomposable KGi-Modules 65

Vi(g - 1) = Vin(g - 1)

and

But g 1 - 1 operates monomorphically on im n, so

Hence

Since ni > 0, there exists a greatest integer k such that k ::; nand

Hence there exists WE Vin such that

and Ck "# 0. Thus

But Vin(gl - 1) S;;; <Vb' .. , vn ), which gives a contradiction. q.e.d.

For cyclic p-groups, however, the situation is quite different. Before we study the cyclic case, we have to prove a general result about grouprings of p-groups.

5.2 Theorem. Let K be a field of characteristic p and (f) a p-group. a) K(f) has precisely one irreducible representation, namely, the unit

representation. b) J(K(f)) is the augmentation ideal;

J(K(f)) is the only maximal right ideal and the only maximal left ideal of K(f).

c) If9t is a proper right ideal of Km, then Km/9t is an indecomposable Km-module. In particular, the free Km-module with one generator is indecomposable.

d) Km has precisely one minimal right ideal, namely K LgE(l; g. e) Every non-zero submodule of Km is indecomposable.

Proof a) This is V, 5.16. . b) By V, 2.1, J(Km) is the intersection of the kernels of all the irreduci

ble representations of Km. It follows from a) that J(Km) is the kernel of the unit representation; but this is the augmentation ideal. By V, 2.2, J(Km) is the intersection of all maximal right ideals of Km. Since dimK Km/J(Km) = 1, it follows that J(Km) is the only maximal right ideal of Km. By symmetry, J(Km) is also the only maximal left ideal of Km.

c) Km/9t has precisely one maximal submodule, namely J(Km)/9t. Hence Km/9t is indecomposable.

d) If 3 is a minimal right ideal of Km, the representation of m on 3 is the unit representation, by a). Thus 3 = Ka for some a E Km and ag = a for all gEm. It is easy to see that a then is a scalar multiple of LgE(l; g.

e) This follows immediately from d). q.e.d.

More detailed results about the group-ring of a p-group over a field of characteristic p will be proved in the next chapter (see VIII, § 2). We shall prove later (in 1003, 10.14 and 11.6) that assertions similar to Theorem 5.2 hold for any projective indecomposable Km-module P; namely, the socle and head of P are irreducible and isomorphic.

5.3 Theorem. Suppose that m = <g) is a cyclic group and K is a field. a) Any indecomposable Km-module is an epimorphic image of Km. b) Suppose that Iml = pn and char K = p. (1) For 1 ::;; i ::;; pn, Km has precisely one Km-submodule of codimen

sion i, namely J(Km)i. J(Km)i is the Km-submodule generated by (g - 1)i. (2) For 1 ::;; i ::;; pn, put Vi = Km/J(Km)i. Then Vi is an indecom

posable Km-module of dimension i, and Vi has a K-basis {v 1 , ... , Vi} such that

(3) Every indecomposable Km-module is isomorphic to Vdor some i.

Proof a) Let V be any indecomposable K(fj-module, and let (J be the linear transformation v ~ vg of V. Let t be an indeterminate over K. Then V becomes a finitely generated K[tJ-module if we put

§ 5. The Number oflndecomposable K(fj-Modules 67

vf(t) = vf(a) (v E V,fE K[tJ).

Since K[tJ is a principal ideal domain, V is the direct sum of cyclic K[tJ-submodules (I, 13.10). But a K[tJ-submodule is the same as a KG)submodule. Since V is indecomposable, it follows that V is a cyclic KG)-module. Thus there is an epimorphism of KG) onto V.

b) There is an K-algebra epimorphism IJ.. of K[tJ onto KG) such that tlJ.. = g, and

kerr:t = (t p" - 1)K[tJ = (t - 1)p"K[t]'

Hence

KG) ~ K[tJj(t - !)p"K[t].

The submodules of KG) correspond to the ideals :5 of K[tJ for which

:5 ;;2 (t - 1)P"K[t].

As K[tJ is a principal ideal domain, :5 = fK[tJ, where II(t - I)P". Hence:5 = (t - 1)iK[tJ for some i (0 ~ i ~ pO) and

By 5.2, the radical J(KG)) of KG) is spanned by the elements

gi _ 1 = (g _ 1)(1 + g + ... + gi-l).

Hence

J(KG)) = (g - 1) KG).

Thus the submodules of KG) are the

and

This proves (1). By 5.2c), Vi = KG)/J(KG))i is indecomposable. If we put


then {v I, ... , Vi} is a K-basis of Vi and

Thus (2) is proved. Suppose that V is any indecomposable Kffi-module. By a), V ~

Kffi/W for some Kffi-submodule W. By (1), W = (g - 1)i Kffi for some i, hence

q.e.d.

5.4 Theorem (D. G. HIGMAN [3], KASCH, KNESER, KUPISCH [1]). Suppose that K is afield of characteristic p and that '-l3 is a Sylow p-subgroup offfi.

a) If'-l3 is not cyclic, there exist indecomposable Kffi-modules of arbitrarily large K-dimension. In particular, there exist infinitely many isomorphism types of indecomposable Kffi-modules.

b) If '-l3 is cyclic, there exist at most I ffil non-isomorphic indecomposable Kffi-modules. Any indecomposable Kffi-module is an epimorphic image ofKffi.

Proof a) Since '-l3 is not cyclic, it follows from 5.1 that for each natural number k there exists an indecomposable K'-l3-module Vk of dimension 2k + 1. By 4.3a), Vk is isomorphic to a direct summand of (Vk(f»)'$. Thus, by the Krull-Schmidt theorem, Vk is isomorphic to a direct summand of W'$ for some indecomposable direct summand W of Vk(f). Hence

b) Now suppose that '-l3 is cyclic. By 5.3, there are 1'-l31 indecomposable K'-l3-modules VI' V2 , ... , where dimK Vi = i, and every indecomposable K'-l3-module is isomorphic to one of them. We show that any indecomposable Kffi-module V is isomorphic to a direct summand of~(f) for some j satisfying j ::;; dimK V. Indeed, Iffi: '-l31 has an inverse in K. Hence by 4.3c), V is isomorphic to a direct summand of (V'$t'. Now Vi! = E8i Ui for certain indecomposable K'-l3-modules Ub so by the Krull-Schmidt theorem, V is isomorphic to a direct summand of Ui(f) for some i and dimK V ~ dimK Ui. Thus V is isomorphic to a direct summand of some Vj(f), where j ::;; dimK V. The number of non-isomorphic indecomposable direct summands of Vj(f) of dimension at least j is at most Iffi: '-l31, since dimKVj(f) = Iffi: '-l3lj· Hence there are at most Iffil non-isomorphic indecomposable Kffi-modules.

§ 5. The Number ofIndecomposable KIfi-Modules 69

By 5.3, Vj is an epimorphic image of K'l3. By 4.2, .ffi is an exact functor. Thus Vjffi is an epimorphic image of (K'l3)ffi ~ Kffi. Thus any direct summand of Vjffi is an epimorphic image of Kffi. q.e.d.

Theorem 5.4 suggests that the theory of representations over a field of characteristic p is particularly simple if the Sylow p-subgroups of ffi are cyclic. For the case in which p2 does not divide 1 ffi I, Brauer developed a deep and rich theory in 1941 (R. BRAUER [3]). This theory was extended only much later to the case of cyclic Sylow p-subgroups (DADE [2]).

We look more closely into a special case which we shall need later.

5.5 Theorem (SRINIVASAN [1]). Let ffi be a group with an Abelian pcomplement ~ and let 'l3 be a Sylow p-subgroup of ffi. Let K be an algebraically closed field of characteristic p.

a) Ifl~1 = k and

k

K~ = EBKei' i=1

where ei' ... , ek are mutually orthogonal idempotents, then

k

Kffi = EBP;, i=1

where P; = (Ket· Also Pi is an indecomposable Kffi-module and (Pi)'ll is isomorphic to K'l3.

b) Suppose that also 'l3 <l ffi. Then each Pi has exactly one maximal submodule Mi, and dimK P;/Mi = 1. If

then P;/Mi is a K~-module with character lJ.i. In particular, Pi if:. FJ for i i= j.

c) If'l3 is cyclic and normal in ffi, thenfor 1 ::; i ::; 1'l31, ffi has exactly Iffi/'l31 non-isomorphic indecomposable Kffi-modules of K-dimension i. These are already indecomposable as K'l3-modules, and they are all the indecomposable Kffi-modules to within isomorphism.

Proof a) Since the order of ~ is not divisible by p and K is algebraically closed, there exist mutually orthogonal idempotents ei (i = 1, ... , k) in K~ such that


k

1 = e1 + ... + ek and KSl = EB Kei· i=l

Hence

k

KG> ~ (KSl)(jj = EB~, i=l

where ~ = (Ke;)(jj. As ~ is a transversal of Sl in G>, we have

(Kei)(jj = EB Kei (8) y = ei ® K~. YE'll

This shows that (~)'ll ~ K~. Hence (~)'ll is indecomposable, by 5.2c). Thus ~ is indecomposable.

b) By 5.2b),

Hence for every x E Sl and Z E J(K~),

(ei ® z)x = eix ® x-1 ZX = cti(x)ei ® x-1 ZX E ei ® J(K~).

Thus Mi = ei ® J(K~) is a K(fj-submodule of ~ of codimension 1. In particular, for x E Sl,

(ei ® l)x = cti(x)(ei ® 1)

and

~ = Mi Ee K(ei ® 1).

By a) and 5.2b), Mi is the only maximal submodule of~. c) For 1 ~ j ~ I~I we consider the KG>-modules

Then dimK Vij = j by 5.3b). Since (Vij)'ll is indecomposable, certainly all Vij are indecomposable K(fj-modules. The representation of Sl on the head (ei ® K~)/(ei ® J(K~)) of Vij belongs to the character cti. Hence

§ 6. Indecomposable and Absolutely Indecomposable Modules 71

Vij 1= Vi'i' if (i,j) # (i/,j'). Thus the Vij (i = 1, ... , /Sl/;j = 1, ... , /~/) are ISlI I~I = 1(f)1 indecomposable non-isomorphic K(f)-modules. By 5Ab) there are no other isomorphism types of indecomposable K(f)modules. q.e.d.

5.6 Remarks. a) Bya theorem of ROITER [1], Theorem 5.1 can be strengthened as follows. Let K be an infinite field of characteristic p and (f) a non-cyclic p-group. Then for infinitely many natural numbers n there are infinitely many types of indecomposable K (f)-modules of dimension n.

b) For the elementary Abelian group of order 4, all indecomposable modules over a field of characteristic 2 have been determined (BASEV [1]; HELLER, REINER [1]). The same has been done for dihedral 2-groups (RINGEL [1]).

c) In contrast to b), the class of indecomposable K(f)-modules for an elementary Abelian group (f) of type (p, p) for p > 2 is extremely large. BRENNER [1] has proved the following result.

Let ~ be any local algebra of finite dimension (cf. 6.1). There exists an indecomposable K(f)-module V such that 6 = HomK(D(V, V) has a nilpotent 2-sided ideal :J such that 6j:J ;;;: ~.

Exercises

8) Let K be a field of characteristic p and (f) an elementary Abelian group of type (p, p). Show that the K(f)-modules mentioned at the beginning of § 5 are indecomposable and pairwise non-isomorphic.

§ 6. Indecomposable and Absolutely Indecomposable Modules

6.1 Definition. a) Let 9i be a ring (with identity element). The units of 9i are those elements in 9i that have left and right inverses in 9i. (On account of the associativity of the multiplication in 9i, both inverses coincide.) If the non-units of 9i form a two-sided ideal of 9i, then 9i is called a local ring.

b) For any ring 9i (with identity element) we define the Jacobson radical J(9i) of 9i as the intersection of the kernels of all irreducible 9i-modules. As in V, 2.2, J(9i) is the intersection of all maximal right ideals of 9i. (We remark that J(9i) in general need not be nilpotent.)


6.2 Lemma. a) If 91. is a local ring, the Jacobson radical J(91.) of 91. is the set of all non-units of 91. and 91.jJ(91.) is a division ring.

b) 1f91.jJ(91.) is a division ring, then 91. is a local ring.

Proof a) We denote by ,3 the two-sided ideal of non-units of the local ring 91.. If m is any maximal right ideal of 91., then m contains only non-units of 91.. Hence m £; ,3, so m = ,3. Thus ,3 is the only maximal right ideal of 91. and ,3 = J(91.).

If a E 91. - J(91.), then a is a unit of 91. and thus has a right inverse in 91.. Then certainly a + J(91.) has a right inverse in 91.jJ(91.). Thus 91.jJ(91.) is a division ring.

b) Suppose now that 91.jJ(91.) is a division ring. We show first that if a has no right inverse in 91., then a E J(91.).

a91. =F 91., so a91. lies in a maximal right ideal m of 91.. But also J(91.) £; m, so a91. + J(91.) £; m. Hence (a91. + J(91.))jJ(91.) is a proper right ideal of the division ring 91.jJ(91.). This forces (a91. + J(91.))jJ(91.) = 0, hence a E J(91.).

Now let b be a non-unit in 91. and suppose b has a right inverse a. If a has a right inverse c, then b = c. But this is impossible as b is not a unit. Hence a has no right inverse and so a E J(91.). But then

1 = ba E J(91.),

a contradiction. Hence b has no right inverse and thus lies in J(91.). Thus J(91.) is the set of non-units of 91., and 9t is a local ring. q.e.d.

6.3 Lemma. 1f9t is a local ring, 0 and I are the only idempotents in 91..

Proof Let 9t be a local ring and e an idempotent of 91.. If e is a unit, then e = 1, for if ea = 1, then

e = e(ea) = e2a = ea = 1.

Similarly, if the idempotent 1 - e is a unit, then e = 0. Thus if e =F ° and e =F 1, then e and 1 - e are non-units. Thus by 6.2a), there follows the contradiction

1 = e + (1 - e) E J(9t). q.e.d.

6.4 Theorem (FITTING). Suppose that 9t is a ring and V an 91.-module with maximal and minimal condition for submodules. The following assertions are equivalent.


a) V is an indecomposable 9{-module. b) 6 = Hom9l(V, V) is a local ring.

Proof a) ::::. b): Suppose IY. E 6. By I, 10.7 either IY. is an automorphism of V or IY. is nilpotent. We show that the non-units of 6 form a two-sided ideal in 6.

If IY. and f3 are non-units of 6, then IY. + f3 is a non-unit by I, 10.10. If IY. is a non-unit of 6, then IY. is nilpotent, so ker IY. ::J 0 and im IY. C V. Hence for any y E 6,

ker (lY.y) ;2 ker IY. ::J 0

and

im(ylY.) £; imlY. C V.

Thus lY.y and ylY. are non-units of 6, and 6 is a local ring. b) ::::. a): If 6 is a local ring, then by 6.3, 0 and 1 are the only idem

potents in 6. But as every direct decomposition V = V1 EB V2 of V gives rise to an idempotent 1t in 6 such that im 1t = V1 and ker 1t = V2 ,

V has to be indecomposable. q.e.d.

The only idem po tents of the ring Z of rational integers are 0 and 1, but Z is not a local ring. We show that this cannot happen for rings with minimum condition for right ideals.

6.5 Theorem. Let 9\ be a ring satisfying the minimum condition for right ideals. If 0 and 1 are the only idempotents in 9\, then 9\ is a local ring.

Proof Let 6 = Hom 91 (9\, 9\). Since the elements of 6 are precisely the left multiplications by elements of 9\, 6 is antiisomorphic to 9\. Thus by hypothesis, 0 and 1 are the only idempotents in 6. Hence 9\ is an indecomposable 9\-module. But by a well-known theorem of Hopkins (see KAPLANSKY [1], p. 134),9\ also satisfies the maximum condition for right ideals. Hence by 6.4, 6 is a local ring. Thus 9\ is a local ring. q.e.d.

Next we introduce the notion of absolutely indecomposable modules. The definition is similar to 2.1.

6.6 Definition. A Kijj-module V is called absolutely indecomposable if

which means that the K-algebra HomK4fi(V, V)/J(HomK4fi(V, V)) is isomorphic to K.

6.7 Lemma. Let V be a K{f}-module. a) If V is absolutely indecomposable, then for any extension L of K,

V L is an absolutely indecomposable L{f}-module. b) If K is algebraically closed and V is an indecomposable K{f}-module,

then V is absolutely indecomposable.

Proof We write 6 = Hom K4fi (V, V). By 1.12, we have

By hypothesis, 6jJ(6) ~ K. Hence by 1.2c),

By 1.2d), J(6)L s; J(6d. As 6 L is a ring with identity element, it follows that J(6)L = J(6d and 6 LjJ(6d ~ L. Hence 6 L is a local ring by 6.2b), and VL is indecomposable by 6.4.

b) As V is indecomposable, Hom K4fi (V, V) is a local ring, by 6.4. Then by 6.2a),

is a division algebra of finite dimension over K. As K is algebraically closed, it follows that

and V is absolutely indecomposable. q.e.d.

It is natural to conjecture by analogy with 2.2 that the indecomposability of VL for every extension L of K implies the absolute indecomposability ofV. But this is not true in general, as is shown by the following example.

6.8 Example (GREEN). Let {f} = <gl) x <g2) be an elementary Abelian group of order p2.

a) Let K be a field of characteristic p and let A be a non-singular n x n matrix such that the characteristic polynomial/of A is irreducible. Let V be a K{!;-module for the matrix representation


of (fj of degree 2n. Then HomKCfi(V, V)/J(HomKCfi(V, V)) is isomorphic to the field K[t]!fK[tJ.

To prove this, we first check that every matrix which commutes with the images of gland g 2 is of the form

where X A = AX. Thus there is an isomorphism a of HomKCfi(V, V) onto the algebra m: of all such matrices. Let

m = {XiX E (K)n, XA = AX},

and let /3 be the epimorphism of m: onto m given by

If 3 = ker a/3, then 3 is nilpotent, since

Also HomKCfi(V, V)/3 ~ m. Since the characteristic polynomial f of A is irreducible,fis also the minimum polynomial of A.

Let W be the space of dimension n on which A operates and o =1= WE W. Then (w, wA, ... , wAn-l) is an A-invariant subspace ofW. As the characteristic polynomial f of A is irreducible,

Suppose X E m and

W = (w, wA, ... , wAn- 1).

n-l wX = I AjwAj (Aj E K).

j=O

If we define the polynomial h by h = Ir::6 Ai, then

wX = wh(A).

Then

WAiX = wXAi = wh(A)Ai = wAih(A),

so X = h(A). Hence

and ~ ~ K[tJ/JK[t]. This is a field, since J is irreducible. Hence HomK(!;(V, V)/'J is a field and J(HomK(!;(V, V» = 'J. Therefore

is isomorphic to the field K[tJ/JK[tJ. b) Now let K = GF(p)(x) be the field of all rational functions in an

indeterminate x over GF(p) and let A be the p x p matrix

o 1 0 0

o 0 1 0

o 0 0 1

x 0 0 0

The characteristic polynomial t P - x of A is irreducible over K (see VAN DER WAERDEN [2J, vol. I, p. 184). Thus if V is the K<f>-module constructed in a), HomK(!;(V, V)/J(HomK(!;(V, V» is not isomorphic to K and V is not absolutely indecomposable.

Nevertheless, for every extension L of K,

is a local ring and hence VL is indecomposable, by 6.4. To show this, we write 6 = HomK(!;(V, V). Then by 1.2c) and 1.4a),

6 L/J(6)L ~ (6/J(6» ®K L ~ K[tJ/(t P - x)K[tJ ®K L

~ L[tJ/(tP - x)L[t].


By 1.2d), J(6)L £ J(6L).1f t P - x is irreducible in L[t], then 6 L/J(6)L is a field, hence J(6)L = J(6L), and 6 L is a local ring by6.2b).

If tP - x is reducible in L[t], then tP - x = (t - z)P for some z E L (see VAN DER WAERDEN, as above). Then

i = (t - z)L[t]/(tP - x)L[t]

is an ideal in L[t]/(tP - x)L[t] for which (L[t]/(tP - x)L[t])/i is isomorphic to Land iP = O. If 3/J(6)L is the corresponding ideal of 6 L/J(6)L' then 3 is a two-sided ideal for which 6 L/3 ~ Land 3P £ J(6)L' As J(6)L is nilpotent, 3 is nilpotent. This shows that 3 = J(6d and 6 L/J(6L) ~ L. SO 6 L is a local ring in this case also, and VL is indecomposable.

The fact that the full analogue of 2.2 does not hold is not very serious for our purpose, since it does hold for perfect fields.

6.9 Theorem. Let K be a perfectfield and V a K{f;-module. Then the following are equivalent.

a) V is absolutely indecomposable. b) For every extension L of K, VL is indecomposable. c) If K is the algebraic closure of K, VR is indecomposable.

Proof By 6.7a), a) implies b)., Trivially, b) implies c). To prove that c) implies a), suppose that VR is indecomposable and write 6 = HomK(fj(V, V). Thus 6 R ~ Homl(lfJ(VR, VR). As VR is indecomposable, certainly V itself is indecomposable, so 6 and 6 R are local rings. Thus 6;(/J(6R) is a division algebra of finite dimension over the algebraically closed field K and is thus isomorphic to K. Write 1) = 6/J(6). It is to be shown that the division ring 1) is isomorphic to K. We have

Let a be an element of J(1) ®K K). Then a is nilpotent and lies in 1) @K L for some finite extension L of K; thus a generates a nilpotent ideal in 1) ®K L. But since K is perfect, J(1) ®K L) = 0 (see VAN DER WAERDEN

[3], vol. II, p. 77). Hence a = 0, and J(1) ®K K) = O. Thus 6 R/J(6)R is a semisimple algebra and J(6R) = J(6k Thus 1) ®K K is isomorphic to 6R/J(6R) and hence to R.. Thus


and 'D is isomorphic to K. Hence V is absolutely indecomposable. q.e.d.

For a fixed KO}-module V, there exists a decomposition of VL into absolutely indecomposable LO}-modules in a finite extension L of K.

6.10 Theorem. Let V be a KO}-module. Then there exists afinite extension L of K such that

for absolutely indecomposable LO}-modules Vi'

Proof For any extension L of K, the number of summands in a direct decomposition of VL is bounded by dimK V. Hence we may choose a finite extension L of K such that

V L = VI EB ... EB Vs,

where the Vi are indecomposable and s is as large as possible. Then for every finite extension L' of L, (Vi)L' is indecomposable. Write 6 i = Homu!'>(Vj, VJ By extending L if necessary, we may suppose that for every i = 1, ... , s, dim LJ(6i) is as large as possible; this means that

for every finite extension L' of L. We have to show that dimL 6jJ(6i) = 1 for all i.

Since Vi is indecomposable, 6jJ(6 i) is a division algebra 'Di. Suppose that dimL 'Di i= 1. Denote by [ the algebraic closure of L. Now l)i ® L [ is not a division ring, so it has zero divisors. Suppose that xy = 0, where

m n ° i= x = L aj ® Ai' ° i= Y = L bk ® A~, j=l k=l

and aj' bk are in 'Di, Ai' A~ are in L Then Aj' A~ are algebraic over L, so they all lie in a finite extension L' of L. Thus x, yare zero divisors in 'Di ®L L'. As (Vi)L' is indecomposable,

is a local ring. But we have

J(6i) <8\ L' = J(6i <8\ L'),

and

is not a division algebra. This contradiction shows that dim L 1)i = 1, so Vi is absolutely indecomposable. q.e.d.

6.11 Example. Suppose that (fj = <gl) x <g2) is elementary Abelian of order p2 and that L is a field of characteristic p. For every a E L we define a representation (la of (fj by

(la is indecomposable, for even its restriction to <gl) in indecomposable.

In which subfield K of L can (la be realized? (la can be realized in K if and only if there exists a non-singular 2 x 2 matrix C = (cij) with coefficients cij in some extension of L such that we have

C ( )C 1 1( * -*d2)E(K)2 (la gl - = -d 2 C22

and

where d = det C. As C12 "# 0 or C22 "# 0, this forces a E K. Hence GF(p)(a) is the smallest field in which (la can be realized.

This observation leads to several remarks. a) If we choose K = GF(p) and a transcendental over K, then Pa

cannot be realized in the algebraic closure of K. b) Choose K = GF(p) and ai such that (K(ai) : K) = i (i = 1, 2, ... ).

Then all (la, can be realized in the algebraic closure of K, but not in any finite extension L of K.

c) Suppose that a is algebraic over K = GF(p) and L = K(a). Let Va = LVI EB LV2 be an L(fj-module for (la' with


We denote Va, regarded as a KQ)-module, by V~. Thus dimK Vao = 2(L: K). Suppose that {II' ... , Is} is a K-basis of Land

s

lia = L ai)j (aij E K). j;1

Thus A = (ai) is the matrix of the K-linear transformation I --+ la of L. Hence the minimum polynomial of A is the minimum polynomial f of a over K and is therefore irreducible. Also, since the degree of f is s = (L: K),f is also the characteristic polynomial of A. Now if Q~ is the matrix representation of Q) relative to the K-basis

ofV~, then

Hence by 6.8a),

Thus by 6.4, Vao is indecomposable. The characteristic roots of the matrix A are the s distinct conjugates

a = aI' ... , as of a over K. Hence there exists a matrix Z E (L)s such that Z-1 AZ = B, where B is the diagonal matrix (bijai)' If we write

then

Thus V~ Q9K L has an L-basis {w 1, ... , w" W'I, •.. , w~} such that

§ 7. Relative Projective and Relative Injective Modules 81

This shows that

where (Wi' W;) ~ Va is an absolutely indecomposable l(f>-module. By the remark, it is clear that for no proper subfield Ll of L the module V~ ®K Ll is a sum of absolutely indecomposable Ll ffi-modules. Hence in the situation of Theorem 6.10, no universal choice of L for all Kffimodules is possible.

Exercises

9) Let 9t be an Artinian ring. Show that the following statements are equivalent.

a) 9t is a local ring. b) Every element of 9t is either nilpotent or invertible.

10) Let V be a Kffi-module and R: the algebraic closure of K. Then the following statements are equivalent.

a) VL is indecomposable for every extension L of K. b) VR is indecomposable. c) VR is absolutely indecomposable.

11) Let ffi be an elementary Abelian group of order p2. Prove that the Kffi-module V of 5.1 is absolutely indecomposable.

§ 7. Relative Projective and Relative Injective Modules

As in § 4, A denotes a commutative ring with identity.

7.1 Definition. Suppose U ~ ffi. a) An Affi-module V is called (ffi, U)-projective, if whenever


is an exact sequence of A(f)-modules and (Trx)u is a direct summand of Uu, then Trx is a direct summand of U.

b) An A(f)-module V is called ((f), U)-injective, if whenever

is an exact sequence of A(f)-modules and (Vrx)u is a direct summand of Uu, then Vrx is a direct summand of U.

If A is a field, the ((f), I)-projective modules are just the projective A(f)-modules, and the ((f), I)-injective modules are the injective A(f)modules. More generally, an A(f)-module V is projective (injective) if and only if it is ((f), I)-projective (((f), I)-injective) and V is a projective (injective) A-module (see Exercise 14).

To study the ((f), U)-projective and ((f), U)-injective modules, we introduce Gaschiitz operators.

7.2 Definition. Suppose that T is a transversal of U in (f). If V is an A(f)module, we call Y a ((f), U)-Gaschiitz operator on V if Y E HomAU(VU, Vu) and

for all v E V. Obviously, this condition is independent of the choice of the transversal T.

7.3 Lemma. Suppose that U ::s; (f).

a) If the A(f)-modules Vi (i E I) all have ((f), U)-Gaschiitz operators, so does E8iEI Vi'

b) If V has a ((f), U)-Gaschiitz operator, so does any direct summand ofV.

c) If I (f) : UI has an inverse in A, any A(f)-module has a ((f), U)-Gaschiitz operator.

d) If W is an AU-module, then WID = W Q9AU A(f) has a ((f), U)Gaschutz operator.

e) The regular A(f)-module A(f) has a ((f), U)-Gaschutz operator.

Proof a) If Yi is a ((f), U)-Gaschiitz operator on Vi' we define Y on E8iE I Vi such that Yi is the restriction of Y to Vi (i E I). Then Y is obviously a ((f), U)Gaschiitz operator on E8iEI Vi'

b) Suppose that V = VI EB V2 , where VI and V2 are A(f)-modules,


and y is a (f), U)-Gaschiitz operator on V. Let 7ri (i = 1, 2) be the projection of V onto Vi defined by the decomposition V = VI EB Vz. Then y7ri induces an AU-homomorphism of Vi into Vi' and for all Vi E Vi we have

since 7ri is an A(f)-homomorphism. Thus y7ri is a (f), U)-Gaschiitz operator on Vi'

c) If V is an A(f)-module and y is defined by

for all V E V, then y is a (f), U)-Gaschiitz operator on V. d) Suppose that T is a transversal of U in (f), so chosen that 1 E T.

Then

wm = ffiW ® t. lET

We define y by

(2: WI ® t)y = WI Q9 1 (WI E W). lET

It is easily checked that y is an AU-homomorphism. For all wE W we have

(W ® t') 2: CIyt = 2: (W ® t'CI)yt = W ® t'. lET lET

Thus y is a (f), U)-Gaschiitz operator on Wm. e) We apply d) to

A(f) ~ AU Q9AU A(f) = (AU)m. q.e.d.

We shall make several applications of the following simple lemma.

7.4 Lemma. Suppose that V and Ware A(f)-modules, that U ::;; (f) and that T is a transversal of U in (f). Suppose that ex E HomAU(VU' Wu). If ex' is the mapping defined by

va' = L Velat (v E V), tET

then a' is independent of the choice ofT and a' E HomACD(V, W).

Proof Suppose that t' = utt for each t E T with Ut E U. Then

L vt,-lat' = L vt-lu;lautt = L vt-lau;lutt = va'. tET tET tET

Thus a' is independent of the choice of T. It follows that for all g E <D,

This shows that a' E HomACD(V, W). q.e.d.

7.5 Theorem (D. G. HIGMAN [4]). Suppose that U ::S; <D and that V is an A<D-module. The following assertions are equivalent.

a) V is (<D, U)-projective. b) V is isomorphic to a direct summand of(Vu)CD. c) There exists an AU-module W such that V is isomorphic to a

direct summand ofWCD. d) V has a (<D, U)-Gaschutz operator.

Proof a) = b): The mapping e of 4.3b) is an A<D-epimorphism of (Vut, onto V and (ker e)u is a direct summand of «VU)CD)U. By hypothesis, V is (<D, U)-projective, so ker e is a direct summand of (VU)CD. Thus

where V' ~ V. b) = c): Take W = Vu. c) = d): This follows at once from 7.3d) and 7.3b). d) = a): Suppose that y is a (<D, U)-Gaschiitz operator on V and

that a is an A<D-epimorphism of an A<D-module U onto V for which (ker a)u is a direct summand of Uu; say

Uu = (ker a)u Etl W,

where W is an AU-module. The restriction IX' of a to W is an AUisomorphism ofW onto Vu. Let fJ = a,-l. Thus fJa = Iv. Also


Let T be a transversal of U in m and define b by

ub = L uC1exYPt (u E U). feT

Then b E HomAIll(U, U), by Lemma 7.4, and

feT feT feT

This shows that U = Ub + ker ex. If ub E Ub n ker ex, then

ub = L ut-1exyPt = L uexC1ypt = O. feT feT

Thus U = Ub ffi ker ex. Hence V is (m, U)-projective. q.e.d.

The dual to 7.5 is the following.

7.6 Theorem (D. G.lflGMAN [4]). Suppose that U ~ m and that V is an Am-module. The following assertions are equivalent.

a) V is (m, U)-injective. b) V is isomorphic to a direct summand Of(Vu)lll. c) There exists an AU-module W such that V is isomorphic to a direct

summand of Will. d) V has a (m, U)-Gaschutz operator.

Proof a) ~ b): The mapping jJ. of 4.3c) is an Am-monomorphism of V into (Vu)lll, and (VjJ.)u is a direct summand of ((Vu)lll)u. By hypothesis, V is (m, U)-injective, so VjJ. is a direct summand of (Vu)lll.

b) ~ c) ~ d): As in 7.5. d) ~ a): Suppose that y is a (m, U)-Gaschiitz operator on V, that V is

an Am-submodule of the Am-module U and that Uu = Vu ffi W for some AU-submodule W. Let n be the projection of Uu onto Vu with kernel W, and define b E HomA (U, V) by

ub = L uC1nyt, feT

where T is a transversal of U in m. If v E V, then vb = v. Thus b is a projection of U onto V, and U = V ffi ker b. By 7.4, b is an Am-homo-


morphism, so ker J is an A(f)-submodule. Hence V is a (f), U)-injective A(f)-module. q.e.d.

7.7 Theorem (D. G. HIGMAN [4]). Suppose that U ::;; (f). a) The (f), U)-projective A(f)-modules are precisely the same as the

(f), U)-injective A(f)-modules. b) If I(f) : UI has an inverse in A, any A(f)-module is (f), U)-projective

and (f), U)-injective. c) Suppose that ~ is a Sylow p-subgroup of (f) and that ~ ::;; U ::;; (f).

Suppose further that A is either a field of characteristic p or a commutative local ring for which the residue class field is of characteristic p. Then any A(f)-module is (f), U)-projective and (f), U)-injective.

Proof a) This assertion follows from 7.5 and 7.6. b) This follows from 7.3c) and 7.5 or 7.6. c) This is a consequence of b). q.e.d.

Now we consider the important case where the ring A is a field, denoted as usual by K.

7.8 Theorem (GASCHUTZ [1]). Let V be a K(f)-module. Then the following assertions are equivalent.

a) V is projective. b) V is injective. c) V has a (f), 1)-Gaschiitz operator.

Proof This follows immediately from 7.5 and 7.6, as (f), 1)-projectivity (injectivity) for K(f)-modules means the same as projectivity (injectivity).

q.e.d.

7.9 Remarks. Those algebras 91 of finite dimension over K, for which 91 is an injective right 91-module, are called quasi-Frobenius algebras. (Such an algebra 91 is also an injective left 91-module.) For quasiFrobenius algebras the notions of projective and injective modules coincide. Conversely, it can be proved that if 91 is a ring for which every projective (injective) module is injective (projective), then 91 satisfies the minimum conditions for left and right ideals and is an injective 91-module.

Quasi-Frobenius algebras can also be characterized by the fact that the formation of annihilators induces a duality of the lattice of right ideals onto the lattice ofleft ideals. In § 11 we shall derive this important property for group-rings over fields from the fact that group-rings are symmetric algebras.

7.10 Lemma. Suppose that U ~ IB ~ (f). 'a) Suppose that Vi (i E I) is an A(f)-nwdule. Then E8iEI Vi is (f), IB)

projective if and only if all the Vi are (f), IB)-projective. b) If V is (f), IB)-projective, then V is also (f), IBg)-projective for all

g E (f). c) If V is (f), U)-projective, then V is also (f), IB)-projective. d) If V is (f), IB)-projective and V<J.l is (IB, U)-projective, then V is also

(f), U)-projective. e) If W is a (IB, U)-projective AlB-module, then any direct summand

ofW(fj is (f), U)-projective.

Proof a) This follows from 7.5 by using 7.3a), b). b) Suppose that T is a transversal of IB in (f). By 7.5, there exists a

(f), IB)-Gaschiitz operator y on V. Then P is a transversal of IBg in (f), and g-lyg is a (f), IBg)-Gaschiitz operator on V. By 7.5, V is (f), IBg)projective.

c) By 7.5, there exists an AU-module W such that V is a direct summand of W(fj. If we put W' = W <J.l, then W(f) ~ W,(fj. Thus V is a direct summand of W,(fj, and V is (f), IB)-projective by 7.5.

d) Suppose that

is an exact sequence of A(f)-modules, and suppose that (Ta)u is a direct summand of Uu. As V<J.l is (IS, U)-projective, (T1J()<J.l is a direct summand of U<J.l. As V is also (f), IB)-projective, it follows that T a is a direct summand of U. Thus V is (f), U)-projective.

e) By 7.5, there exists an AU-module V such that V<J.l ~ W EE> W' for some AlB-module W'. Thus

V(f) ~ W(fj EE> W,(fj.

Hence W(fj is (f), U)-projective. By a), so is any direct summand of W(fj. q.e.d.

7.11 Theorem. a) If V is a projective A(f)-module and U ~ (f), then Vu is a projective AU-module.

b) Suppose that V is an A (f)-module and that U is a subgroup of (f) for which I (f): UI has an inverse in A. If Vu is a (U, 1)-projective AU-module, then V is a (f), 1 )-projective A(f)-module.

Proof a) Since (A(f))u is a free AU-module, the restriction of any free Affi-module to U is a free AU-module. Thus the restrictions of projective A(f)-modules are projective AU-modules.

b) By 7.7b), V is (f), U)-projective. By hypothesis Vu is (U, I)-pro-jective. Hence by 7.lOd), V is also (f), I)-projective. q.e.d.

Before we consider the consequences of 7.11, we remark that the relative statement corresponding to 7. 11 a) is in general not true.

7.12 Example. Let (f) = <a, b, c> be the non-Abelian group of order p3 and exponent p for p > 2, where

aP = bP = cP = 1, [b, a] = c.

Let K be any field of characteristic p. We take U = <a>, 5D = <a, c>. We regard K as a module for the trivial representation of U. Then by 7.3d), KG; is (f), U)-projective. We have

p-l

KG; = EB KXij' i.j=O

where Xu = 1 ® bic j • Thus the module-structure of (KG;))2l is described by

(1)

and

If we write

Xuc = Xi.j+l (j modulo p)

p-l

Vi = EB KXij' j=O

we obtain the decomposition

as a direct sum of K5D-modules. To show that (KG;)!ll is not (5D, U)-projective, it suffices to show that V1 is not (5D, U)-projective, by 7.3. Suppose that y is a (5D, U)-Gaschiitz operator on V1 • Then

§ 7. Relative Projective and Relative lnjective Modules

p-l

(3) " -I I Xlj = L. XljC YC 1=0

for all j = 0, ... , p - 1. Suppose that

(4) p-l

xljY = L ajkXlk k=O

with ajk E K. As Y is a KU-homomorphism, we have

p-l

(xlja)y = XI,j+1 Y = L aj+l,kxlk k=O

p-l p-l

= (xljy)a = L ajkxI,k+1 = L aj,k-Ixlk' k=O k=O

This shows that

(5)

Now (3), (1) and (4) imply that

p-l p-l

xlj = L XI,j_IYC ' = L aj-I,kXlkC'

This shows that

1=0 k,I=O

p-l p-l

= L aj-I,kxl,k+1 = L aj-I,k-Ix!k' k,I=O k,I=O

p-l

tJjk = L aj-I,k-I' 1=0

But now (5) gives the contradiction

p-l

1 = L aj_I,j_1 = paOO = O. 1=0

Hence K(I; is «f), U)-projective, but (K(I;)!l is not (m, U)-projective.

89

7.13 Lemma. Suppose that U :::;; 58 :::;; G> and U <l G>. If V is a (G>, U)projective AG>-module, then V'1l is (58, U)-projective.

Proof As V is (G>, U)-projective, V is a direct summand of wm for some AU-module W, by 7.5. By Mackey's lemma, V'1l is a direct summand of

k k

(W(f))'1l = EB((W (8) gihtg'n'1l)'1l = EB((W @ gi)U)'1l i=l i=l

for some gi E G>. Thus V'1l is (58, U)-projective, by 7.3d) and b). q.e.d.

7.14 Theorem. Suppose that char K = p and that ~ is a Sylow p-subgroup of G>. Let V be a K G>-modu Ie. Then V is projective if and only if V'll is a projective K~-module.

Proof This follows trivially from 7.11. q.e.d.

We see from 7.14 that the projective K~-modules, where ~ is a p-group and K is a field of characteristic p, are particularly important. Fortunately these are easy to describe.

7.15 Theorem. Let ~ be a p-group and K any field of characteristic p. Then any finitely generated projective K~-module is free.

Proof Let P be a finitely generated projective K~-module. Then there exists a finitely generated free K~-module F such that F = P Ef> P'. Since P and P' are finite-dimensional vector spaces over K, there exist decompositions

P = EB Pi and P' = EB Pi, i j

where P; and ~ are indecomposable K~-modules. But by 5.2, K~ is an indecomposable K~-module. By the Krull-Schmidt theorem (I, 12.4), each Pi must be a free K~-module with one generator. Thus P is a free K~-module. q.e.d.

The assertion of Theorem 7.15 is also valid for not necessarily finitely generated projective modules. Indeed, Kaplansky has shown that any projective module over a local ring is free, and under the assumptions of Theorem 7.15, K~ is a local ring by 5.2.

We shall often use the following consequence of 7.15.

7.16 Corollary (DICKSON). Suppose that char K = p and that P is afinitely generated projective K(fj-module. If ~ is a Sylow p-subgroup of (fj, then I~I divides dimK P.

Proof By 7.11a), P'Il is a projective K~-module. By 7.15, P'Il is free. Hence I~I divides dimK P. q.e.d.

How does the process of inducing influence projectivity of modules?

7.17 Theorem. Suppose that U ::; (fj and that V is an AU-module. Then the induced module V<fJ is a (fj, 1)-projective A(fj-module if and only if V is a (U, 1)-projective AU-module.

Proof a) Suppose first that V<fJ is a (fj, 1)-projective A(fj-module. By 4.3 a), (V<fJ)u = VY/ EB V', where Viis an AU-module and Y/ is an AUmonomorphism. But (V<fJ)u is a (U,1)-projective AU-module by 7.13, hence VY/ is also (U, 1)-projective. As V is isomorphic to VY/, V is a (U, 1)projective AU-module.

b) Suppose now that V is a (U, 1)-projective AU-module. By 7.10e), V<fJ is (fj, 1)-projective. q.e.d.

Besides the induction process we occasionally consider the "inflation process". We prove the following as a further application of Gaschiitz operators.

7.18 Theorem. Suppose that 91 <I (fj and that V is a non-zero K(fjj91)module. Thus V becomes a K(fj-module, if vg (v E V, g E (fj) is defined to be v(91g). Then the following assertions are equivalent.

a) V is a projective K(fj-module. b) The characteristic of K does not divide 1911 and V is a projective

K(fjj91)-module.

Proof If f3 is any K-linear mapping of V into V, then

v L g-lf3g = v L (91gt 1f3(91g) = 1911v L h-1f3h gE<fJ gE<fJ hE<fJ/91

for all v E V. a) => b): Suppose first that V is a projective K(fj-module. By 7.8,

there exists a (fj, l)-Gaschiitz operator y on V. Thus

v = v L g-lyg = 1911v L h-1yh gE<fJ hE<fJ/91

for all v E V. Since V #- 0, it follows that char K does not divide 1911 and that y' = 191IY is a ((f)/91, l)-Gaschiitz operator on the K((f)/91)-module V.

b) = a): If char K does not divide 1911 and y' is a ((f)/91, l)-Gaschiitz operator on the K((f)/91)-module V, then from

v = v L h-1y'h = 1911-1v L g-ly'g, hEG;j91 gEG;

it follows that 1911-1y' is a ((f), l)-Gaschiitz operator on the K(f)-module ~ q~

7.19 Theorem. Let V, W be K(f)-modules. a) IfF is afree K(f)-module of rank r, then V ®K F is afree K(f)-module

of rank r(dimK V). b) Suppose that U :::; m :::; (f). If V is ((f), m)-projective and WID is

(m, U)-projective, then V ®K W is ((f), U)-projective. c) If W is projective, then V ®K W is also projective.

Proof a) We have F ~ F~, where Fo is the K-module of dimension r. Hence by 4.15a),

The assertion follows at once, since dimdV ®K Fo) = r(dimK V). b) Let S be a transversal of m in (f) and let T be a transversal of U

in m. Then TS is a transversal of U in (f). Let

be a ((f), m)-Gaschiitz operator on V and let

be a (m, U)-Gaschiitz operator on WID' Then

is a ((f), U)-Gaschiitz operator on V ®K W; for if v E V, W E W,

(v ® w) L s-lt-1(y ® <5)ts 5ES,tET


L vs-le l yts ® ws-lelbts 5ES,tET

L vs-lys ® ws-lt-lbtS, 5ES,tET

5ES,tET

= (L VS-lyS) ® w = v ® w. 5ES

c) This is an immediate consequence of b). q.e.d.

Before considering consequences for the radical of a group-ring, we make the following remark.

7.20 Theorem. Let U be a subgroup of ffi and K a field the characteristic of which does not divide I ffi : UI. Let V be a Kffi-module.

a) IfW is a Kffi-submodule of V such that Wu is a direct summand of vu , then W is a direct summand ofV.

b) IfVu is a completely reducible KU-module, then V is a completely reducible Kffi-module.

Proof a) By 7.7b), W is (ffi, U)-injective, hence by Definition 7.1b) W is a direct summand of V.

b) Now a) applies to every Kffi-submodule of V. Hence V is com-pletely reducible. q.e.d.

7.21 Theorem. Suppose that 91 is a normal subgroup offfi. a) J(K91) s; J(Kffi). b) (GREEN, STONEHEWER [1], VILLAMAYOR [1]). If char K does not

divide Iffi/91I, then J(Kffi) = J(K91)Kffi and

dimKJ(Kffi) = (dimKJ(K91))lffi/91 I·

c) (WILLEMS [1]). If J(Kffi) = J(K91)Kffi, then char K does not divide I ffi/91 I·

Proof a) Suppose that x E J(Km) and let V be an irreducible Kffi-module. By Clifford's theorem (V, 17.3a)), V<Jl is a completely reducible Kffimodule, say

where the Vi are irreducible Km-modules. Since x E J(Km), Vix = 0, so also Vx = 0. As this is the case for all irreducible Kffi-modules V, it follows that x E J(Kffi).

b) Suppose that x E J(Kffi).IfT is a transversal ofm in ffi containing 1, then since Kffi = EBtET(Km)t, we can write x in the form x = LtET xtt with uniquely determined elements X t of Km. We show that X t E J(Km).

To do this, let W be an irreducible Km-module. We form the induced Kffi-module

V = Wffi = EB W ® t. tET

If W E Wand U E 91, then

Thus W ® t is an irreducible Km-module and V<Jl is a completely reducible Km-module. Since char K does not divide /ffi/m/, it follows from 7.20 that V is also a completely reducible Kffi-module. Since x E J(Kffi), we conclude that Vx = 0. In particular, we see that for all WE W,

° = (W ® l)x = L WX t ® t. tE T

Thus WX t = 0, so WX t = 0. As this holds for all irreducible Km-modules W, we see that X t E J(Km).

It follows from this together with a) that

J(Kffi) = EBJ(Km)t = J(Km)(Kffi) tET

and

c) We consider K(ffi/m) as a Kffi-module. Take x E J(Km). As J(Km) is contained in the augmentation ideal


of K91, we have x = LUE9IaU(U - 1) for some au E K. Hence for all g E Gl,

(g91)x = L au(g(u - 1)91) = O. UE9I

Thus by our assumption

K(Gl/91)J(KGl) = K(Gl/91)J(K91)KGl = O.

So K(Gl/91) is a completely reducible KGl-module by 1.6. Hence K(Gl/91) is also a completely reducible K(Gl/91)-module. Thus char K does not divide IGl/91I. q.e.d.

7.22 Theorem. Suppose that Gl has a normal Sylow p-subgroup ~ and that char K = p. Then J(KGl) is spanned over K by all x - y with x, y E Gl and xy-t E ~. Also

Proof This follows immediately from 5.2 and 7.21. q.e.d.

Finally, we mention a fact about regular submodules.

7.23 Theorem. If V is a KGl-module, L is an extension of K and V L has an LGl-submodule isomorphic to LGl, then V has a KGl-submodule isomorphic to KGl.

Proof By 7.8, LGl is an injective LGl-module, so V L has a direct summand LGl-isomorphic to LGl = (KGl)L' By 1.21, V has a direct summand isomorphic to KGl. q.e.d.

Exercises

12) Suppose that Ut , ... , Um are subgroups of Gl and that the greatest common divisor of the orders IUd is 1. If V is an AGl-module which is (Gl, UJprojective for all i, then V is also (Gl, I)-projective.

13) Suppose that D is a Dedekind domain. Show that a finitely generated DGJ-module V is projective if and only if for every prime ideal V of D the localization V~ is a projective D9 GJ-module. (Hint: a) Show that if V~ is (GJ, i)-projective for all V, then V is (GJ, 1)projective.

b) Apply the following theorem: If V is a finitely generated Dmodule, then V is a projective D-module if and only if for every prime ideal V of D, Vp is a projective Dp-module. (See BOURBAKI [2], p. 138)).

14) Suppose that A is a commutative ring and that V is an AGJ-module. Then V is an injective AGJ-module if and only if V is (GJ, i)-injective and V is an injective A-module. And V is a projective AGJ-module if and only if V is (GJ, i)-projective and V is a projective A-module.

15) a) Prove 7.10c), d), e) by constructing Gaschiitz operators. b) Prove 7.11a) by constructing a (U, l)-Gaschiitz operator for Vu.

Why does a similar construction not work in the relative case?

16) Let V = V(n, q) be an n-dimensional vector space over the field K = GF(q) and let GJ = GL(n, q) act naturally on V. For which pairs (n, q) is V a projective KGJ-module?

17) Suppose that U ~ GJ and that V is a KU-module. a) If dimK V = 1, then Vm is a projective KGJ-module if and only if

char K does not divide lUI. b) If char K does not divide lUI and V is an irreducible KU-module,

then Vm is isomorphic to a direct summand of KGJ.

18) Suppose that 91 <J GJ and that K is an arbitrary field. a) If V is a projective KGJ-module, then

Vo = {VIVEV,vg = v forallgE91}

is a projective K(GJj91)-module, where the module structure of Vo is defined by v(91g) = vg for v E Vo and 9 E GJ.

b) If V is a projective KGJ-module and we put

then VjV1 is a projective K(GJj91)-module, where the module structure of V jV1 is defined by (v + V1)(91g) = vg + V1.

§ 8. The Dual Module 97

19) Suppose that P is a projective Kffi-module. a) If L is an extension of K, then PLis a projective Lffi-module. b) If Ko is a subfield of K such that (K: Ko) < 00, then the Koffi

module Po, obtained as in 1.16 from P, is projective. c) If in addition to b), K is a Galois extension of Ko, then for any

automorphism 11 of Kover Ko, the algebraic conjugate P" of P is projective.

§ 8. The Dual Module

In this section we collect some elementary results about the dual module, many of which will be used later on.

8.1 Definition. Suppose that K is an arbitrary field and V a Kffi-module. The K-vector space V* = HomK(V, K) dual to V becomes a Kffi-module if we put

v(fg) = (vg-l)f

for all v E V,JE V* and g E ffi. For obviously fg E V* andfl = f; also if v E V, f E V* and gl' g2 are in ffi, then

We give a more explicit description of the dual module.

8.2 Lemma. Let V be a Kffi-module with K-basis {Vl"'" vn}. Let {fl' ... ,fn} be the dual basis of V* .. thus vdj = Dij. If

then

n

Vig = L aij(g)Vj (g E ffi, aij(g) E K), j=l

n

hg = L aji(g-l}fj. j=l

(In terms of matrices, this shows that if (! is a matrix representation


corresponding to V, then the representation Q*, where Q*(g) = Q(g-Iy, corresponds to V*).

Proof By definition of V*,

This shows that

n

jjg = L akig-I)k q.e.d. k=i

In this section we want to develop a duality theory for K{f}-modules. We use the corresponding statements for K-modules without proof. These statements are either standard results of the duality theory of finite-dimensional vector spaces or follow easily from these.

8.3 Lemma. Suppose that V is afinitely generated K{f}-module. a) IfW is a K{f}-submodule of V, then

W"- = {Ilf E V*, wf = 0 for all w E W}

is a K{f}-submodule of V*. If Wz s; Wi s; V, where WI' Wz are K{f}submodules, (WdWz)* and Wt!wt are isomorphic K{f}-modules.

b) If V and Ware finitely generated K{f}-modules and rI. E HomKID(V, W), then the mapping rI.* of W* into V* defined by

v(frl.*) = (vrI.)f (v E V, f E W*)

is a K{f}-homomorphism. Ifrl. E HomKID(V, W) and f3 E HomKID(W, X), then (rl.f3)* = f3*rI.*. If

is an exact sequence of K{f}-modules, then also

P* 11* o -+ X* -+ W* -+ V* -+ 0


is exact. (* is a contravariant and exact functor on the category of finitely generated K(fJ-modules.)

c) Let! be the canonical isomorphism of V onto V** defined by

f(v!) = vf (v E V, f E V*).

Then! is a K(fJ-isomorphism of V onto V**. d) If V = Vl EB Vz is a decomposition of V into K(fJ-submodules Vl'

V1 , then V* = vt EB Vi is a decomposition of V* into K(fJ-submodules. We have vt ~ Vi and Vi ~ Vi-

e) V is irreducible if and only ifV* is irreducible. f) V is indecomposable if and only ifV* is indecomposable. g) If U is a K(fJ-submodule of V, [U, (fJ].L jU.L = CV*/u" «fJ) and

[U.L, (fJ] = W.L, where WjU = Cv/u«fJ).

Proof a) Let a be the restriction mapping of V* into W*. From

w«fg)a) = w(fg) = (wg-l)f = (wg-l)(fa) = w«fa)g)

for w E W,J E V* and 9 E (fJ, it follows that a is a K(fJ-homomorphism. Then a is an epimorphism of V* onto W* with kernel W.L. Thus W.L is a K(fJ-submodule of V*, and W* ~ V* jW.L.

Now (WtlWl)* is isomorphic to the K(fJ-submodule

{Ilf E Wi, wf = 0 for all WE W1 }

of Wi, and in the isomorphism Wi ~ V*jwt, this corresponds to Wt!W{

b) It is well-known that 0:* E HomK(W*, V*) and that the resulting functor is contravariant and exact. Also, IX* is a K(fJ-homomorphism, since

v«(fg)IX*) = (vlX)(fg) = «vlX)g-l)f = «vg-l)o:)f

= (Vg-l)(fIX*) = (v«(fo:*)g)

for all v E V, f E W* and 9 E (fJ. c) For v E V, f E V* and 9 E (fJ, we have

f«vr)g) = (fg-l)(V!) = v(fg-I) = (vg)f = f«vg)r).

This shows that r E HomK(!; (V, V**), and it is well-known that r is an isomorphism.


d) The direct decomposition V* = vt EB Vi of the K-space V* is well-known. By a), it is a direct decomposition ofV* into K(fj-submodules.

e) If V* is irreducible, then V is irreducible by a). If V is irreducible, then V** is irreducible by c), hence V* is irreducible by a).

f) This follows similarly from d) and c). g) If f E V*, f lies in [U, (fj]l. if and only if (u(g - l»f = 0 for all

u E U, g E (fj. Since (u(g - 1»f = u(f(g-t - 1», this condition is equivalent to f(g-t - 1) E Ul. for all g E (fj. Hence [U, (fj]l./Ul. = Cvw,(fj).

Applying this to the subspace Ul. of V*, we find that for v E V, f(vr) = 0 for all f E [Ul., (fj] if and only if f'((vr)(g - 1» = 0 for all f' E Ul.,g E (fj. Thus vf = o for allf E [Ul., (fj] ifand onlyifv(g - 1) E U for all g E (fj. Thus if W/U = Cv/u(fj),

W = {vlv E V, vf = 0 for all f E [Ul., (fj]}.

q.e.d.

8.4 Lemma. a) Let V be a finitely generated K(fj-module and let L be an extension of K. Then (V ®K L)* and V* ®K L are isomorphic L(fj-modules.

b) If Vt and V2 are finitely generated K(fj-modules, then (Vt ®K V2)* and Vt ®K Vt are isomorphic K(fj-modules.

Proof a) As is well-known, there exists an L-isomorphism a of V* ®K L onto (V ®K L)* such that

for all v E V, f E V* and It, 12 in L. For g E (fj,

(v ® It)((f ® 12)ga) = (v ® It)((fg ® 12)a) = lt/2(v(fg»

= ltI2((vg-t)f) = (vg- t (8) It)((f (8) 12)a)

= ((v (8) It)g-t)((f ® 12)a)

= (v ® It)(((f (8) 12)a)g).

Thus a is an L(fj-isomorphism. b) There exists a K-isomorphism f3 of Vt (8)K Vt onto (Vt ®K V2)*

such that for Vi E Vi and /; E Vi* (i = 1, 2)

For every g E (fj we have

(VI ® V2)((f1 ® f2)gfJ) = (VI ® V2)((flg ® f2g)fJ) = (V I (flg))(V2(f2g))

= ((V lg- l )fl)((V2g-l)f2)

= (VIg- I ® V2g- I )((f1 ® f2)fJ)

= (VI ® V2)((f1 ® f2)fJg)·

Hence fJ is a K<f;-isomorphism.

8.5 Lemma. Let V be a finitely generated K<f;-module. a) The mapping afor which

(V ® f)a = vf (v E V,f E V*)

q.e.d.

is a K<f;-epimorphism of V ®K V* onto K, where K is regarded as a module for the trivial representation of <f;.

b) Let {VI' ... , vn} be a K-basis of V and let {fl' ... , fn} be the dual basis of V*; thus vdj = Jij. Then

for all g E <f;. Hence V ®K V* contains a submodule Vo isomorphic to K.

c) If char K does not divide dimK V, then Vo is a direct summand of the K<f;-module V ®K V*.

d) If char K divides dim K V, then K appears as a composition factor in V ® K V* with multiplicity at least 2.

Proof a) We have for every V E V,j E V* and g E <f;,

(vg ® fg)a = (vg)(fg) = (vgg-l)f = vf = (vf)g.

Hence a is a K<f;-epimorphism of V ®K V* onto K. b) By 8.2,

n n

Vig = L aij(g)vj and j;g = L aji(g-l )jj. j=l j=!


Hence

n n

= I I aki(g-l)aij(g)vj ® fk j,k=l i=l

n n

= I bkjVj ® fk = I Vj ® jj. hk=l j=l

As {V 1, ... , Vn} is a K-basis of V and}; =F 0, we have L?=1 Vi ® }; =F 0. Hence

n

Vo = K I Vi ® }; i=l

is a Kffi-submodule isomorphic to K, c) In the notation of a) and b),

If char K does not divide n, we conclude that V ®K V* = ker II. $ Vo. d) If char K divides n, then (L;i'=1 Vi ® };)II. = O. From

and

the assertion follows. q.e.d.

8.6 Theorem. Suppose that V and Ware irreducible Kffi-modules, Then the following assertions, in which K is regarded as the module for the trivial representation of ffi, are equivalent.

a) V* ~ W. b) V ®K W has afactor module isomorphic to K. c) V ® K W has a submodule isomorphic to K.

Proof By 8.5, a) implies b) and c).


To show that b) implies a), suppose that rx is a Kffi-epimorphism of V (8) K W onto K. We define a mapping f3 of W into V* by putting

v(wf3) = (v ® w)rx (v E V, W E W).

Then for all g E ffi,

v((wf3)g) = (vg- 1)(wf3) (by the module structure ofV*)

= (vg- 1 (8) w)rx = (v (8) wg)g-lrx = (v ® wg)rxg-l

= (v ® wg)rx (since K is a trivial Kffi-module)

= v((wg)f3).

Thus f3 is a Kffi-homomorphism of W into V*. As Wand V* are irreducible and f3 :j:. 0, it follows that W ~ V*.

To prove that c) implies a), suppose that

is an exact sequence of Kffi-modules. By 8.3b), the sequence

(V ®K W)* -+ K* -+ 0

is also exact. By 8Ab), (V (8)K W)* ~ V* (8)K W*. Also K* ~ K, so since b) implies a), V** ~ W*. Hence V* ~ W by 8.3c). q.e.d.

We now give an example to show that for irreducible Km-modules V and W, the existence of a composition factor of V ®K W isomorphic to K does not imply the duality of V and W.

8.7 Example. Let ffi = SL(2, 5). We consider the Kffi-modules, where K is a field of characteristic 5. By 3.10, there are five irreducible Kffi-modules to within isomorphism, namely

Vi = KXi EB Kxi-ly EB ... EB Kyi (i = 0, 1,2,3,4).

Thus - I operates non-trivially on Vi and V3 , but trivially on Yo, V2

and V4• Hence every composition factor of V2 (8)K V4 is isomorphic to one of the modules Yo, V2 or V4 • If ni is the multiplicity of Vi as a composition factor of V2 (8)K V4, then

Also, if (Ii denotes the representation of ffi on Vi and

xM) = tr (];(g) for all g E ffi,

then

Applying this with

gives the congruences

no - nz + n4 == - 1 and no - n4 == 0 (mod 5).

It follows that no = n4 = 1 and nz = 3. Thus Vz <8\ V4 has a composition factor isomorphic to K, but Vz and V4 are not dual since they have different dimensions.

8.8 Lemma. Let V and W be finitely generated Kffi-modules. a) HomK(V, W) becomes a Kffi-module ifwe put

for all v E V, a E HomK (V, W) and g E ffi. Then

HomK(!;(V, W) = {ala E HomK(V, W), ag = afor all g E ffi}

is the set of all elements ofHomK(V, W) which are fixed under ffi. b) The mapping fJ ofV* ®K W into HomK(V, W) defined by putting

v((f ® w)fJ) = (vJ)w (v E V, f E V*, W E W)

is a Kffi-isomorphism ofV* ®K W onto the Kffi-module HomK(V, W).

Proof a) It is easy to check that IY.g E HomK(V, W), a1 = a and (ag1)gz = a(glgZ) for alIa E HomK(V, W) and gi E ffi. Thus HomK(V, W) is a Kffimodule. Also ag = a if and only if va = ((vg-1)a)g for all v E V, that is, (V()()g = (vg)()( for all v e V. Thus HomK(!;(V, W) is the set of elements of Hom K(V, W) left fixed by all elements of ffi.


b) It is well-known that fJ is a K-isomorphism of V* ®K W onto HomK(V, W). For v E V, f E V*, wE Wand 9 E G>, we have

v((f ® w)gfJ) = v((fg ® wg)fJ) = (v(fg))(wg) = ((vg-l)f)(wg)

and

v((f ® w)fJg) = ((vg-I)((f ® w)fJ))g = ((vg-I)f)(wg).

Thus fJ is a KG>-isomorphism. q.e.d.

8.9 Lemma. Let VI and V2 be finitely generated KG>-modules. We denote by B(VI' V2) the K-vector space ofK-bilinear forms on VI x V2 •

a) B(Vl' V2) becomes a KG>-module ifwe put

for Vi E Vi (i = 1,2), f E B(VI' V2) and 9 E G>. b) Vt ®K Vt and B(VI' V2) are isomorphic KG>-modules; infact, the

mapping y defined by

is a KG>-isomorphism.

Proof a) is easily checked. b) It is well-known that y is a K-isomorphism of Vt ®K Vt onto

B(Vl' V2). For Vi E Vi' fi E Vi* and 9 E G>, we have

(((fl ® f2)g)y)(VI' v2) = ((fIg ® f2g)y)(V l , v2)

= (VI (flg))(V2(f2g))

Thus y is a KG>-isomorphism.

= (V lg- l )fl(V2g- l )f2

= ((fl ® f2)y)(V lg- l, V2g- l )

= (((fl ® f2)y)g)(VI, V2)'

8.10 Lemma. Let VI' V2 be finitely generated KG>-modules.

q.e.d.

a) Suppose that r:t. E Hom K(V2 , Vi), f is a K-bilinear form on VI x V2

and


Jor all Vi E Vi' Then rx is a K{f)-homomorphism if and only if J is {f)-invariant; (that is, J(V1g, V2g) = J(V1' v2)Jor all Vi E Vi and all g E (f)).

b) Vz ~ Vt if and only if there exists a non-singular (f)-invariant bilinear Jorm on VI x V2 •

Proof a) This follows from

J(V1g, vzg) - J(v1, V2) = (V1g)(V2grx) - (V1gg- 1)(V2 rx)

= (V1g)(V2grx - vzrxg).

b) It is clear that in a), rx is an isomorphism if and only if J is non-singular. The assertion thus follows from a). q.e.d.

8.11 Theorem (Gow). Suppose that char K i= 2 and that V is an indecomposable K{f)-module Jor which V ~ V*.

a) There exists a non-singular symmetric or symplectic (f)-invariant Jormon V.

b) [JV is absolutely indecomposable, there cannot exist both symmetric and symplectic non-singular (f)-invariant bilinear Jorms on V.

Proof Let 6 = HomK(!;(V, V) and 3 = J(6). a) By hypothesis, there exists a K(f)-isomorphism rx of V onto V*.

Let rx* be the dual linear mapping of V** = V onto V*, defined by

w(vrx) = v(wrx*)

for all v, wE V. By 8.3b), c) rx* is a K(f)-isomorphism. Put f3 = rx*rx-1. Then f3 E 6 and

w(vrx) = v(wf3rx)

for all v, W E V. For B = ± 1, a bilinear form Ie on V is defined by putting

Ie(v, w) = v(wrx) + BW(Vrx).

By 8.10, Ie is (f)-invariant, J1 is symmetric and J-1 is symplectic. Suppose that J1' J-1 are both singular. Since

Je(v, w) = v«w t BWf3)rx),


it follows that 1v + [3 and 1v - [3 are both singular. They are therefore non-units of the local ring 6 and thus lie in ,3. Hence 2· 1v E ,3 and then 1 vE,3 since char K "# 2. This is a contradiction. So either fl or f-l is non-singular.

b) Suppose that fl' f2 are non-singular (f)-invariant bilinear forms on V, where fl is symmetric and f2 is symplectic. Then there exist Kisomorphisms IXl' a2 of V onto V* such that

Thus ·a1 , a2 are K(f)-isomorphisms, IXi = al and IX~ = -IX2. Put [3 = a1 a21, then [3 E 6. From a1 = [3a 2 we obtain IXi = a~[3*, so IXl = - a2 [3*· Thus

Since V is absolutely indecomposable, 6/,3 ;;; K, so [3 = a1 v + y for some a E K and y E,3. Then [3* = a1v* + y* and by 8.3b), y* E HomK(!;(V*, V*). Thus a1 y*a11 E 6. Since y is singular, so is y*, whence a1 y*al1 E ,3. Thus

1 - [3 - [3* -1 - 1 + *-1 -a v - Y - - - a1 IXl - a v IX 1Y IXl

and hence 2alv E,3. Since char K i= 2, this implies that a = O. Thus [3 E ,3, [3 is singular and IXl = [3IX2 is not an isomorphism, a contradiction.

q.e.d.

Theorem 8.11 is wrong for fields of characteristic 2. Let (f) = SL(2, 5) and let K be an algebraically closed field of characteristic 2. Then there exists an indecomposable projective K(f)-module P ;;; p* with two composition factors which are both isomorphic to an irreducible module of dimension 4; also P does not carry a (f)-invariant non-singular symmetric form, and, since char K = 2, this implies that P carries no non-singular symplectic form (WILLEMS [2J).

8.12 Theorem. Suppose that V is an absolutely irreducible K(f)-module. If V ;;; V*, then to within a scalar multiple, there exists only one non-zero (f)-invariant bilinear form f on V, and this form is non-singular. If char K is different from 2, f is symmetric or symplectic.

Proof We denote by Bo(V) the space of all (f)-invariant bilinear forms on V; thus f E Bo(V) if and only if


for all Vi E V and all 9 E (f). By 8.8b) and 8.9b),

HomK (V, V) ~ V* @K V ~ V* Q9K V* ~ B(V, V).

Hence by 8.8a), HomK(li(V, V) ~ Bo(V). Since V is absolutely irreducible, it follows that

The existence of a (f)-invariant non-singular form follows from 8.10, and that the form is symmetric or symplectic follows from 8.11a). q.e.d.

A corresponding result can be proved in the case of characteristic 2.

8.13 Theorem (FONG [2]). Let K be a perfect field of characteristic 2 and let V be an irreducible K(f)-module. If V ~ V* and V is not a module for the trivial representation of (f), then there exists a non-singular (f)-invariant symplectic form on V. In particular, dimK V is even.

Proof If dimK V = 1, say V = Kv, then V ~ V* implies vg = vg- i for all 9 E (f). Since vg = av for some a E K x, we conclude that a2 = 1, hence a = 1 as char K = 2. Hence dimK V ~ 2.

By 8.10, there exists a non-singular (f)-invariant bilinear form f on V. Put q(v) = f(v, v). Thus

where

We distinguish two cases.

Case 1. Suppose that

for all Vi E V. Choose linearly independent elements Vi' V2 in V, as we may since dim K V ~ 2. If q(v i ) i= 0, there exists an a in the perfect field


K such that

Thus the set

U = {v/v E V, q(v) = O}

is a non-zero (f)-invariant subspace of V. Since V is irreducible, U = V. Thus

o = q(v) = f(v, v)

for all v E V, andfis symplectic.

Case 2. If the hypothesis of case 1 does not hold, [,] is a non-zero (f)invariant bilinear form on V. Since char K = 2, [,] is symplectic. Let R be the radical of [,], defined by

R = {V/VEV,[V,W] = 0 forallwEV}.

Since [,] is non-zero, R i= V. But R is (f)-invariant and V is irreducible, so R = 0 and [ , ] is non-singular. q.e.d.

8.14 Defmition. Let V be a vector space of finite dimension over the field K.

a) A quadratic form q on V is a mapping of V into K such that

for all v, V1, V2 in V and k E K, where f is a bilinear form on V. b) A quadratic form q on V is called non-singular if the corresponding

bilinear form f is non-singular. c) Let V be a K(f)-module. A quadratic form q on V is called (f)

invariant if q(vg) = q(v) for all v E V, g E (f). (In this case, the corresponding bilinear form f is obviously also (f)invariant.)

8.15 Remark. a) If char K i= 2 and q is a quadratic form on V, then the bilinear form f defined by

is obviously symmetric and f(v, v) = 2q(v). Hence we can forget about q and work only with the bilinear form f

But if char K = 2, then f(v, v) = ° for all v E V, hence f is symplectic.

b) If char K =j:. 2 and V is an absolutely irreducible self-dual K(f)module, then V carries a (f)-invariant bilinear form, which is either symmetric or symplectic (8.12). If K is a perfect field of characteristic 2 and V =j:. K, then by 8.13 V carries a non-singular (f)-invariant symplectic form. We can ask whether V carries in addition a (f)-invariant quadratic form.

8.16 Defmition. Let K be a field of characteristic 2 and V a K(f)-module. a) The K-linear mapping a defined by

(v ® w)a = w ® v

is a K(f)-isomorphism of V ®K V. b) We put

S(V) = {tit E V ®K V, ta = t}

and

A(V) = {t(a - 1)lt E V ®K V} = im(a - 1).

Then S(V) and A(V) are K(f)-submodules of V ®K V. As

(a - 1V = a2 - 1 = 0,

we obtain A(V) s;: S(V). (If char K =j:. 2, then obviously V ®K V = S(V) EB A(V).)

8.17 Lenuna. Let K be a perfect field of characteristic 2 and V a finitely generated K(f)-module.

a) (V ®K V)/S(V) ~ A(V). b) S(V)jA(V) ~ V (2), where V(2) is the K(f)-module algebraically con

jugate to V under the automorphism a ~ a2 of the perfect field K. c) If V carries a (f)-invariant symplectic form f which is not identically

zero, then there exists a K(f)-epimorphism <5 of A(V) onto the trivial K(f)module K such that

(v ® w + w ® v)<5 = f(v, w).

Proof. a) () - 1 is a KG>-epimorphism of V ®K V onto A(V), and the kernel of () - 1 is obviously S(V).

b) Let {Vt, ... , vn} be a K-basis ofV. Then

A(V) = <Vi ® Vj + Vj ® v;ji,j = 1, ... , n; i =1= j)

and

S(V) = <Vi ® Vi' A(V)li = 1, ... , n),

where <T) denotes the vector space over K spanned by T. Hence

{Vi ® Vi + A(V)li = 1, .. . ,n}

is a K-basis of S(V)/ A(V). If

then

n

Vig = L aij(g)vj (aij(g) E K), j=t

(Vi ® Vi + A(V))g = Vig ® Vig + A(V)

n

= L aij(g)aik(g)Vj ® Vk + A(V) j,k=t

n .

= L aij(g)2vj ® Vj + A(V). j=t

This shows that S(V)j A(V) ;;;; V(2). c) The mapping () - 1 induces a KG>-isomorphism r of (V ®K V)/S(V)

onto A(V) such that

(V ® w + S(V))r = V ® w + w ® v.

The mapping ¢ E HomK(V ®K V, K) defined by

(V ® w)¢ = f(v, w)

is a KG>-homomorphism of V ®K V into the trivial KG>-module K, as f is G>-invariant. As f is symplectic, we obtain


(v ® w + w ® v)4> = f(v, w) + f(w, v) = 0 and

(v ® v)4> = f(v, v) = o.

This shows S(V) s;; ker 4>. Hence there exists a KG)-epimorphism ¢ of (V ®K V)/S(V) onto K such that

(v ® w + S(V))¢ = f(v, w).

If we define (j by (j = r- 1 ¢, then

(v ® w + w ® v)(j = (v ® w + S(V))¢ = f(v, w). q.e.d.

8.18 Lemma. Let K be afield of characteristic 2 and V a KG)-module. Let {VI' ... , vn} be a K-basis of V and q a quadratic form on V. We put qii = q(vi) and qij = f(v i, Vj) (i =/; j), where f is defined by

f(v, Vi) = q(v + Vi) - q(v) - q(v').

We define (J( E HomK(S(V), K) by

Then q is G)-invariant if and only if (J( E HomK(!;(S(V), K).

Proof. For v = Ii=1 XiVi (Xi E K), we have

q(v) = L XiXj%' i"'j

Now q is G)-invariant if and only if

(1)

(2)

q(Vig) = q(Vi) (i = 1, ... ,n), and

for all g E G). Suppose that

n

Vig = I aij(g)Vj. j=1

§ 8. The Dual Module

Then (1) is equivalent to

(1')

n

= L aij(g)2 qjj + L aij(g)aik(g)qjk' j=l j<k

Also (2) is equivalent to

(2') qij = f(vi' v) = f(vig, Vjg)

= L: aik(g)ajl(g)qkl (i =/; j). k;fl

On the other hand, r:J. is in HomK(l;(S(V), K) if and only if

and for i =/; j

qii = qiig = (Vi ® Vi)r:J.g = (Vig ® Vig)r:J.

= (~aij(g)aik(g)Vj ® Vk)r:J. },k

n

= L: aij(g)2qjj + L: aiig)aik(g)qjk j=l j<k

% = %g = (Vi ® Vj + Vj ® Vi)r:J.g

= (Vig ® Vjg + Vjg ® Vig)r:J.

= Laik(g)ajM)(Vk ® VI + VI ® Vk)r:J. k,1

= L: aik(g)ajl(g)qkl' k,.,

Hence the assertion follows.

113

q.e.d.

8.19 Theorem. Let K be a perfect field of characteristic 2 and V an irreducible K(fJ-module not isomorphic to the trivial K(fJ-module. Then thefollowing assertions are equivalent.

a) There exists a non-singular (fJ-invariant quadratic form on V. b) There exists a K(fJ-epimorphism ofS(V) onto the trivial K(fJ-module

K.

Proof Let {v 1, ... , vn} be a K-basis of V.

a) ~ b): Let q be a non-singular (}}-invariant quadratic form on V and f the corresponding symplectic form. We put

qii = q(Vi) and % = f(Vi' v) (i =f. j).

If we define a E HomK(S(V), K) by

then a E HomK(l;(S(V), K) by 8.18, and obviously a =f. O. b) ~ a): Suppose that 0 =f. a E HomK(l;(S(V), K). We put

By 8.18, this defines a (}}-invariant quadratic form q on V. We have to show that q is non-singular.

By 8.17b),

S(V)I A(V) ~ V(2) ;t K.

As V is irreducible, so is V(2). Thus A(V) ~ ker a and hence

for some i =f. j. The <f>-invariant symplectic form f on V is thus not the zero form. Therefore

W = {wlw E V,f(v, w) = 0 for all v E V}

is a proper K<f>-submodule of V. As V is irreducible, we obtain W = O. Hence f is non-singular, and q is a non-singular {}}-invariant quadratic ~oo~ ~~

8.20 Theorem (WILLEMS [2]). Let K be a perfect field of characteristic 2 and let V be an irreducible self-dual K<f>-module which is not isomorphic to the trivial module.

a) IfV does not carry a non-singular (}}-invariant quadratic form, then there exists an indecomposable K{}}-module M and a submodule N such that MIN is the trivial module and N ~ V(2). (By 10.5 this means that V(2) is a composition factor of P1 J(K{}})/P1 J(K{}})2, where P1 denotes the indecomposable projective K<f>-module for which PdPl J(K<f» ~ K.)

b) If{}} is soluble, then V carries a non-singular {}}-invariant quadratic form.

Proof a) As V ~ V* i K, by 8.13 there exists a non-singular symplectic GJ-invariant form on V. Thus by 8.17c), there exists a KGJ-epimorphism b of A(V) onto K. We put W = ker band U = S(V)/W. Then

U /(A(V)/W) ~ S(V)/ A(V) ~ V(2) (by 8.17b))

and

A(V)/W ~ K.

By hypothesis, V does not carry a GJ-invariant non-singular quadratic form, so the trivial module K is not an epimorphic image of S(V), by 8.19. Thus U is indecomposable. Then u* is also indecomposable and has a composition series

o eX c u*

such that U * /X ~ K and

b) Obviously, we can assume that V is a faithful KGJ-module. We put

f = IOz,(GJ)I-1 I g, gE02 ·((!;)

As Oz,(GJ) <J GJ, f is a central idempotent of KGJ. Suppose now that V does not carry a non-singular GJ-invariant

quadratic form. Then by a) there exists an indecomposable KGJ-module U* such that

As multiplication by f induces the identity on U*/X, certainly U*f '# O. As U* is indecomposable, U*f = U*. But then for every u E U* and every g E Oz,(GJ) we obtain

ug = (uf)g = u(fg) = uf = u.

So Oz,(GJ) operates trivially on U*, hence also on X ~ V(Z). As X is irreducible, the normal 2-subgroup 0Z'.2(GJ)l02,(GJ) of GJ/02 ,(GJ) operates trivially on X, by V, 5.17). But as (f) is soluble, Oz,.z«f) > 1.


This is a contradiction, for (f) operates faithfully on V and on X ~ V(2). Hence V carries a non-singular (f)-invariant quadratic form. q.e.d.

8.21 Application. a) Suppose that K = GF(21 ), (f) = Sp(2m, 21) and that V is the natural K(f)-module of dimension 2m over K. If (2m, 21) =1= (2, 2), then

for both types of the orthogonal groups O± (see II, 10.16d)). Thus there is no (f)-invariant quadratic form on V of the kind described in Theorem 8.20. Hence by 8.20a), there exists an indecomposable K(f)-module M such that MIN ~ K and N is an algebraic conjugate of V. Using Exercise 34 we see that Hl((f), N) =1= O. Thus also Hl((f), V) =1= O. (In fact, iHl((f), V)i = 21 (POLLATSEK [1]).

b) A similar argument applies for the natural module V = V(4, q) for the Suzuki group Sz(q) (see XI, § 3). There does exist on V a non-singular Sz(q)-invariant symplectic form (XI, 3.8). If there were an invariant quadratic form, we would obtain a monomorphism of the simple group Sz(q) into some orthogonal group 0(4, q) and indeed into PSO'(4, q). But if the quadratic form is equivalent to X 1X 2 + X3X4, then

PSO'(4, q) ~ PSL(2, q) x PSL(2, q);

and in the other case,

PSO'(4, q) ~ PSL(2, q2)

(see VAN DER WAERDEN [1], p. 20 and p. 26; cf. exerc. 33). Hence PSO'(4, q) has Abelian Sylow 2-subgroups, whereas Sz(q) does not.

By the same argument as in a), we obtain HI (Sz(q), V) =1= O.

8.22 Remark. Let K be a splitting field for (f). It follows easily from V, 13.7 that if char K = 0, the number of irreducible self-dual K(f)modules is equal to the number of real conjugacy classes in (f), that is, of classes fixed under formation of inverses. It can be proved by using Brauer characters that if char K = p, the number of isomorphism types of irreducible self-dual K(f)-modules is equal to the number of real p' -classes.

8.23 Lemma. a) K(f) ~ (KGj)*. b) If P is a finitely generated projective KGj-module, then p* is also

projective.

Proof a) We define a K-bilinear form f on KGJ x KGJ by the rule

Then f is non-singular and

f(ag, bg) = f(a, b)

for all a, b E KGJ, g E GJ. Hence by 8.10, KGJ ~ (KGJ)*. b) There exists a finitely generated free KGJ-module F such that

F = PI EB P2 , where PI ~ P. By 8.3d),

As F = KGJ EB ',;,' EB KGJ for some m, we find by repeated application of 8.3d) and by a) that

F* ~ (KGJ)* EB ',;,' EB (KGJ)* ~ KGJ EB . m' EB KGJ ~ F.

From F ~ P! EB PI* and PI ~ P, it follows that p* is projective. (We could also easily construct a (GJ, 1)-Gaschiitz operator for P*.) q.e.d.

8.24 Applications. a) Let (f; = SL(2, p) and, as in 3.10, let Vi be the K(f;module of homogeneous polynomials in two variables of degree i over K = GF(p). By 3.10, every absolutely irreducible KGJ-module is isomorphic to one of the Vi for ° ::; i ::; p - 1. By 8.3e) and 8.4a), Vi is absolutely irreducible of dimension 3, so Vi ~ V2 . It follows from 8.12 that if p is odd, there exists a non-singular GJ-invariant symmetric or symplectic form f on V2 • As dimK V2 = 3 is odd, f has to be symmetric. Hence we obtain a homomorphism of SL(2, p) into the orthogonal group 0(3, p) on V2 belonging to the formf The kernel of the representation of SL(2, p) on V2 is obviously < - I). So PSL(2, p) is isomorphic to a subgroup of 0(3, p). Comparing orders, we easily derive the isomorphism PSL(2, p) ~ 0(3, p)' of II, 10.11.

The same method works if p is replaced by pI (p odd). The proof that V2 ~ Vi requires additional arguments in this case, as SL(2, pI) has more than one absolutely irreducible module of dimension 3, namely the modules conjugate to V2 under automorphisms of the field GF(pI).

b) Let K = GF(q), where q is odd. Let V be a vector space of dimension 4 over K and let GJ be the group of all linear transformations of V onto itself of determinant 1; thus GJ = SL(4, q). Then the components

V(i) (i = 0, ... ,4) of the exterior algebra of V are K(f)-modules; in particular, V(4) is one-dimensional, and if V(4) = Kw, then wg = (det g)w = w for all g E (f). For WI' Wz E V(2) we have WI 1\ Wz = Wz 1\ WI' If we put

then f is a symmetric bilinear form on V(2). It is easy to see that f is non-singular. Then for all g E SL(4, q),

Hence f is invariant under SL(4, q). (Compare also Exercise 31). As the kernel of the representation of SL(4, q) on V(2) is precisely (- I), we obtain a monomorphism of SL(4, q)/( - I) into an orthogonal group 0(6, q). Now

and

where 8 = 1 or 8 = - 1. Hence 8 = 1 and

SL(4, q)/( - I) ~ 0(6, q)',

where the orthogonal group 0(6, q) belongs to a space of index 3. (Of course it may be seen directly that V(Z) has index 3.) Thus

PO(6, q)' ~ PSL(4, q),

the isomorphism connected with Plucker's geometry of lines.

We conclude this section with a lemma of "Clifford type" involving the K(f)-structure of HomK(V, W). It is related to V, 17.5.

8.25 Lemma. Suppose that 91 <J (f). Suppose that V and Ware K(f)modules, W'1I is absolutely irreducible and that

V'1I = WI Ei7 ... Ei7 Wr ,

where Wi ~ W'1I (i = 1, ... , r). Then HomK'1I(W'1I' V'1I) is a K(fj-submodule ofHomK(W, V) and

Proof. Hom K'1I(W'1I' V'1I) is a K(fj-submodule of HomK(W, V), since by 8.8 it is the set of elements of HomK(W, V) left fixed by all elements of the normal subgroup 91 of (fj.

We define a K-linear mapping <5 of W ®K HomK'1I(W'1I, V'1I) into V by putting

It is easily checked that <5 is well-defined. Let rxi be a K91-isomorphism of W'1I onto Wi' Then

Hence im <5 = V. From

r r

dim KHomK'1I(W'1I, V'1I) = dimKE8HomK'1I(W'1I, Wi) = dimKE8K = r, j=l j=l

we obtain

Hence <5 is an isomorphism. It is a K(fj-isomorphism, since if wE W, rx E HomK'1I (W'1I' VjJl) and g E (fj, then

((w ® rx)g)<5 = (wg ® rxg)<5 = (wg)(rxg) = ((wgg- 1)rx)g = (wrx)g

= ((w ® rx)<5)g. q.e.d.

Exercises

20) Suppose that V is a finitely generated K(fj-module. By 8.8, HomK (V, V) is a K(fj-module. Show that the trace mapping on HomK(V, V) is a K(fj-homomorphism of HomK(V, V) onto the trivial K(fj-module K. Compare this result with 8.5a).


21) Give a "natural" proof of 8.5b) using 8.8b).

22) Let P be a projective K(f)-module. Show that the dual module p* of P is projective by constructing a ((f), l)-Gaschiitz operator on P*.

23) Let K be a perfect field of characteristic 2 and V an absolutely irreducible K(f)-module. Show that the K-space of (f)-invariant quadratic forms on V has dimension at most l. (Hint: Let ql and q2 be (f)-invariant quadratic forms on V. We can assume that the corresponding symplectic forms, defined by

are the same. Show that there exists v E V, v =1= 0, such that q 1 (v) = q2(V) and calculate qi(va) for a E K(f).)

24) (WILLEMS) Let Bo be the K-space of (f)-invariant bilinear forms on K(f).

a) The forms fh (h E (f)) defined by

{1 if g2 = hg1 fh(g 1, g 2) = 0 otherwise

form a K-basis of Bo-b) If we define a multiplication * on Bo by

(fl * f2)(gl' g2) = L fl(gl' X)f2(X, g2)' XE(jj

then Bo becomes a K-algebra with unit element. c) The K-linear mapping r:x defined by gr:x = fg is an antiisomorphism

of K(f) onto Bo. d) Suppose a E K(f). Then ar:x is a non-singular bilinear form on K(f)

if and only if a is a unit of K(f). e) Suppose a = L9E(jj agg E K(f) and char K =1= 2. Then ar:x is sym

metric (symplectic) if and only if ag-1 = ag (ag- t = -ag) for all g E (f). What is the corresponding result if char K = 2?

f) Now suppose that char K = p and that p divides 1(f)1. Show that there exist "many" linearly independent non-singular (f)-invariant symmetric bilinear forms on K(f). If p =1= 2, there does not exist a nonsingular symplectic (f)-invariant form on K(f).


25) Let V be a finitely generated Km-module and Q a representation of m on V. If V ~ V* and 8 is an eigen-value of Q(g) (g E m) in some extension field of K, then 8-1 is also an eigen-value of Q(g).

26) Consider the n-dimensional vector space V = V(n, q) over K = GF(q) as a module for m = GL(n, q) in the natural way. Suppose that n>1.

a) V ~ V* as Km-modules if q > 2 or n > 2. b) Consider now V as a module for f> = SL(n, q). Then for any q,

V ~ V* only if n = 2.

27) Take K = GF(q) and L = GF(q"). Let the multiplicative group LX of L operate on the K-vector space L by x -+ ax. Show that L is a se1fdual LX-module if and only if q" = 2, 3.

28) Suppose that we have the same situation as in Exercise 27. Let U be a subgroup of LX and suppose that L is an irreducible KU-module. If L is a self-dual KU-module, then either n = 1 and lUI ~ 2 or n = 2m is even and lUi divides qm + 1.

29) Let V = V(2m, q) be a non-singular symplectic space. a) If 3 is a cyclic subgroup of the symplectic group Sp(2m, q) operat

ing irreducibly on V, then 131 divides qm + 1. b) Sp(2m, q) has a cyclic subgroup of order qm + 1 which operates

irreducibly on V. (See II, 9.23.)

30) Let V = V(n, q) be a vector space over GF(q) with a non-singular symmetric bilinear form. Suppose that q is odd and n > 1.

a) If the orthogonal group O(n, q) of isometries of V has a cyclic subgroup 3 which operates irreducibly on V, then n = 2m is even, 131 divides qm + 1 and V has index m - 1. (For the last statement observe that in the case of index m, V has exactly (qm - 1)(qm-1 - 1) vectors v '" 0 for which (v, v) = 0, and 3 operates fixed point freely on these vectors.)

b) If n = 2m and V has index m - 1, there is a cyclic subgroup 3 of O(2m, q) of order qm + 1 operating irreducibly on V. (See HUPPERT

[3]).

31) Let V be a Km-module with dimK V = n. Let Q be the representation of m on V and assume that det Q(g) = 1 for all gEm. Consider the homogeneous components V(i) of the exterior algebra of V as KG>-


modules and prove that V(i)* ~ v(n-i). (Construct a non-singular (f)invariant bilinear form on V(i) X v(n-i).)

32) Suppose that V is a K(f)-module and that L is an extension field of K. We denote by Bo(V) the space of (f)-invariant bilinear forms on V. Interpret and prove the formula BO(VL) ~ Bo(V) ®K L.

33) Prove the isomorphism PSL(2, q2) ~ PSO(4, q)', where 0(4, q) is the orthogonal group of a space of dimension 4 and index 1 over GF(q) (q odd), by the following argument.

Write K = GF(q), L = GF(q2) and let a be the only non-trivial automorphism of Lover K. Let V = V(2, q2) be the vector space of dimension 2 over L, on which (f) = SL(2, q2) operates naturally. We denote by V~ the L(f)-module conjugate to V under a and put W = V ®K V~.

a) As SL(2, q2) = Sp(2, q2), there is a (f)-invariant non-singular symplectic form on V. Construct from this a non-singular (f)-invariant symmetric bilinear form on W.

b) Prove Hom ujj (W, W) = L and show that W is an absolutely irreducible L(f)-module. (Observe first that W restricted to a Sylow p-subgroup of SL(2, q2) has only one irreducible submodule if char K = p.)

c) Show that there exists a K(f)-module Wo and a non-singular symmetric (f)-invariant bilinear form fo on Wo such that W ~ Wo ®K L.

d) Show that this gives a monomorphism of PSL(2, q2) into PSO(4, q)'. Show by comparing orders that this is an isomorphism and that the form fo on Wo has index 1.

34) Let W be a K(f)-module and V a submodule of W such that W/V ~ K is the trivial K(f)-module. Suppose W = V EB Kw as a vector space over K.

a) If the mappingffrom (f) to V is defined by

wg = w + f(g),

then f is a l-cocycle in Zl((f), V). b) f E Bl((f), V) if and only if V is a direct summand of W. c) Suppose in addition that V is irreducible. Then Hl((f), V) -# 0 if

and only if V is isomorphic to a composition factor of P1 J(K(f))jP1 J(K(f))2 where P1 is the projective K(f)-module characterized by PdPl J(K(f)) ~ K.

§ 9. Representations of Normal Subgroups 123

§ 9. Representations of Normal Subgroups

Clifford's theorem (V, 17.3) states that if 91 <l (fj, K is any field and V is an irreducible K(fj-module, the restriction V'lI is a completely reducible K91-module and its composition factors form a class of (f)-conjugate K91-modules. These results are very general; in particular, there is no restriction on the characteristic of K. In this section, we shall consider the extent to which the same is true for indecomposable K(fj-modules, and we shall attempt to extend the results of V, § 17 to the modular case. We begin with a counterexample which shows that if V is an indecomposable K(fj-module, the indecomposable components of V'lI may have different dimensions.

9.1 Example. Let (f) = <gl) x <g2) be an elementary Abelian group of order p2 and let K be a field of characteristic p. Let V be a 3-dimensional vector space over Kandlet {VI' V2, V3} bea K-basisofV. Then V becomes a K(fj-module if we put

An easy calculation shows that

Thus HomKm(V, V)jJ(HomKm(V, V)) is isomorphic to K, hence HomKm(V, V) is a local ring and V is absolutely indecomposable. If we put 91 = <g2), then

is a decomposition of V'lI into indecomposable K91-modules of different dimensions.

This example shows that there is no analogue of Clifford's theorem (V, 17.3) for indecomposable modules. But an analogue will be proved

under special conditions. Before doing so, we introduce the following notation, which will be used frequently in this section.

9.2 Definition. Suppose that 91 <J ffi. If V1 and V2 are K91-modules, V1

and V2 are said to be ffi-conjugate if there exists a K-isomorphism e of V1 onto V2 and an element g of ffi such that

(ve)x = (vxg)e

for all v E V1, X E 91. Thus if W is a Kffi-module and V a K91-submodule of W'11> Vg is a K91-submodule of Wm ffi-conjugate to V. Also, if V is a K91-module and g E ffi, then V ® g is a K91-submodule of (V<!i)m fficonjugate to V, for V ® 1 is a K91-submodule of (V<!i)m K91-isomorphic to V. The inertia subgroup of V in ffi is defined by

{gig E ffi, V ® g ~ V} (V, 17.6).

9.3 Theorem (NAKAYAMA [3]). Suppose that 91 <J ffi, that K is a field and V is an indecomposable Kffi-module. If V is a (ffi, 91)-projective Kffimodule, then Vm is a direct sum of ffi-conjugate indecomposable K91-modules, each isomorphism type occurring with the same multiplicity in the sense of the Krull-Schmidt theorem. (By 7.7b), this theorem can be applied if char K does not divide I ffi/91 I·)

Proof By 7.5, V is isomorphic to a direct summand of (Vm)m. By the KrullSchmidt theorem, there is an indecomposable direct summand W of Vm such that V is isomorphic to a direct summand of Wm. 1fT is a transversal of 91 in ffi, then

Wm = ffiW ® t. IE T

Since W ® t is a K91-module conjugate under ffi to W, W ® t is indecomposable. A further application of the Krull-Schmidt theorem shows that

(1)

where each Vi is K91-isomorphic to W ® t for some t E T. Thus Vi is K91-isomorphic to V1g for some g E ffi. But

(2)

is also a decomposition of V')l into indecomposable K91-submodules, so by the Krull-Schmidt theorem, the multiplicity of Vi in (1) is equal to that of V1g in (2) and hence to that of V1 in (1). q.e.d.

The following statement, which is dual to Clifford's theorem, is sometimes useful.

9.4 Theorem. Suppose that 91 <l <», that K is a field of characteristic p and p does not divide 1<»/911. If V is a completely reducible K91-module, then V<» is a completely reducible K<»-module.

Proof If T is a transversal of 91 in <», we have

(V<»)')l = EBV ® t, lET

where each V ® t is a completely reducible K91-module. Hence (V<»k is completely reducible. As char Kdoes not divide 1<»/911, V<» is completely reducible by 7.20b). q.e.d.

9.5 Remark. The conditions in 9.4 are necessary. The fact that 91 <l <» alone is not sufficient to imply the conclusion of 9.4 is shown by the example where <» is a non-identity p-group, 91 = 1, char K = p and V is the trivial K91-module. Then V<» ;;;; K<» is indecomposable by 5.2, but V<» is certainly not completely reducible.

Also in the case when 91 *' <» and char K does not divide 1<»: 911, the conclusion of 9.4 need not hold. To scc this, let <» be the group SL(2, 3) of order 24, let 91 be a Sylow 3-suhgroup of <» and let K be an algebraically closed field of characteristic 3. Let V be the trivial K91-module of K-dimension 1. We consider the K<»-module V<».

To within isomorphism, <» has exactly three irreducible modules Vi (i = 1, 2, 3), and dimK Vi = i, by 3.10. By Nakayama's reciprocity theorem (4.13), the multiplicity ni of Vi in the head V = V <»/V<» J(K<») of V<» is equal to the multiplicity of V in the socle of (Vi)')l' By 5.2, V is the only irreducible K91-module, so nl = 1, nz ?: 1 and n3 ?: 1. Now V2 is a module for the natural representation of SL(2, 3) of degree 2, so (Vz)')l is not a trivial K91-module. Thus nz = 1. From

if follows that n3 = 1. Thus dimK V = 6, so V<»J(K<») i= O. Hence V<» is not completely reducible.

Next we prove a generalization of V, 17.11.

9.6 Theorem (H. N. WARD [2], WILLEMS [1]). Suppose that 9l <J (fj, K is an arbitrary field and V is an indecomposable K9l-module. Let 3 be the inertia group of V in (fj and suppose that

V'" = VI EB ... EB V"

where the Vi are indecomposable K3-modules. a) v: is an indecomposable K(fj-module (i = 1, ... , r). b) If Vi is irreducible for some i, then V~ and V are irreducible. c) V~ ~ VXi if and only if Vi ~ Vj .

Proof a) Let T be a transversal of 3 in (fj containing 1. Since 9l <J 3, (V"')ill is the direct sum of K9l-modules of the form

V ® s (s E 3). Since 3 is the inertia group of V, V ® s ~ V. Hence by the Krull-Schmidt theorem,

(3)

for some ni' Thus for any g E (fj,

Vi ® g ~ V ® g EB . ~i • EB V ® g.

This implies that

(4) (V/D)ill ~ (EB Vi ® t) ~ EB(V <8l t EB '~i' EB V ® t~. tET ill tET

Suppose now that ViD = W EB X is a direct decomposition of the K(fj-module vt'. By 4.3a), Vi is isomorphic to a direct summand of (ViDh. By the Krull-Schmidt theorem, we may suppose that Vi is isomorphic to a direct summand of W",. Suppose that

Restricting to 9l, we see by (3) that

where WI ~ ... ~ Wni ~ V. As W is a K(fj-module,

and Wjt, V ® t are isomorphic K91-modules. Hence V ® t appears as a direct summand of W'!! with multiplicity at least ni' If t, t' are distinct elements of T, then V ® t and V ® t' are non-isomorphic K91-modules. Hence we see from (4) that (V?'),!! is isomorphic to a direct summand of W'!!. Since dimK vt, ~ dimK W, we obtain X = O. Thus vt, is indecom-posable. .

b) Suppose now that Vi is irreducible. Let W be any irreducible factor module of vtD• Then by Nakayama's reciprocity theorem (4.13), Vi is isomorphic to a K.3-submodule of W3. Since (V3)'!! is the direct sum of K91-modules V ® S (s E .3) and .3 is the inertia group of V, (Vi)'!! is the direct sum of ni submodules isomorphic to V for some ni' Hence Wn contains a submodule X which is the direct sum of ni submodules isomorphic to V. As W'!! is completely reducible, the indecomposable K91-module V is irreducible. Thus if g E (f), W'!! contains the K91-submodule Xg, which is the direct sum of ni submodules isomorphic to V(8)g.

Again, let T be a transversal of.3 in (f). Then the K91-modules V (8) t (t E T) are irreducible and non-isomorphic, and they all appear as direct summands of W'!! with multiplicity at least ni' Thus

Hence vtD = W is irreducible. c) Suppose that V?' ~ vt', but Vi * Vj' By 4.3a),

(Vth = V; E8 V;' and (v;"h = Vi E8 Vi',

where V;, Vi, V;', Vi' are K.3-modules, V; ~ Vi and Vi ~ Vj' By the KrullSchmidt theorem, Vi is isomorphic to a K.3-direct summand of Vi'. Thus (V;)'!! is isomorphic to a K91-direct summand of (Vj')'!!. Using (3) and (4), it follows that (Vi)'!! is isomorphic to a K91-direct summand of

E8 (V ® t E8 . ~. E8 V ® t). l#IET J

Hence by (3), V is K91-isomorphic to V ® t for some t E T, t "# 1. This contradicts the definition of the inertia group .3, since t ~ .3. q.e.d.

9.7 Remark. The following is a generalization of V, 17.12. Suppose that 91 <J (f) and that (1911, 1(f)/911) = 1. Let K be an alge-

braically closed field of any characteristic. Let W be an irreducible K91-module the inertia group of which is Gl. Then there exists an irreducible KGl-module V such that V'1I ~ W.

A short proof of this result, using Brauer characters, was recently given by GOW [1].1 We shall prove a special case which will be used later. For this we need the following lemma.

9.8 Lemma. Suppose that 91 <J Gl and Gl = 91(g). Let Q be a homomorphism of 91 into a group f). Suppose that there exists an element z of f) such that

and gnQ = zn, where n = I Gl/91 I· Then there exists a homomorphism (j of Gl into f) such that Q is the restriction of (j to 91 and g(j = z.

Proof. We define (j by putting (giX)(j = Zi(XQ) for all x E 91 and o ~ i < n. Then (gix)(j = zi(xQ) for any integer j, since if j = nq + i and 0 ~ i < n, then

(giX)(j = (gignqx)(j = zi((gnqx)Q) = zi(gnQ)q(xQ)

= zi+nq(xQ) = zi(xQ).

Thus if x, yare elements of 91 and i, j are positive integers,

(gixgiy)(j = (gi+iX9iy)(j = zi+i(X9iY)Q = Zi+i(X9iQ)(YQ)

= Zi+iz-i(xQ)zi(YQ) = Zi(XQ)zi(YQ)

= (giX)(j(giY)(j·

Hence (j is a homomorphism of Gl into f) with the required properties. q.e.d.

9.9 Theorem. Let K be an algebraically closed field of characteristic p and let 91 be a normal subgroup of Gl for which Gl/91 is cyclic. Let V be a K91-module the inertia group in Gl of which is Gl itself.

a) (SRINIVASAN [1]) If V is an irreducible K91-module, there exists a (necessarily irreducible) KGl-module W such that W'1I ~ V.

b) (WILLEMS [1]) If Gl/91 is a p'-group and V an indecomposable K91-module, there exists a KGl-module W such that W'1I ~ V.

1 See also ISAACS [1].

Proof Let n = 1(fj/91I. By hypothesis, (fj = 91<h) for some element h. We denote the representation of 91 on V by Q. Since (fj is the inertia group of Q, there exists a non-singular K-linear mapping y of V onto itself such that for all x E 91

(5)

Thus

This shows that

(6)

a) Now V is irreducible. Since K is algebraically closed, it follows from (6) and Schur's lemma that

Q(hn)y-n = a1

for some a E KX. Choose bE KX such that bft = a and put z = by. Then

for all x E 91 and Q(hft) = zn. By 9.8, the assertion follows. b) Since K is algebraically closed, it follows that

Hence it follows from (6) that

(7) Q(hft)y-n = a1 + IY., where a E KX and IY. E J(HomK91 (V, V».

Again we choose b E KX such that bft = a and put z = by. Thus

for all x E 91 and

(8) e(hft)z-n = 1 + f3

where P is a nilpotent element of HomK9I(V, V). It follows from (8) that

z-"e(h") = e(h"r1e(h")z-"e(h")

= e(h"rl(l + P)e(h") = 1 + p,

since h" E 91 and 1 + P E HomK9I(V, V). Thus

(9)

Since P is nilpotent, there exists an integer k such that ppt = O. Hence it follows from (8) and (9) that

Write hP' = g. Since 1<»/911 is not divisible by p, <» = 91(g). Also

e(xg) = e(xhi ) = z-pte(x)zpt

and e(g") = z"pt. Thus the assertions again follow from 9.8. q.e.d.

We turn now to analogues of V, 17.12b). For this we need the following lemmas.

9.10 Lemma. Suppose that 91 <l ffi and that V is a Kffi-module. For oc E HomK!Il(V9I, V!Il) and g E <», we define ocg by

Then ocg E HomK9I(V9I, ~), and oc --. oc g is an algebra automorphism of HomK9I(V9I, V9I).

Proof Suppose v E V, g E <», h E 91 and oc E HomK9I (V9I , V9I). Then

(vh)ocg = (Vh)g-locg = ((vg-l)(ghg-l»ocg = ((vg-1)oc)(ghg-1)g

= vg-locgh = (vocg)h.

Thus oc g E HomK9I(V9I , V9I). It is obvious that oc --. ocg is K-linear and bijective. If oc, P E HomK9I(V9I, V9I) and v E V, then

9.11 Lemma. Suppose 91 <J 05. Let V be a K05-module and let WI' W2

be K05-modules of equal dimension m, on which 91 acts trivially. a) There is a K-monomorphism ! of

into the complete matrix ring (HomK91(V91 , V91 ))m defined as follows. Let {w I, ... , wm} be a K-basis of WI and let {W'I' ... , w~} be a K-basis of W2 • IflY.. E Hom K91(V91 (8\ (WI)91' V91 ®K (W2)91) and

m

(v ® Wi)1Y.. = L VlY..ij ® wj (VlY..ij E V), j=1

we put IY..! = (lY.. i). b) If WI = W2 and w; = Wi' then! is an algebra monomorphism of

into the complete matrix ring (HomK91 (V91 , V91))m' c) Suppose that

m m

Wig = L aij(g)wj and w;g = L bij(g)wj. j=1 j=1

m m

L aij(g)lY..jk = L lY..~bjk(g)· j=1 j=1

Proof a) For every IY.. E HomK(V ®K WI' V ®K W2), there exist uniquely defined lY..ij E HomK(V, V) such that

m

(v ® Wi)1Y.. = L VlY..ij ® wj (i = 1, ... , m). j=1

Obviously, the mapping! defined by IY..! = (lY..i) is a K-monomorphism. If x E 91 and IY.. E HomK91 (V91 ®K (WI )91, V91 ®K (W2k), then

m

L (vx)lY..ij ® wj = (vx ® w;}1Y.. = (vx ® WiX)1Y.. = (v ® Wi)XIY.. j=l


= (v ® Wi)IXX = (t VlXij ® Wi)X J=1

m

= L (VlXij)X (8) wi· j=1

Comparing coefficients of wi, we obtain lXij E HomK91 (V91 , V91)' b) Suppose that IX, f3 are elements of

HomK91 (V91 ®K (W1)91' V91 ®K (W1)91)' If IXr = (IX i) and f3r = (f3ij), then

(v (8) Wi)lXf3 = ((v ® w;}IX)f3 = Ct1 VlXij (8) Wj ) f3

Hence (1Xf3)r = (IXr)(f3r). c) If 9 E (fj and IX E HomKIli(V ®K W1, V ®K W2), then

m m = L L (Vg)lXjk (8) aij(g)w~

j=1 k=1

and

m m

= L L (vlXij)g ® bjk(g)w£. k=1 j=1

This implies that

m m

L (Vg)lXjkai/g) = L (VlXi)gbjk(g)· j=1 j=1

Substituting v for vg, we obtain

m m

V L aij(g)ajk = L (vg- 1 )aijgbjk(g) j=l j=l

m

= V L a~bjk(g)· q.e.d. j=l

9.12 Theorem (H. N. WARD [2]; HUPPERT, WILLEMS [1]). Suppose that 91 <l (fj, K is an arbitrary field and V is a K(fj-module.

a) If V'JI is absolutely indecomposable and W is an indecomposable K(fj/91)-module, then V ®K W is an indecomposable K(fj-module.

b) If V'JI is absolutely irreducible and W is an irreducible K(fj/91)module, then V ®K W is an irreducible K(fj-module.

c) IfV'JI is absolutely indecomposable and W1, W2 are K(fj/91)-modules such that V ®K W1 ~ V ®K W2 , then W1 ~ W2 ·

Proof a) Suppose that n is an idempotent in HomKffi(V ®K W, V ®K W). We want to show that n = 0 or n = 1. Let {w 1, ••• , wm } be a K-basis of Wand

m

Wig = L aij(g)wj (g E (fj). j=l

We apply Lemma 9.11 with W1 = W2 = Wand Wi = wi (i = 1, ... , m). Thus we obtain nr = (ni), where

m m (10) L aij(g)njk = I n~ajk(g)·

j=l j=l

By 9.11b), nr = (ni) is an idempotent of (HomK'JI(V'JI' V'JI))m' As V'JI is absolutely indecomposable, we have the decomposition

Suppose that

nij = Pij1V + Qij (i,j = 1, ... , m),


where Q0 E J(HomK91(Vm. V91)) by 9.10. Comparing the coefficients of Iv in (10), we obtain

m m

L aij(g)pjk = L pijajk(g)· j=l j=l

Hence the mapping n', defined by

m

win' = L PijWj (i = 1, ... , m) j=l

is a KG}-homomorphism ofW. As (nij) is idempotent, n' is also idempotent. Hence the indecomposability of W implies n' = 0 or n' = 1. Replacing n by 1 - n if necessary, we can assume that n' = 0, hence

for all i,j. As J(HomK\ll(Vm. V91)) is a nilpotent ideal, this implies that

for sufficiently large t. Hence n = nt = 0, and V ®K W is indecomposable.

b) Let U be a non-zero KG}-submodule of V ®K Wand let u be a non-zero element of U. Write

k

U = L Vi ® Wi (Vi E V, Wi E W), i=l

where W 1 # ° and V 1, ••• , Vk are linearly independent over K. Since '4t is absolutely irreducible, K = HomK\ll(V91, V91 ). Hence by Jacobson's density lemma (V, 4.2), given V E V, there exists

a = L axx E K91 (ax E K) XE91

such that V1 a = v and Via = ° for 2 ::; i ::; k. Since 91 operates trivially on W, we obtain

(Vi ® w;}a = L aX(vi ® wJx = L ax(vix ® Wi) = Via ® Wi XE\ll XE\ll

and ua = v ® WI' Thus V ® WI ~ U. Hence for all gEm,

Since WI =I 0 and W is irreducible, it follows that U = V ®K W. Hence V ®K W is irreducible.

c) Now suppose that a is a Km-isomorphism of V ®K WI onto V ®K W2 • Then certainly dimKWI = dimKW2 , and we can apply 9.11. Thus in the notation of 9.11c), we have aT = (ai) and

m m

(11) L aij(g)ajk = L afjbjk(g)· j=1 j=1

Again we obtain

where cij E K and Yij E J(HomK91(V91, V91»' Hence (11) implies that

m m

L aij(g)Cjk = L cijbjk(g)· j=1 j=1

Thus the mapping a', defined by

(i = 1, ... , m)

is in HomK(!;(WI, W2). Application of the same procedure to the inverse of a shows that a' has an inverse. Hence WI and W2 are isomorphic Km-modules. q.e.d.

9.13 Corollary. Suppose that m <J m and that VI' V2 are Km-modules for which (VI)91 and (V2)91 are isomorphic absolutely irreducible Km-modules. Then there exists a K(m/m)-module W of K-dimension 1 such that V2 ;;:;

VI ®K W.

Proof By 4.15b),

Since K(m/m) has the trivial module as a quotient module,


V2 ®K K(GJ/91) has a quotient module isomorphic to V2. Hence V2 is isomorphic to a composition factor of Vl ®K K(GJ/91). Let

be a composition series of the KGJ-module K(GJ/91). Then by 9.12b),

is an irreducible KGJ-module. Thus by the Jordan-Holder theorem, V2 ;::: Vl ®K W for some irreducible K(GJ/91)-module W. Since dimK Vl

= dimK V2, certainly dimK W = 1. q.e.d.

From Theorem 9.12 we can easily deduce results about the irreducible and indecomposable modules of direct products.

9.14 Theorem. Let K be a splitting field for GJ l and GJ 2. Let Vl' ... , Vs be all the irreducible K GJ l-modules and let Wl , ... , Wt be all the irreducible KGJ 2-modules (to within isomorphism). Then the Vi ®K Wj' regarded as K(GJ l x GJ 2)-modules with

are all the irreducible K(fjl x (fj2)-modules, to within isomorphism.

Proof We apply 9.12b) to the normal subgroup (fjl of (fjl x (fj2 = (fj. Vi can be regarded as a K(fj-module on which (fj2 operates trivially. By hypothesis, the restriction of this module to (fj 1 is absolutely irreducible. Hence by 9.12b), the Vi ®K Wj are irreducible K(fj-modules.

Suppose that Vi ®K Wj and Vk ®K WI are isomorphic KGJ-modules. Restriction to (fj 1 shows that

By the Jordan-Holder theorem, Vi ;::: Vk, so i = k. Similarly j = I. Finally, it must be shown that the Vi ®K Wj are all the irreducible

K(fj-modules to within isomorphism. We have s = I(GJ l ) and t = I(GJ 2),

where I(X) is the number of conjugacy classes of X if char K = 0 and I(X) is the number of pi-classes of X if char K = p, by V, 5.1 and 3.9. Thus

is the number of isomorphism types of irreducible K({fj1 x (fj2)-modules. q.e.d.

The statement corresponding to 9.14 for indecomposable modules is unfortunately not as complete.

9.15 Theorem. Let K be an algebraically closed field. a) If Vi is an indecomposable K{fj;-module (i = 1, 2), then V1 (8\ V2 is

an indecomposable K({fjl x (fj2)-module. b) If Vi' Wi are indecomposable K{fj;-modules (i = 1,2) and V1 ®K V2

~ W1 ®K W2 , then Vi ~ Wi (i = 1, 2). c) Every indecomposable K(m1 x ( 2 )-module is of the type V1 ®K V2

for suitable Kmi-modules Vi if and only ifat least one of the orders Imll, I (fj 21 is prime to char K.

Proof a), b) The proofs of these assertions are the same as those of 9.14, using 9.12a), c) instead of 9.12b) and the Krull-Schmidt theorem instead of the Jordan-Holder theorem.

c) In the case when char K = 0, the assertion follows from 9.14, since in this case there is no difference between irreducibility and indecomposibility. Suppose that char K = p.

First suppose that p does not divide Im21. We show (following CONLON [1 J) that any indecomposable K(m1 x ( 2)-module V is of the type Vl ®K V2 · Since p does not divide Im l x m2 : m11, V is isomorphic to a direct summand of (V'S> )'S>, x'S>2, by 4.3c). By the Krull-Schmidt , theorem, there exists an indecomposable direct summand W of V'S> such that V is isomorphic to a direct summand of W<!i, x <!i2 • Now '

W<!i,X<!i 2 = EB W ® g, gE <!i2

This shows that W<!i, x <!i2 is isomorphic to W @K Km2. Let

be a decomposition of Km2 into indecomposable Km2-modules Ui .

(Since p does not divide Im21, the Ui are in fact irreducible.) Then V is


isomorphic to a direct summand of EBi(W ®K UJ By a), each W ®K Ui is an indecomposable K(GJ 1 x GJ2)-module. Hence by the Krull-Schmidt theorem, V ~ W ®K Ui for some i.

Now suppose that char K = p divides IGJ 11 and IGJ21. Let ~i be a Sylow p-subgroup of GJ i• We put GJ = GJ1 X GJ2 and ~ = ~1 X ~2' Let illi be a subgroup of ~i of index p. By 5.1, there exists an indecomposable K~-module W such that dimK W > p2 and ill1 x ill2 operates trivially on W. If W is of the type W1 ®K W2 for K~i-modules Wi' then Wi is indecomposable and illi operates trivially on Wi' Thus dimK Wi ::; p by 5.3. This gives the contradiction dim K W ::; p2. Hence W is not of the type W1 ®K W2 for K~i-modules Wi'

Put V = W m. Suppose that

V = V1 EEl ... EEl Vs'

where the Vi are indecomposable KGJ-modules. By 4.3, W is isomorphic to a direct summand of (Wm)'ll' By the Krull-Schmidt theorem, W is isomorphic to a direct summand of (Vk)'ll for some k. Suppose that Vk = Xl ®K X2, where Xi is an indecomposable KGJi-module (i = 1,2). If

(Xi)'ll, = EB Xij (i = 1, 2), j

where the Xij are indecomposable K~i-modules, then

(Vk)'ll ~ (X1)'ll1 ®K (X2)'ll2 ~ EBX1j ®K X21· j,l

By a), each X1j ®K X21 is an indecomposable K~-module. Thus by the Krull-Schmidt theorem, we get the contradiction W ~ Xlj ®K X21 for some j, I. Thus Vk is an indecomposable KGJ-module not of the form Xl ®K X2 for indecomposable KGJ;-modules Xi' q.e.d.

Proof a) If {WI' ... , wn } is a K-basis of V2, then 2::'=1 Vi ® Wi (Vi E Vd lies in CV(fjl x 1) if and only if Vi E CV1(fjl) for all i = 1, ... , n. Thus

§ 9. Representations of Normal Subgroups

cv(m) ~ CV,(m 1) ®K Vz· Application of the same argument to Cv, (m 1) ®K VZ yields

Thus

139

b) We put Hi = HomK(V;, Vi) and H = HomK(V, V). By V,9.14, there exists a K-algebra isomorphism, of HI ®KHz onto H such that

for all Vi E Vi' lY.i E Hi (i = 1,2). We regard HI' Hz, H as Km 1-, Kmz-, Km-modules respectively in the sense of8.8. Then, is a Km-isomorphism, for if Vi E Vb lY.i E Hi and gi E mi, then

(VI ® VZ)(((1Y. 1 ® IY. Z)glgZ)') = (VI ® VZ)((IY. Igl ® IY.zg z)')

= VI(1Y. 1gd ® Vz(lY.zgz)

= ((VIgIl )IY. I)gl ® ((Vzg2 1 )lY.z)gz

= ((VIgIl ® v2 g2"1)(1Y.1 ® IY.Z),)glgZ

= (VI ® VZ)(((IY. I ® IY.Z)')glgZ)·

So CH, (m l ) ®K CH2 (mz) and CH(m) are isomorphic K-algebras. The assertion now follows from the observation that

9.17 Theorem. Let m = m l x mz, let Vi be a Kmi-module and let V = VI ®K Vz·

a) If VI is indecomposable and Vz is absolutely indecomposable, then V is an indecomposable Km-module.

b) If VI and Vz are absolutely indecomposable, then V is absolutely indecomposable.


Proof. Let 6 i = HomK(!;,(Vi , V;) (i = 1,2). We consider the mapping (X

from 6 1 ®K 6 2 onto 6dJ(6 1) ®K 6 2/J(62) defined by

for (J"i E 6 i (i = 1, 2). Then r:t. is an algebra epimorphism, and the kernel of r:t. is the ideal

Thus

Now 6dJ(6 1) is a division algebra since VI is indecomposable and 6 2/J(6 2) = K since V2 is absolutely indecomposable. Hence

is a division algebra, and so 3 :2 J(61 @K 6 2), But 3 is nilpotent, so 3 = J(6 1 ®K 6 2 ) and

If we put 6 = HomKo;(V, V), then by 9.16b)

Thus 6 is a local ring and hence V is indecomposable. In case b), we obtain

and V is absolutely indecomposable. q.e.d.

9.18 Theorem (ROTH [1J, SCHWARZ [1J). Let K be a splitting field of Gl. Suppose that 91 is a normal subgroup of Gl such that Gl/91 is Abelian and char K does not divide I Gl/91I. Let V be an irreducible KGl-module. By Clifford's theorem, there is an irreducible K91-module W such that

m

V'JI ~ eEBW @ gi, i=1

where {g l' ... , gn} is a transversal of the inertia subgroup 3 of W in (fj. Then

r

W<f> ~ eE8V ®K Uj , j;l

where the Uj are irreducible K(fj-modules of K-dimension 1 on which 91 operates trivially. The V ®K Uj are irreducible and pairwise non-isomorphic. Also

Proof It follows from the hypotheses on K and (fj/91 that

n

K(fj/91) = E8 Uj , j;l

where n = I (fj/911 and U1, ... , Un are all the irreducible K(fj/91)-modules to within isomorphism; further dimK Ui = 1 for all i and the Ui form a group under the tensor product operation. Now W Q9 gi is a K91-module and (W Q9 gJ<f> ~ W<f>. Hence by 4.15b),

so

Let

n

(V<Jt)<f> ~ V ®K K(fj/91) ~ E8(V Q9K U), j;l

n

emW<f> ~ (V<Jt)<f> ~ E8 V Q9K Uj .

j;l

Then 6 is a subgroup of the group formed by the Ui . Let {U1 , ... , Ur} be a transversal of 6 in this group. Then

r

emW<f> ~ 161 E8V Q9K Uj . j;l

Since rl61 = I (fj/91I, this gives


In particular, Wm is completely reducible and the socle of Wm is Wm

itself. Hence by Nakayama's lemma (4.13b)), the multiplicity of V in Wm

is equal to the multiplicity of W in the head of V~, and this is e. Hence

q.e.d.

By repeated application of 9.18, we can easily prove that if m is soluble and K is a splitting field of m such that char K does not divide Iml, then the dimension of any irreducible Km-module divides Iml. Of course, this is true for any group m, by V, 12.11. But this method can be used to obtain a result even in the case when char K does divide Iml. To do so, we need the following lemma, which is of independent interest.

9.19 Lemma. Let K be afinite or algebraically closed field of characteristic p. Let 91 be a normal subgroup of m of index a power of p. Let V be an irreducible Km-module for which all the irreducible K91-submodules of V~ are isomorphic. Then V is an irreducible K91-module.

Proof (GREEN). First suppose that K is finite. Let W be an irreducible K91-submodule of V~ and let q = I HomK91(W, W)I. Put

f(j) = 1 + q + ... + qi- I (j ~ 1).

We prove by induction on j that f(j) is the number of irreducible K91-submodules of any K91-submodule U of V91 of composition length j. This is obvious ifj = 1. For j > 1, we observe that by Clifford's theorem, V91 is completely reducible. Hence so is U, and we can write U = WI EB UI , where WI ~ W. By the inductive hypothesis, the number of irreducible K91-submodules of UI is f(j - 1). Now if W* is an irreducible K91-submodule of U not contained in UI , then W* n UI = 0, so W* is of the form

for some lI. E HomK91 (WI , UI). Thus the number of irreducible K91-submodules of U is

Since U1 is the direct sum of j - 1 Kin-submodules isomorphic to W, this number is

J(j - 1) + I HomK\Jl(W, W)ji-l = J(j - 1) + qi- 1 = J(j).

Hence the number of irreducible Kin-submodules of V\Jl is J(I), where I is the composition length of V\Jl' Since q is a power of p,J(l) == 1 (p). But a Sylow p-subgroup ~ of m permutes the irreducible Kin-submodules of V\Jl among themselves, so there exists an irreducible Kin-submodule W of V\Jl invariant under ~. Hence W is invariant under ~in = m. Since V is an irreducible Km-module, W = V, and V is an irreducible Kin-module.

Now suppose that K is algebraically closed, and let k be the number of pi -conjugacy classes of m. By 2.6, there exists a finite splitting field L for all the subgroups of m, and since K is algebraically closed, we may suppose that L s; K. Then by 3.9, there are k non-isomorphic irreducible Lm-modules V1, ... , Vk • By 2.4, V ~ (Vi)K for some i. Similarly, the irreducible Kin-submodules of V\Jl are isomorphic to WK for some irreducible Lin-module W. Hence by 1.22 the irreducible submodules of (Vi )9/ are all isomorphic to W. By the finite case, (Vi)\Jl is an irreducible Lin-module, and V\Jl is irreducible. q.e.d.

Combining 9.18 and 9.19, we obtain the following.

9.20 Theorem (SWAN [1]). Suppose that m/in is soluble and that K is an algebraically closed Jield. Let V be an irreducible Km-module. Then the composition length OJV91 divides Im/inl.

Proof This is proved by induction on Im/inl. Suppose that in < m. As m/in is soluble, there exists IDl <J m such that IDl ~ in and IDl/in is a non-identity Abelian group of prime-power order. By Clifford's theorem,

for certain m-conjugate irreducible KIDl-modules U1, ... , Us. By the inductive hypothesis, s divides I m/IDlI. We distinguish two cases.

Case 1. Suppose that char K does not divide I IDl/in I· By V, 17.3[) (Clifford's theorem),

m

(U1)\Jl ~ e EBW ® gil i~l


where {g I' ... ,gm} is a transversal of the inertia subgroup of W. By 9.18, em divides IIDl/91I. As Uj is (V-conjugate to UI , each (U)91 also splits into em irreducible parts. Hence the composition length of V91 is sem, and this divides I (V/IDlI I IDl/91 I = I (v/91 I·

Case 2. Suppose that char K = p and that IDl/91 is a p-group. By Clifford's theorem,

where the Xi are the homogeneous components of (U1)91' and Xl' ... , Xr are transitively permuted by IDl. Let '1: be the stabiliser of Xl; thus IIDl: '1:1 = r. By V, 17.3e), Xl is an irreducible K'1:-module. Hence by 9.19, Xl is an irreducible K91-module. Thus the composition length of (UI )91 is r, as is that of each (U)91' Hence the composition length of V91 is sr, and sr divides

Since 91 :::;'1:, sr divides 1(v/91I. q.e.d.

9.21 Theorem. Suppose that (V is soluble and that K is an algebraically closed field. Let V be an irreducible K(V-module and let m be a normal Abelian subgroup of (V. Then dimK V divides I (V/ml (cf V,17.lO).

Proof As m is Abelian,

where dimK Wi = 1. Thus the composition length m of V'H is equal to dimK V, and by 9.2?, this divides I(V/ml. q.e.d.

9.22 Remark. Theorem 9.21 remains valid under the weaker hypothesis that K is algebraically closed, char K = p and (V is p-soluble ( DADE [3], SWAN [1]).

9.23 Theorem. Let K be an algebraically closed field of characteristic p, let (V be a soluble group and pa the highest power of p dividing I (VI. If V is an irreducible K(V-module and pa divides dimK V, then V is a projective K(V-module. (This is a partial converse of Dickson's theorem 7.16.)

Proof We proceed by induction on I(VI. Let 91 be a maximal normal subgroup of (V. Then I (V/911 is a prime.

Case 1. Suppose that p # I ffi/91I. By 9.20, we obtain

V'JI = WI EB ... EB W"

where the ~ are irreducible K91-modules of the same dimension and r divides Iffi/91I. Hence p does not divide r, and so pa divides 1911 and dimK~' By induction, each ~ is a projective K91-module. Hence V is a projective Kffi-module by 7.11b).

Case 2. Now suppose Iffi/911 = p. By Clifford's theorem we have

m

V'JI = eEB~, j;1

where the ~ are irreducible, non-isomorphic ffi-conjugate K91-modules. By 9.20, em divides I ffi/911 = p.

If V'JI is irreducible, then pa divides dimK V'JI but not 1911. This contradicts 9.21. Hence em = p. By 9.19, e = p is impossible. Hence e = 1 and m = p. Then V ~ W? by V, 17.3e). Now pa-I divides dimK WI and is the highest power of p dividing 1911. Hence by induction WI is a projective K91-module. Then V is a projective Kffi-module by 7.17.

q.e.d.

9.24 Remark. 9.23 is not true for arbitrary finite groups. Let ffi be the first simple group of lanko of order

175560 = 23 .3'5.7.11.19.

Let K be an algebraically closed field of characteristic 2. Then there are two types of irreducible Kffi-modules of dimension 56 = 23 . 7. These are not projective; the corresponding projective modules have dimension 552 = 23 . 3· 23 (LANDROCK, MICHLER [1J).

Exercises

35) Suppose that 91 <J ffi and that V is an irreducible K91-module. Let .3 be the inertia group of V in ffi and suppose that

where the Vi are irreducible K3-modules. Show that if W is an irreducible KGj-module and V is isomorphic to a direct summand of W'JI' then W ~ vtSJ for some i.

a) Prove this by using characters in the case K = C. b) Prove it in the general case by arguments similar to those in the

proof of 9.6.

36) Show that the statement in Exercise 35 does not remain true if the word "irreducible" is replaced by "indecomposable". (Consider a group of type (p, p) and use representations as in 6.11.)

37) (WILLEMS [1]) Let K be an algebraically closed field and suppose 91 <J Gj. Let V be an irreducible K91-module and let 3 be the inertia group of V in Gj. Then the following statements are equivalent.

a) V(fj is an irreducible KGj-module. b) V:J is an irreducible K3-module. c) 3 = 91.

(In the proof of b) ~ c), assume that 91 < 3 and take a subgroup U such that 91 < U ::5: 3 and U/91 is cyclic. By 9.9, there is a KU-module W such that W'JI ~ V. Consider VU.)

38) a) Show that there is no analogue of Corollary 9.13 for indecomposable modules.

b) Suppose that K is an algebraically closed field and 91 <J Gj. Let V1 , V2 be KGj-modules such that (V1)91 and (V2)91 are indecomposable and isomorphic. If V2 is (Gj, 91)-projective, there exists a K(Gj/91)-module W such that V2 ~ V1 ®K Wand dimK W = 1.

39) Let K be a field of characteristic p and Gj = (g) a cyclic group of order pn. Which indecomposable representations of (gP) can be extended to representations of Gj?

40) Suppose that K is algebraically closed and that char K does not divide I Gj/91 I· Let V be a KGj-module and let W1 , ... , Wk be all the irreducible K(Gj/91)-modules (to within isomorphism). Suppose that the V ®K Wi are irreducible non-isomorphic KGj-modules. Show that V91 is irreducible

a) by character calculations for char K = 0, b) by module-theoretic considerations in the general case.

(Hint: a) Let (] be the regular character of Gj/91, lifted to Gj, and calculate (X, X(])(fj, where X is the character of V.

b) On the one hand

§ 10. One-Sided Decompositions of the Group-Ring 147

k

(V'JIyfJ ~ V ®K K(G3/91) ~ EB (dimK Wi)(V ®K Wi), i;1

where V itself appears with multiplicity only 1. On the other hand, if V'JI = V1 EB ... EB Vs for irreducible K91-modules Vi' then V appears in each (completely reducible) vt, with multiplicity at least 1.)

41) Give an example to show that Theorem 9.9b) is not true in general without the assumption that G3/91 is a p'-group.

42) Let K = GF(2) and consider V = GF(4) as an irreducible K3-module, where 3 is the cyclic group of order 3. Show that V (8\ V is not an indecomposable K(3 x 3)-module.

43) Suppose that KG3 = EBi P" where the Pi are indecomposable projective Km-modules. Show that

and

dimK EB HOmK(l;(Pi ®K Pj' Pk ®K P') = Iml3

i,j,k,l

dimK EB HomK(l;(Pi , Pk ) ®K HOmK(l;(Pj, P') = Im12• i,j.k,l

Hence there exist indices i,j,k,l such that

44) Suppose 91 <J m. Let V be a Km-module such that V'JI is absolutely indecomposable. Then the following assertions are equivalent.

a) V'JI is a projective K91-module and char K does not divide 1m/mi. b) V is a projective Km-module.

(Use 9.12.)

§ 10, One-Sided Decompositions of the Group-Ring

Let ~ be an algebra of finite dimension over a field K. In the sequel we write ~ = ~/J(~). As ~ is semisimple, it follows from V, 3.3 that there is a decomposition


(1) I = e1 + ... + en

of the unit element I of 21 into mutually orthogonal idempotents ei' such that

is a direct decomposition of 21 into irreducible 2l~modules ei 21. We now wish to obtain from (1) a decomposition

1 = e1 + ... + en,

where e l' ... , en are mutually orthogonal idempotents of 21 such that ei = ei + J(21). This process is known as "lifting" the idempotents ei' With an application in mind in which 21 will not be a ring with minimum condition but the group-ring of a finite group over a ring of p-adic integers, we proceed at first rather more generally than is necessary for the purpose of this section.

10.1 Lemma. Let 9i be a ring and let 3 be a two-sided ideal of 9ifor which n~1 3n = 0. We put 3° = 9i. For a, b E 9i we define

{o ifa=b d(a, b) = 2-k ifa _ bE3\buta _ b~3k+1.

a) d is a metric on 9i and the so-called ultrametric inequality

d(a, b) ~ max(d(a, c), d(c, b))

holds for all a, b, c E 9i. We call the Hausdorff topology defined on 9i by d the 3-adic topology on 9i.

b) With the topology defined in a), 9i is a topological ring; that is, the mappings (a, b) -+ a + b, a - b, ab are continuous. We call the topological ring 9i an 3-adic ring.

c) The sequence (an) of elements an E 9i is a Cauchy sequence if and only if(an+1 - an) is a null sequence.

Proof. a) It is obvious that for a, bE 9i, we have d(a, b) 2: 0, d(a, b) = d(b, a); and d(a, b) = ° is equivalent to

a - bEn 3" = 0. n=1


Suppose that d(a, c) = 2-m and d(c, b) = 2-", where n :::;; m. Then a - c E 3 m and c - bE 3". Since 3 m ~ 3", it follows that

a - b = (a - c) + (c - b) E 3 m + 3" = 3".

Hence

d(a, b) :::;; 2-" = max(d(a, c), d(c, b)) :::;; d(a, c) + d(c, b)

and thus d is a metric on 9l. b) If d(a, ao) :::;; 2-" and d(b, bo) :::;; 2-", then

and

Thus addition, subtraction and multiplication are continuous operations, and 9l is a topological ring.

c) Suppose that (a,,+! - a,,) is a null sequence, that is,

For n ~ n(k) and m ~ 0 we obtain

m+n-l

a,,+m - a" = L (ai+l - ai) E 3\ i=n

so d(a,,+m, a,,) :::;; 2- k• Thus (a,,) is a Cauchy sequence in 9l. q.e.d.

We come now to the lifting of idempotents.

10.2 Theorem. Let 9l be an 3-adic ring which is complete with respect to the 3-adic topology. We put 9l = 9l/3.

a) There exist polynomials p" (n = 0, 1, ... ) with rational integral coefficients such that p,,(O) = 0, having the following properties.

If e = eo + 3 is an idempotent in 9l, the limit e of the sequence (p,,(eo» exists, and e is an idempotent in 9l for which e - eo E 3. If a E 9l and eoa = aeo (or eoa = 0 or aeo = 0), then ea = ae (or ea = 0 or ae = 0).

b) Suppose that I = e1 + ... + en' where ei E 91, er = ei and


eiej = 0 for i =f. j. Then there exist mutually orthogonal idempotents ei in 91 such that 1 = e 1 + . . . + en and ei + 3 = ei.

Proof. a) We define polynomials Pn in Z[t] recursively by Po(t) = t and

for n > O. Then Pn(O) = 0 for all n. We have

We prove by induction on n that

This is true for n = O. For n > 0, we have

and

Thus

Pn(eo) - eo = 3Pn-l (eo)2 - 2Pn-l (eo)3 - eo

== 3e6 - 2e6 - eo == 0 mod 3,

Pn(eo)2 - Pn(eo)

= 4(Pn-l (eo)2 - Pn-l (eo»3 - 3(Pn-l (eo)2 - Pn-l (eo»2

E (32'-'? + (32'-'? = 3 2'.

Hence by 10.1, (Pn(eo)) is a Cauchy sequence. Since by hypothesis 91 is complete, the limit e of this sequence exists. Since 91 is a topological ring and

we obtain

Thus e2 = e. Since the set

§ to. One-Sided Decompositions of the Group-Ring 151

3 = {aJd(a,O) ::;; t}

is closed, it follows from Pn(eO) - eo E 3 that e - eo E 3. If eoa = aeo, then Pn(eo)a = aPn(eo) for all n, so

ea = lim Pn(eo)a = lim aPn(eo) = ae. n-oo n-oo

Since Pn(O) = 0, it follows similarly from eoa = 0 that ea = 0 and from aeo = 0 that ae = O.

b) We construct a set el, ... , em of mutually orthogonal idempotents ei in 9l such that ei + 3 = ei for 1 ::;; m < n by induction on m. For m = 1 this is done by applying a). For m > 1, there exists a set of mutually orthogonal idempotents e I, ... , em-I for which ei + 3 =

ei, by the inductive hypothesis. Hem = a + 3 (with a E 9l), put

b = a - a(el + ... + em-I) - (el + ... + em-da

+ (el + ... + em-I)a(el + ... + em-I)'

Then ejb = bej = 0 for 1 ::;; j ::;; m - 1 on account of the mutual orthogonality of the ei' Since eim = emej = 0 for 1 ::;; j ::;; m - 1, we have eja E 3 and aej E 3. Thus b + 3 = a + 3 = em' Hence by a), there exists an indempotent em in 9l for which em - b E 3 and ejem = emej = 0 for 1 ::;; j ::;; m - 1.

We apply this result with m = n - 1 and then put

From 10.2 we deduce a basic theorem about direct decompositions of not necessarily semisimple algebras.

10.3 Theorem. Let m: be an algebra of finite dimension over a field K. We put m: = m:/J(m:). Let

be a direct decomposition of 21 into minimal right ideals ei 21, where the ei are mutually orthogonal idempotents in m:for which T = el + ... + en'


a) There exist mutually orthogonal idempotents ei in 21 such that

ei + J(21) = ei and 1 = e1 + ... + en'

b) 21 = e121 ED ... ED en 21. c) eiJ(21) is the only maximal submodule of ei 21, and ei 21jeiJ(21) ~

ei21. Hence eim is an indecomposable projective m-module the head of which is irreducible and isomorphic to ei m.

Proof By V, 2.4, J(m) is nilpotent. Thus the J(m)-adic topology of m introduced in 10.1 is discrete, and 21 is trivially complete. Hence a) follows from 10.2. The assertion b) follows at once from a).

If a E ei m n J(21), then a = eib for some b E m. Thus

a = eib = efb = eia E eiJ(m).

Hence ei21 n J(m) = eiJ(21). Therefore

eimjeiJ(m) = eimj(eim n J(m)) ~ (eim + J(m))jJ(m)

= (ei + J(m))(mjJ(m)) = eim.

As eim is an irreducible m-module, eiJ(m) is certainly a maximal submodule of ei 21. If Wl is any maximal submodule of ei m, then ei 21jWl is irreducible and is therefore annihilated by J(m). It follows that

so eiJ(m) is the only maximal submodlile of eim. This implies the indecomposability of eim, since a decomposable module possesses more than one maximal sub module. q.e.d.

We now need an elementary lemma about finitely generated modules.

10.4 Lemma. Let 9i be any ring with unit element and V a finitely generated 9i-module.

a) Every proper submodule W of V is contained in a maximal submodule ofV.

b) (Nakayama's lemma; cf vol. I, p. 643) If W is an 9i-submodule of V for which V = W + VJ(9i), then W = V.

Proof a) Suppose that V = 2:7=1 Vi9i. Consider the set


Y = {XIX is a submodule of V and W s; X c V}.

As WE Y, [f is not empty. We show that [f is inductive. To do this, let :If be a chain in Y. Then Y = UlEJf' X is a submodule of V and W s; Y. Suppose that Y = V. Then for every i = 1, ... , k, there exists X; E :If such that Vi E Xi' As :If is a chain, some Xj contains Xi for all i = 1, ... , k, so V s; Xi' a contradiction. Hence Y is inductive. Thus by Zorn's lemma, [f has maximal elements, and these are maximal submodules of V containing W.

b) Suppose that W c V. By a) there exists a maximal submodule X of V such that W S; X. Then V IX is an irreducible 9i-module and is therefore annihilated by J(9i). Hence VJ(9i) S; X. Thus

X ;;2 W + VJ(9i) = V.

This is a contradiction, hence W = V. q.e.d.

10.5 Lemma. Let 9i be an arbitrary ring with unit element. a) Suppose that P is a projective 9i-module and V is a finitely gene

rated 9i-module. If there exists an 9i-epimorphism rt. of P/PJ(9i) onto VjVJ(9i), then there is an 9i-epimorphism y ofP onto V such that

xy + VJ(9i) = (x + PJ(9i))rt.

for all x E P. b) If P is a finitely generated projective 9i-module, f3 E Hom91(P, P)

and xf3 -x E PJ(9i)for all x E P, then f3 is an automorphism ofP.

Proof a) Let Vy, Vp denote the natural epimorphisms of V onto V IV J (9i) and P onto PIP J (9i) respectively; thus

VVit = v + VJ(9i) (v E V) and yVp = Y + PJ(9i) (y E P).

Vp

,f~;J)------->'O v • v/VJ(\R)---_. 0

Vv

Then f3 = Vprt. is an epimorphism of Ponto VjVJ(9i). Since P is projective, there exists y E Hom91 (P, V) such that yVy = f3 = vprt.. Now f3 is an epimorphism; thus, given v E V, there exists yEP such that


v + VJ(9l) = yf3 = YYVv = YY + VJ(9l).

Thus V = PI' + VJ(9l). Since V is finitely generated, it follows from 10.4b) that V = PI'. Thus I' is an epimorphism of Ponto V.

b) It follows from the hypothesis that P = Pf3 + PJ(9l). Since P is finitely generated, we get P = Pf3 by lO.4b). Thus f3 is an epimorphism. Hence P /ker f3 ~ P is projective and we obtain P = ker f3 EB P' for some submodule P' isomorphic to P. Thus

PJ(9l) = (ker f3)J(9l) EB P/J(9l).

Since ker f3 n P' = 0, it follows by Dedekind's identity that

P J (9l) n ker f3 = (ker f3)J (9l).

Hence for x E ker f3 we have

x = x - xf3 E PJ(9l) n ker f3 = (ker f3)J(9l).

Thus ker f3 = (ker f3)J(9l). But ker f3 is finitely generated since it is a direct summand of P. Thus by lO.4b), ker f3 = 0 and f3 is a monomorphism. Hence f3 is an isomorphism of Ponto P. q.e.d.

10.6 Theorem. Let 9l be an arbitrary ring with unit element and suppose that PI' P2 are finitely generated projective 9l-modules. Then

if and only if PI ~ P2 •

Proof If PI ~ P2 , then certainly PdPI J(9l) ~ P2/P2 J(9l). Now suppose that a is an isomorphism of PdPI J(9l) onto P2/P2 J(9l).

We apply 10.5a) to a and to a-I. Thus there exists an epimorphism f3 of PI onto P2 such that

for all x E PI , and there exists an epimorphism I' of P2 onto PI such that

for all y E P2 . Thus if x E PI'


It follows from 1O.5b) that fJy is an automorphism of Pl' Hence fJ is a monomorphism and indeed an isomorphism of P1 onto Pz. q.e.d.

10.7 Definition. a) Let 9i be a ring and Van 9i-module. If S is a subset of V, we define the annihilator of S by

An\J!(S) = {rlr E 9i, Sr = a}.

Obviously An\J!(S) is a right ideal in 9i. If 6 is a subset of 9i, we put

Anv(6) = {vlv E V, v6 = a}.

If 6 is a left ideal of 9i, then Anv(6) is an 9i-submodule of V. b) If 6 is a subset of the ring 9i, we put

R(6) = {rlr E 9i, 6r = o}

and

L(6) = {rlr E 91, r6 = a}.

Obviously R(6) is a right ideal and L(6) is a left ideal in 9i. If 6 is a right ideal of 9i, then R( 6) is a two-sided ideal of 91; if 6 is a left ideal of 9i, then L(6) is a two-sided ideal of 9i. If 6 1,62 are additive subgroups of 9i, then

and

analogous statements hold for the left annihilators. c) Let m: be an algebra of finite dimension over a field K. We denote

by STem:) the socle of the right m:-module m: and by Sl(m:) the socle of the left m:-module m:. Thus by 1.6, Sl(m:) = R(J(m:» and Sr(m:) = L(J(m:». By the remarks in b), SAm:) and Sl(m:) are both two-sided ideals of m:.


10.8 Remark. If ~ is an algebra of finite dimension, then ~/J(~) is the largest completely reducible (right or left) factor module of~. But it is not true in general that Sr(~) = Sl(~); this is shown by the example of the ring of all triangular matrices over a field (see Exercise 45). If however ~ is the group-ring of a finite group over a field, then Sr(~) = Sl(~)' as we shall see in 11. 6b ).

We now study the projective modules over an algebra.

10.9 Theorem. Let ~ be an algebra of finite dimension over a field K. a) Any finitely generated projective ~-module P is determined to

within isomorphism by its head P/PJ(~). b) If V is a finitely generated completely reducible ~-module, there

exists afinitely generated projective ~-module P such that V ~ P/PJ(~). c) A finitely generated projective ~-module P is indecomposable if

and only if P/PJ(~) is irreducible. Any finitely generated projective indecomposable ~-module is isomorphic to one oj the right ideals ei ~ of 10.3.

Proof a) This is 10.6. b) Since

we can assume that V is irreducible. Since VJ(~) = 0, V may be regarded as an irreducible ~-module, where ~ = ~/J(~). Let

be a direct decomposition of ~ into minimal right ideals ei ~, where the ei

are mutually orthogonal idempotents in ~ for which I = e I + ... + en. Thus V ~ ei~ for some i. But by 10.3c), ei~ ~ ei~/eiJ(~) for the indecomposable projective ~-module ei~.

c) Suppose that P/PJ(~) is reducible. Since P/PJ(~) is completely reducible, there exist non-zero completely reducible modules VI' V2

such that

By b), Vi ~ P;/p;J(~) for some projective module p;. Hence

By a), P ~ PI EB P2 , so P is decomposable.

If P/PJ(m:) is irreducible, then PJ(m:) is the only maximal submodule of P, so P is indecomposable.

If P is a projective indecomposable finitely generated m:-module, P/PJ(m:) is irreducible. As in b), we have

for some i, so P ~ ei m: by a). q.e.d.

The iteration of the formation of head and soc1e of a module brings us to the Loewy series.

10.10 Definition. Suppose that m: is an algebra of finite dimension and V an m:-module.

a) The lower Loewy series

of V is defined by Vn = V J (m:t (n = 0, 1, ... ). b) Dually, the upper Loewy series

of V is defined by 5n/5n - 1 = S(V /5n - 1) for n > 0.

10.11 Theorem. Let .m: be an algebra of finite dimension over K and V an m:-module.

a) There exists an integer k (called the Loewy length of V) such that

and

b) Also

° = 50 C 51 C ... C 5k = V,

Proof a) The existence of k follows from the fact that J(m:) is nilpotent. If Vi = Vi +1 and i < k, then

a contradiction. b) From (5n+d5n)J(21) = 0, it follows immediately that 5n S;;

Anv(J(21t). We establish equality by induction on n. For n = 0, it is trivial. For n > 0, we have

by the inductive hypothesis. It follows that Anv(J(21t) S;; 5n • Thus Anv(J(21t) = 5n and in particular 5k = V.

If there exists i < k such that AnV(J(21)i) = Anv(J(21)i+1), then it follows from (VJ(21)k-i-1)J(21)i+1 = VJ(21)k = ° that

a contradiction. Thus

° = 50 C 51 C ... C 5k = V. q.e.d.

In many cases, the projective indecomposable KG>-module the head of which yields the unit representation of G> can be described grouptheoretically.

10.12 Theorem. Let K be afield of characteristic p. a) Suppose that ~ is a subgroup of G> of order prime to p. If P is a

projective indecomposable KG>-module the head of which yields the unit representation ofG>, then P is isomorphic to a direct summand ofVC», where V is a module for the unit representation of ~. In particular, dimK P :::; IG>: ~I·

b) Suppose further that G> possesses a p-complement .0. If V is the module for the unit representation of.o, then VC» is the projective indecomposable KG>-module' the head of which yields the unit representation of G>. The element

is an idempotent and eKG> ~ VC».

Proof a) Since p tl~l, V is a projective K~-module. By 7.17, vC» is a projective KG>-module.

Let W be the i-dimensional K<fi-module corresponding to the unit representation of <fi. We denote by k the multiplicity of W in the head of Vffi and by I the multiplicity of V in the soc1e of Wf>. Then I = 1, and by 4.13a),

If Vffi = P1 EEl ... EEl Pr with indecomposable projective K<fi-modules ~, then there exists some i such that ~/~J(K<fi) ~ W. Hence ~ ~ P. Finally,

b) By Dickson's theorem (7.16) we know that the order I~I ofa Sylow p-subgroup ~ of <fi divides dimK P. By a),

Thus Vffi ~ P. Since char K f 1.01, e is an idempotent in K<fi. Clearly ex = e for all

x E .0. If T is a transversal of .0 in <fi, then

eK<fi = E8 K(et). tET

If g E <fi and tg = xt' with x E .0 and t' E T, then

(et)g = ext' = et'.

But if V = Kv, then also (v ® t)g = v (8) t'. Hence eK<fi ~ Vffi . q.e.d.

10.13 Theorem. Suppose that m: is an algebra offinite dimension over K and that m: is an injective m:-module. (By 7.8, the group-ring of afinite group over any field has this property.) Let P be a finitely generated indecomposable projective m:-module.

a) Then the socle S(P) of P is the unique minimal submodule of P and PSr(m:) = S(P). In particular, S(P) is irreducible.

b) IfP1 and P2 are indecomposable projective m:-modules, both finitely generated, then P1 ~ P2 if and only ifS(P1) ~ S(P2)'

c) If V is an irreducible m:-module, there exists an indecomposable projective m:-module P such that S(P) ~ V.


Proof a) Since ill: is an injective ill:-module, so is any finitely generated free ill:-module and hence also any finitely generated projective ill:-module.

We consider the socle S = S(P) of P. Suppose IX E Homm(S, S). If we denote by I the embedding of S in P, we have the following diagram.

As P is injective, there exists [3 E Homm(p, P) such that for all S E S,

s[3 = SI[3 = SIXI = SIX.

Hence restriction to S is an algebra epimorphism of Homm(p, P) onto Homm(S, S). If S is reducible, it is decomposable and then Homm(S, S) is not a local ring. As every epimorphic image of a local ring is again local, we conclude that Homm(p, P) is not a local ring, in contradiction to the indecomposability of P. Thus S is irreducible.

As P is a direct summand of a free ill:-module F, we obtain

S = S(P) = S(F) II P = FSr(ill:) II P = PSr(ill:).

b) Now let y be an isomorphism ofS(pl) onto S(P2). From the diagram

it follows that there exists a homomorphism 15 of P1 into P2 such that y is the restriction of 15 to S(Pl)' As

kerJ II S(Pl) = kery = 0

and S(Pl) is the only minimal submodule of P1, we conclude that ker 15 = O. Hence P1 15 ~ P1 is an injective submodule of P2 • Thus P2 = P1J EB P~ for some P~. The indecomposability of P2 forces P2 = P1 15 ~ Pl'

c) By 10.9, the number of isomorphism types of indecomposable


projective finitely generated m:-modules is equal to the number oftypes of irreducible m:-modules. Thus by b) all isomorphism types of irreducible m:-modules must appear among the soc1es of indecomposable projective m:-modules. q.e.d.

We remark that in a) P is the injective envelope of S(P) in the sense of the theorem of Eckmann and Schopf (see MACLANE [1], p. 103). Then b) follows from the fact that the injective envelope is uniquely determined.

Every module is an epimorphic image of a projective module. But in general an indecomposable module need not be an epimorphic image of an indecomposable projective module. We can state the following lemma.

10.14 Lemma. Let V =f. 0 be a Kffi-module of finite dimension over K. a) V is an epimorphic image of an indecomposable projective Kffi

module P if and only if the head H(V) of V is irreducible. Then H(V) ~ H(P).

b) V is isomorphic to a submodule of an indecomposable projective Kffi-module P if and only if the socle S(V) of V is irreducible. Then S(V) ~ S(P).

Proof a) Suppose first that V = Po: for an indecomposable projective module P and 0: E HomKo;(p, V). By 10.9c), H(P) is irreducible, hence H(V) ~ H(P) is also irreducible.

Now suppose conversely that H(V) = V /VJ(Kffi) is irreducible. Let P be the indecomposable projective Kffi-module for which H(P) ~ H(V) (1O.9b)). Let vp denote the natural epimorphism of Ponto H(P), Vv the natural epimorphism of V onto H(V) and 0: an isomorphism of H(P) onto H(V).

Vp

pi 'T V-----+.H(V)-----+.°

Vv

As P is projective, there exists a mapping (J E HomKo;(p, V) such that

Hence im f3 ~ ker vv. As ker Vv is the only maximal submodule of V, this implies that im f3 = v.


b) As the socle of P is irreducible (10.13a)), the existence of a monomorphism of V into P implies that S(V) = S(P) is irreducible.

Suppose conversely that S(V) is irreducible. By 7.8, K(fj is injective. Hence by 10.13c), there exists an indecomposable projective module P for which there is an isomorphism IX of S(V) onto S(P).

SiP) -----to P

,I 0------->. S(V) -----to V

As P is injective (7.8), there exists a f3 E HomKm(V, P) such that VIX = vf3 for all v E S(V). Hence

ker f3 n S(V) = 0

and thus ker f3 = O. Hence f3 is a monomorphism of V into P. q.e.d.

If K is a splitting field for (fj and char K does not divide I (fj I, the structure of the free K(fj-module K(fj is very simple, and we know from the Wedderburn theory that any irreducible module V occurs in K(fj with multiplicity dimK V. If on the other hand char K divides l(fjl, this holds only for the multiplicity in K(fjjJ(K(fj). Our next steps are directed towards a determination of the multiplicities of the composition factors in K(fj and in the indecomposable projective K(fj-modules.

10.15 Definition. Let ~ be an algebra of finite dimension and let

be a decomposition of ~ into indecomposable right ideals ei~' where the ei are mutually orthogonal idem po tents such that 1 = e1 + ... + en' We denote by cij the multiplicity of the irreducible ~-module ej~ as a composition factor of the indecomposable projective ~-module ei~' Since

certainly Cii ~ 1. Suppose the numbering is so chosen that e 1 ~, .•• , ek ~ are representatives of the isomorphism types of the irreducible ~modules. The k x k matrix C = (Ci) is called the Cartan matrix of ~.


Clearly C is the unit matrix if and only if each ei ~ is irreducible, that is, if ~ is a completely reducible ~-module and therefore a semisimple algebra.

10.16 Lemma. Suppose that ~ is a K-algebra, that e is an idempotent in ~ and V an ~-module. Then there is a K-isomorphism of Hom~(e~, V) onto Ve.

Proof We define a mapping fJ of Hom~(e~, V) into V by r:xfJ = er:x for r:J. E Hom~(e~, V). Since er:x E V, we have

Thus fJ is a K-linear mapping of Hom~(e~, V) into Ve. Given an element ve EVe, define the mapping r:J. of e~ into V by

ear:J. = vea (a E ~).

Then r:x E Hom'!l(e~, V) and r:J.fJ = er:x = ve. Thus fJ is an epimorphism of Hom~(e~, V) onto Ve. If r:J.fJ = 0, then er:J. = 0 and

(ea)r:x = (er:x)a = 0

for all a E ~, so r:J. = o. Thus fJ is an isomorphism. q.e.d.

10.17 Lemma. Suppose that ~ is an algebra offinite dimension over K. Let

be a decomposition of ~ into indecomposable right ideals ei~' If V is a finitely generated ~-module, the multiplicity m of the irreducible module ei~ (where ~ = ~/J(~)) as a composition factor of V is determined by

Proof We prove this by induction on dim V. Let VI be a maximal submodule of V. As ei ~ is projective, we obtain from the exact sequence

an exact sequence


This shows that

We have

Ifml is the multiplicity ofeim as a composition factor ofVl , then by the inductive hypothesis,

This proves the first part of the assertion; the second part follows by 10.16. q.e.d.

Lemma 10.17 gives us a formula for the Cartan numbers Cij'

10.18 Theorem. Suppose that m is a K-algebra offinite dimension and that K is a splittingfieldfor m = mjJ(m). Suppose further that m = EBi=l eim is a direct decomposition of m into indecomposable m-modules ei m as in 10.3.

a) cij = dimKeimej' b) The multiplicity of eim as a direct summand ofm in the sense of the

Krull-Schmidt theorem is dimKeim. c) The multiplicity of ei m as a composition factor of m is dimK mei' d) If Cij = cjJor all i, j, then dimK ei m = dimK mei' (This hypothesis

about the symmetry of the Cartan matrix is always satisfied in grouprings, as we shall see in 1l.6d).)

Proof a) By Definition 10.15, cij is the multiplicity ofejm as a composition factor of ei m, and by 10.17 this is dimK ei mej.

b) By the Wedderburn theory, the multiplicity of eim in m is dimKeim. By 10.9a), the indecomposable projective m-module eim is determined to within isomorphism by its head eim, so dimKeim is also the multiplicity of ei m as a direct summand of m.

c) This follows at once from 10.17. d) We count the multiplicity of ei m as a composition factor of ~ in

two ways.

§ 11. Frobenius Algebras and Symmetric Algebras 165

On the one hand this multiplicity is dimK ~ei' by c). On the other hand, the multiplicity of ej~ as a direct summand of ~ is dimKej~ by b), and the multiplicity ofei~ as a composition factor of ej~ is Cji. Thus if as in 10.15 el ~, ... , ek2I are the isomorphism types of irreducible 2I-modules, then

k k

dimK 2Iei = L Cji dimKej2I = L cij dimKej2I = dimKei~· q.e.d. j=l j=l

Exercises

45) Let 2I be the ring of all n x n triangular matrices with coefficients in a field K. Find J(2I), Sr(2I) and Sz(2I). Compare the right ~-modules 2IjJ(2I) and Sr(2I) and the left modules 2IjJ(2I) and S,(~).

46) Let ~ be an algebra of finite dimension and P a projective finitely generated 2I-module. We put 15 = PjPJ(2I). If rI. E Hom'II(P, P), then the mapping a defined by

(x + PJ(2I))a = Xrl. + PJ(2I) (x E P)

is an element of Hom2l(P, 15). Show that the mapping rI. -t a is an algebraepimorphism of Hom'II(P, P) onto Homm(P, 15) with kernel Hom'II(P, PJ(2I)) = J(Hom'II(P, P)).

47) Let 2I be an algebra of finite dimension and let 9l be a right ideal of 2I such that 9l 5li: J(2I). Show that 9l contains an idempotent e "# o.

§ 11. Frobenius Algebras and Symmetric Algebras

In several places we have already encountered a certain duality of grouprings. We shall now study this duality in the more general context of Frobenius algebras and symmetric algebras.

11.1 Definition. Let ~ be a K-algebra of finite dimension. a) ~ is called a Frobenius algebra if there exists a non-singular K

bilinear form (,) on ~ such that for all a, b, C E 2I we have

(ab, c) = (a, bc).


b) ~ is called a symmetric algebra if there exists a symmetric nonsingular K-bilinear form (,) on ~ such that for all a, b, c E ~

(ab, c) = (a, be).

For Lie algebras over a field of characteristic 0, the existence of a bilinear form with the properties introduced in a) is equivalent to semisimplicity; this is an essential point of departure in E. Cartan and Killing's determination of all simple Lie algebras over the field of complex numbers. For associative algebras, however, the class of Frobenius algebras is larger than the class of semisimple algebras even in characteristic 0 (see Exercise 50).

11.2 Theorem. If K is an arbitrary field, K(f; is a symmetric algebra.

Proof For gl> g2 E (f;, we put

and extend this to a bilinear form on K(f;. Obviously this is a symmetric K-bilinear form. From

it follows easily that

(ab, c) = (a, be)

for all a, b, c E K(f;. It remains only to show that (,) is non-singular. Suppose that a = LgE(!; agg and (a, b) = 0 for all bE K(f;. Then for all h E (f; we have

Thus a = 0 and (,) is non-singular. q.e.d.

In the following theorem we collect some basic properties of Froben ius algebras and use the notation of 10.7 for annihilators.

11.3 Theorem. Let 21 be a Frobenius algebra with bilinear form (,). a) If'R is a right ideal of 21, then

§ II. Frobenius Algebras and Symmetric Algebras 167

L(9\) = raja E 21, (a, r) = 0 for all r E 9\}.

If,B is a left ideal of 21, then

R(,B) = raja E 21, (I, a) = 0 for alii E ,B}.

b) For every right ideal 9\ of21 we have

(1)

and

(2) R(L(9\» = 9\.

Similarly, if,B is a left ideal of 21, we have

(1')

and

(2') L(R(,B» = ,B.

c) The mapping 9\ --+ L(9\) is a duality of the lattice of right ideals of 21 onto the lattice of left ideals of21, that is,

and

for all right ideals 9\1' 9\2 of21. Similarly ,B --+ R(,B) is a duality of the lattice of left ideals of 21 onto the lattice of right ideals of 21.

Proof a) Let 9\ be a right ideal of 21 and put

w- = raja E 21, (a, r) = 0 for all r E 9\}.

Suppose x E 9\1.. Then for all r E 9\ and a E 21 we have

o = (x, ra) = (xr, a).


Since (,) is non-singular, it follows that xr = ° for all r E 9l. Thus x E L(9l) and hence 9l.L £; L(9l). If conversely x E L(9l), then for all r E 9l,

° = (xr, 1) = (x, r).

This shows that L(9l) s 9l.L. Together we obtain 9l.L = L(9l). The assertion for left ideals follows similarly.

b) Since (,) is non-singular, it follows from well-known assertions about bilinear forms that

The assertion (1') is proved similarly. Since L(9l)9l = 0, we have certainly 9l £; R(L(9l)). We obtain from (1) and (1')

dimK R(L(9l)) = dimK ill: - dimK L(9l)

= dimK ill: - (dimK ill: - dimK 9l) = dimK 9l.

This shows that R(L(9l)) = 9l for all right ideals 9l of ill:. Similarly L(R(£)) = £ for all left ideals £ of ill:.

c) Trivially we have

and

With b), it follows that

L(9l 1 n 9l2) = L(R(L(9l 1)) n R(L(9l 2))) = L(R(L(9l 1) + L(9l 2)))

= L(9l 1) + L(9l2)·

The corresponding assertion for left ideals follows similarly. q.e.d.

11.4 Remarks. Here we cannot go into the many relations between the assertions made in 11.3; for this we must refer to the theory of Frobenius and quasi-Frobenius algebras (see CURTIS and REINER [1], chap. IX.) We confine ourselves to the following remarks.

a) In 7.9, quasi-Frobenius algebras were defined to be those algebras

§ II. Frobenius Algebras and Symmetric Algebras 169

m: of finite dimension for which m: is an injective right m:-module. It can be shown that this is equivalent to the assertions (2) and (2') in 1l.3b) for all left ideals j! and all right ideals 9l. Also the assertions of 1l.3c) characterize quasi-Frobenius algebras; indeed, any algebra of finite dimension which admits any duality of the lattice of right ideals onto the lattice of left ideals is a quasi-Frobenius algebra.

b) The assertions of 1l.3b) characterize Frobenius algebras. Another characterization in terms of dual modules can be given. If V is a right m:-module, then V* = HomdV, K) can be made into a left m:-module by putting

v(af) = (va)f

for v E V, f E V* and a E m:. Then Frobenius algebras are precisely those algebras m: of finite dimension for which the dual module m:* of the free m:-module m: on one generator is a free left m:-module with one generator.

We prove only one of the facts mentioned in 11.4.

11.5 Theorem. If m: is a Frobenius algebra, then m: is an injective right m:-module.

Proof. By Baer's lemma, it suffices to prove that if 9l is a right ideal of m: and rx E Hom2((9l, m:), then rx can be extended to an element of Hom2((m:, m:).

F or each a E m:, the mapping !a of 9l into m: given by r!a = ar for r E 9l is an m:-homomorphism of 9l into m:, and the mapping a --+ !a is a K-homomorphism of m: into Hom2((9l, m:) with kernel L(9l). Obviously, each !a can be extended to an element !~ of Hom2((m:, m:) simply by putting x!~ = ax for all x E m:. We show that

Let (,) be a bilinear form defined on the Frobenius algebra according to Definition 11.1a). We then define ¢ E HomK(m:, K) by putting a¢ = (a, 1) for a E m:. For rx E Hom2((9l, m:) we put rx' = rx¢. Thus rx' E HomK(9l, K). If rx ' = 0, then for all r E 9l and all a E m: we have

° = (ra)rx' = (ra)rx¢ = ((rrx)a)¢ = ((rrx)a, 1) = (rrx, a).

As (,) is non-singular, this implies rrx = 0 and thus rx = O. Hence a -+ a' is a K-monomorphism of Hom2((9l, m:) into Homd9l, K). Thus


By 11.3b),

dim K9t = dimKm: - dimKL(9t) = dimKm:jL(9t)

= dimdrala E m:}.

Hence Hom~(9t, m:) = {rala E m:}, and so each a in Hom~(9t, m:) can be extended to an ~ in Hom~(m:, m:). q.e.d.

We now collect those properties of symmetric algebras which are of interest for group-rings.

11.6 Theorem. Suppose that m: is a symmetric algebra. a) If.3 is a two-sided ideal in m:, then L(.3) = R(.3). b) Sl(m:) = Sr(m:). c) If m: = EBi=l ei m: is a direct decomposition of m: into indecom

posable right ideals ei m: according to 10.3, then eiSr(m:) is the socle of ei m: and

Thus head and socle of ei m: are isomorphic. d) Suppose that K is a splitting field for m:/J(m:). Then cij = Cji. The

Cartan matrix is thus symmetric. e) Either (i) ei m: is irreducible, Cii = 1 and cij = 0 for all j i: i or (ii) Cii ~ 2.

Proof a) On account of the symmetry of the bilinear form, it follows from 11.3a) that

L(.3) = {ala E m:, (a, i) = 0 for all i E .3}

= {ala E m:, (i, a) = 0 for all i E.3} = R(.3).

b) By lO.7c), Sr(m:) = L(J(m:)) and Sl(m:) = R(J(m:)). Thus Sr(m:) = Sl(m:), by a).

c) By 11.5, m: is an injective m:-module. Hence by lO.13, S(ei m:) = eiSr(m:) is an irreducible m:-module.

Put m: = m:/J(m:) and let m: = Ef)7=1 J;m: be the direct decomposition of m: into simple two-sided ideals J;m:, where T = 11 + ... + lk is a


decomposition ofthe unit element into mutually orthogonal central idempotents]; of m. We decompose the ];m into irreducible right ideals, say

];m = E!jeijm, j

where]; = Lj eij is a decomposition of the unit element]; of the algebra ];m into mutually orthogonal idempotents eij' By 10.2b), there exist mutually orthogonal idempotents eij in m such that 1 = Li,jeij and eij + J(m) = eij' Put

The /; are then mutually orthogonal idempotents in m, but they are not necessarily central. If V is an irreducible m-module and V/; :f= 0, then V/; = V]; = V and Vlj = V];lj = ° for allj :f= i; thus V may be regarded as an irreducible mkmodule and is therefore isomorphic to eil m, by V, 4.1. Thus to prove that eil Sr(m) ~ eil m, we show that eil Sr(m)/; :f= 0.

Suppose that eil Sr(m)/; = 0. We observe first that Sr(m)/; is a twosided ideal of m. To see this, observe that/;x - x/; E J(m) for all x E m, since]; is central in m = m/J(m). Since also Sr(m)J(m) = 0,

Since Sr(m) is a left ideal, it follows that Sr(m)/; is a two-sided ideal. Hence by a),

which shows that

But meil is a non-zero left ideal of m and therefore contains a minimal left ideal E. But then J(m)E = ° and

This is a contradiction, so c) is proved. d) From the decomposition

172

we obtain

VII. Elements of General Representation Theory

n n

m = EBeim = EBmej i=1 j=l

n

m = EB eimej, i,j=1

which is a direct decomposition of m as a vector space over K. For j =f. k and all a, b E m, we have

and from the symmetry of the form (,), it follows that for i =f. /,

The K-subspace (eime).L orthogonal to eimej therefore contains

It now follows from 1O.18a) that

Cij = dimKeimej = dimK m - dimK(eime).L

:::; dimK m - I dimK ek mel = dimK ej mei = Cji' (k,l) '" U, i)

Similarly we get Cji :::; Cij' so Cij = Cji' e) If eim is not irreducible, then eiSr(m) s; eiJ(m) by lO.3c). But also

eiSr(m) ~ eimjeiJ(m) by c), and so Cii ;2 2. q.e.d.

11.7 Remarks. a) It can be shown that the two socles Sr(m) and Sl(m) coincide even for quasi-Frobenius algebras.

b) In VIII, 2.13 it will be proved that if 3 is the augmentation ideal of the group-ring of a p-group over a field of characteristic p and 3 8 ::::> 3 8 +1 = 0, then 3 8 +1- i = L(3i). The proof uses the symmetry of the group-ring. By 10.11b), this shows that

o C 3 8 C 38 - 1 C ... c 3° = KG)

is the upper Loewy series of KG).


Exercises

48) If m is a symmetric algebra, then the full matrix ring (m)m is also a symmetric algebra.

49) Show that the tensor product of two Frobenius (or symmetric) algebras is again a Frobenius (or symmetric) algebra.

50) Let V be a finite dimensional vector space. Show that the exterior algebra of V is a Frobenius algebra. When is it a symmetric algebra?

51) If 0 =1= fE K[t], then K[t]/.{K[t] is a symmetric algebra.

52) Give an example of a symmetric algebra m and a two-sided ideal 3 in m such that m/3 is not a Frobenius algebra.

53) Let m be a K-algebra of finite dimension. Show that the following assertions are equivalent.

a) m is a Frobenius algebra. b) There exists f E HomK(m, K) such that the kernel of f contains no

proper right ideal of m.

54) Let m be a K-algebra of finite dimension. Show that the following assertions are equivalent.

a) m is a symmetric algebra. b) There exists f E HomK(m, K) such that the kernel of f contains

no proper right ideal of m andf(ab) = f(ba) for all a, b E m.

55) If an algebra m of finite dimension contains only one irreducible right ideal, then m is a Frobenius algebra.

56) (TSUSHIMA [1]) Let m be a symmetric algebra and choose f E HomK(m, K) as in exercise 54).

a) For any g E HomK(m, K), there exists a E m such thatg(x) = f(ax) for all x E m.

b) Let 3 be a two-sided ideal in m such that m/3 is a symmetric algebra and let g be an element of HomK(m/3, K) as in exercise 54. Then there exists an element a in the centre of m such that g(x + 3) = f(ax) for all x E m. Further L(3) = R(3) = am.


§ 12. Two-Sided Decompositions of Algebras

In § 10 we considered the direct decomposition of the group-ring 2l = KQj into indecomposable right ideals. By lifting idempotents it was shown that these decompositions correspond exactly to the direct decompositions of 2l = 2ljJ(2l) into irreducible right ideals. Unfortunately the situation is not so simple for decompositions of 2l into two-sided ideals, for a central idempotent of 2l cannot in general be lifted to a central idempotent of 2l.

First we prove a generalization of V, 3.8 about direct decompositions of algebras into two-sided ideals. '

12.1 Theorem. Let 2l be an algebra of finite dimension over a field K, a) There exists a direct decomposition

of2l into two-sided ideals ~i =1= 0, which are indecomposable as two-sided ideals. For i =1= j we have ~i~j = 0.

b) Suppose that 1 = fl + ... + ft, where Ji E ~i' Then the Ji are mutually orthogonal idempotents in the centre Z(2l) of 2l and ~i = Ji2l = 2lJi.

c) Z(2l) = fl Z(2l) EEl ... EEl ftZ(2l) is a direct decomposition of the commutative algebra Z(21) into indecomposable ideals JiZ(21) = ~i n Z(2l). EachJiZ(2l) is a local ring and so JiZ(2l)jJ(JiZ(2l» is afield.

d) For every 21-module V,

t

V = EBVJi i=1

is a direct decomposition of V into 2l-modules.Ifin particular V is indecomposable, then V = Vjj for some j.

e) For each right ideal 91 of2l we have

t

91 = EB (91 n ~;). i=1

In particular, each indecomposable right ideal of 2l lies in one of the ~i' f) If 211 and 212 are two-sided ideals of 2l and 2l = 211 EEl 2l2' then

211 and 212 are both direct sums of certain ~i' In particular, the ~i in a) are uniquely determined.

§ 12. Two-Sided Decompositions of Algebras 175

Proof a) The existence of a direct decomposition of the algebra m: into indecomposable two-sided ideals is clear, since m: has finite dimension over K. For i =I- j we have

b) For i =I- j we have fdj E mimj = O. Hence

t

ji = lji = L Jjji = ji2. j=1

Further for all a E m: we have

t t

L aji = al = la = L jia. i=1 i=1

Since aji and jia lie in mi, we see by comparing components that aji = jia. Thus ji E Z(m:). Since Jjmi £; mjmi = 0 for i =I- j, we obtain

Hence mi = jim: = m:fi. c) Obviously Z(m:) = EB:=1jiZ(m:) is a direct decomposition of

Z(m:). Suppose thatji = I' +!", whereI',!" are non-zero idempotents of Z(m:). Then

and since I', !" lie in Z(m:), I'm: and!"m: are two-sided ideals of m:. However this contradicts the indecomposability of mi = ji m: as twosided ideal.

jiZ(m:)jJ(jiZ(m:)) is a commutative semisimple algebra. Hence by Wedderburn's theorem

for certain fields K 1, ... , Kn. The Kj are minimal (right) ideals, so by 10.3, there exists a decomposition of jiZ(m:) as the direct sum of n ideals. Since jiZ(m:) is indecomposable, n = 1 and jiZ(m:)/J(jiZ(m:)) is a field. Thus jiZ(m:) is a local ring.


d) We have

t t

V = V1 = V L Ji £ L VJi £ V. i=1 i=1

Hence V = EBl=1 V fi. As Ji is central, V Ji is an m-rnodule.

e) By d), 9l = EBl=1 9lJi and

Thus

t

9l = EB (9l n ~i)' i=1

f) Bye),

t t

m = m1 ffi m2 = EB(m1 n ~i) ffi EB(m2 n ~i)' i=1 i=1

Hence for all i,

The indecomposability of ~i forces ~i £ m1 or ~i £ m2 . Thus

mj = EB ~i (j = 1, 2). !Il,~91J

A different aspect of two-sided decompositions is often useful.

q.e.d.

12.2 Theorem. Suppose that m is a K-algebra of finite dimension and that K is a splitting field for mjJ(m). With the notation of 12.1, the following assertions hold.

a) J(Z(m:)) = J(m:) n Z(m:). b) The ring JiZ(m:)jJ(JiZ(m:)) is isomorphic to K. Thus Z(m)jJ(Z(m:))

is the direct sum of t ideals each isomorphic to K. c) There are precisely t algebra homomorphisms C\:1' ... , C\:t of Z(m:)

onto K. With appropriate numbering we have

JiC\:i = 1 and jjC\:i = 0 (j =1= i).

Further

ker IX; = J Ct;Z(m:)) ffi EB .fjZ(m:). j,';

Proof a) J(m:) (") Z(m:) is a nilpotent ideal of Z(m:) and is therefore contained in J(Z(m:)). Conversely, if x E J(Z(m:)), then x is nilpotent. Since x E Z(m:), xm: is a nilpotent ideal of m:, and it follows that x E J(m:) (") Z(m:).

b) Put L; = .fiZ(m:)/J(.fiZ(m:)). By 12.1c), L; is a field. Since Z(m:) = EBl;l.fiZ(m:), we have

Z(m:)/J(Z(m:)) ~ Ll ffi ... ffi Lt·

Using a), we obtain

Now (Z(m:) + J(m:))/J(m:) is a subalgebra of Z(m:/J(m:)), and since K is a splitting field for m:/J(m:), we conclude that m:/J(m:) is a direct sum of complete matrix rings over K. Hence Z(m:/J(m:)) is isomorphic to a direct sum of fields each isomorphic to K. Thus there is a K-algebra monomorphism (j; of L; (i = 1, ... , t) into a direct sum K ffi ... ffi K. We denote by 1tj the projection of K ffi ... ffi K onto the j-th summand. Then there existsj such that (j;TCj #- O. Since ker (J;TCj is an ideal in Li, (JiTCj is a K-linear isomorphism of Li onto K. Thus Z(m:)/J(Z(m:)) is a direct sum of t ideals each isomorphic to K.

c) By b), .fiZ(m:) is spanned over K by J(.fiZ(m:)) and.fi. Thus Z(m:) is spanned over K by fl' ... ,j; and J(Z(m:)). Also,fl' ... ,j; are linearly independent modulo J(Z(m:)). Thus for 1 :$ i :$ t, there exists a Klinear mapping lXi of Z(m:) onto K such that .fjlXi = 0 for j #- i, .filXi = 1 and J(Z(m:))lXi = O. Clearly lXi is an algebra homomorphism and ker lXi is as stated.

Conversely, suppose that IX is an arbitrary algebra homomorphism of Z(m:) onto K. Since 1 = fl + ... + j;, there exists an i such that .fi1X #- o. Since for j #- i we have

o = (.fi.fj)1X = (.fiIX}(.fjIX),

we obtain.fj1X = O. Obviously J(Z(m:)) 5; ker IX, so IX = IX;. q.e.d.

12.3 Remark. If K<ll is semisimple and K is a splitting field for K<ll, the number t in 12.1 is the class-number h(<ll) of <ll (V, 5.1). A corresponding

description of t is unfortunately completely lacking in the case when char K divides l(fjl. It happens sometimes that t = 1, which means that K(fj is an indecomposable algebra. This seems to be rare for simple groups, but if (fj is isomorphic to the Mathieu group Wl22 and K is an algebraically closed field of characteristic 2, then K(fj is indecomposable (JAMES [1]). A necessary and sufficient condition for t = 1 is known only for p-constrained groups; this condition is that (fj has no non-identity normal p'-subgroups, that is, Op,(fj) = 1 (see 13.5). The existence of group-rings K(fj with t = 1 shows that central idempotents of K(fj/J(K(fj) cannot in general be lifted to central idempotents of K(fj.

If 21 is a semisimple algebra, the two-sided ideals ~i of 12.1 are just the sums of all the irreducible right ideals of a given isomorphism type. In general the connection between the two-sided decomposition in 12.1 and the decomposition of 21 into right ideals in 10.3 is not so simple.

12.4 Theorem. Let 21 be an algebra of finite dimension over a field K. Let 21 = EBi=l ei 21 be a decomposition of 21 into indecomposable right ideals. Put ei21 '" ej 21 if there exists a sequence

ei 21 = ei 21, ei 21, ... ,ei 21 = eJ.21, 12m

such that any two neighbouring right ideals in this sequence have at least one composition factor in common. Obviously'" is an equivalence relation. If d l , ... , d s are the equivalence classes of"', then s = t. If we put

21j = EB ei 21, ei~E .<JIj

then 21 = EBi=l 21j is the decomposition of 21 into indecomposable twosided ideals of 12.1.

Proof If ei21 and ej21 have the composition factor ek21 in common, then by 10.17, ei21ek i= 0 and ej21ek '# O. By 12.1e), there exist ideals ~, ~', ~" in the set {~l' ... , ~t} such that ei21 ~ ~, ej 21 ~ ~' and ek21 ~ ~". It follows that 0 i= ei21ek ~ ~~", so ~ = ~"; also 0 i= ej21ek ~ ~'~", so ~' = ~". Hence ei21 + ej21 ~~. Repeated application of this argument shows that the right ideal

21j = EB ei 21 ej'!lEdj

is contained in a certain ~j" Obviously 21 = EBl=121j'


We show now that 2lj is a two-sided ideal of 21. To do this, suppose that ei 21 ~ 2lj and ek 21 ~ 21" where j -=1= 1. Then ei 21 and ek 21 have no composition factor in common. In particular, ek 21 has no composition factor isomorphic to ei21, so it follows from 10.17 that ek21ei = O. This shows that 2l/21j = 0 for j -=1= 1. Thus

and so 2lj is a two-sided ideal in 21. Since 2lj is also a direct summand of 21, it follows from 12.1f) that 2lj is a direct sum of certain ~i' Since 2lj ~ ~j" we obtain 2lj = ~j" q.e.d.

12.5 Theorem. Let 21 be a K-algebra of finite dimension and suppose that K is a splittingfieldfor 2l/J(21). Further let

be a direct decomposition of 21 into indecomposable right ideals. a) If ei21 and ej 21 do not lie in the same two-sided indecomposable

direct summand ~k of 21, then cij = O. Thus the Cartan matrix can be written in the form

C = (~' ~, ~ ~), o 0 0··· Ct

where Ci corresponds to the two-sided ideal ~i (i = 1, ... , t). b) The matrices Ci in a) are indecomposable in the sense that the right

ideals ej 21 in ~i cannot be ordered such that Ci has the form

with square matrices Ci1 , Ci2 .

Proof a) Suppose that ei21 ~ ~i' and ej 21 ~ ~j" where ( -=1= j'. Then


Hence by 10.17

b) Suppose that an ordering is possible for which Ci is decomposable. Then the ejm which belong to Cil form a union of equivalence classes for the relation", of 12.4. But this contradicts 12.4. q.e.d.

We now introduce the important concept of blocks.

12.6 Definition. Let K be a field of characteristic p. Let

t t

Kffi = EB~i = EBfiKffi i=l i=l

be the decomposition of Kffi defined in 12.1. a) ~i is called a block ideal andfi a block idempotent of Kffi. If K is a

splitting field for Kffi/J(Kffi), we call the algebra homomorphism (Xi of Z(Kffi) into K for whichfi(Xi = 1 (cf. 12.2c)) the block character of ~i'

b) For i = 1, ... , t, the block 14i is defined to be the class of all Kffi-modules V for which Vfi = V. Sincefijj = Ofori # j,everynon-zero Kffi-module lies in at most one block. If V is indecomposable, then V belongs to a block by 12.1d).

A Kffi-module belongs to a block if and only if all its indecomposable direct summands belong to the same block. A Kffi-module belongs to a block if and only if all its composition factors belong to the same block.

Every block contains irreducible modules, for 14i contains the composition factors of ~i'

c) Let V be a Kffi-module for the trivial representation of ffi of degree 1. By b), V belongs to a block, and we choose the numbering so that this blockis8i1 • ThusVfl = V. Wecal18i1 theprincipalblockofKffi.

For the proof of an important property of central idempotents in group-rings, we need a lemma which is centnil for some deeper results in modular representation theory.

12.7 Lemma (R. BRAUER [6]). Suppose that K is a field of characteristic p, that '13 is a p-subgroup of ffi and that ~ is a subgroup for which C(fj('13) ::;: ~ ::;: N(fj('13). We define a K-linear mapping y of Z(Kffi) into K~ by putting

C~c x)Y = XEC~Cm('fl~

for each conjugacy class C of ffi. Then Y is an algebra homomorphism of Z(Kffi) into Z(K~), and the kernel of Y is spanned by the class-sums of those conjugacy classes C of ffi for which C ('\ C(!j(~) = 0.

Proof. Since C(!j (~) <:J ~, the set C ('\ C(!j(~) is ~-invariant for any conjugacy class C of (fj. Thus C ('\ C(!j(~) is a union of conjugacy classes of ~. Hence Y is indeed a K-linear mapping of Z(K(fj) into Z(K~). For ~ = 1, we have ~ = (fj and the assertion is trivial. Suppose henceforth that ~ '# 1.

Let Cl , ... , Ch be the conjugacy classes of (fj. We put C; = Ci - Ci ('\ C(!j(~), Ci = LXEC, x and c; = LXEC; x. As {C1' ... , Ch} is a K-basis of Z(Kffi), we have

h h

(CiY + c;)(CjY + c)) = CiCj = L aijlc/ = L aij/(c/Y + C;) /=1 /=1

for certain elements aij/ in K. The elements of (fj appearing in (CiY)c) and in c;(CjY) obviously all lie outside C(!j(~), and all elements appearing in (CiY)(CjY) lie in C(lj(~). We show that no element of C(lj(~) appears in c; ci; it will then follow by comparing coefficients that

h

(CiY)(CjY) = L aij/(cIY) = (CiCj)Y' /=1

Suppose that gEC(!j(~), and for fixed i,j write

m = {(x,y)lxEC;,YECi,xy = g}.

We have to show that Iml == 0 (p), for g appears in c;ci with coefficient Iml. Since g E Cm(~), a permutation representation of ~ on m is defined by putting (x, y) ~ (XU, yU) for u E ~. Since C; ('\ C(!j(~) = 0, there is no orbit oflength one. The length of any orbit of ~ on m is thus a power of p larger than 1. Hence Iml == 0 (p).

We still have to prove the assertion about the kernel of y. If C is a conjugacy class in (fj for which C ('\ C(!j(~) = 0, then obviously (LXECX)Y = O. Since no element of C(!j(~) appears in more than one CiY, it follows from

that each aiciy = O. Thus ai = 0 unless C i ('\ C(!j(~) = 0. q.e.d.

12.8 Theorem (OSIMA [1J). Suppose that char K = p and that f = L9E(jj agg is an idempotent in Z(Kffi). If ag i= 0, then 9 is a pi-element.

Proof(PAsSMAN [5J). Suppose that ag i= 0, but that p divides the order of g. Suppose that 9 = uv = vu, where u is a p-element and v is a p' -element. We apply 12.7 with ~ = (u) and f, = Cf5j(u). Then l' = fy is an idempotent in Z(Kf,). Iffy = LhEf> bhh with bh E K, then from 9 E C(jj(u) it follows that bg = ag i= O.

Let n be a natural number such that pft ~ jf,j and pn - 1 is divisible by the order of v. Put c = v-If' and write c = LhEf>chh. Then since f E Z(Kf,), we have

P" -p"j'p" -If' C = V = V = c.

We use the subspace Q(Kf,) of Kf, defined as in 3.1 by

By 3.3,

Q(Kf,) = <ab - ba j a, b E Kf,).

c = cpo = L cf' hP" + r hEf>

for some r E Q(Kf,). Since pn ~ jf,j, all the hP" are pi-elements. Since l' = VC, we get

Hence if we write r = LhEf>rhh, then ru i= 0. But by definition of Q(Kf,), r is a linear combination of elements of the form xy - yx with x, y E f,. Thus there exist x, y E f, such that xy = u, but yx i= u. However this is impossible since u lies in the centre of f,. q.e.d.

12.9 Theorem. Let m be a normal subgroup of ffi and let K be a field of characteristic p.

a) Suppose that m is a p'-group and put

Then f is a central idempotent of Kffi and fx = f for all x E m. Further,

Kffi = fKffi EB (1 - f) Kffi


is a direct decomposition of KG) into two-sided ideals. The mapping rx of fKG) onto K(G)jill) defined by (fg)rx = illg (g E G)) is a K-algebra isomorphism and also an isomorphism of KG)-modules.

b) The following assertions are equivalent. (i) ill is a pi_group.

(ii) There is an idempotent e in KG) such that eKG) and K(G)jill) are isomorphic KG)-modules.

(iii) There is a right ideal 9t in KG) such that KG) = ker 13 EB 9t, where 13 is the natural epimorphism of KG) onto K(G)jill)for whichgf3 = gill for all g E G).

Proof a) Obviously fx = f for all x E ill, so f is an idempotent. Since ill is the union of certain conjugacy classes of G), f lies in the centre of KG). If T is a transversal of ill in G), then since fx = f for all x E ill, we obtain

fKG) = EB K(ft). IE T

If g E G) and tg = yt' (t, t' E T, y E ill), then

(ft)g = ft ' and (illt)g = illt'.

Hence there is a KG)-isomorphism rx of fKG) onto K(G)jill) such that (ft)rx = illt. It follows that (fg)rx = illg for all g E G). From

we conclude that rx is also an algebra isomorphism. b) By a), (i) implies (ii). If (ii) holds, then since

KG) = eKG) EB (1 - e)KG)

is a decomposition of KG) into right ideals, eKG) is a projective KG)module. Since

eKG) ~ K(G)jill) ~ KG)jker 13,

it follows that KG) = ker 13 EB 9t for some right ideal 9t. If (iii) holds, 9t is a projective right ideal of KG) and

9t ~ K(f)jker 13 ~ K((f)jill).

It follows from 7.18 that ill is a pi_group. q.e.d.


Exercise

57) Let V be an indecomposable K(f)-module. If V does not belong to the principal block PAl' then Hi«f), V) = 0 for all i ~ 1. (For i = 1 use Exercise 34. For i > 1 consider an exact sequence

where P is projective, and P and W belong to the same block as V. Then use W«f), V) ~. Hi- 1«f), W).)

§ 13. Blocks of p-Constrained Groups

13.1 Lemma. Let K be afield of characteristic p. a) Let f1 be the block idempotent for the principal block of K(f). Then

f1 (1 - g) = 0 for all g E Op,«f)). b) If K(f) is an indecomposable algebra, then Op,«f)) = 1.

Proof a) Let

f = IOp,«f))1-1 L g. 9 EO,.(Ii;)

By 12.9a), f is a central idempotent of K(f), so fd and f1 (1 - f) are central idempotents of K(f). Sincef1 K(f) is an indecomposable two-sided ideal of K(f), either fd = 0 or f1 (1 - f) = O. But if Kv is the module for the trivial representation of (f), then vf = v, so vfd = Vf1 =! 0 and hence fd=! O.Thusf1(1.-f) = Oandfd = f1·Sincefg = ffor all g EOp,«f)), it follows that f1g = fdg = fd = f1 .

b) Since

is a decomposition of K(f) into two-sided ideals and K(f) is indecomposable, 1 - f1 = 0 andf1 = 1. Hence by a), Op,«f)) = 1. q.e.d.

The converse of 13.1b) is in general false, for it can be shown that if (f) is the alternating group of degree 5 and p = 3, then Op,«f)) = 1 but (f) has 3 blocks (Exercise 69). However, the converse is true for p-constrained groups. To prove this, we need the following lemma.

§ 13. Blocks of p-Constrained Groups 185

13.2 Lemma (COSSEY, GASCHVTZ [1]). Let K be afield of characteristic p and let V be an irreducible K(fj-module. Then the following assertions are equivalent.

a) V belongs to the principal block. b) If C is any conjugacy class of (fj and c = Lgecg, then

V(c - ICi) = O.

Proof Let 3 be the K-space spanned by all c - I C I, where C runs through the conjugacy classes of (fj and c = Lgecg. Thus dimK 3 = h - 1, where h is the class number of (fj. So 3 is a subspace of Z(K(fj) of codimension 1. Condition b) is equivalent to V3 = O.

Let

be the augmentation ideal of K(fj. Then 3 ~ 21, so 3 ~ 21 II Z(K(fj). But Z(K(fj) !l;f 21, since 1 E Z(K(fj) and 1 ~ 21. Thus 3 = Z(K(fj) II 21. Suppose

t

K(fj = EBJiK(fj i=1

as in 12.1. Now K(fjj2l is a module for the trivial representation of (fj, so (K(fjj2l)Ji = 0 for all i > 1. Thus Ji E 21 and so Ji E 3 (i > 1). Since

t

Z(K(fj) = EBJiZ(K(fj), i=1

it follows that

t

3 = 31 E9 EBJiZ(K(fj), i=2

where 31 = f1 Z(K(fj) II 3 = f1 Z(K(fj) II 21. Hence dirndl Z(K(fj)/31 = 1 and 31 is a maximal ideal of the ringf1Z(K(fj). But by 12.1c),JIZ(K(fj) is a local ring, so 31 = J(f1 Z(K(fj)) and the elements of 31 are nilpotent elements of Z(K(fj). Hence 31 ~ J(K(fj), and since V is irreducible, V31 = O. Thus

t t

V3 = V31 + L VJiZ(K(fj) = L Vfi. i=2 i=2


Thus V3 = 0 if and only if Vj; = 0 for all i 2': 2. Since V necessarily lies in a block, it follows that V3 = 0 if and only if Vfl = V. q.e.d.

13.3 Definition. A group (f) is called p-constrained if

We remark that by VI, 6.5 every p-soluble group is p-constrained.

13.4 Theorem. Let K be a field of characteristic p. Then the intersection of the kernels of all irreducible K(f)-modules is the largest normal p-subgroup Op(f) of (f).

Proof We denote by Sl the intersection of all the kernels of irreducible K(f)-modules. Then Op(f) ::; Sl by V, 5.17. Conversely, it follows from VI, 7.20 that Sl consists only of p-elements, hence Sl ::; Op(f). q.e.d.

13.5 Theorem (COSSEY, FONG, GASCHUrz). Let (f) be a p-constrained group and let K be a field of characteristic p. Then K(f) is an indecomposable algebra if and only ifOp,(f) = 1.

Proof If K(f) is indecomposable, then Op,(f) = 1 by 13.1b). Suppose that Op,(f) = 1. It has to be shown that (f) has only one

block. Since every block contains irreducible K(f)-modules, it is sufficient to show that any irreducible K(f)-module V belongs to the principal block. We shall do this by using 13.2. Thus we show that if C is any conjugacy class of (f) and c = Lg EC g, then V (c - Ie/) = o.

If C £; Op(f), then V(1 - x) = 0 for all x E C by 13.4, so V(c - lei) = o. Suppose that C ~ Op(f). Choose gEe, then g ~ Op(f). Since Op,(f) = 1, it follows from the assumption that (f) is p-constrained that C(fj(Op(f)) ::; Op(f). Thus Op(f) ~ C(fj(g). Let

pn = IC(fj(g)Op(f): C(fj(g)1 = IOp(f): Op(f) II C(fj(g)l,

so pn > 1. Let 5 be a transversal of Op(f) II C(fj(g) in Op(f) and let T be a transversal of C(fj(g)Op(f) in (f). Thus 151 = pn and 5T is a transversal of C(fj(g) in (f). Hence

As

§ 13. Blocks of p-Constrained Groups 187

and Op(G)) operates trivially on the irreducible KG)-module V (by 13.4), we obtain

vc = I I vgSf = lSI L vg f = pn I vg f = O. fET SES fET fET

But also ICI = IG): Co;(g)1 == O(p), so V(c - ICI) = o. q.e.d.

13.6 Theorem. Let G) be a p-constrained group and put 91 = Op,(G)). Let K be a field of characteristic p. Then

is the central idempotent fl of KG) belonging to the principal block of G).

Proof We write

By 12.9, f is a central idempotent in KG) and there is an algebra isomorphism (3 off KG) onto K(G)/91) such that (fg){3 = 91g for all g E G). But since Op,(G)/91) = 1 and G)/91 is also p-constrained, K(G)/91) is an indecomposable algebra by 13.5. ThusfKG) is indecomposable, sofKG) is a block ideal of KG). If Kv is the module for the trivial representation of G), then vf = v, so Kv belongs to the block defined by f Thus f belongs to the principal block of G). q.e.d.

13.7 Theorem (FONG, GASCHUTZ [1]). Let G) be a p-constrained group and let K be afield of characteristic p. Let V be an irreducible KG)-module. Then the following conditions are equivalent.

a) V belongs to the principal block of G). b) Op,(G)) operates trivially on V. c) 0p',p(G)) operates trivially on V.

Proof Let fl be the central idempotent belonging to the principal block of G). If a) holds, then Vfl = V, so Op,(G)) operates trivially on V by 13.1. Thus a) implies b).

If b) holds, V can be regarded as an irreducible K(G)/Op,(G)))-module. It follows from 13.4 that 0p',p(G)) operates trivially on V. Thus b) implies c).

And if c) holds, then by 13.6, Vfl = v for all v E V. Thus V belongs to the principal block. Hence c) implies a). q.e.d.


From 13.7 one of the rare statements about tensor products follows.

13.8 Theorem (FONG, GASCHlJTZ [1]). Let (f) be a p-constrained group and let K be a field of characteristic p. If VI' V2 are irreducible K(f)modules in the principal block fJB1 of (f), then all the composition factors of VI ®K V2 belong to fJBI ·

Proof By 13.7, Op,(f)) operates trivially on VI and on V2 , hence also on VI ®K V2 · Thus Op,(f)) operates trivially on all the composition factors of VI ®K V2 and these therefore belong to the principal block by 13.7.

q.e.d.

We derive another result from 13.2.

13.9 Theorem (COSSEY, FONG, GASCHDTZ~. Let K = GF(p) and let V be an irreducible K(f)-module isomorphic to some chief factor IDl/91 of (f). Then V belongs to the principal block.

Proof By 13.2, we have to show that

(1)

for all x E IDl and for every conjugacy class C of (f). If x g X-I E 91 for every x E IDl and gEe, this is clear. Suppose then that there exists gEe such that IDl/91 is not centralized by g91. Let :1)/91 = CID/91 (g91); thus IDl i :1).

Let S be a transversal of:1) (\ IDl in IDl; thus lSI is a power ofp greater than 1. Let R be a transversal of IDl:1) in (f). Then S R is a transversal of:1) in (f). Finally let T be a transversal of Cm(g) in :1). Then TSR is a transversal of Cm (g) in (f) and

C = {lsrlt E T, s E S, r E R}.

Since t E :1), gt == g mod 91. Since also sEIDl and IDl/91 is Abelian,

for all x E IDl, so

mod 91,

§ 14. Kernels of Blocks 189

since I S I is a power of p greater than 1. And again, since I C I = IT II S II R I, also x 1c == 1 mod 91. Thus (1) is proved. q.e.d.

13.10 Remarks. a) Theorem 13.8 is not true for insoluble groups. Let (l) be the alternating group m:s and K an algebraically closed field

of characteristic 2. Then K(l) has two blocks. The principal block PA 1

contains the trivial module V1 and two irreducible modules V2 , V3 each of dimension 2, while PA2 contains only one irreducible module, namely, V4 ~ V2 ®K V3 (see Exercises 64, 65). Hence the statement of 13.8 is not true for K(l).

b) Let (l) be the symmetric group 6 4 and K an algebraically closed field of characteristic 3. Then K(l) has three blocks :?A1,:?A2 and :?A3. Here PAl contains the trivial module V1 and the module V of dimension 1 corresponding to the sign representation of 6 4 , The blocks PA2 and :?A3 each contain only one irreducible module, say V2 and V3 . Then V3 ~ V2 ® K V (see 15.10b)). Hence tensoring the module V2 with the module V in PA 1 does not leave V2 in its block. So any extension of 13.8 beyond the principal block seems impossible, even for soluble groups.

Exercises

58) Find the number t of indecomposable two-sided direct summands of K(l) in the case when (l) is the alternating group of degree 4 and char K is 2 or 3.

59) Let (l) = SL(n, q), where q = pl. Let K = GF(q) and let V be the natural module for (l) of dimension n over K.

a) If V E :?A1, then (n, q - 1) = 1. b) If(n, q - 1) = 1 and n is not divisible by p, then V E :?A 1.

(Hint for the proof of b): From Schur's lemma, it follows that for every conjugacy class C of (l), L9EC 9 = a1, where a E K and na = ICltrg. Calculate the action of LXEdx - 1) on V, distinguishing between the cases when p divides I C I and when it does not.)

§ 14. Kernels of Blocks

14.1 Lemma (WILLEMS [3]). Suppose that 91 <J (l). We put

a = a(91) = L X. XE91

Let V be an irreducible K<f>-module and P(!j(V) the uniquely determined projective K<f>-module such that

P(!j(V)/P(!j(V)J(K<f» ~ V (cf 10.9).

P V {P(!j/91(V) if 91 operates trivially on V, a) (!j()a ~ . o otherwise.

(Here P(!j/91(V) is the uniquely determined projective K(<f>/91)-module for which P(!j/91(V)/P(!j/91(V)J(K(<f>/91)) ~ V.)

b) P(!j(V)a is the maximal submodule of P(!j(V) on which 91 operates trivially.

c) P (V)/Anp(j;(v)a is the maximal factor module of P(!j(V) on which 91 operates trivially.

d) Ifchar K divides 1911, then a2 = 0 and P(!j(V)a ~ Anp(j;(v)a.

Proof a) Obviously, xa = a E Z(K<f» for all x E 91. Hence the K-linear mapping a, defined by

(g91)a = ga (g E <f»

is well-defined and is a K<f>-epimorphism of K(<f>/91) onto K<f>a. If <f> = UtET ~Ht, where T is a transversal of 91 in <f>, then the elements ta (t E T) are linearly independent over K. Hence we obtain

This yields the K<f>-isomorphism

From

with a suitable K<f>-module W we obtain

for

Thus Po;(V)a is isomorphic to a direct summand of K«fj/91) and hence is a projective K«fj/91)-module. As Po;(V) has the irreducible head V, either Po;(V)a = 0 or else Po;(V)a has the irreducible head V and is hence indecomposable. If 91 does not operate trivially on V, we conclude that Po;(V)a = O. Hence we obtain

EB d(V)Po;j\Jl(V) ~ K«fj/91) ~ K(fja = EB e(V) Po;(V) a, v v

where d(V) and e(V) are certain multiplicities and both sums run over the irreducible K«fj/91)-modules V, considered in the right-hand sum as K(fj-modules. On the left side all these V appear as heads. On the right side V appears as a head if and only if Po;(V)a # O. The Krull-Schmidt theorem now shows that

b) We put

U = {uJu E Po;(V), ux = u for all x Em}.

As 91 <J (fj, certainly U is a K(fj-submodule of Po;(V). From ax = a for all x E 91 we conclude that Po;(V)a s; U.

If 91 does not operate trivially on V, we have by a) Po;(V)a = O. As the socle ofPo;(V) is isomorphic to V (see 11.6c)), we obtain U n S(p(ij(V)) = 0, hence U = O.

Now suppose that 91 operates trivially on V. Then by 10.14 we conclude from S(U) ~ V that U is isomorphic to a submodule of the uniquely determined projective (and injective) K«fj/91)-module Po;j\Jl(V) with socle V. By a) we have

Hence P o;(V)a = U in this case also. c) Now let Po;(V)/W be the largest factor module ofPo;(V) on which 91

operates trivially. As

for all Z E Po;(V), x E 91, we certainly have


If 91 does not operate trivially on V, then Pffi(V)jW = 0, for V is the head of Pffi(V). In this case we have

W = AOp/S(Vl a = Pffi(V).

Suppose now that 91 does operate trivially on V. Then Pffi(V)/W has the irreducible head V. Thus by 1O.14a), Pffi(V)/W is an epimorphic image of Pffi/91(V). Using a), we obtain

dim KPffi/91(V) ;;:: dimKPffi(V)jW ;;:: dimKPffi(V)/Aop/S(vl a

= dimKPffi(V)a = dim KPffi/91(V).

This proves W = AOP/S(Vl a. d) As xa = a for all x E 91, we have a2 = Imla. If char K divides

1911, then a2 = ° and so Pffi(V)a s; AOp/S(Vla. q.e.d.

14.2 Lemma (WILLEMS [3]). Let V be an irreducible K(f)/m)-module. Then

where K is to be considered as the trivial K(f)-module.

Proof By 9.3,

where the ~ are projective, indecomposable (f)-conjugate Km-modules. As 91 operates trivially on V, we must have S(~) = K for somej. But then ~ is (f)-conjugate only to ~ itself, so

for some natural number e. As 91 operates trivially on K, we have by 14.1a) (with (91, 91) in place

of (f), 91))

Hence

§ 14. Kernels of Blocks

Therefore by 14.1,

edimKP9l(K) = dimKPm(V) = dimKAopo;(v)a + dimKPm(V)a

= e(dimKP9l(K) - 1) + dimKPm/9l(V).

This shows that e = dimK Pm/9l(V).

193

q.e.d.

Let Pl = P(!j(K) denote the indecomposable projective Kffi-module for which PdPl J(Kffi) ~ K. By Dickson's theorem (7.16), we know that the highest power Iffilp of p = char K which divides Iffil also divides dimK Pl· If ffi has a p-complement, then I ffi I P = dimK Pl by 1 O.12b). We show

14.3 Theorem (WILLEMS [3]). The following assertions are equivalent. a) dimKP(!j(K) = Iffil p. b) For every 91 <I ffi, we have

c) For every composition factor Sl of ffi we have dimKPSI(K) = ISlip"

Proof. a) ~ b): By 14.2, we have

dimKP9l(K)dimKPmm(K) = dimKPm(K) = 1(f)1p"

As 1911p divides dim KP9l(K) and Iffi/91lp divides dim KPmm(K), we conclude that

b) ~ a): This follows immediately. The equivalence of b) and c) is obvious. q.e.d.

We remark that the class of groups ffi for which dim K Pm(K) = Iffilp is definitely larger than the class of groups with a p-complement. The simple groups PSL(2, p) have no p-complement for p ~ 13 (see II, 8.27), but it can be shown that dimKPl = p in all groups PSL(2, p)(BURKHARDT [1]).

14.4 Definition. a) Let V be a Kffi-module. We put

Sl(V) = {gig E ffi and vg = v for all v E V}


and

91(V) = n ft(W), w

where the intersection runs through all composition factors W of V. b) Let ~ be a block of GJ and f the corresponding idempotent in

Z(KGJ). We put

and

91(&6) = 91(fKGJ).

14.5 Lemma. a) For every KGJ-module V in the block ~ and every 9 E ft(&6) we have V(l - g) = O.

b) Let f be the idempotent in Z(KGJ) belonging to &6. Then 91(&6) = n~~r~ ft(V) = {gig E GJ,j(1 - g) E fJ(KGJ)}.

Proof a) If V is a module in &6, then V = Vf and hence

V(l - g) = Vf(1 - g) = 0

for every 9 E ft(&6). b) The first statement is obvious. Suppose that f(1 - g) E f J(KGJ). If V is a composition factor of

fKGJ, then V = Vf and hence

V(l - g) = Vf(l - g) ~ VfJ(KGJ) = O.

This implies 9 E 91(&6). Suppose conversely that 9 E 91(~) and that

s

fKGJ = EBeiKGJ i=l

is a decomposition of fKGJ into indecomposable right ideals ei KGJ and f = Ii=l ei' Then by 10.9c), eiKGJjeiJ(KGJ) is an irreducible module in &6. Thus


and

s

f(1 - g) = f L e;(1 - g) E fJ(Km). q.e.d. ;=1

14.6 Theorem (WILLEMS [3J). Let V be an irreducible Km-module and char K = p.

a) 5l(pCJj(V)) = Op,(91(PCJj(V))) = Op,(5l(V)). b) 91(PCJj(V)) = Op',p(5l(V)). c) If K is the trivial Km-module, then 5l(P(lj(K)) = Op,(m) and

91(P(lj(K)) = 0p',p(m).

Proof a) We put 91 = Op,(5l(V)). As p does not divide 1911, the trivial K91-module K is projective and hence P\ll(K) = K. We conclude from 14.2 that

Lemma 14.1a) now implies

thus PCJj(V) = P(lj(V)a and hence 91 operates trivially on P(lj(V). This shows 91 ~ 5l(PCJj(V)).

We put 9Jl = 5l(Po;(V)). Then 9Jl operates trivially on P(lj(V), Hence

and so K is a projective K9Jl-module. By Dickson's theorem (7.16), 9Jl must be a pi_group. So we have proved

This shows

We also have obviously

hence


So

b) m(Pm(V))/Sl(Pm(V)) is faithfully represented on Pm (V) and operates trivially on all the Ol-composition factors. Hence it is a p-group. This shows that

thus

As the normal p-subgroup

of Ol/Sl(Pm(V)) operates trivially on every composition factor of Pm(V), we obtain

c) This is the special case V = K. q.e.d.

14.7 Theorem. Let V be an irreducible KOl-module in the block f!4. a) Sl(p(jj(V)) = Sl(gH). So all the projective modules belonging to the

block gH have the same kernel. b) (BRAUER [8], MICHLER [1], WILLEMS [3])

and

In particular, m(gH) is p-nilpotent.

Proof a) If f is the idempotent corresponding to f!4 in Z(KOl), then

s

fKOl = EBeiKOl, i=1


where the ej Kffi are just the indecomposable projective Kffi-modules in ~. Thus they are the P(fj(V), where V runs through the irreducible modules in~. Thus

Sl(~) = n Sl(P(fj(V». Veal Virr

Suppose g E Sl(P(fj(V». Then by 14.6a),

Thus for every composition factor W of P(fj(V),

Repeating this procedure, we see from 12.4 that g E Sl(P(fj(X» for every irreducible module X in ~. Hence

Sl(P(fj(V» ~ n Sl(P(fj(W» = Sl(~) and Sl(P(fj(V» = Sl(~). Weal Wirr

b) We have 91(~) ~ 91(p(fj(V» for all irreducible modules V in ~. As normal p-subgroups operate trivially on all irreducible modules, we also have

91(~)/Sl(~) ~ 91(P(fj(V»/Sl(P(fi(V» (by 14.7a»

This implies

and

= 0p',p(Sl(V»)!Op,(Sl(V» (by 14.6a), b»

= Op(Sl(V)/Op,(Sl(V)))

= Op(Sl(V)!Sl(P(fi(V)))

= Op(Sl(V)!Sl(~»

(by 14.6a»

~ Op(ffi/Sl(~» ~ 91(~)/Sl(~).


Finally, for every irreducible module V in f1l we have

(14.7a))

= Op,(91(Pm(V))) (14.6a))

= Op,(91(f1l)).

We mention an important special case.

q.e.d.

14.8 Theorem (BRAUER [7]. If f1l1 is the principal block of K(fj, then

Proof If K denotes the trivial K(fj-module, then

Also

~(f1l1) = ~(Pm(K)) (14.7a))

= Op,(fj) (14.6c)).

91(f1l1) = 91(Pm{K)) (14.7b))

= 0p',p(fj) (14.6c)). q.e.d.

As an application, we derive a representation-theoretic characterization of p-nilpotent groups.

14.9 Theorem. Let K be afield of characteristic p. The following assertions are equivalent.

a) The principal block [lJ 1 of(fj contains just one irreducible K(fj-module. b) (fj is p-nilpotent. c) Every block of (fj contains just one irreducible K (fj-module. d) The Cartan matrix of K(fj is a diagonal matrix.

Proof a) => b): If [lJ1 contains only one irreducible K(fj-module, this is necessarily the module for the trivial representation of (fj. So by 14.5, 91(f1l1) = (fj. Hence 0p',p(fj) = (fj by 14.8, and (fj is p-nilpotent.

b) => c): Suppose now that (fj is p-nilpotent. By 14.8, 91(f1l1) = (fj, so by 14.5, the only irreducible K(fj-module in f1l1 is the module V1 for the trivial representation of (fj.

Let V1, ..• , Vk be all the irreducible K(fj-modules to within isomorphism. By 10.9, there exists a projective indecomposable K(fj-module P;

such that ~/PiJ(K(f») ~ Vi' We show that all the K(f)-composition factors of Pi are isomorphic. This is clearly the case for PI' since all composition factors of PI belong to f!ll and are therefore isomorphic to VI' If

is a K(f)-composition series of PI' then ~/Wj-l ~ VI (j = 1, ... , m). Since

all the composition factors of PI Q9K Vi are isomorphic to Vi' By 7.19, PI Q9K Vi is projective, so by 10.9, PI Q9K Vi is the direct sum of certain Pj' Thus any composition factor of any of these Pj is isomorphic to Vi' It follows that j = i and that any composition factor of ~ is isomorphic to

Vi' We conclude from 12.4 that Pi and Pj lie in distinct blocks for i i= j.

Hence each block contains only one Vi' c) => d): This is true by definition of the Cartan matrix and 12.4. d) => a): This is trivial. q.e.d.

We apply the preceding results in the following example.

14.10 Example. Let q be a prime and let (f) = (x, y) be a metacyclic group of order qnr, defined by the relations

where k' == 1 (qn) and (r, q) = 1. We make the assumption that C(!j(x) = (x). This means that ki =1= 1 (qn) for 0 < i < r. Observe that r divides q - 1.

We shall describe the irreducible modules, the indecomposable projective modules and the Cartan matrix of (f) over a field K of characteristic p.

a) Suppose first that p = q. We construct the principal indecomposable projective module

Po by using 10.12b). (For technical reasons we start the numbering of the indecomposable projective modules with Po rather than the usual Pl') Thus Po = K(!j, where K is the trivial module for the p-complement (y) of (f). As the Sylow p-subgroup ~ = (x) of (f) is a transversal of (y) in (f),

(Po)'ll is isomorphic to the indecomposable K'P-module K'P. Hence by 5.3 its only submodules are

The action of y on these vector spaces is described by

(1 ® (1 - X)i)y = 1 ® (1 - X)iy = 1 ® y(1 - XY)i = ly ® (1 - Xk)i

= 1 ® ((1 - x)(1 + x + ... + Xk-I))i

= 1 ® (1 - x)i{k + (x - 1) + (x 2 - 1) + ... + (X k- I - l)r

== 1 ® (1 - x)ik i mod Vi+I.

Hence V;/Vi+1 is the I-dimensional irreducible K(fj-module on which x operates trivially and y induces multiplication by ki.

Let a denote the character of (fj given by a(x) = 1 and a(i) = ki. Let Mj be an irreducible K(fj-module corresponding to aj • By the assumption that C(!j(x) = <x), it follows that a is faithful on <y). As <x) = Op(fj), Mo, M I , ... , Mr - I are all the irreducible K(fj-modules.

We determine the multiplicity of Mj as a composition factor in Po. The calculation above shows that

V;/Vi + 1 ~ Mi (i modulo r).

Hence we obtain the composition series of Po illustrated in the following diagram.

Vo = Po

Mo

VI

MI V2

Vp"_l

Mo

Vp" = 0

If pn - 1 = rs, then the multiplicity of Mj in Po is s for j i= 0 and s + 1 for j = o.

We still have to find the other indecomposable projective K(f)modules. In this example it is very simple. Put Pi = Po ®K Mi(i = 0, ... ,r - 1). Then by 7.19, Pi is projective. As

P; is indecomposable by Dickson's theorem (7.16). The epimorphism rx of Po onto Mo gives rise to an epimorphism rx ® 1 of Pi = Po ®K Mi onto Mo ®K Mi ~ Mi' Hence Pi is the indecomposable projective K(f)-module with head Mi' We see immediately that Pi has a composition series described by the following diagram

Mi

Thus the multiplicity of Mj in P; is s for i :f= j and s + 1 for i = j. Hence the Cartan matrix of K(f) is the r x r matrix

(

s + 1 s

s s + 1 c=. . . . . .

s s

As this matrix is obviously indecomposable in the sense of 12.5, (fj has only one block for the prime p. Since Op,(f») = 1, this fact also follows from 13.5,

b) Now we assume that p divides r. Then (fj is p-nilpotent and we can apply Theorem 14.9. Hence every block contains exactly one irreducible K(fj-module. We assume now that K is algebraically closed.

Let V be an irreducible K <x)-module with character A. We denote by :J = :J(A) the inertia subgroup of V in (f). Let 9t/<x) be the normal p-complement of :J/<x) and write t = 19t/<x)l. Then A has exactly t extensions A1 , ... , At to 9t, which correspond to irreducible K9t-modules V1 , ... , Vt of K-dimension 1. If K denotes the trivial K<x)-module, then

3 = 3(A)

<x) qn

V'Jl ~ ((V1)(x»)'Jl ~ V1 ®K K(m/<x») (4.l5b))

~ V1 E!1 ... E!1 Vt,

as K(m/<x») is a direct sum of modules of dimension 1. We first construct indecomposable projective K(fj-modules. We

obtain

As the Sylow p-subgroup 6 of 3 is a transversal of m in 3, we have

Hence V/J is indecomposable, and thus by 9.6 the V/fJ are indecomposable. As p t 19i 1, certainly Vi is a projective K9i-module, and hence P; = vt) is a projective K(fj-module by 7.17.

Next we determine the head of p;. As 31m is a cyclic p-group, each of the characters Ai of m has a unique extension f.li to 3. As 3 <l (fj, we certainly have 3(f.li) ~ 3. But as (f.lJ<x) = A, we see that :5 is the inertia group of f.li' Let Mi be a K3-module belonging to f.li' By 9.6b) (with 3 in the place of 91 and 3 of 9.6), Ni = M? is an irreducible K(fj-module. Hence we obtain

dimKHomKffi(Pi, Ni ) = dimKHomKffi(Viffi, Ni) = dimKHomK'Jl(Vi, (Ni)'Jl)

= dimK HomK'Jl(V;, Vi E!1 ... ) ~ 1.

Thus Ni is the head of Pi' As by 14.9 all composition factors of Pi are isomorphic to Ni , the multiplicity of Ni in P; is

Finally we have to show that the Ni are all the irreducible K(fjmodules (to within isomorphism). Let W be any irreducible K(fj-module

§ 15. p-Chief Factors of p-Soluble Groups 203

and let A be a character of <x) such that the K<x)-module VA belonging to A is a direct summand of W(x)' If we apply our construction to this A, we obtain

Hence

o "# dimKHomK<x)(W(x), V;) = dimKHomK(jj(W, V~)

= dimKHomK(jj(W, P1 EB ... EB Pt)·

Thus W is isomorphic to an irreducible submodule of some Pi and hence to Ni .

Exercise

60) (WILLEMS) Suppose char K = p. Let 91 be a normal p-subgroup of ffi and V an irreducible Kffi-module. Then

L(P(jj/\Jl(V)) + L(K91) - 1 ~ L(P(jj(V)),

where L denotes the Loewy length.

§ 15. p-Chief Factors of p-Soluble Groups

If ffi is a p-soluble group, any p-chieffactor IDl/91 of ffi may be regarded as an irreducible Kffi-module, where K = GF(p), by putting

Clearly, Op,(ffi) operates trivially on such a factor, so by 13.7, this module lies in the principal block. In this section we study relations between the Loewy series of the principal indecomposable projective Kffi-module P1

and the p-chief factors of ffi.

15.1 Definitions. Throughout this section we use the following notation. Let U be a subgroup of ffi and K = GF(p). The K(fj-module induced

from the trivial KU-module K is denoted by K~. We write

A~ = <x(g - l)lx E K~, g E (!j) = K~3,

where 3 is the augmentation ideal of K(!j. If we regard K~ as being the K-vector space with K-basis the set of right co sets of U in (!j, then {Ug - Uig E (!j - U} is a K-basis of A~.

15.2 Lemma (GASCHUTZ). Suppose that (!j = [f) and [ n f) = :D, where [ and :D are normal in (!j and [/:D is an elementary Abelian p-group; thus [/:D may be regarded as a K(!j-module. Let V be a K(!j-module. IfU ~ f), there exists a monomorphism a of HOmK<!i([/:D, V) into HOmK<!i(A~, V) such that

(Uhc - U)(cxa) = (:Dc)cx

for all cx E HomK<!i([/:D, V), h E f) and c E [.

Proo!SupposethatUhlCl = Uh2ClwithhiEf),ciE[.Thenf)cl = f)Cl since U ~ f), so C1C2 l E f) n [ =:D and :DCl = :DCl' Thus if cx E HomK<!i([/:D, V), there exists a K-linear mapping cxa of A~ into V such that

(Uhc - U)(cxa) = (:Dc)cx,

since (f) = f)[ and rUg - Ulg E (f) - U} is a K-basis of A~. !Y.a is a K(f)-homomorphism, for if g = h'c' with h' E f), c' E [,

((Uhc - U)g)(cxa) = ((Uhh'ch'c' - U) - (Uh'c' - U))(cxa)

= (:Dch' c')cx - (:Dc')cx = (:Dch')cx

= (:Dc9)cx = ((:Dc)cx)g

= ((Uhc - U)(cxa))g,

Now a is a homomorphism of HomK<!i([/:D, V) into HOmK<!i(A~, V), since if CX i E HOmK<!i([/:D, V) (i = 1, 2), then

(Uhc - U)((cx l + c(2)a) = (:Dc)(cxl + cx l )

= (:Dc)cx l + (:Dc)cx2

= (Uhc - U)(cx l a + cxla).

Finally a is a monomorphism, since if cxa = 0, then (:Dc)cx = ° for all c E [ and so cx = O. q.e.d.


15.3 Lemma (GASCHlJTZ). Suppose that G} is p-soluble and IG}: UI is a power of p. Let V be an irreducible KG}-module and let

[ = Cm(V) = {gig E G}, vg = v for all v E V}.

If we put :D = ['[P, there is a monomorphism r of HomKm(A~, V) into HomKm([/:D, V) such that

(:Dc)([3r) = (Uc - U)[3

for all [3 E HomKm(A~, V), c E [.

Proof Given [3 E HomKm(A~, V), put cp = (Uc - U)[3 for all c E [.

Then if Ci E [(i = 1, 2),

(clc2f13 = (UC IC2 - U)[3 = ((UCI - U)c2 + (UC2 - U))[3

= ((UCI - U)[3)c2 + (UC2 - U)[3

= (UCI - U)[3 + (UC2 - U)[3,

since [ = Cm(V). Hence

and 71 is a homomorphism of [ into V. Since V is an elementary Abelian p-group, :D ~ ker 71. So there exists [3r E HomK([/:D, V) such that

(:Dc)([3r) = (Uc - U)[3

for all c E [. In fact, [3r is a KG}-homomorphism, for if C E [ and g E G),

then

((:Dc)g)([3r) = (:Dcg)([3r) = (Ug-1cg - U)[3

= ((Ug-1 - U)cg + (Uc - U)g + (Ug - U))[3

= ((Ug- 1 - U)[3)cg + ((Uc - U)[3)g + (Ug - U)[3

= ((Ug-1 - U)g)[3 + ((:Dc)([3r))g + (Ug - U)[3

= ((:Dc)([3r))g.

It is easy to see that r is a homomorphism. To verify that it is a monomorphism, suppose that 0 1= [3 E HomKm(A~, V) and let

~ = {YIY E (fj, (Uy - U)f3 = OJ.

For Yl' Y2 E~, we have

(UYIY2 - U)f3 = (UYl - U)Y2f3 + (UY2 - U)f3

= (UYI - U)f3Y2 = O.

Thus U :::;; ~ :::;; (fj and so I (fj : ~I is a power of p. We show that

Suppose that Y E ngem ~g. Then for all g E (fj,

(Ug - U)f3y = (Ugy - U)f3 - (Uy - U)f3 = (Ugyg-lg - U)f3

= «Ugyg- 1 - U)g + (Ug - U»f3

= (Ugyg- 1 - U)f3g + (Ug - U)f3 = (Ug - U)f3,

since gyg- 1 ERAs 13 :f. 0 and V is irreducible,

«Ug - U)f3lg E (fj) = V.

Thus y E Co;(V) = G:. Since (fj/(i; is faithfully represented on the irreducible K(fj-module V,

we have Op«fj/(i;) = 1. Now suppose that f3! = O. Then (Uc - U)f3 = 0 for all c E (i;, so (i;:::;; R Since ngeo; ~g :::;; (i;, this implies that (i; = ngem ~g.1f 91/(i; is a minimal normal subgroup of (fj/(i;, then 91/(i; is a pi -group (as Op«fj/(i;) = 1), hence 91 :::;; ~ and thus 91 :::;; ngeo; ~g = (i;. This proves (fj = (i; = R Hence (Ug - U)f3 = 0 for all g E (fj, and 13 = 0, a contradiction. q.e.d.

15.4 Lemma. Let (fj be a p-soluble group. a) IfWl is a minimal normal p-subgroup of(fj and Wl = Cm(Wl), then Wl

has a complement in (fj (cf II,3.3). b) Suppose that V is an irreducible K(fj-module, where K = GF(p) and

put (i; = Cm(V). Suppose also that !l is a normal subgroup of (fj such that !l :::;; (i;, that (i;/!l is an elementary Abelian p-group and, as a K(fj-module, (i;/!l is the direct sum of submodules isomorphic to V. Then there exists a subgroup f) of (fj such that (fj = (i;f) and (i; n f) = !l.

Proof a) We may assume that Wl < (fj. Let 91/Wl be a minimal normal

subgroup of (fj/IDI. As (fj/9Jl is faithfully and irreducibly represented on 9Jl, certainly 91/9Jl is not a p-group. Thus 91/9Jl is a p' -group. By I, 18.1, 9Jl has a complement .Q in 91, and by I, 18.2, all such complements are conjugate in 91. Hence by the Frattini argument, (fj = 91No;(.Q) = 9JlNo;(.Q). Thus 9Jl n N(f;(.Q) <l (fj. But since 9Jl is minimal and.Q does not centralize 9Jl, we have 9Jl n N(f;(.Q) = 1 and N(f;(.Q) is therefore a complement of 9Jl in (fj.

b) If V is the trivial module, (£ = (fj and we may take ~ = :1). Otherwise (£ < (fj. Choose a minimal subgroup ~ of (fj for which (fj = (£~ and (£ n ~ ~ :1). Then (£ n ~ <l (fj, so if (£ n ~ > :1), we may choose a normal subgroup 91 of (fj, maximal with respect to :1) ::; 91 < (£ n ~. Then by hypothesis, the K(fj-module (£ n ~)/91 is the direct sum of K(fj-modules isomorphic to V. Hence

Thus by a), (£ n ~)/91 has a complement ft/91 in N91. Hence

and (£ n ft ~ 91 ~:1). This contradicts the minimality of ~, so (£ n ~ = :1). q.e.d.

15.5 Theorem (GASCHUTZ). Let (fj be a p-soluble group, K = GF(p), let V be an irreducible K(fj-module and put (£ = C(f;(V).

a) For any p-complement U of (fj, HOmK(f;(A~, V) and HomK(f;(£/(£'(£P, V) are isomorphic.

b) Let PI be the principal indecomposable projective K(fj-module. Then V occurs as a composition factor in the completely reducible K(fj-module PI J(K(fj)/Pl J(K(fj)Z if and only if V is isomorphic to a complemented p-chief factor of (fj, and the multiplicity of V in PI J(K(fj)/Pl J(K(fj)2 is equal to its multiplicity in the maximal completely reducible factor module of (£/(£'(£P.

Proof a) Let :1) be the intersection of all normal subgroups X of (fj for which (£'(£P::; X < (£ and (£/X is K(fj-isomorphic to V. Then HomKo;(£/(£'(£P, V) is isomorphic to HomKo;(£/:1), V) and (£/:1) is the direct sum of K(fj-submodules isomorphic to V. By 15.4, there exists· a subgroup ~ of (fj such that (fj = (£~ and (£ n ~ = :1). Hence I (fj : ~I = 1(£::1)1 is a power of p. Thus there exists a p-complement U of (fj such that U ::; ~. By 15.3, there is a monomorphism 't" of HomKo;(A~, V) into HomK(f;(£/:1), V) such that

(l)c)(f37:) = (Uc - U)f3

for all c E (t, f3 E HomKm(A~, V). And by 15.2, there is a monomorphism (J

of HomKm((tjl), V) into HomKm(A~, V) such that (J7: is the identity mapping. Hence 7: is an isomorphism.

Since the p-complements of <I; are conjugate (VI, 1.7), a) holds for any p-complement U of <I;.

b) Let r1' r 2 be the multiplicities of V in P1 J(K<I;)jP1 J(K<I;)2 and in the maximal completely reducible factor module of (tj(t'(tp respectively. By 4.12,

and

By 10.12, P1 = K~, so P1 J(K<I;) = A~. Hence by a), HomKm(P1 J(K<I;), V) and HomKm((tj(t'(tP, V) are isomorphic and r1 = r2'

Let mjlE be a p-chief factor of <I; complemented by ~l' If (t1 = Cm(mjlE), then (ttl((t1 n ~l) is a self-centralizing complemented p-chief factor of <I; K<I;-isomorphic to mjlE. Thus (ttl((tl n ~1) is an irreducible K<I;-module which occurs as a composition factor in the maximal completely reducible factor module of (ttl(t~ (t~ and hence in P1 J(K<I;)jP1 J(K<I;)2.

Conversely, by a), a composition factor of P1 J(K<I;)jP1 J(K<I;f is of the form (tjl), where l) <J <I; and (tjl) is self-centralizing. By 15.4a), (tjl) is complemented. q.e.d.

15.6 Remark. The multiplicity r of V in P1 J(K<I;)jP1 J(K<I;)2 can also be found in the following way. In any chief series of <I;, r is the number of complemented p-chief factors in this series which are K<I;-isomorphic to V. (This number is indeed the same for every chief series.)

Theorem 15.5 gives a complete description of the second Loewy factor P1 J(K<I;)jP1 J(K<I;)2 of Pl' Much less is known about the composition factors of P1 J(K<I;)2. We shall however show that every p-chief factor of <I; is isomorphic to some composition factor of a certain factor module of Pl' We first make the following elementary remarks.

15.7 Lemma. Suppose that U :::; <I;. a) If 91. <J <I;, then K~~91 (regarded by means of inflation as a K<I;

module) is isomorphic to K~91'

b) If U :::; ~ :::; (fj the K-linear mapping {3 of K~ onto K~ given by (Ug){3 = lEg (g E (fj) is a K(fj-epimorphism of K~ onto K~.

Proof a) If v is the natural homomorphism of (fj onto (fj/91, there exists a K(fj-isomorphism r:J. of K~91 onto K~~~91 such that

«U91)g)r:J. = (Uv)(gv)

for all g E (fj. b) This is obvious. q.e.d.

15.8 Theorem (GREEN, HILL [1]). Suppose that (fj is a p-soluble group, U is a p-complement of (fj and e = Nr;(U). Then K~ is an indecomposable factor module of the principal indecomposable projective K(fj-module PI = K~. Every p-chieffactor of(fj is isomorphic to a composition factor of K~.

Proof By 1S.7b), K~ is an epimorphic image of the K(fj-module PI = K~. By 1O.12b), PI is the projective K(fj-module for which PI/PI J(K(fj) is the module for the trivial representation of (fj. By 1O.9a), K~ has only one maximal submodule and is therefore indecomposable.

Suppose that 91/Wl is a p-chief factor of (fj. If 91/Wl is central in (fj, then 91/Wl is K(fj-isomorphic to K~/K~J(K(fj). Hence we may assume that 91/Wl is not central.

We show first that it is sufficient to consider the case 9Jl = 1. For if the theorem is proved in this case, then the K«fj/Wl)-module 91/Wl is isomorphic to a composition factor of K~{;:, where :t/Wl = N(f;f.Dl(UWl/Wl). Hence by 1S.7a), the K(fj-module 91/Wl is isomorphic to a composition factor of K~. Now e :::; :t and hence 91/Wl is isomorphic to a composition factor of K~ by 1S.7b).

Suppose then that Wl = 1. Then 91 is a minimal normal non-central p-subgroup of (fj. Let ~ be a Sylow p-subgroup of (fj. Then 91 :::; ~ and (fj = U~. Now [91, U] :::; 91 and [e, U] :::; U since e = N(f;(U), Thus

[91 ('\ e, U] :::; 91 ('\ U = 1.

Hence

[91 ('\ e, (fj] = [91 ('\ e, U~] = [91 ('\ e, ~] < 91

by III, 1.2 and 2.6. But by III, 1.6b), [91 ('\ e, (fj] <:J (fj, so [91 ('\ e, (fj] = 1. Thus 91 ('\ e :::; 91 ('\ Z«fj) = 1.


By 15.7b), the K-linear mapping p of K~ onto K~91' given by (6g)P = 691g (g E G» is a KG>-epimorphism of K~ onto K~91' We have

= IG>: 61 - IG>: 6 911 = (1911 - 1) IG>: 691 1,

since 6 n 91 = 1. But if T is a transversal of 691 in G>, then 6yt - 6t E ker P for all y E 91 - {1}, t E T, and these elements 6yt - 6t are linearly independent. Since the number of them is dimK ker p, it follows that they form a K-basis of ker p. Hence there exists a K-linear mapping y of ker pinto 91 given by

((6y - 6)t)y = yI (y E 91, t E T).

This is a KG>-homomorphism, since if g E G>, y E 91, t E T and tg = sy't' with s E 6, y' E 91 and t' E T, then

((6y - 6)tg)y = ((6ysy' - 6y')t')y = ((6ySy' - 6y')t')y

= (ysy't(y't'rl = yst'

(since 91 is Abelian)

Thus Y is a KG>-epimorphism of ker f3 onto 91, and so 91 is isomorphic to a composition factor of K~. q.e.d.

15.9 Examples. a) Suppose G> = .03 is the split extension of the quaternion group.o of order 8 by a cyclic group 3 of order 3,"where 3 operates non-trivially on .0. Let K = GF(2) and L = GF(4).

By 10.12, the principal indecomposable projective KG>-module P{ is induced from the trivial module K for the 2-complement 3 of G>. Hence dimKP~ = 1G>:31 = 8.

Let V be the irreducible KG>-module isomorphic to the complemented 2-chief factor .o/Z(.o) of G>. Then C(f;(V) = .0, and by 15.5, V appears as a composition factor of P~ J(KG>)/P~ J(KG»2 with multiplicity 1. As .o/Z(.o) is the only complemented 2-chief factor of G>, by 15.5,

By 11.6c) the socle S of P{ is isomorphic to the module K for the

trivial representation of K(f). Thus by 8.3a),

K ~ S* ~ P'*/S.l - - 1 ,

and P~* is indecomposable by 8.3f). By 8.23b), P~* is also projective, so P{* ~ P{. From dimdP{ J(K(f))).l = 1 we conclude that

(P{ J(K(f)))* ~ P{*/(P{ J(K(f))).l (8.3a))

~ PUS.

As P{ J(K(f)) has only one irreducible factor module and this is isomorphic to V, PUS has only one irreducible submodule U/S, and this is isomorphic to V*. Hence U ~ P{ J(K(f)f But as 02(f)) = .Q centralizes all irreducible K(f)-modules, V is the only irreducible K(f)-module of dimension 2 and hence V* ~ V. Therefore we have the following diagram of sub modules of P{ .

P' 1

K

P; J(Kffi)

v P; J(Kffi)2

W

u v

s K

o

We still have to determine the 2-dimensional module

As 3 < 3 x Z(.Q) < (f) and I (f) : 3 x Z(.Q) I = 4, P{ has a factor module PUW of dimension 4 by 15.7. Hence

and


We claim that P~J(Kffi)2/U is isomorphic to KEEl K. Otherwise by 10.14a) it would be an epimorphic image of P~. But this is impossible, as PUP; J(Kffi)2 has only one composition factor isomorphic to K. Thus the lattice of submodules of P; is the following.

K

v

v

K

P' I

P; J(K(fj)

s

o

Obviously V ®K L ~ V2 EEl V3, where V2 , V3 are irreducible non-isomorphic Lffi-modules, and L, V2 , V3 represent all types of irreducible Lffi-modules. As complete reducibility is not affected by extension of the ground-field (1.8), we see that the factor modules in the Loewy series of Pi = P; ®K L are

As in Example 14.10, we see that ~ = Pi ®L Vi (i = 2,3) is the indecomposable projective Ul'}-module with head Vi. As V2 Q9L V2 ~ V3 and V2 Q9L V3 ~ L, we see that P2 has a descending series of submodules with factor modules (from above) isomorphic to

To show that these factors are the Loewy factors of P2 , we have only to observe that

and that every completely reducible Kffi-module remains completely reducible by tensoring with the module V3 of dimension 1.


Similarly, P3 has the Loewy factors

Hence the Cartan matrix is

2 2) 4 2 .

2 4

b) It is much simpler to study the group-ring KGJ, where GJ = ,03 is the same group as in a), but K is an algebraically closed field of characteristic 3. Now GJ is 3-nilpotent and has three 3'-classes, hence there are three irreducible KGJ-modules Vi' V2 , V3. We assume Vi = K.

As ,0 is a 3-complement of GJ, by lO.12b) Pi = W~ is the principal indecomposable projective KGJ-module, where Wi denotes the trivial K,Q-module. The Cartan matrix of KGJ is a diagonal matrix by 14.9. Hence Pi has the Loewy factors Vi' Vi' Vi' So ell = 3 and eli = 0 for i = 2,3.

Let W2 be an irreducible, non-trivial K,Q-module of dimension 1. Then ,0 is the inertia group of W2 , hence P2 = W~ is irreducible, by 9.6. As W2 is a projective K,Q-module, P2 is a projective KGJ-module, by 7.17. We put V2 = P2 •

Let W3 be the irreducible K,Q-module of dimension 2. Its inertia group is GJ. Hence by 9.9a) there exists an irreducible KGJ-module V3 of dimension 2 such that (V3)o ~ W3. We put P3 = Wf. Then P3 is projective by 7.17. We have

Now Z(,Q) is represented faithfully on W3, but Z(,Q) is represented trivially on (Vi)o for i = 1, 2. Thus

{o for i = 1, 2, dimK HomKo;(P3, Vi) = 1

for i = 3.

Hence P3 = W: is the projective indecomposable KGJ-module with head V3. As

the Loewy factors of P3 are V3, V3, V3. Hence the Cartan matrix of GJ is


(3 0 0) c = 0 1 O.

003

15.10 Examples. a) We now consider the case (fj = 6 4 and K any field of characteristic 2. As (fjj02«fj) ~ 6 3 , K(fj has two irreducible modules K and V, where V is absolutely irreducible of dimension 2. Thus K is a splitting field for K(fjjJ(K(fj). We denote by Pl , P2 the indecomposable projective K(fj-modules for which

As Pl is induced from the trivial module for a 2-complement of (fj, dimK Pl = 8. By 10.18b),

hence dimK P2 = 8. Next we determine the Cartan matrix of (fj. Using the symmetry of

C, we obtain

8 = dimK Pl = Cll + C12 dimK V = Cll + 2C12'

8 = dimK P2 = C12 + 2C22'

Thus C12 is even. By 15.5, Pl J(K(fj)jPl J(K(fj)Z is isomorphic to K EEl v. As head and socle of Pl are isomorphic to K, we conclude that C 11 :?: 3 and C12 :?: 1. As C12 is even, this forces Cll = 4, C12 = C21 = 2, C22 = 3. Thus

(As (fj is not 2-nilpotent, we could have deduced C12 #- 0 from 14.9.) It is considerably more difficult to obtain the Loewy series of Pl and

P2. We start by looking at several K(fj-modules. (1) V ®K V ~ P{ EEl V, where P{ is indecomposable and has two

composition factors isomorphic to K. As the Klein subgroup operates trivially on V, we can regard V as an

irreducible K6 3-module. If P{, P~ are the indecomposable projective K6 3-modules with heads K, V respectively, then


This implies that dim KP'1 = 2 and P~ = V. Hence V is a projective K6 3-

module. (It is not projective as a Km-module I). By 7.19), V ®K V is a projective K6 3-module. We consider the operation of an element g of order 3 on V ®K V. The eigen-values of g on V are 6, 6- 1, where 6 3 =

1 =I 6. Hence on V ®K V the eigen-values of g are 1, 1,6,6-1. This rules out the possibilities V ®K V ~ P{ EEl P{ and V ®K V ~ V EEl V. Hence

(2) Let W be the Km-module of dimension 4 induced from the trivial K6 3-module. Then the head and socle ofW are isomorphic to K, and the composition factors of Ware K, V, K.

By Nakayama's theorem, the multiplicity of K in the head of W is

The multiplicity of V in the head of W is

The statement about the socle of W follows similarly. If all composition factors of Ware isomorphic to K, the elements of order 3 of m operate trivially on W, by Maschke's theorem. But this is not true, for W is the module for the natural permutation representation of m = 6 4 of degree 4.

(3) There exists a Km-module U1 such that dimK U1 = 4, the head of U1 is isomorphic to V, the socle of U1 is isomorphic to K and the composition factors of U1 are V, K, K.

Let

be a composition series of W. Then by (2),

We consider T = W ®K V and put Ti = Wi ®K V. Then


and by (1),

We have

As

T has no factor module isomorphic to K. Thus T has a factor module U1

with the required properties. (4) The dual Ut of U1 is a module U2 with only one composition

series

where

(5) Now we can determine the Loewy series of Pl' By 15.5,

As Wand U2 are isomorphic to factor modules of PI (lO.14a)), there are sub modules ZI and Z2 of PI for which

Hence

and

Therefore PI J(Kffi)2 /P1 J(Kffi? has composition factors K and V. As S(P1) ~ K, this implies

and

To determine the Loewy series of the indecomposable projective K<f>-module Pz with head V, we show first that

(6) PI ®K V ~ Pz EB Pz· By 7.19c), PI ®K V is projective and hence is isomorphic to a direct

sum of copies of PI and Pz. The composition factors of PI ®K V are those of

K ®K V = V, taken 4 times,

and those of

V ®K V ~ P{ EB V (cf. (1)), taken twice.

Hence the multiplicities of K, V in PI ®K V are 4, 6 respectively. Using the Cartan matrix, this implies that

(7) From the Loewy series of PI we see that PI ®K V has submodules RI, Rl such that

Hence (Pl EB PZ)/(Pl EB Pz)J(K<f»l has at least the composition factors V, V, V, K. Thus P1J(K<f»/P1J(K<f»1 must have composition factors V and K. Hence dimKP1J(K<f»1 :$ 3. This still allows the possibilities that

(i)

or

(ii)

We show that case (ii) is not possible. In (3) we constructed a K<f>-module UI with head V, socle K and of

dimension 4. By lO.14a), this module is isomorphic to a factor module


of Pz, say Pz/Y ~ UI . But this implies

Thus case (i) holds. The structure of Pz is illustrated in the following diagram.

v

y

V P2J(K(»j3 = S(P2).

V

Hence

{

V

P,J(KffiY/P,J(KffiYH '" K! V

The upper Loewy series of Pz is

for i = 0,

for i = 1,

for i = 2,

for i = 3.

the terms of the upper and lower Loewy series are therefore not the same. b) We consider K{f;, where again {f; = 6 4 , but K is an algebraically

closed field of characteristic 3. Now {f; has four 3'-classes. Hence there are four irreducible K{f;

modules VI' Vz, V3 , V4 . We assume that VI is the trivial module and Vz the module corresponding to the sign representation of {f;. {f; has only one 3-chief factor and this is complemented. Vz is the module obtained from this by extension of the ground-field.

Let WI be the module for the trivial representation of a Sylow 2-subgroup ~ of (f;. Then PI = WT is by 1O.12b) the principal indecomposable projective K{f;-module. As


by 15.5, the Loewy factors of PI are VI' Vz, VI' Certainly Pz = PI @K Vz is projective and indecomposable. As Pz

has a factor module isomorphic to VI @K Vz ~ Vz, Pz is the indecomposable projective K(fj-module with head Vz· From Vz @K Vz ~ VI we conclude easily that Pz has the Loewy factors Vz, VI' Vz.

Let T be the trivial module for K6 3 and consider To;, on which (fj induces the natural permutation representation of degree 4. If {VI' vz, V3' V4} is a K-basis of To; the elements of which are permuted by (fj, then as char K = 3, we have the direct decomposition

where

w = {.t aivij.t ai = o}. ,=1 ,=1

This is a decomposition into K(fj-submodules. By Nakayama's lemma,

Hence W has no factor module isomorphic to VI or Vz. A similar calculation shows that W has no submodule isomorphic to VI or Vz. As VI and Vz are obviously all the irreducible K(fj-modules of dimension 1, we see that W must be irreducible. We put W = V3 •

Finally put V4 = V3 @K Vz. As Vz has dimension 1, V4 is irreducible. If g = (1, 2) E 6 4 and Xi denotes the character of (fj belonging to Vi' then X3(g) = 1, hence

This proves that V3 *' V4 •

From 10.18b), we conclude that

4

24 = dimK K(fj = L dimK Vi dimK Pi i=I


This shows that Vi ~ Pi for i = 3, 4. The Cartan matrix is hence

Thus K(fj has three blocks. We remark that V3 and V4 can also be constructed in a different way.

Let 'l' be a Sylow 2-subgroup of (fj and U an irreducible K'l'-module of dimension 1. Then

As (V3)'ll is a direct sum of irreducible K'l'-modules of dimensions 1 or 2, there exists a U such that HomK(l)(U(l), V3) i= o. Thus U(l) has a factor module isomorphic to V3, hence U(l) ~ V3. As U is a projective K'l'module, this construction once more makes it evident, by 7.17, that V3 is a projective K(fj-module.

Exercises

61) Let K be a field of characteristic p and P1 the principal indecomposable projective K(fj-module. Then the following assertions are equivalent.

a) (fj has a p-factor group different from 1. b) P1 J(K(fj)/P1 J(K(fj)2 has a composition factor isomorphic to the

trivial module.

62) In examples 15.9a) and b), determine a) the composition factors of all Vi ®K Vj' b) the decomposition of the projective modules ~ ®K Vj and

Pi ®K Pj into indecomposable projective modules.

63) In example 15.10a), prove the following decompositions. P1 ®K V ~ P2 EB P2; P2 ®K V ~ P1 EB P2; P1 ®K P1 ~ 4P1 EB 4P2; P1 ®K P2 ~ 2P1 EB 6P2; P2 ®K P2 ~ 3P1 EB 5P2· (Here, 4P1 stands for P1 EB P1 EB P1 EB P1 , etc.)

§ 15. p-Chief Factors of p-Soluble Groups 221 64) Let (fj be the alternating group 215 and K an algebraically closed field of characteristic 2. Determine all irreducible K(fj-modules by the following procedure. Let V1 = K be the module for the trivial representation of (fj.

a) As (fj ~ SL(2, 4) (II, 6.14), there are irreducible non-isomorphic modules Vz, V3 of dimension 2. Show that Vi* ~ Vi (i = 2, 3).

b) Show that V4 = Vz ®K V3 is an indecomposable projective K(fjmodule by considering (V4hJ for a Sylow 2-subgroup ~ of (fj.

c) Show that V4 is irreducible. d) V1 , Vz, V3, V4 are all irreducible K(fj-modules (to within isomor

phism). e) Let W be the trivial module for the subgroup 214 of (fj. Show that

W(fj ~ V1 E8 V4 . Use W(fj to construct V4 independently of b) and c). '

65) In the case of Exercise 64, determine the Cart an matrix by the following arguments.

a) Let K be the module for the trivial representation of a Sylow 5-subgroup of (fj. Show by Nakayama's reciprocity theorem that K(fj is the principal indecomposable projective K(fj-module Pl'

b) Use Nakayama's reciprocity theorem and Mackey's theorem to prove that C 11 = 4.

c) Show that the Cartan matrix is of the form

(4 2 2 0)

C = 2 Czz CZ3 0

2 CZ3 C33 0 o 0 0 1

where either Czz = C33 = 2, CZ3 = 1 or Czz = C33 = 3, CZ3 = O. d) Show that Vz ®K Vz has the composition factors V1 , V1 , V3 .

e) If Pi denotes the indecomposable projective K(fj-module with head '1;, show that P1 ®K Vz ~ 2Pz E8 2P4 ·

f) Conclude that Czz = C33 = 2 and CZ3 = C32 = 1.

66) In the case of Exercise 64, determine a) the composition factors of all the tensor products Vi ®K Vj' b) the decomposition of all the modules Pi ®K Vj and Pi ®K Pj as

direct sums of indecomposable projective modules.

67) Consider the example of exercise 64 and show the following. a) P1 has Loewy factors V1 , Vz E8 V3 , V1 E8 V1 , V2 E8 V3, V1 . (Use

Exercise 61.)


b) Show that there does not exist an indecomposable K(f)-module W with socle S ~ VI EB VI and W/S ~ V2· (SupposethatW/S ~ V2andS ~ VI EB VI' Let 'l3bea Sylow 2-subgroup of (f). Show first that ifU is a K-subspace ofS such that U (') WJ(K'l3) = 0, then U is a K'l3-direct summand of W'fl and a K(f)-submodule, hence a K(f)-direct summand of W, a contradiction. Hence S s:: WJ(K'l3). Sho~ that dim K WJ(K'l3) = 3 and conclude that W'fl is projective. Hence W is a projective K(f)-module, a contradiction.)

c) P2 has Loewy factors V2, VI' V3, VI' V2. (Use the fact that P2, and hence also P2J(K(f»)jS(P2)' is self-dual.)

d) Show that there does not exist an indecomposable K(f)-module W with socle S, where W/S ~ V2 and S ~ V2 or S ~ V3.

68) Consider K(f), where (f) = m:s and K is an algebraically closed field of characteristic 5.

a) As (f) ~ PSL(2, 5) (II, 6.14), there exist irreducible K(f)-modules VI' V2, V3 of dimensions 1,3,5, and no others.

b) Let ~ again denote the projective K(f)-module with head '1;. Show that dimK PI = 5, therefore ell = 2 and c 12 = 1. Conclude that dimK P2 = 10 and P3 = V3 .

c) Show that the Cart an matrix is

(2 1

C = 1 3

° ° d) The Loewy factors of P2 are V2, VI EB V2, V2. (Use the fact that P2

is self-dual.)

69) Finally consider K(f), where (f) = m:s and K is an algebraically closed field of characteristic 3.

a) There is no irreducible K(f)-module of dimension 2. (Consider the restriction to a Sylow 2-subgroup.)

b) Let 91 denote the normalizer of a Sylow 5-subgroup of (f) and consider K91, where K denotes the trivial K91-module. Show that K91 is projective and has a factor module isomorphic to the module VI for the trivial representation of (f). Show finally that Kffi = PI and that PI has Loewy factors VI' V2, VI' where V2 is an irreducible K(f)-module of dimension 4.

c) Show that dimK P2 = 9 and that there are irreducible projective K(f)-modules Vi = Pi (i = 3, 4) of dimension 3.

d) The Cartan matrix of K(f) is

§ 16. Green's Indecomposability Theorem 223

and KG> has three blocks. e) Let K be the trivial module for the subgroup m:4 of G>. Show that

K(fj ~ V1 EB V2 .

f) Let W be the non-trivial K91-module of dimension 1. Show that W(fj ~ V3 EB V4 •

(Exclude the possibility that W(fj ~ V3 EB V3 by using KG> ~ (K91)(fj and the decomposition of K91 into irreducible modules.)

§ 16. Green's Indecomposability Theorem

16.1 Definition. A subgroup 91 of a group G> is called subnormal if there exists a series

of subgroups such that 91i <J 91i+1 (1 ::;; i < n).

The following theorem plays a part in many investigations and is the main theorem of this section.

16.2 Theorem (GREEN [1]). /.jet K be a perfect field of characteristic p and let 91 be a subnormal subgroup ofG> of index a power ofp. If V is an absolutely indecomposable K91-module, then V(fj is an absolutely indecomposable KG>-module.

For the proof we need a lemma and some preliminary remarks.

16.3 Remarks. a) We may suppose that K is algebraically closed. For let K be the algebraic closure of K. By 6.7, V K is absolutely

indecomposable. Thus, assuming the theorem in the algebraically closed case, (VR)(fj is absolutely indecomposable. Thus (V(fjh is indecomposable and by 6.9, V(fj is absolutely indecomposable.

b) We may suppose that IG>: 911 = p.

For by 16.1, there is a series

of subgroups such that 91i <J 91 i +1 (1 ~ i < k). By adding more terms if necessary, we may suppose that 191i+1 : 91i l = p (i = 1,2, ... ,k - 1). Thus if the result is known for index p, the absolute indecomposability of V 9l;+l follows from that of V9\ since

c) We may suppose that the inertia subgroup of V in m is m itself. For otherwise, by b), the inertia subgroup is 91. Then VfD is an

indecomposable Km-module by 9.6a). As K is algebraically closed, VfD is absolutely indecomposable.

16.4 Lemma. Suppose that 91 <J m and that V is a K91-module such that the inertia group of V in m is m itself. ltTite S = HomKfD(VfD, VfD) and m = m/91. Then, corresponding to each h E m, there is a subspace Sh of S with the following properties.

a) If gEm and g = g91,

Sg = {ala E S, (V ® 1)a £; V ® g-1}.

b) S = E8hE® Sh' c) Sh Sh = Sh h flor all h1' h2 inm. Further, there exist units t/lh in

1 2 1 2

S such that

(Thus S is an algebra graded by m/91). d) S1 and Hom K91(V, V) are isomorphic K-algebras.

Proof a) Let T be a transversal of 91 in m. For each t E T, define

Then SI is a subspace of S. If gEm, g = g91 and g = xt with x E 91, t E T, then V ® g-1 = V ® t- 1 ; hence Sg is well-defined.

To prove the remaining assertions, observe that for any gEm, V ® g is a Kin-submodule of Vffi • We shall use the following.

(*) If ~ E HomK91(V ® 1, V ® g-l), there exists~' E Sgsuch that


(v ® g')~' = (v ® 1)~g' (v E V, g' Em).

To see this, observe that we may define ~'E HomK(V(jj, V(jj) by

(v ® t)~' = (v ® 1)~t (v E V, t E T).

Then if g' E m and g' = xt with x E 91, t E T,

(v ® g')~' = (vx ® t)~' = (vx ® l)~t = (v ® 1)~xt = (v ® 1)~g'.

It follows at once that ~'E HomK(jj(V(jj, V(jj). And since (V ® 19' =

(V ® 1)~ £; V ® g-l, we obtain ~' E 6 g•

b) It is to be proved that

First suppose that 0 = ItET Yt, with Yt E 6 t. Then for any v E V, gEm,

I (v ® g)Yt = o. tET

But (v ® g)Yt = (v ® 1)Ytg £; V ® t- 1g and V(jj = EBtET V ® t- 1g. Hence (v ® g)Yt = 0 for all v E V, gEm, and Yt = O.

Next suppose that a E 6. Let 7rt be the projection ofV(jj onto V ® t- 1 ;

thus 7rt E HomK\Jl(V(jj, V ® t- 1) and

By (*), applied to the restriction of a7rt to V ® 1, there exists Pt E 6 t such that

(v ® g)Pt = (v ® 1)a7rtg (v E V, gEm).

Hence

(v ® g) I Pt = I (v ® 1)a1tB = (v ® 1)ag = (v ® g)a, tET tET

so

Thus

c) If hi = gi91 and rii E e;h, (i = 1,2) then by a),

(V ® l)ril ri2 £; (V ® g1 1)ri2 = ((V ® l)ri2)g11

£; V ® g"2 1g1 1 = V ® (glg2tl,

SO ril ri2 E e;h h . Thus e;h e;h £; e;h h . 1 2 1 2 1 2

Suppose that t E T. By hypothesis, there exists a K91-isomorphism Pt

of V @ 1 onto V ® t -1. By (*), there exists l/II E e;I such that

It is clear that l/It is surjective and hence non-singular. Thus l/Ii1 E e; and l/Ii1 E e;r', since (V ® t)l/It = V ® 1. In general, we have

Thus

so

In particular, e;h = e; 1 l/Ih' and similarly, e;h = l/Ih e; l' d) If ri E Hom K91(V, V), then by (*), there exists p E e;1 such that

(v ® g)P = (Vri ® l)g = Vri ® g.

The mapping ri -+ P is an algebra monomorphism. It is in fact an isomorphism, since if P' E e;1'

(v @ 1)p' = Vri' ® 1

for some ri' E HomK;n(V, V). q.e.d.

16.5 Proof of 16.2. By the remarks in 16.3, we may suppose that 91 <J (f), (f)/91 is a p-group and the inertia subgroup of V in (f) is (f) itself. Put

§ 16. Green's Indecomposabi1ity Theorem 227

(!j = (!j/91 and 6 = HomK<»(V<», V<»). By 16.4, 6 has a subalgebra 6 1

isomorphic to Hom K91 (V, V), and there exist units l/Ih (h E (!j) of 6 such that l/Ih61 = 6 1 l/Ih = 6h, 6 h 6h = 6 h hand

1 2 1 2

hE(f; hE(f; hE (f;

Hence 3 = J(61)6 is a two-sided ideal of 6 and

(1) 3 = EB l/IhJ(61), hE (f;

Thus 3 s;;; J(6). Since V is an absolutely indecomposable K91-module,

Thus the elements of 6 1 are expressible as Al + a with A E K, a E J(6d. Hence the elements of 6 are all of the form

with b E 3. It follows from this and (1) that if eh {ehlh E (!j} is a K-basis of the algebra 6/3. Now

so


Then

so

(2)

(Thus we have an element of Z2(m, K x ).) Put

It follows from (2) that

Since K is perfect and I m I is a power of p, there is an element Vh E K x such that v~(fil = J.lh (h E m). Since char K = p, it follows that

Ah hVhh =VhVh· l' 2 1 2 2 1

Also {Ji,lh E m} is a K-basis of 6/3, so the algebra 6/3 is isomorphic to Km. Since m is a p-group,

Km/J(Km) ~ K.

Since 3 ~ J(6), J(6/3) = J(6)/3. It follows that

6/J(6) ~ (6/3)/J(6/3) ~ Km/J(Km) ~ K.

Hence VID is absolutely indecomposable. q.e.d.

We mention a special case of 16.2.

16.6 Theorem. Suppose that m is a p-group and that U ::; m. Let K be a perfect field of characteristic p and let V be an absolutely indecomposable KU-module. Then VI» is absolutely indecomposable.

Proof This follows at once from 16.2, since U is subnormal in m (see II1,2.3). q.e.d.

16.7 Theorem (WILLEMS [1]). Let K be an algebraically closed field of characteristic p. Suppose that 91 <l m and let V be an indecomposable K91-module with inertia subgroup 3. Then the following are equivalent.

a) VI» is an indecomposable Km-module. b) V3 is an indecomposable K3-module. c) 31m is a p-group.

Proof a) => b): Since VI» = (V3 tf>, this is trivial. b) => c): If 31m is not a p-group, it contains a non-identity cyclic

p'-subgroup U/91. By 9.9b), there exists a KU-module W such that W'Jt ~ V. Hence by 4.15b),

But U/91 is a non-identity p' -group, so

K(U/91) ~ K EEl U

where K is the trivial KU-module and U is a non-zero KU-module. Hence

and

Thus V3 is decomposable, contrary to b). c) => a): As 31m is a p-group, V'J is indecomposable, by 16.2. Hence

Vff> is indecomposable by 9.6. q.e.d.

In preparation for the proof of a theorem of Fong, we introduce the projective envelope.


16.8 Lemma. Let V be a K(fj-module. a) To within isomorphism, there is exactly one projective K(fj-module

P for which PIP J(K(fj) ~ V IV J(K(fj). We put P = P(V); P(V) is called the projective envelope of V.

b) There exists an epimorphism Av of P(V) onto V such that ker Av ~ P(V)J(K(fj).

c) Suppose that V, Ware K(fj-modules and a is an epimorphism of V onto W. Then peW) is isomorphic to a direct summand of P(V).

Proof a) By 1O.9b), there exists a projective K(fj-module P such that P/PJ(K(fj) ~ VjVJ(K(fj), since VjVJ(K(fj) is completely reducible. By 10.6, P is uniquely determined by this condition.

b) Let a be an isomorphism of P/PJ(K(fj) onto VjVJ(K(fj). By 10.Sa), there is an epimorphism A of P onto V for which

XA + VJ(K(fj) = (x + P J(K(fj))a.

If x E P and XA = 0, then (x + PJ(K(fj))a = 0 and x E PJ(K(fj). c) Consider the commutative diagram

V IJ. • W ------I. 0

'·1 p '·1 P(V) -------~ P(W)

Since Aw is an epimorphism, there exists fJ E HomKo;(P(V), peW)) such that fJAw = Ava. Since Ava is an epimorphism,

peW) = P(V)fJ + ked w

= P(V)fJ + P(W)J(K(fj).

Hence by nilpotency of J(K(fj), peW) = P(V)fJ and fJ is an epimorphism. Since peW) is projective, peW) is isomorphic to a direct summand of P(V). q.e.d.

Now we can prove an interesting theorem about the projective indecomposable modules for soluble groups.

16.9 Theorem (FONG [lJ). Suppose that (fj is soluble and that K is an algebraically closed field of characteristic p. Let V be an irreducible Kffimodule and let P(V) be its projective envelope. Then

ProoJ(w. SCHWARZ [1]). This is proved by induction on 1<»1. Let 91 be a maximal normal subgroup of <». Then either 1<»/911 = p or 1<»/911 is not divisible by p.

a) Suppose first that 1 <»/911 = p. By Clifford's theorem, there exists a decomposition

V'JI = WI E9 ... E9 Wr,

where WI' ... , Wr are irreducible K91-modules of equal dimensions. By 9.20, r divides 1 <»/911 = p. Thus

By Nakayama's reciprocity theorem, the multiplicity of V in the head of W~is

Thus there is an epimorphism of W~ onto V. Hence by 16.8c), P(V) is isomorphic to a direct summand of P(W~). Since WI is irreducible, P(W1) is indecomposable; hence by 16.2, P(WSfJ is indecomposable. By 16.8b), there is an epimorphism of P(W1) onto WI' so by 4.2, there is an epimorphism of P(WSfJ onto W~. By 16.8c), P(W~) is a direct summand of P(P(WSll). By 7.17, P(WSfJ is projective, so P(P(WSfJ) = P(Wd(fj and P(W~) is a direct summand of P(W1)(fj. Since P(W1)(fj is indecomposable, so is P(W~), so P(V) ~ P(W~) = P(W1)(fj. Thus

dimK P(V) = 1 <»/911 dimK P(W1)

= 1 <»/91II 91 I/dimK W1)p"

by the inductive hypothesis. Hence

as required. b) Now suppose that 1<»/911 is prime to p. By 7.21b), J(K<») =

J(K91)K<».1f g E <», g-I(K91)g = K91, so g-IJ(K91)g = J(K91). Thus

J(K<») = K<»J(K91).


Consider K(f) as a Kin-module; then the greatest completely reducible factor module of (K(f))91 is

Write

K(f)/J(K(f)) = EBdiVi i

with non-isomorphic irreducible K(f)-modules Vi' Then

K(f) = EBdiP(Vi) i

and

(K(f))91 = EB diP(Vi)91' i

But since K(f) is a projective Kin-module,

(K(f))91 = P((K(f))91)

= P((K(f))91/(K(f))91J (Kin))

= P((K(f)/J(K(f)))91)

= EBdiP((Vi)91)' i

But since there is an epimorphism of P(Vi) onto Vi and hence of P(Vi)91 onto (Vi)91, P((VJ91) is a direct summand of P(P(Vi)91) = P(Vi)91' Hence

and

But by Clifford's theorem,

(Vi )91 = EB cij Wij j


with non-isomorphic irreducible K91-modules Wij of equal dimension. Thus

and

P((Vj>m) = EBcijP(Wij) j

dimK P(VJ = I Cij dimK P(Wjj). j

By the inductive hypothesis,

so

dimK P(VJ = (~Cjj) I 91lp(dimK Wil)p"

By 9.20, Ij cij divides I (fj/911 and is therefore a pi-number. Thus

dimKP(VJ = 1911p (ICijdimK Wij}'

= I (fjlp(dimK Vj)p" q.e.d.

Theorem 16.9 holds also for p-soluble groups; the proof is similar and uses the extension of 9.20 to p-soluble groups.

With our methods, we can deal with p-nilpotent groups.

16.10 Theorem (MICHLER [2], SRINIVASAN [1]). Suppose that (fj is p-nilpotent and 91 = 0 p,(fj). Let K be a perfect splitting field for 91 of characteristic p. Let f1' ... ,fm be a complete set of representatives of the (fjconjugacy classes of indecomposable central idempotents in Z(K91). Let

be the inertia group of ii, and let

p"

(fj = U 'Xjgij j=l

be a decomposition of (fj into right eosets of'Xj.


a) The elements

pOI

/;' = L gi"/ figij (i = 1, ... , m) j=l

are the indecomposable central idempotents of K(f). In particular, (f) has exactly m blocks.

b) Let VI' ... , Vm be a complete set of representatives of the (f)conjugacy classes of irreducible K9l-modules. Write Pi = V/Jj. Then to within isomorphism, PI' ... , Pm are all the indecomposable projective K(f)modules.

c) Wi = PJPiJ(K(f)) is an absolutely irreducible K(f)-module. In particular, K is a splitting field for (f).

d) (Wi)m ~ EB!~i1 Vi ® %. In particular, dimK Wi = I (f) : :!d dimK Vi' Also, the composition length of P; is I:!J9lI·

e) If I (f)/9l1 = pa,

and

Proof (1) (f) has at least m blocks. The indecomposable central indempotents of K9l are the elements

-11' gij Jigij, so

m pQi m

1 = L L gi-/figij = L fi'· i=l j=l i=l

Clearly /;' E Z(K(f)), so (f) has at least m blocks. (2) As K is a splitting field of 9l and 9l is a p' -group,

r

K9l = EBVi i=l

for absolutely irreducible K9l-modules Vi' Then

r

K(f) ~ (K9lY" = EB Vim, i=l

and by 16.2, Vim is absolutely indecomposable. If Vi is (f)-conjugate to Vj' Vim ~ VJ. Thus the decomposition of K(f) contains at most m isomorphism types of indecomposable direct summands. Being direct summands of K(f), the V?' are of course projective.

Combining (1) and (2), we obtain the following. (3) (f) has exactly m blocks and the idempotents 1/ are the block

idempotents of K(f). Supposing (as we may) that no two of VI' ... , Vm are isomorphic, PI' ... , Pm are all the types of projective indecomposable K(f)-modules. In particular, all the indecomposable projective modules in a given block are isomorphic. The Cartan matrix is thus a diagonal matrix (cf. 14.9).

Thus a), b) are proved. (4) Since the Pi are absolutely indecomposable, it follows that for any

extension L of K,

(PJPiJ(K(f»)) ®K L ~ (~ ®K L)/(~ ®K L)(J(K(f») ®K L)

~ (Pi ®K L)/(~ ®K L)J(L(f») (by 1.5)

is an irreducible L(f)-module. Hence Wi is absolutely irreducible. (5) By Clifford's theorem, (Wi)<Jl is a completely reducible K91-module.

It follows from Nakayama's reciprocity theorem that

This is 1 if Vjm ~ Vr» and 0 otherwise. But VJ ~ V? if and only if Vj' Vi are (f)-conjugate. Thus

and

pO;

(Wi)<Jl ~ EB Vi ® gij j=1

Since the Cart an matrix is diagonal, all the composition factors of Pi are isomorphic to Wi' The composition length of Pi is thus

(6) Since the block ideal fiK(f) is the direct sum of dimK Wi indecomposable K(f)-modules each isomorphic to Pi'

dimKf;K(f) = (dimK Wj)(dimK P;)

= pa'(dimK Vj)pa(dimK Vj)

= pa+a'(dimK Vjf.

Further, f;K(f)/JU;K(f») is isomorphic to the direct sum of dimK W; modules isomorphic to Wj, so

Thus

A number of similar facts can also be proved for p-soluble groups (see SCHWARZ [1]).

Exercises

70) Suppose that K = GF(2), (f) = 6 3 and 91 = m:3. Then there is an irreducible K91-module V of dimension 2 such that Vcr. is decomposable. (Thus in 16.2, absolute indecomposability cannot be replaced by indecomposability.)

71) Suppose that K is algebraically closed of characteristic 5, (f) = m:5 and U = m:4. Then there is an irreducible KU-module V of dimension 3 for which vcr. is decomposable. (Thus in 16.2, the condition that 91 be subnormal cannot be replaced by 91 ::s;; (f»).

72) Is Theorem 16.9 valid for arbitrary groups?

73) (WILLEMS [1]) Suppose that K is an algebraically closed field of characteristic p. Suppose that 91 <l (f) and that V is an irreducible K91-module with inertia group 3. Then the following are equivalent.

a) 3/91 is a p'-group. b) vcr. is completely reducible. c) If ~/91 E Sp(f)/91), V'fl is completely reducible.

74) a) Formulate and prove the assertions about the injective envelope dual to 16.8.

b) Give an example of a module for which the injective and projective envelopes are not isomorphic.


Notes on Chapter VII

§ 1: 1.17 is stated in BRAUER [4], but essentially, it is already contained in BRAUER [I] on page 101. The proof of 1.20 was communicated to us by A. Dress. 1.22 was first proved by E. Noether and published in DEURING [I]. § 3: 3.9 was first proved in BRAUER [2]. The proof given here is taken from BRAUER [6]. For an interesting different treatment of 3.10, we refer to GLOVER [I]. § 4: This section follows mainly two fundamental papers of D. G. HIGMAN [2, 3]. § 5: The central result 5.4 of this section comes from D. G. HIGMAN [3], with some improvements from KASCH, KNESER and KUPISCH [I]. § 6: After 6.6, the treatment follows HUPPERT [4]. The result in 6.11 is already contained in the unpublished thesis of D. VOIGT (Kie1 1965). § 7: The use of Gaschiitz operators to characterize projective and injective modules first appears in GASCHUTZ [I]. Relative projective and relative injective modules are introduced in D. G. HIGMAN [4]. § 8: 8.11 was communicated to us in a letter by R. Gow. § 9: 9.18 was proved for characteristic 0 in ROTH [I] by character calculations, and the general case is from SCHWARZ [I]. The Exercises 40,41,43 and 44 are taken from WILLEMS

[I ]. § 10: Most of the content of this section appears in the work of Brauer, Nakayama and Nesbitt on algebras. In NAKAYAMA [1],10.13-10.18, among other results, will be found. The concepts of projective and injective modules, only introduced some years later, did much to clarify the situation. In particular, compare 10.13 with 11.5. The basic result 10.12 is essentially in BRAUER and NESBITT [I]. § 11: Frobenius and symmetric algebras are introduced in NESBITT [I]; for instance, most of 11.6 is there. 11.3 is from NAKAYAMA [2]. § 12: Many results in this section are from the basic paper BRAUER [6]. § 13: 13.5, 13.7 and 13.8 were first proved using character theory in FONG and GASCHUTZ

[I]. The treatment in this section follows COSSEY and GASCHUTZ [I]. § 14: Most of this section follows WILLEMS [3]. Most of 14.9 is already in BRAUER and NESBITT [1]. § 15: 15.1-15.6 are due to oral communications by Gaschiitz. § 16: The proof of Lemma 16.4 is strongly influenced by a lecture of E. Dade at Aarhus in 1978.

Chapter VIII

Linear Methods in Nilpotent Groups

The subject of this chapter is commutator calculation. It will be recalled that the commutator [x, y] of two elements x, y of a group is defined by the relation

We then have

[xy, z] = [x, z ]Y[y, Z ], [x, yz] = [x, z ] [x, y y.

These relations are rather similar to the conditions for bilinearity of forms, and there are a number of ways of formalizing this similarity. Once this is done, commutator calculations can be done by linear methods. Several examples of theorems proved by this method will be given in this chapter.

One of these is the determination of the Suzuki 2-groups (see 7.1). This is accomplished by applying the idea sketched above to express the relevant commutation in terms of bilinear forms. No other structure of linear algebra is used in this problem. But the Witt identity

is remarkably like the Jacobi identity

[x,y,z] + [y,z,x] + [z,x,y] = 0

of Lie rings. This can be formalized by associating a Lie ring with any nilpotent group, and this will be used in the discussion of other problems.

One of the best-known such problems is the following. Suppose that Am is the group-ring of the group m over the commutative ring A, and let 3 be the augmentation ideal of Am. If n is a positive integer, define mn to be the set of elements x of m such that x-I E 3n• It is easy to see

§ l. Central Series with Elementary Abelian Factors 239

that (f)n is a normal subgroup of (f). The dimension subgroup problem is to characterize (f)n within (f) itself.

Magnus and Witt used the procedure mentioned above to solve this problem in the case when A is the ring 7L of integers and (f) is a free group; this is presented in § 11. No definitive answer is known for a general group (f) when A is the ring of integers. On the other hand, the problem was solved by Jennings for any group (f) in the case when A is a field. This is presented in § 2 in the case of prime characteristic. The Lie ring method is not needed for this.

The simplification effected when commutation is replaced by a bilinear form is, of course, not very great in itself, but an additional advantage of doing so is that operations on the linear structures involved may be used, even when no corresponding operation exists for the group. The operation most successfully used in this connection is extension of the ground-ring. An application involving this is given in §§ 9-10, where it is shown that the class of a nilpotent group having a fixed point free automorphism of prime order p is bounded by a function of p.

Another method for applying linear methods to nilpotent groups is the use of the Baker-Hausdorff formula and the corresponding inversion formula (see, for example, AMAYO and STEWART [1]). This, however, will not be described in this chapter.

§ 1. Central Series with Elementary Abelian Factors

Throughoutthis section let p be a fixed prime. We shall give a construction of two central series Kn(f)), An(f)) (n ~ 1) of a group (f), for which Kn(f))/Kn+1 (f)) and An(f))j An+1 (f)) are elementary Abelian p-groups. The lower central series of (f) will be denoted by

here Yn(f)) = [Yn-1 (f)), (f)] for n > 1. If m is a positive integer, (f)m = <xmlx E (f). Thus (f)mn ~ (f)m)n and (f)j91)" = (f)n91j91 for 91 <J (f).

1.1 Lemma. a) Ifx, yare elements of a group (f),

n

(xy)p" := xP"yP" mod Y2(f))P" TI YP,(f))P"-' r=1

for all n ~ 1.

240 VIII. Linear Methods in Nilpotent Groups

b) If x, yare elements of a group ffi, S ::;; ffi and x, [x, y] belong to S,

n

[xP",y] == [x,y]P" modY2(S)P"flyp '(S)P"-' r=1

for all n ~ 1.

Proof a) Let

n

91 = Y2(ffi)P" fl Yp,(ffi)P"-' . r=1

By the Hall-Petrescu formula (III, 9.4), there exist elements Ci E Yi(ffi) (i = 2, ... , pn) such that

If (i, p) = 1, then (1") is divisible by pn, and for i > 1, c~~') E Y2(ffi)P" ::;; 91. And if i = prj, where r ~ 1 and (p, j) = 1, then (1") is divisible by pn-r and i ~ pr, so c:~') E Yp,(ffi)P"-' ::;; 91. Hence

b) By a),

n

(x [x, y])P" == xP"[x, y]P" mod Y2(S)P" fl YP,(S)P"-'. r=1

But

so the assertion follows at once. q.e.d.

1.2 Lemma. If n ~ 0,

n

[Yi(ffi)P", Yiffi)] ::;; fl Yj+ip,(ffi)P"-' . r=O

Proof Let 91 = fl~=o Yj+iP,(G'»P"-'. Since 91 <l (f), it is sufficient to show that expo, y] Em for any x E Yi(f)), Y E Yj(f)). By l.lb),

§ 1. Central Series with Elementary Abelian Factors 241

n

[xP",y] == [x,y]P" modY2(f»)P"IlYp .(f»)p"-', r=1

where f) = <x, [x, y]). Then Y2(f») is the normal closure of [x, y, x] in f), by 111,1.11, so Y2(f») ~ Y2i+i(f»)· Since f) ~ Yi«f»), it follows that Ym(f») ~ Ymi+ i(f») for all m ~ 2. Thus YP,(f»)P"-' ~ Yp'i+ i(f»)P"-' ~ 91 for r = 1, ... ,n.AlsoY2(f»)P" ~ Yi+i(f»)P" ~ 91 and [x, y]P"E Yi+j«f»)P" ~ 91. Hence [x P", y] E 91, as required. q.e.d.

1.3 Lemma. For i ~ 1, k ~ 0, put

Proof By 1.2,

k-h [Yi+h«f»)Pk-h, Yj«f»)] ~ Il Yj+(i+h)p,«f»)pk-h-· ~ 91i+j,k'

r=O

since j + (i + h)pr + (k - h - r) ~ j + ipr + k - r ~ j + i + k. Hence

(1)

mp m' (lC'.)pH1 ••• (IC'.)P m :J~ik ~ :J~ikYi w Yi+k w ~ :J~i,k+1'

It follows at once that 91fkm ~ 91i,k+m' Now suppose that x E Yi«f»), y E 91jl . By (1), [x, y] E 91i+j,1 and

[x, y, x] E 912i+j,I' Thus if f) = <x, [x, y]), Y2(f») ~ 912i+j,I' Again by (1), it follows that Yn(f») ~ 91ni+j ,I' Thus for 1 ~ r ~ k,

Also, [x, y]pk and Y2(f»)P' are both contained in


Hence by 1.1 b),

[x P', Y] E 91i+j ,k+I'

It follows that

and

Hence

q.e.d.

1.4 Definition. For any group Gl, put

Thus An(Gl) is a characteristic subgroup of Gl and

1.5 Theorem. a) [Am(Gl), An(Gl)] ~ Am+n(Gl). b) An(Glt ~ An+)Gl). c) For n > 1, An(Gl) = [An-1(Gl), Gl]An-l(Gl)P.

Proof In the notation of 1.3, An(Gl) = 911,n-l, so a), b) follow at once. It follows from a) and b) that

so it only remains to prove that

for i = 1, ... , n. But for i = 1, ... , n - 1,

and

1.6 Corollary. Suppose that

where (f)i <1 (f), (f)/(f)i+1 ~ Z((f)/(f)i+1) and (f)/(f)i+1 is an elementary Abelian p-group (i = 1, 2, ... ). Then An((f) ~ (f)n.

Proof We use induction on n. For n > 1, An- 1((f) ~ (f)n-1' by the inductive hypothesis. Hence by 1.5,

1.7 Theorem. a) Suppose that ex, p are homomorphisms of (f) into GJ which induce the same homomorphism on (f)/Az((f). Then ex, p induce the same homomorphism on An((f)/An+1((f) for all n 2:: 1.

b) Suppose that ~ ~ (f) and ~AZ((f) = (f). Then ~An+1((f) = (f) for all n 2:: 1.

Proof a) This is proved by induction on n and is trivial for n = 1. For n > 1, it follows from the inductive hypothesis that if u E An-1 ((f), then uex = (uP)v for some v E An(GJ). Thus

since by 1.5, An(GJ)/ An +1 (GJ) is central and elementary Abelian. Also, if x E (f),

[u,x]ex = [uex,xex] = [(uP)v,xex] == [uP,xex] mod An+1(GJ).

But by hypothesis, xex = (xP)y for some y E AZ(GJ), so

[u, x]ex == [uP, (xP)y] == [uP, y] [uP, xP]Y == [uP, y] [uP, xP] mod An+1 (GJ).

Using 1.5a),

so


By 1.5c),

so iX, [3 induce the same homomorphism on An ((f))/An + 1 ((f)). b) Again, we use induction on n. For n > 1, the inductive hypothesis

yields (f) = ~An((f)). Thus by 1.5,

An((f)) = [An- 1 ((f)), (f)] An-l ((f))P

~ [~An((f)), ~An((f))] (~An((f)))P

~ ~An+l ((f))

q.e.d.

To investigate further the factors of this series, we prove the following.

1.8 Theorem. For i ~ 1, k ~ 0, let

a) Any element of'Rik can be written in the form

for suitable aj E Yj((f)). b) If x E Yi((f)) and y E Yi((f))PYi+l ((f)), then

( )p' - p' d m xy = x mo :rli.k+l'

c) There is a surjective mapping [3 of the direct product

onto 'Rid'Ri,k+l given by

(2) (- - -)[3 p' p.- 1 m ai' ai+l, ... , ai+k = ai ai+l ... ai+k;lli,k+l,

where ai+ j E Yi+ i(f)) and ai+ j = ai+ jYi+ i(f))PYi+ j+1 ((f)). Except in the case when p = 2 and i = 1, [3 is an epimorphism.

Proof Note first that 91ik :s; 91jl whenever i ~ j and i + k ~ j + I.

Suppose that x, yare elements of y;(m) and ~ = <x, y). Thus ~ :$; Yi(m) and Ym(~) :$; Yim(m). Hence by 1.1,

k (3) (xy)pk == XPkypk mod Y2i(m)pkYp(~)pk-l n YiP,(m)pk-'.

,=2

a) This is proved by induction on k. For k = 0, it is trivial. For k > 0, observe that

Y2i(m)pk :$; 91i+l,k-I,

Yp(~)pk-l :$; YPi(m)pk-l :$; 91 i +I ,k-I'

Yip,(m)pH:$; 91i+I,k-l,

since ip' + k - r ~ i(r + 1) + k - r ~ i + k. Thus by (3), x ~ X Pk91i+I,k_1 is a homomorphism of Yi(m).

Now suppose that U E 91ik • Then there exist elements x 10 ... , Xm in y;(m) and v E 91 i + I ,k-1 such that

k • u = xf ... x! v.

Since x ~ x P'91i+I ,k-l is a homomorphism,

xf' ... x!' = af'v'

for ai = Xl ... x m, v' E 91i+I,k-I' Thus u = ar'v'v, and the assertion follows by applying the inductive hypothesis to V'V.

b) By (3),

k

(xy)P' == xP'yP' mod Y2i(m)p'Yp(~)pk-l n YiP,(m)p'-', ,=2

where ~ = <x, y). Now Y E'91il , so by 1.3, yP' E 9li ,k+I' Also Y2i(m)p' :$; 9l i ,k+1 and, for r ~ 2, Yip,(m)p'-' :$; 91 i ,k+I' since

ip' + k - r ~ i + k + i(p' - 1) - r ~ i + k + (p' - r - 1)

~i+k+1.

Finally,

by 1.3. Thus

(,,)pk-l mpk-l m m Yp ~ ~ ~l2i,l ~ n2i,k ~ ~li,k+1'

Hence

( )pk _ pk d m xy = x mo ~li,k+1'

c) It follows from b) that there exists a mapping f3 given by (2). For if ai+ j E Yi+ i(f) and x E Yi+ i(f))PYi+ j+1 (f),

also 91i+j,k-j+1 ~ 91i,k+1' By a), f3 is surjective. Suppose that x, yare elements of Yi(f). Since ~ = <x, y) ~ Yi(f),

Yp(~) ~ Yip (f). Thus by(3),

k (xy)pk == XPkypk mod Y2i(f)pk n Yip,(f)pk-'.

,=1

Now Y2;(f))P' ~ 91i,k+1' And except in the case when i = 1, p = 2 and r = 1, i(p' - 1) ~ r + 1, so

ip' + k - r ~ i + k + 1

and YiP,(f))pk-' ~ 91i,k+ l' Thus except when i = 1 and p = 2,

( )pk _ pk pk d m xy = X Y mo ni,k+1'

Since 91i+j,k-j+1 ~ 91i,k+1' it follows that f3 is an epimorphism. q.e.d.

In § 13, we shall apply the following theorem to free groups.

1.9 Theorem. Suppose that (f) is a group and that for i = 1, ... ,n + 1, no non-identity element of (f)/Yi(f) is of finite order; (that is, (f)!Yi(f) is torsion1reej.

a) For i = 1, ... , n + 1,

b) There is a bijective mapping IY. of


where aj E Yi((f;) and Qi = aiYi((f;)PYi+l((f;)· c) For p odd, (J. is an isomorphism. For p = 2, the restriction of (J. to

is an isomorphism of this group onto ()'n((f;) (') Y2((f;))A.n+1((f;)!An+l((f;).

Proof a) This is proved by induction on i and is trivial for i = 1. Suppose that i > 1 and put

Clearly 91i ,n-i+l ~ A.n+l ((f;) (') Yi((f;). Suppose that x E A.n+l ((f;) (') Yi((f;). Then

by the inductive hypothesis. It follows from 1.8a) that x = aP·- i+ 2 y, where a E Yi-l((f;) and y E 91i,n-i+l' Thus aP·-'+2 = xy-l E Yi((f;). Since

(f;/Yi((f;) is torsion-free, a E Yi((f;). Hence aP·-'+2 E 91i,n-i+l and x E 91t,n-i+l' Therefore

b) By 1.8c), (J. is surjective. To show that (J. is injective, suppose that ai' bi are elements of Yi((f;) for which

mod A.n+1 ((f;).

We prove that ai == bi mod Yi((f;)PYi+l ((f;) (i = 1, ... , n) by induction on i. If i > 1, the inductive hypothesis gives aj == bj mod y/(f;)PYj+1 ((f;) for j = 1, ... , i - 1. Hence by 1.8b), a{-J == br-J mod A.n+ 1 ((f;). Thus

this also holds for i = 1. By a),

Hence

and

for some C E y;(fj). Since Yi(fj)/Yi+1 (fj) is a torsion-free Abelian group, it follows that

and ai == hi mod Yi(fj)PYi+1 (fj). Thus a is injective. c) By b) and 1.8c), a is an isomorphism for p odd. For p = 2, the

restriction a1 ora to

is /3y, where /3 is the epimorphism of g; onto 912,n-2/912,n-l given in 1.8c) and Y is the natural epimorphism of 912,n-2/912,n-1 onto 912,n-zAn+1(fj)/An+1(fj). By a) and b), a 1 is an isomorphism onto (An(fj) (') Y2(fj))An+1 (fj)/An+1 (fj). q.e.d.

We turn now to our second central series.

1.10 Definition. For any positive integer n, write

Then Kn(fj) is a characteristic subgroup of (fj, and

(fj = K (fj) > K (fj) > ... > K (fj) > .... 1 -2 - -n-

Note that it is possible to have Kn-l (fj) = Kn(fj) > Kn+1 (fj). For if (fj is Abelian, Kn(fj) = (fjP" where k is the smallest integer for which pk ~ n. Thus Kn(fj) = (fjpk if pk-l < n ::;; pk.

If (fj is a finite p-group, K 2(fj) = A2(fj) is the Frattini subgroup of (fj. If (fj is a group of exponent p, Kn(fj) = Yn(fj) for all n.

To prove that this series is central, we need the following lemmas.

1.11 Lemma. Suppose that i ~ 1,j ~ 1, h ~ O. Let

h

91 = n YH jp,(fWh~'. r=O

ffx E Yi(Q)) and n ~ 2,

h

Yn«x, 91») ::;; n Yin+jp,(Q))P'~'. r=O

Proof This is proved by induction on n. For n = 2, we have

Yz( <x, 91») = [91, <x, 91)]

by III, 1.11; since <x, 91) ::;; Yi(Q)), it follows that

For n > 2, we have

Using the definition of 91 for n = 2 and the inductive hypothesis for n > 2, it follows that for n ~ 2,

Applying III, 1.l0a) and 1.2,

h

Yn( <x, 91») ::;; n [Yin-H jp,(Q))ph~', Yi(Q))] r=O

h h-r n n (Q))P h~,~' ::;; YHp'(in-Hjp')

r=O s=O

::;; n Yin+ jp,+,(Q))ph~'~', r+s,;;h

since i + pSin - pSi ~ in. The lemma follows at once. q.e.d.

1.12 Lemma. IJi ;::: 1,j ;::: 1, h ;::: 0, k ;::: 0,

[Yi(OW', Yj(fj)P'] ~ Kip'+ jph(fj).

Proof Suppose that x E Yi(fj) and y E Yj(fj). Let z = [x, yph] and f) = <x, z). By 1.1b),

k [x P', yph] == [x, yph]pk = Zpk mod Y2(f))P' TI Ypm(f))pk~m.

m=l

For n = 1, ... , pk, define

h

f)n = TI Yin+ jp,(fj)ph~: r=O

By 1.2, z E f)1. Thus f) ~ <x, f)1), and by 1.11, Yn(f)) ~ f)n. Hence

n

[x P" yph] E TI f)~~~m. m=O

Now let 91 = Kipk+jph(fj).lfm + h - r;::: k,thenYipm+jp,(fj)Ph~' ~ 91, by 1.10. Thus if s = max(m + h - k + 1, 0),

h f)pm ~ 91 n Yi~+ jp,(fj)ph~' ~ 91Yipm+ jp,(fj).

r==s

Hence

since ipk + jps+k-m ;::: ipk + jph.

Thus [x P" yph] E 91 and x pk91 commutes with y ph91. Hence each element of Yi(fj)pk91/91 commutes with each element of y)(fj)Ph91/91, and

[Yi(fj)P" Yj(fj)ph] ~ 91. q.e.d.

1.13 Theorem. a) [Km(fj), Kn(fj)] ~ Km+n(fj). b) Kn(fj)P ~ Kpn(fj). c) For n > 1, Kn(fj) = [Kn- 1(fj), (fj]Km(fj)P, where m is the least

integer Jar which pm ;::: n.

Proof a) follows at once from 1.10 and 1.12.

b) By a), Yp(Kn(<»)) ~ Kpn(<»), so Kn(<»)/Kpn(<») is regular (III, 10.2). But Kn(<»)/Kpn(<») is generated by elements of order p since if ipk ~ nand x E Yi(<»)' (Xpky E Kipk+1(<») ~ Kpn(<»). Hence (Kn(<»)/Kpn(<»))P = 1 by III, 10.5, and Kn(<»)P ~ Kpn(<»).

c) By a) and b), [Kn- 1(<»), <»]Km(<»Y ~ Kn(<»). Suppose that ipk ~ n. If k = 0, Yi(<»yk = Y;(<») ~ Yn(<») = [Yn-1(<»)' <»] ~ [Kn- 1(<»), <»]. If k > 0, then ipk-1 ~ rn by definition of rn, and Yi(<»)pk ~ (Yi(<»)Pk-ly ~ Kipk-l(<»)P ~ Km(<»)P. Thus Yi(<»yk ~ [Kn- 1 (<»), <»] Km(<»y in any case, and the assertion follows from 1.10. q.e.d.

1.14 Corollary. Suppose that

where Sln <:l <», [Sln' <»] ~ Sln+l and Sl~ ~ Slnp for all n ~ 1. Then Sln ~ Kn(<») for all n ~ 1.

Proof This follows from l.13c) by induction on n (cf. 1.6). q.e.d.

1.15 Example. Let <» be a p-group of maximal class of order pm, where rn ~ p + 1. For n > 1, define the integers I, a as follows:

Then Kn(<D) = Ya+1+I(p-l)(<D). Ifrn - 1 = (p - l)q + rand 2 ~ r ~ p, Krp.(<») > K 1 +rp'(<») = 1.

For such a group Yi(<D)pk = Yi+k(p-1)(<D) (III, 14.16). Hence Ya+1+I(p-1)(<D) = Ya+l (<D)P' ~ K(a+1)p'(<») ~ Kn(<D), as (a + 1)pl ~ n. Conversely, if ipk ~ nand k ~ I, then i ~ 1 + apl-\ so that

i + k(p - 1) ~ 1 + k(p - 1) + apl-k ~ 1 + k(p - 1) + a + pH - 1

~ 1 + k(p - 1) + a + (I - k)(p - 1) = 1 + I(p - 1) + a.

Also if k > I, 1 + k(p - 1) ~ a + 1 + I(p - 1). Thus Kn(<») ~ Ya+1+I(p-1)(<»)·

Of course An(<») = Yn(<»), by 1.6.

Exercise

1) Prove that for any group <D,

Deduce that a) [,{n((}», (})] = Y2((}»)P"-1 ... Yn+1 ((}»; and b) for n > 1, ,{n((}» = [,{n-1 ((}», (})](})P·-'.

§ 2. Jennings' Theorem

There is a relation between the central series "n((}» i!ltroduced in § 1 and the group-ring over a field of characteristic p. This was first proved by JENNINGS [1]. Jennings proved his theorem only for finite p-groups; however it is true for all groups. For the rather easy extension to the general case we need the following lemma.

2.1 Lemma. Suppose that (}) is a nilpotent group, (}) = <Xl> ... , xn) and each Xi is of finite order. Then (}) is finite.

Proof We use induction on the class c of (}). By 111,1.11, yA(}» = <X), where X = {[Xi" .•. , xd 11 ::; ij ::; n}. X is finite and by III, 6.8, each element of X is of finite order. Hence the Abelian group (c((}» is finite. This is the assertion if c = 1. If c > 1, (})fyA(}» is finite by the inductive hypothesis, so (}) is finite. q.e.d.

2.2 Definitions. If A is any commutative ring with identity and (}) is a group, the group-ring (I, 16.6) of (}) over A will be denoted by A(}). If J1 , J2 are A-submodules of A(}), J1 J2 denotes the A-module spanned by all products ab, with aEJ1 , bEJ2 • Thus (J1J2)J3 = J1(J2J3), and we can form the powers J~ of J 1 . If ~h, 32 are 2-sided ideals of the ring A(}), so are 3132 and 31. The augmentation ideal 3 of A(}) (III, 18.3) is defined by

3 has A-basis {g - Ilg E (}), g #- I}.

2.3 Lemma. Let 31' 32 be two-sided ideals of A(}) and let

fti = {xix E (}), x-I E 3;} (i = 1,2).

§ 2. Jennings' Theorem 253

b) IfSl = {xix E ffi, x - 1 E 31~h + 3231}' then [Sll' Sl2] ::s;; R c) Ifx E Sll' Y E Sl2' then

[x, y] - 1 - {(x - 1)(y - 1) - (y - l)(x - I)} E 3(3132 + 3231)'

where 3 is the augmentation ideal of Affi.

Proof If Vi is the natural homomorphism of the ring Affi onto Affi/3i' the restriction f.li of Vi to ffi is a group-homomorphism of ffi into the group of units of Affi/3i. Since Sli = ker f.l;, a) is clear. Suppose that x E Sll and y E Sl2. Then

where z = (x - l)(y - 1) - (y - 1)(x - 1). But x - 1 E 31 and y - 1 E 32' so Z E 3132 + 3231. Hence [x, y] - 1 E 3132 + 3231 and [x, y] E R By a), Sl is a subgroup, so b) is clear. Also

[x, y] - 1 - Z = (X- 1y-1 - 1)z.

Since x-1 y-1 - 1 E 3, c) is proved. q.e.d.

2.4 Dermition. For any group ffi and any prime p, Kn(ffi)/Kn+ 1 (ffi) is an elementary Abelian p-group by 1.13. We may therefore think of Kn(ffi)/Kn+1 (ffi) as a vector space over GF(p) and choose a GF(p)basis 8n of it. For each Ii E 8n, choose a fixed b E Kn(ffi) such that Ii = bKn+1 (ffi), and denote by 8n the set of all such elements b. Let 8 = Un~l 8n, Such a set 8 will be called a K-net on ffi. Clearly 8n = (8 n Kn(ffi» - (8 n Kn+1 (ffi».

By an ordered K-net is meant a K-net 8 fully ordered in such a way that (i) a < b if a E 8 m, bE 8 n, m < n; and (ii) every non-empty subset of 8 n has a greatest element.

By the well-ordering principle, every group ffi possesses an ordered K-net.

2.5 Lemma. Let 8 be a K-net on ffi and let n be a positive integer. For each d = 1, ... , n, let Bd be a finite subset of (8 n Kiffi» -(8 n Kd+1(ffi». Let F be a finite subset of ffi. Then there exists a subgroup f) of ffi and a normal subgroup .R of f) with the following properties.

a) 0')1 = f)/.R is ajinite p-group and Kn +1(0')1) = 1. b) Bd S;; Kif) and F S;; f).

c) There exists a K-net C on (f)1 such that for d = 1, ... , n, (C n KA(f)1)) - (C n Kd+1(f)1)) :2 {b51lb E Bd}'

Proof If x E Kd(f)), there is a finitely generated subgroup (f)* of (f) such that x E KA(f)*); this is clear from 1.13c). Hence there exists a finitely generated subgroup ~ of (f) containing F such that b E KA~) for all bE Bd (d = 1, ... ,n), since F and the Bd are finite. Let 5l = Kn+1 (~). By 1.10, 5l ~ ~P"+\ so every element of (f)1 = N51 is of finite order. Also 5l ~ Yn+1(~)' so (f)1 is finitely generated and nilpotent. By 2.1, (f)1 is finite. Since (f)r+! = 1, (f)1 is a finite p-group. Thus a) and b) are clear.

Now {bKd+1(f))lb E Bd} is a linearly independent set of elements of KA(f))/Kd+1(f)). Hence {bKd+1(~)lb E Bd} is a linearly independent set of elements of KA~)/Kd+1(~)' Hence {bKd+1(~)lb E Bd} may be embedded in a GF(p)-basis Cd of KA~)/Kd+1 (~). Clearly a K-net C on (f) 1 may be formed from the Cd so that c) holds. q.e.d.

2.6 Theorem (JENNINGS [1]). Let (f) be a group and let p be a prime. Let B be an ordered K-net on (f), and Bn = (B n Kn(f))) - (B n Kn+1 (f))) (n ~ 1). Let K be a field of characteristic p and let ,3 be the augmentation ideal of K(f); If x E Kn(f)), then x - 1 E ,3n. For each n ~ 1, define

1:-:; ei < p,b1 < ... < bY' iteidi = n}.

Then {b* + ,3n+1lb* E B:} is a K-basis of the vector space ,3n/,3n+1.

Proof. The proof will be carried out in several steps. a) If x E Kn(f)), then x - 1 E ,3n. Also B: ~ ,3n. Let

is a central series of (f). If x E (f)n' xP - 1 = (x - 1)P E ,3pn, since K is of characteristic p, whence x P E (f)Pn" Thus (f)~ :-:; (f)pn" By 1.14, (f)n ~ Kn(f))

for all n ~ 1. Thus if x E 1<n(6)), X - 1 E ,3n. In particular b - 1 E ,3n if b E Bn. It follows that B: ~ ,3n.

b) If b1, ... , bm are elements of Bn and x E 1<n+1(6)),

m

b1 ... bmx - 1 == L (bi - 1) mod,3n+1. i=1

To see this, expand

b1 .•• bmx = {1 + (b 1 - 1)} ... {1 + (bm - 1)} {1 + (x - 1)}.

By a), bi - 1 E ,3n, so any product (b i - l)(bj - 1) lies in ,3n+1. Also x - 1 E ,3n+1 by a); hence the assertion.

c) ,3 is spanned over K by ,32 and all b - 1 (b E B 1)' The augmentation ideal,3 is spanned over K by all x - 1 with x E 6>.

By definition of B1, x = b1 .•• bmy, where bi E B1, Y E 1<2(6)). By b), x - 1 lies in the space spanned by ,32 and all b - 1 (b E Bl)'

d) ,3n is spanned over K by B: and ,3n+1. Suppose that this is false. Let Vn be the vector space spanned

over K by B: and ,3n+1. By a), Vn ~ ,3n. Also V1 = ,3 by c), for Bt = {b - lib E B1}. Let n be the smallest integer for which Vn "# ,3n. Thus n > 1. It follows from c) that ,3n is spanned over K by ,3n+1 and all (b1 - 1)· .. (bn - 1) with bi E B1. Since Vn "# ,3n, some such element does not lie in Vn. Let X be the set of finite sequences (b 1 , ... , bm) for which bi E Bd" d 1 + ... + dm = nand (b i - 1)· .. (bm - 1) 1: Vn. Thus X is non-empty.

Choose (b 1 , ••• , bm) E X with bi E Bd" d1 + ... + dm = nand bm

as large as possible in the ordering of B; this is possible since dm ::;; n and every non-empty subset of Bd has a greatest element. If dm = n, we have bm - 1 E B: ~ Vn, a contradiction. Thus dm < n. Write d = dm, b = bm .Now(b1 - 1) .. ·(bm- 1 - 1)E,3n-d bya).Bychoiceofn,Vn_d = ,3n-d; it follows that (b 1 - 1)··· (bm - 1) lies in the vector space spanned over K by ,3n+1 and {u(b - l)lu E B:-d}. Hence there exists u E B:-d such that u(b - 1) ¢ Vn. Write

u = (al - l}e, ... (ar - It"

where aiEBe, 1::;; ei < p, a1 < ... < ar and elcl + ... + ercr =

n - d. Since ~(b - 1) ¢ Vn, we have u(b - 1) ¢ B:. Hence either ar > b, or ar = band er = P - 1.

First suppose that ar > b. Since ar - 1 E ,3e, and b - 1 E ,3d,

[aY' b] - 1 == (ar - l)(b - 1) - (b - l)(ar - 1) mod ,3C+1,


where C = Cr + d, by 2.3c). By 1. 13a), [ar , b] E KJ(fj). Hence we may write

Car, b] = a'l ... a~a',

where aj E Be (j = 1, ... , s) and a' E Ke+1«(fj). By b),

Hence

s [a" b] - 1 == L (aj - 1) mod3C+l.

j=1

s

(ar - 1)(b - 1) == (b - 1)(ar - 1) + L (aj - 1) mod jC+1. j=1

Multiplying on the left by u' = (a1 - 1)e l ••• (ar - 1)e,-1,

s

u(b - 1) == u'(b - 1)(ar - 1) + L u'(aj - 1) mod3n+l, j=1

. , ""'n-c B ( b ) d Slnceu E;J . utnow al , ... ,al , ... ,a" ... ,ar, ,ar an e 1 e,-l

(a l , ... , al' ... , ar, ... , a" aj) cannot lie in X on account of the defini-e 1 e,-l

tion of b, for ar > band aj > b. Hence u'(b - 1)(ar - 1) E Vn and u'(aj - 1) E Vn . Since 3 n+l s Vn , it follows that u(b - 1) E Vn, a contradiction.

Thus ar = band er = P - 1. Hence u(b - 1) = v(b - 1)1', where v = (a l - 1)e l ••• (ar - 1 - 1)e'-1 E 3n- pd (since Cr = d). Since K is of characteristic p, (b - 1)P = bP - 1. By 1.13b), Kd«(fj)P S KpA(fj), so bP E KpA(fj). Hence we may write

with bj E Bpd (j = 1, ... , t) and b' E Kpd+l((fj). By b),

Since v E 3n - pd,

t

bP - 1 == L (bj - 1) mod 3Pd+ l . j=l

t

u(b - 1) = v(b P - 1) == L v(bj - 1) mod3n+1• j=l

Since bi > b, (ai' ... , ai' ... , ar - I , ... , ar - I , bi) ¢ X. Thus e1 e r- 1

v(bi - 1) E Vn. Hence u(b - 1) E Vn, a contradiction. Thus d) is proved. e) {b* + 3n+l lb* E Bn is linearly independent over K. Suppose that this is false. Then there is a non-trivial relation L Akbt

= u E 3n+1. Write each bt E B:inthisrelationas(b l - l)e l '" (b r - 1)'" with bi E B, bl < ... < b" 1 :::; ei < p; write the element u of 3n+1 as a K-linear combination of elements (Xl - 1)'" (Xn+l - 1) with Xi E G>. The set Bd of elements of Bd occurring in this relation is finite (d = 1, ... , n); also the set F of elements Xi of G> occurring in it is finite. By 2.5, there exists a subgroup i) and a normal subgroup.R of i) such that G>l = i)j.R is a finite p-group; further there exists a ,,-net Con G>l such that Cd = (C n "iG>l)) - (C n "d+l (G>l)) ::2 {b.Rlb E Bd}. We may extend the ordering on Bd to an ordering of Cd and thus make C into an ordered ,,-net. Let C = {ai' ... , am}, al < az < ... < am' Since I"AG>I):"d+I«(fit)1 = plcdl, 1G>11 = pm.

We wish to apply d) to the ordered ,,-net C on G>I' Thus for each d ~ 1, let

where a i E Ce, . 1f:J is the augmentation ideal of KG>l' :Jd is spanned by C: and :Jd+l; thus

But the above relation implies a non-trivial relation between the elements of C: modulo :In+\ so

Hence if C* = U d~l C:,

IC*I > L dim K (:Jdj:Jd + I). d~l

But

thus IC*I :::; pm - 1. Now by Y, 5.16,:J is the radical of KG>l' By Y, 2.4, :J is nilpotent. Hence

pm _ 1 ;::: IC*I > L dimK(gdlgd+1) = dimKg. d~1

But dimK 9 = I (fj 11 - 1 = pm - 1, a contradiction. With d) and e), 2.6 is proved. q.e.d.

2.7 Theorem (JENNINGS [1]). Suppose that (fj is a group, K is a field of characteristic p, and 3 is the augmentation ideal of K(fj. Suppose that x E (fj. Then x E Kn(fj) if and only ifx - 1 E 3n•

Proof If x E Kn(fj), then x-I E 3n by 2.6. If x ~ Kn(fj), there is a least integer m such that x ~ Km(fj) and 1 < m ::; n. Thus x E Km- 1(fj). There exists a K-net B on (fj such that x E B. By 2.6, there exists a K-basis of 3m- 113m containing x - 1. Thus x-I ~ 3m. Since n ;::: m, x-I ~ 3n.

q.e.d.

2.8 Corollary. Let (fj be a finite p-group, K a field of characteristic p and 3 the augmentation ideal of K(fj. Suppose that K/(fj) > Kl+1 (fj) = 1,

/

S = (p - 1) L ndn• n=1

Then 35 +1 = O. 35 is of dimension lover K and is spanned by LXE(lj X.

Proof Let B be an ordered K-net on (fj. By 2.6, 35 + 1 = 0 and 35 is spanned by nbEB (b - IV- 1 , where the factors are taken in increasing order of bE B. But

n (b - IV- 1 = n (1 + b + ... + bP- 1) = L x. q.e.d. bEB bEB XE(lj

2.9 Examples. a) Suppose that (fj is an Abelian p-group of exponent pm and that I (fjpH: (fjpil = pWi (i = 1, ... , m). By 1.10, Kn+1 (fj) = Kn(fj) if nisnot a power ofp, and Kpi(fj) = (fjpi, Kpi+1(fj) = (fjpi+1. Thusdpi-l = Wi (i = 1, ... , m), and

m

S = (p - 1) L pi-1Wi · i=1

b) Let ID be ap-group of maximal class of order pm, where m 2: p + 1, m = (p - l)q + r + 1 and 1 ::; r ::; p - 1. It follows from 1.15 that

§ 2. Jennings'Theorem 259

ifn = 1;

if n = ap' with 1 ~ a ~ p - 1, 0 ~ 1 ~ q, (a,l) i= (1, 0) and a ~ rifl = q;

otherwise.

The integer s of 2.8 is given by

s = (p - 1) + tp(p - l)(pq - 1) + t(p - l)pQr(r + 1).

For p = 2, this reduces to s = 2m- 1.

2.10 Theorem (Jennings' formula). Suppose that GJ is a finite p-group, K is a field of characteristic p and 3 is the augmentation ideal of KGJ. SupposefurtherthatK,(GJ) > Kl+1(GJ) = landthatIKn(GJ):Kn+1(GJ)1 = pd. (n = 1, ... , 1). Let s = (p - 1) L~=l ndn, and define the integers Cn by

, s n (1 + t i + t2i + ... + t(P-1)it, = L cntn. i=l n=O

Proof Let B be a K-net on GJ, and let Bn = (B n Kn(GJ)) - (B n Kn+1 (GJ)). Thus IBnl = dn- Write Bn = {bn1 , ... , bndJ. By 2.6, dimK3n/3n+1 is the number of lexicographically ordered products

n (bij - l)eij i,j

with 0 ~ eij < p and Li,j ieij = n. This number is obviously cn. q.e.d.

2.11 Remarks. a) If K = GF(p), 3/32 is isomorphic to <»/K2(GJ) since c 1 = d l' More explicitly, there is an isomorphism between them in which x-I + 3 2 corresponds to XK2(GJ) for all x E GJ.

b) If <» is a 2-group of maximal class of order 2m, 2.10 yields Cn = 2 for 1 ~ n < 2m - 1 and C2m-1 = 1.

A remarkable symmetry of the Cn follows from 2.10.

2.12 Theorem. Suppose that GJ is a finite p-group, K is a field of characteristic p and 3 is the augmentation ideal of KGJ. If3s ::::> 3s+1 = 0,


and

Proof In the notation of 2.10, s = (p - 1) L:=1 idi and

s I

L cntn = n (1 + t i + t 2i + ... + t(P-1)i)d,

n=O i=1

I = n t(p-1)id,(t-(P-1)i + C(p-2)i + ... + 1)d,

i=1

I = t S n (1 + C i + ... + C(p-1)i)d,

i=1

Comparing coefficients, Cn = cs- n (n = 0, ... , s). Thus if 1 :::; n :::; s - 1,

Also Cs = Co = 1 (cf. 2.8), so

Hence if 1 :::; n :::; s,

n-1

dim K K(f)/3n = dimK K(f)/3 + L dimK3 m/3m+1 m=1

n-1

= dimK3s + L dimK3s-m/3s-m+1 m=1

- d' ,,!,s-n+1 - ImK " .

Thus


The following theorem shows that the right and left annihilators of powers of 3 are powers of 3. In the terminology of Loewy series (see VII, 10.10), the upper and lower Loewy series have the same terms.

2.13 Theorem (E. T. HILL [1]). Suppose that G) is afinite p-group, K is a field of characteristic p and 3 is the augmentation ideal of KG). Suppose 3s ::::l 3s+1 = O. For any element a of KG) and any integer n such that 1 ::;; n ::;; s, the following statements are equivalent:

a) a E 3s+1-n.

b) The coefficient of 10; in ab is 0 for every b E 3n. c) ab = 0 for every b E 3n. d) The coefficient of 10; in ba is 0 for every b E 3n. e) ba = 0 for every b E 3n.

Proof For 1 ::;; n ::;; s, we define

U = {ala E KG), ab = 0 for all b E 3n},

v = {ala E KG), the coefficient of 10; in ab is 0 for all b E ,3n}.

Clearly, ,3s+1-n £; U £; V. For any elements a, b of KG), denote by (a, b) the coefficient of 10; in

abo Then ( , ) is a K-bilinear form on KG). We show that this form is non-singular. Suppose that a E KG) and a :f. O. Then a = LXE6'> Axx, where Ax E K and Ay :f. 0 for some y E m, and (a, y-1) = (y-l, a) = Ay :f. O.

Now

v = {ala E KG), (a, b) = 0 for all b E ,3n}.

Since ( , ) is non-singular,

Using2.12,itfollowsthatdimKV = dimK,3s-n+1.Since,3s+1-n £; U £; V, it follows that 3s+1- n = U = V. This establishes the equivalence of a), b), c). Similarly a), d), e) are equivalent. q.e.d.

It follows from 2.13 that, in the notation of VII, 10. 7b),


thus 3s- n+1 is the right and left annihilator of 3n. The lower Loewy series (VII, 10.10) of the Kffi-module Kffi is

It follows from VII, 10.11b) that the upper Loewy series is the same series in the reverse order.

We conclude this section by showing how 2.7 can be used to deduce a slight generalization of III, 9.7 from the following elementary ringtheoretical formula.

2.14 Lemma. If a, b are elements of an associative algebra over a field of odd characteristic p,

p-l p-l L (a + ib)P + L bjabp- 1- j = 0. i=O j=O

Proof We have

p-l (a + b)P = aP + bP + L Sj'

j=l

where Sj is the sum of all products Cl C2 ... cp in which j of the Ci are b and the remainder are a. Replacing b by ib (i = 0, 1, ... ), we see that

p-l (a + ib)p = aP + iPbP + L ijsj.

j=l

Summing over i = 0, 1, ... ,p - 1 and using Fermat's theorem,

p-l (P-l ) p-2 (P-l ) i~O (a + ib)p = i~ i bP + j~l i~ ij Sj + (p - 1)Sp_l'

But for 1 ::;; j ::;; p - 2, Lf:l i j == ° (p), so since p is odd,

p-l L (a + ib)P = -Sp-l'

i=O

The assertion follows from this since by definition of Sp-l,

p-l - " bi bP - 1 - j Sp-l - L... a .

j=O q.e.d.


To interpret the second of the two sums in 2.14 group-theoretically, we need the following lemma.

2.15 Lemma. Let (f) be a group, K a field of characteristic p and 3 the augmentation ideal of K(f). Let x, y be elements of (f) and write a = x - 1, b = y - 1. Define the mapping e of K(f) into itself by

ce = cb - bc (c E K(f)).

a) For n ~ 0,

aen == [x,y, ... ,y] -1 mod3n+2• n

b) For all c E K(f),

p-I

cep - I = L bicbP- I - i. i=O

Proof a) This is proved by induction on n. It is trivial for n = 0. If n > 0, there exists c E 3 n+1 such that

ae n- I = [x,y, ... ,y] - 1 + c, n-I

by the inductive hypothesis. Now since b E 3, e carries 3 m into 3 m+l ;

hence em carries 3 into 3 m+l . Thus aen- I E 3 n and

[x, y, ... , y] - 1 E 3n• n-l

By 2.3c),

[[x,y, ... ,y],y] -1 - ([x,y, ... ,y] -1)eE3n+2• n-I n-I

Hence

[x, y, ... , y] - 1 - (ae n- I - c)e E 3n+2• n

Since e is linear and ce E 3n+2, the assertion follows. b) Define the mappings el , e2 of K(f) into itself by

ce l = cb, ce2 = bc (c E K(fj).


Thus ~ = ~1 - ~2 and ~1~2 = ~2~1' Since char K = p, it follows that

p-1

~p-1 = L ~~~f-1-i, i=O

which is the assertion. q.e.d.

2.16 Theorem (ZASSENHAUS). Suppose that p is an odd prime. If x, yare elements of a group G),

(This holds for any order of the factors in the product).

Proof Let K = GF(p) and let a = x - 1, b = y - 1 in the group-ring KG). If 3 is the augmentation ideal of KG), we have

xi = (1 + a)(1 + b)i = 1 + a + ib + Ci'

where Ci E 3 2• Therefore

(xi)P = 1 + (a + ib + c;)p == 1 + (a + ib)P mod3p +1.

Hence

p-1 p-1 TI (xi)P == 1 + L (a + ib)P mod 3P+1. i=O i=O

But also by 2.15,

p-1

[x,y, ... ,y] == 1 + a~p-1 == 1 + L biabp- 1- i mod3p +1. p-1 i=O

Multiplying,

p-1 p-1 p-1 [x, y, ... , y] TI (xi)P == 1 + L (a + ib)p + L biabp- 1- i mod3 P+\

p-1 i=O i=O i=O

since a E 3, b E 3. It follows from 2.14 that

p-l [ ] n ( i)p 1 ""p+1 X, y, ... , y xy - E ~ . p-1 i=O

By 2.7,

p-1 [x, y, ... , y] n (xi)p E Kp+1 (<D).

p-1 i=O

By 1.10, Kp+1(<D) = <D P2 yz (<D)pyp+ 1 (<D), which yields the assertion. q.e.d.

The following can be used to prove III, 14.21.

2.17 Theorem. If p is a prime and x, yare elements of a group <D, [x, y, ... , y] E <D(P)<D,pYP+1(<D), where <D(P) is the group generated by all

p-1 elements of the form (uvtPuPv P with U, v in <D.

Proof Let {s: be a free group for which there is an epimorphism of {s: onto <D and let a, b be preimages of x, y respectively. It is sufficient to prove that c = [a, b, ... , b] E (s:(P){s:'PyP+1({s:)' By 2.16, c E (s:PYp+1({s:)' so p-1

for certain}; E {s: (i = 1, ... , k), g E Yp+1 ({s:). Thusf! ... f{ = cg-1 E {s:'. Since {s:j{s:' is Abelian, (fl ... fk)P E {s:'. But {s:j{s:' is a free Abelian group and is therefore torsion-free. Hence f1 ... fk E {s:'. Now (uv)P == uPvP mod (s:(P) for all u, v in {S:, so

Therefore ff ... f{ E {s:(P){s:'P and c E {s:(P){s:'PYP+1 ((s:). q.e.d.

Exercises

2) For any finite p-group <D, write cn(<D) for the integer denoted in 2.10 by Cn' Prove that if <D, ~ are finite p-groups,

3) Let K be a field of characteristic p. Let ~ be a K-algebra having K-basis {ail'" aiJO ~ n < q} for elements a1' ... , ad' in which any

product of q of the ai is zero. Show that 1 + ai is a unit of m, and let GJ be the group generated by 1 + aI' ... ,1 + ad. Let Ij be a free group of rank d. Prove that m:: ~ KIj/3Q, where 3 is the augmentation ideal of KIj, and deduce that GJ ~ Ij/Kilj).

§ 3. Transitive Linear Groups

Our usual assumption that all groups mentioned are finite unless otherwise stated comes into force again here.

In § 3-§ 7, the structure of a 2-group GJ in which the involutions are permuted transitively by a soluble subgroup m of Aut GJ will be discussed. In this situation, QI(Z(GJ)) can be regarded as a vector space over GF(2) and m induces a soluble group of linear transformations of this vector space which permutes the one-dimensional subspaces transitively. The corresponding condition over GF(p) will be studied in general in XII, § 7; here we shall only deal with the case p = 2. We say that a group m of operators on a group lID is irreducible if lID =I 1 and there is no m::invariant subgroup U of lID such that 1 < U < lID. Otherwise m is reducible. If lID is soluble, it is of course necessary for irreducibility that lID be elementary Abelian.

The first theorem is concerned with tn-transitivity (II, 1.14).

3.1 Theorem (WIELANDT [4]). a) If(f) is a !-transitive permutation group, then either (f) is primitive or (f) is a Frobenius group.

b) Let m be a reducible group of automorphisms of a finite group lID. Suppose that the orbits of m on lID - {I} are all of the same length. Then the stabiliser of any non-identity element of lID is {I}.

Proof a) Suppose that (f) is !-transitive and imprimitive on Q. Then Q = lJ'1 U ... u lJ'r and lJ'i n 'lj = 0 (i =I j), where 1 < r < /Q/ and GJ permutes the lJ'i. Write

and put all = 1. Then the orbits of the stabiliser GJ I on Q - {1} are all of the same length m > 1.

Since 1 E lJ'1' lJ'1 remains fixed under GJ I . Hence lJ'1 is a union of orbits of GJ I , among them {I}. Hence s = /lJ'I/ == 1 (m); in particular, (m, s) = 1. For i =I 1, lJ'i (f) I remains fixed under (f) I but does not contain 1; hence lJ'i(f)t is a union of orbits of (f)t of length m. Hence / lJ'i(f)l! == 0 (m). On the other hand, 'l'i(f)1 is the disjoint union of certain

§ 3. Transitive Linear Groups 267

'l'j' so l'l'i(fj11 == 0 (s). Since (m, s) = 1, it follows that l'l'i(fj11 == 0 (ms). Further

(1) s

'l'i(fj 1 = U aij(fj l' j=l

and I aij(fj 1 I = m since aij # 1. Therefore (1) is a disjoint union. In particular,

(2)

for i # 1 and allj. We show that if a E 'l'i with i # 1, then (fjea leaves invariant every

element of 'l'1' For there exists h E (fj such that ah = 1. Then 'l'ih = 'l'1

and 'l'lh = 'l'j for some j # 1. Suppose that g E (fj1,a' Then gh E (fj1h,1 and 'l'jl = 'l'j. Also, if bE 'l'1, then bh E 'l'j and by (2),

Hence

It follows that if a ¢ 'l'1 and 1 # b E 'l'1' then (fj 1.a = (fj l,b' For (fj1,a:::; (fj1,b and 1(fj1: (fj1,al = m = 1(fj1: (fjl,bl· Hence if 1 # bE 'l'1, (fj1,b = (fj1,c for all 1 # CEQ and (fj1,b = 1. Thus (fj is a Frobenius group.

b) Let p be the regular representation of lID; thus ~ normalises p(IID). Let (fj = ~p(IID). Then (fj is transitive on lID and the stabiliser of 1 is ~. It follows from the hypothesis that (fj is ~-transitive. But since ~ is reducible, ~ is not a maximal subgroup of (fj and (fj is not primitive. By a), (fj is a Frobenius group. Thus if v E lID - {1}, ~v = (fj1,v = 1.

q.e.d.

3.2 Theorem. lfthe group (fj possesses afaitJiful irreducible representation, the centre 3 of (fj is cyclic.

Proof Let p be a faithful, irreducible representation of (fj on a vector space V. Since 3 is the centre of (fj, p(3) s;; HomK(!l(V, V). By I, 10.5, HomK(!J(V, V) is a division algebra. Thus p(3) generates a field contained in HomK(!J(V, V). Since finite multiplicative subgroups of fields are cyclic, p(3) is cyclic. Since p is faithful, 3 is cyclic. q.e.d.

3.3 Lemma. Let 21 be a p-group and let V be a faithful, irreducible K21-module for any field K. Suppose that 21 has a normal subgroup in which is elementary Abelian of order p2. Let ~ = C'1l(in). Then 121: ~I = p and V = VI EEl ... EB Vp, where each Vi is ~-invariant. For any a E 21, there exists a permutation 1C of {1, ... , p} such that Via = Vi'"

Proof Since in :::; ~, the centre of ~ is not cyclic. Hence by 3.2, V is a reducible K~-module. Thus ~ i= 21. Since p2 does not divide the order of the group of automorphisms of in, 121 : ~ 1 = p. By V, 17.3,

where each Vi is a completely reducible K~-module, all irreducible K~-submodules of Vi are isomorphic and an irreducible K~-submodule of Vi is not isomorphic to an irreducible K~-submodule of ~ if i i= j. It follows from 3.2 that no Vi is a faithful K~-module, so t > 1. By V, 17.3d), there exists a transitive permutation representation p of 21 on {VI' ... , V,} such that ViP(a) = Via for alIa E 21. Thus ~ is contained in the kernel of p and t = p, by I, 5.11. q.e.d.

3.4 Lemma. Let V be an elementary Abelian q-group and let 21 be a p-group of automorphisms of V, where p is an odd prime. Suppose that all the orbits of 21 on V - {1} are of the same length m. Then 21 is cyclic and the stabiliser of any non-identity element of V is 1.

Proof We observe first that the two conclusions of the lemma are equivalent. For if 21 is cyclic, the stabiliser of any non-identity element of V is the unique subgroup of 21 of index m, which is thus 1. And if 21 is not cyclic, then by V, 8.7, the semidirect product V21 is not a Frobenius group, since p is odd. Hence the stabiliser in 21 of some non-identity element of V is not 1.

Suppose that the lemma is false. Then the stabiliser of some nonidentity element of V is not 1, so 21 is irreducible by 3.1. Also 21 is not cyclic and p is odd, so by III, 7.5, 21 possesses a normal subgroup in which is elementary Abelian of order p2. Let ~ = C'1l(in). By 3.3, 121 : ~I = p and V is the direct product of ~-invariant subgroups VI' ... , Vp. Suppose 1 :::; j < k :::; p and choose non-identity elements u, v in Vj' Vk

respectively. Let U, m, W be the stabilisers of u, v, uv respectively in 21. If W ;t, ~,choose a E W, a ~ ~; by 3.3, Vp = V" Vka = Vs for suitable r, s. Since (ua)(va) = uv, {r, s} = {j, k}. Thus Vp2 = Vj and Vj is invariant under <a2 , ~> = ~, contrary to the irreducibility of ~. Hence W :::; ~.If a E W, the equation (ua)(va) = uv now yields ua = u, va = v, so a E U n m. Thus W :::; U n m. But by hypothesis, U, m, Ware all

§ 3. Transitive Linear Groups 269

of the same order. Hence U = m. Thus all the non-identity elements of all the Vi have the same stabiliser, which is thus 1. This is a contradiction.

q.e.d.

3.5 Theorem. Let V be an elementary Abelian 2-group of order 2", and let ~ be a soluble group of automorphisms of V of odd order. Suppose that ~ permutes the set of non-identity elements of V transitively. Then ~ is similar to a subgroup of the group of permutations of K = GF(2") of the form x ~ a(x8) (a E KX , 8 E Aut K). If IB is the subgroup of ~ consisting of elements which correspond to permutations of K of the form x ~ ax, then IB <l ~ and IB is cyclic of order k, where k divides 2" - 1 and 2" - 1 divides kn. Also IB is irreducible on V, and IB permutes the set of nonidentity elements of V in orbits of length k.

Proof Let 0: be the Fitting subgroup of ~. Each Sylow subgroup 6 of 0: is normal in ~ and thus permutes the set of non-identity elements of V in orbits of equal length (II, 1.5). Since I~I is odd, it follows from 3.4 that 6 is cyclic and that the stabiliser in 6 of any non-identity element of V is {1}. Since 0: is the direct product of its Sylow subgroups, 0: is cyclic. Also the stabiliser in 0: of any non-identity element of V is {1}, since such a stabiliser would otherwise contain an element of prime order. Thus ~ permutes the set of non-identity elements of V in orbits of length

10:1· Since an automorphism of a cyclic group is determined by the image

of a generator, I Aut 0:1 < 10:1· Hence I~: C'll(~)1 < I ~I. But C'll(~) = ~, by III, 4.2b). Thus I ~ I ~ 10:1 2 - 1. Since ~ permutes the set of nonidentity elements of V transitively, IVI - 1 divides I~I. Thus IVI ~ 10:1 2.

Now suppose that U is an irreducible O:-invariant subgroup of V of minimal order. Since the orbits of ~ on V - {1} are of length 10:1, lUI - 1 is divisible by 10:1. Thus lUI> 10:1. Thus IUI2 > 10:12 ;;:: IVI· By the Maschke-Schur theorem (1,17.7), V = U x W for some 0:invariant subgroup W of V. Hence IWI < lUI. By minimality of lUI, W = 1. Thus V = U and V is an irreducible O:-group. By II, 3.11, ~ is similar to a subgroup of the group of permutations of K = GF(2") of the form x ~ a(x8) (a E KX

, 8 E Aut K); also, if IB is the subgroup of ~ consisting of elements which correspond to permutations of K of the form x ~ ax, IB = C'll(m. Hence IB = 0: <l ~, IB is cyclic and IB is irreducible on V. Let k = IIBI. Since ~/IB is isomorphic to a subgroup of Aut K, I~: IBI divides n. Hence I~I divides kn. Since IVI - 1 divides I~I, 2" - 1 divides kn. Since IB is isomorphic to a subgroup of KX, k divides 2" - 1 and IB permutes the set of non-identity elements of V in orbits of length k. q.e.d.


3.6 Lemma. Let V be an elementary Abelian q-group of order qn, and let IX be an automorphism of V of order k, where (k, q - 1) = 1 and qn - 1 ~ kn. Suppose that (IX) is irreducible, that V = V1 X ... X V, and that IX permutes the Vi' Then r = 1.

Proof By 11,3.10, k divides qn - 1. Hence (k, q(q - 1)) = 1 and k is odd. Since (IX) is irreducible, (IX) permutes the Vi transitively. Hence r divides k, and the Vi are all of the same order qm, where rm = n. Further, the stabiliser of Vi under the permutation group induced by (IX) is (IX'), and (IX') operates faithfully and irreducibly on each Vi' Hence qm - 1

is divisible by~, by 11,3.10. Hence r

k qn - 1 q'm - 1 qm _ 1 > _ > __ = . - r - rn r2m

Thus

r2m ~ 1 + qm + ... + qm(,-l)

~ 1 + 2m + ... + 2m(,-1).

Since 2m- 1 ~ m, it follows that

and r2 > 2' - 2. Hence r ~ 4. Since mr2 ~ (2m, - 1)j(2m - 1), it is easy to check that if r = 4 or r = 3, then m = 1. Since r divides k and k is odd, r #- 2. Hence either r = 1 or m = 1. But if m = 1, IVil = q. Since (k, q - 1) = 1, a' is identity on each Vi' Thus a' = 1. If v E V1 and v #- 1, v(va)··· (va,-l) is then invariant under IX. Since (a) is irreducible, n = 1; hence r = 1 in any case. q.e.d.

§ 4. Some Number-Theoretical Lemmas

We shall need some number-theoretical properties of the integers k, n occurring in 3.5.

As far as 4.5 inclusive, let k, n denote positive integers such that

(1) n > 1, k divides 2n - 1 and 2n - 1 divides nk.

§ 4. Some Number-Theoretical Lemmas 271

Thus if d = (n,2n - 1), the possible values of k are c 2n ; 1, where c is

any divisor of d. Hence if d = 1, k = 2n - 1. If d > 1, there is an odd prime divisor p of n such that n is divisible by the order of 2 modulo p. Thus it follows, for instance, that if d > 1, n cannot be a power of a prime; again, if n = 2p for some prime p, then p = 3. Using these facts it is easy to check that the following table gives all values of n ::;; 55 for which d > 1.

12 18 20 21 24 30 36 40 42 48 54 3 9 5 7 3 3 9 5 21 3 27

We need to solve certain congruences of the following kind.

4.1 Lemma. Suppose that iI, ... , ir (r ~ 2) are non-negative integers, 0::;; I::;; rand

If the ij are distinct modulo n, then

n

and 2r < 2n.

Proof Since 2n == l(k), we may suppose that 0 ::;; ij < n for all j. Thus the ij are distinct. Suppose that the ij' when written in increasing order, are hI, ... , hr ; thus

hI < h2 < ... < hr·

If the given congruence is multiplied by 2n-\ a congruence is obtained which involves

Denote the largest exponent occurring here by dj ; thus dj = hj - I + n - hj if) > 1 and d l = hr - hI' If d = min(d l , ... , dr), then

(2) rd ::;; d I + . . . + dr = (r - 1 )n.

Now one of these congruences is of the form L == R (k), where L, Rare sums of powers of 2 such that 2d appears in L, and all the other exponents which occur in either L or R are distinct and less than d.

Thus

Hence L ~ R + k. Thus

k ::;; L ::;; 2d + 2d- 1 + ... + 1 ::;; 2d+1 - 1.

Combined with (2), this is the first of the stated inequalities. By (1), nk ~ 2n - 1, so

n- 21+n(1-+) ~ n(k + 1) ~ 2n - 1 + n > 2n. n

Hence 2n > 2'. q.e.d.

This gives a bound for n in terms of r, which we give explicitly for 1 < r ::;; 8.

4.2 Lemma. Suppose that n, r are integers greater than 1 and that 2n1r < 2n. Then n ::;; f(r), where f(2) = 7, f(3) = 14, f(4) = 21, f(5) = 29, f(6) = 37, f(7) = 45 and f(8) = 54.

Thus in solving the congruence of 4.1 for a particular value of r, only a finite number of cases need be considered. We consider first some cases with r = 2 and r = 3.

4.3 Lemma. a) If 2i == 1 (k), then i == 0 (n). b) If 2i == -1 (k), then n = 2 and i == 1 (2). c) If 2i + 2j == 1 (k), then i == j == n - 1 (n). d) If 2i + 2j == - 1 (k), then either (i) n = 2, i == j == 0 (2), (ii) n = 3,

{i,j} == {t, 2} (3), or (iii) n = 6, k = 21 and {i,j} == {2, 4} (6). (A statement such as {i,j} == {1,2} (n) means of course that either

i == 1, j == 2 (n) or i == 2, j == 1 (n)).

Proof In a) and b) we may suppose 0 < i < n. In c) it is clear that i ¢ 0 (n) and j $. 0 (n); thus we may suppose that 0 < i < j < n since a) may be used in the case i = j. In d), if i == 0 (n),j == 0 (n) or i == j (n), we may use b), and this gives the solution (i); thus we may suppose that o < i < j < n in this case also.

§ 4. Some Number-Theoretical Lemmas 273

Under these suppositions the conditions of 4.1 are satisfied with r = 2 in a), b) and r = 3 in c), d). By 4.1, 2n1r < 2n. By 4.2, n ~ 7 in a), b) and n ~ 14 in c), d).

In a), b) the situation is clear if n = 2. If n > 2, suppose 2i + 1 = kl. Then kl ~ 2n- 1 + 1 < 2n - 1. Hence k < 2n - 1 and from the above table n = 6, k = 21. Then 211 < 63, I < 3, I = 1 and 2i + 1 = k = 21, which is false.

In c), d) the situation is clear if n = 3, since ° 3, suppose t + 2i + 1 = kl. Then kl ~ 2n- 1 + 2n- 2 + 1 < 2n - 1. From the list n = 6 or n = 12 and 2n - 1 = 3k. Hence I = 1, and 2i + 2i + 1 = k. We use the fact that every non-negative integer is uniquely expressible as a sum of distinct powers of 2. For n = 12,

is impossible. For n = 6, 1 + 22 + 24 = k = 2i + 2i + 1 implies that i = 2,j = 4ind).

Next we handle some cases when the number of powers of 2 involved is 4.

4.4 Lemma. a) If 2h + 2i + 2i == 1 (k), then either (i) {h, i, j} == {-1, -2, -2} (n), or (ii) n = 6, k = 21, {h, i, j} == {1, 2, 4} (6).

b) If 2h + 2i + 2i == -1 (k), then either (i)n = 2, {h, i, j} == {O, 1, 1} (2); (ii) n = 3, {h, i, j} == {O, 0, 2} or {I, 1, I} (3); (iii) n = 4, {h, i, j} == {I, 2, 3} (4); or (iv) n = 6, k = 21, {h, i,j} == {O, 3, 5}, {I, 1, 4} or {2, 3, 3} (6).

c) If 2i + 2i == 1 + 2h (k), either (i) {i,j} == {O, h} (n), (ii) n = 6, k = 21,{i,j} == {3,4}(6),h == 1 (6),or (iii) n = 6,k = 21,{i,j} == {2,3} (6), h == 5 (6).

Proof These results are easily deduced from 4.3 except in the case when the exponents are distinct and non-zero modulo n. In this case 2n/4 < 2n by 4.1 and n ~ 21 by 4.2. Clearly we may suppose n > 4. We also suppose that h, i, j lie between ° and n, and write

Then Ik/l ~ 2n- 1 + 2n- 2 + 2n- 3 + 1 < 2n - 1. Thus k < 2n - 1 and the possible values of n are 6, 12, 18, 20, 21. Of course I is odd and

III < 2n ; 1.


If k = (2n - 1)/3, then I = ± 1. Also n is even and k = 1 + 22 + ... + 2n- 2 . In a), b), I = 1. Now k = 2h + 2i + 2j + 1 implies {h, i,j} = {2, 4, 6} and n = 8, a contradiction. And k = 2h + 2; + 2j - 1 implies {h, i, j} = {I, 2, 4} and n = 6, which is one of the stated solutions. In c), I = ± 1. But 2; + 2j = 1 + 2h + k implies {i,j} == {3,4} (6), h == 1, n = 6, k = 21; and 2i + 2i + k = 1 + 2h implies {i, j} ==

-{2, 3} (6), h == 5, n = 6, k = 21. If k '# (2n - 1)/3, n = 18, 20 or 21. These cases have to be treated

individually. One writes out all the possible values of kl as sums of powers of 2, and it is very easy to convince oneself that no such kl can satisfy the above equations. q.e.d.

4.5 Lemma. a) If 29 + 2h + 2; + 2i == 1 (k), either (i) {g, h, i, j} == {-2, -2, -2, -2} or {-3, -3, -2, -l}{n), or (ii) n = 6, k = 21 and {g, h, i, j} == {O, 0, 2, 4}, {O, 1,3, 5}, {l, 1, 1, 4} or {I, 2, 3, 3} (n).

b) If 2h + 2; + 2j == 1 + 29 (k), either (i) {h, i, j} == {O,g - 1,g - 1} or {-I, -l,g} (n), or (ii) n = 6, k = 21 and one of the following holds:

g == 0, {h, i,j} == {2, 3, 5} (6);

g == 1, {h, i,j} == {2, 2, 4} or {3, 3, 3} (6);

g == 2, {h, i,j} == {I, 3, 4} (6);

g == 4, {h, i,j} == {l, 2, 5} (6);

g == 5, {h, i,j} == {l, 1, 3} or {2, 2, 2} (6).

Proof This is proved along the same lines. First we prove the results in the case when some of the exponents are congruent modulo n or when one ofthem is zero modulo n by appealing to 4.3 and 4.4. In the remaining cases n ::; 29 by 4.1 and 4.2. There are no solutions for n = 5, so n is 6, 12, 18, 20, 21 or 24. The case k = (2n - 1)/3 can be handled rather easily as in 4.4; the other cases must be treated individually. q.e.d.

On account of the exceptional role of n = 6, we shall need a certain fact about GF(26).

4.6 Lemma. Let F = GF(26 ). If £ is any non-zero element of F, there exists elements a, bin F, not both zero, such that

§ 5. Lemmas on 2-Groups 275

Proof Let P be the mapping of F into itself for which xp = x(1 + eX). Then P is an additive homomorphism, and I ker PI = 2 since e =1= O. Hence limPI = 25. If K is the subfield of F having 4 elements, K n (imp) =1= O. Thus there exists x E F such that x(1 + eX) is a non-zero element of K. Now K consists of the 21st powers of the elements of F; thus x(1 + eX) = a21 for some a E F, a =1= O. Define bE F by X = a16b2.

Then b2(1 + ea l6b2 ) = b2a2lx- 1 = a5, and a5 + b2 + sa l6b4 = O. q.e.d.

§ 5. Lemmas on 2-Groups

In this section, we present some results on groups of prime-power order which are useful for the determination of the Suzuki 2-groups.

We begin with a counting argument. In 5.1 and 5.2, p is an arbitrary prime, and if 1 is a finite p-group, j(l) is the number of elements of 1 of order at most p.

5.1 Theorem (BLACKBURN [1]). Let <D be a p-group and let 91 be a nonidentity subgroup of exponent p in the centre of <D. If Yf is the set of maximal subgroups of 91,

L j(<D/9Jl) = (IYf I - l)j(<D/91) + j(<D). WiE.l!"

Proof Let k = IYfI. Thus 1911 = (p - 1)k + 1. Note that if 3 is a subgroup of 91 of order p, the number of maximal subgroups of 91/3 is (k - 1)/p.

Let

![IJ= {(x, 9Jl)lx E <D - 91, 9Jl E Yf, x P E 9Jl}.

If x E <D - 91 and x P E 91, let

Clearly, l![IJxl is the number of maximal subgroups of 91/<x P). Thus l![IJxl = kif x P = 1 and I ![IJx I = (k - 1)/p if x P =1= 1. Hence


where x runs through the set of elements of (1) - ill for which x P E ill. Now x P E 91 if and· only if (xm)p = 1; thus the number of x E (1) - 91 for which x P Em is (j(1)jm) - 1)1911. Hence

I~I = (j(1») - 1m!) (k - k ~ 1) + (j(1)jm) _ 1)l mlk ~ 1

= j((f;)(k _ k ~ 1) + j((f;jm)lmlk ~ 1 _ klml.

On the other hand, suppose that 9Jl E :Yf. The number of elements x E (f; - 91 for which x P E 9Jl is (j(1)j9Jl) - p)I9JlI, so

Comparing the two expressions for I~I,

j((f;) + j((f;jm)k - 1 _ k = I (j((f;j9Jl) _ p)!. p p ~E~ P

Multiplication by p yields the stated result. q.e.d.

To apply this, we shall need to know j(l) if 1 is a 2-group and 14>(1)1 = 2. First we prove the following.

5.2 Lemma. Suppose that (f; is a p-group and that (f; = ,!US, where ['!I, ~] = 1. Let ::D = '!I (\~. Suppose that I::D I = p and that '!Ij::D, ~j::D are of exponent p. Then

Proof We have (f; ~ ('!I x ~)j«x, X-I) for a generator x of ::D. Thus

If i = 0, the number of solutions of aP = Xi is j('!I); otherwise it is (I'!II - j('!I))j(p - 1), since '!I/::D is of exponent p. Similar statements hold in ~, and the assertion follows at once. q.e.d.

5.3 Lemma. Suppose that 6) is a 2-group, 16): 4>(6))1 = 2d and 14>(6))1 = 2. If j = I {xix E 6), x 2 = 1} I, j is either 2d, 2d + 2d- r or 2d - 2d- r , whpre r is a positive integer satisfying 2r ~ d. Further, if j is 2d + 2d- 2d or 2d - 2d- td, 6) is extraspecial.

Proof This is proved by induction on 16)1. If 6) is Abelian, 4>(6)) = 6)2 and j = 16): 6)21 = 2d. If 6) is non-Abelian, there exist non-commuting elements x, y of 6). If ~ = <x, y), ~ is non-Abelian of order 8, so the number i of elements of ~ of order at most 2 is either 2 or 6. Since <x,4>(6))) and <y,4>(6))) are non-central normal subgroups of 6) of order 4, their centralizers are of index 2, for the order of the automorphism group of a group of order 4 is not divisible by 4. Hence if (t = C(9(~), 16): (tl ~ 4. But (t n ~ = 4>(~) = 4>(6)), so 1(t~1 = 41(t1 ~ 16)1 and 6) = (t~. Thus by 5.2,

j = !(il + (8 - i)(2d- 1 - I)) = il + 2d+1 - 41 - 2d- 2 i,

where I is the number of elements of (t of order at most 2. If (t is elementary Abelian, I = 2d-1, so j = 2d- 2i.Ifi = 2, j = 2d - 2d- 1, and if i = 6,j = 2d + 2d- 1• If (t is not elementary Abelian, then 4>«(t) = 4>(6)) and we can apply the inductive hypothesis to (t. Thus I is either 2d- 2 ,

2d- 2 + 2d- s- 2 or 2d- 2 - 2d- s- 2, where s is a positive integer satisfying 2s ~ d - 2. If I = 2d- 2, j = 2d, and in the other cases, j is 2d - 2d- r or 2d + 2d- r , where r = s + 1, so 2r ~ d.

Suppose that 6) is non-Abelian and is not extraspecial. If (t is elementary Abelian, (so that r = 1), then l(tl > 2, so d > 2 = 2r. In the remaining case, (t is not extraspecial, so the inductive hypothesis gives 2s < d - 2. Thus 2r < d in any case. q.e.d.

The following theorem is an application of this counting argument.

5.4 Theorem. Suppose that 6) is a 2-group of class at most 2 and exponent at most 4. Suppose that {x I x E 6), x2 = 1} is a subgroup 91 of 6). Then 16)1 ~ 19113 ; indeed 16)1 < 1911 3 if6) is not special.

Proof We prove this by induction on 1911. If 1911 = 2, 6) is either cyclic of order at most 4 or the quaternion group of order 8, and the assertion is clear. Suppose then that 1911 > 2. Clearly 4>(6)) ~ 91. .

First suppose that 6)' < 91. We have 91 ~ £, where £16)' = Q1 (6)/6)'). Since £16)' is elementary Abelian,

£/6)' = (91/6)') x (~/6)')

for some subgroup st ~ i!. Since (fj' < 91, st < i!, so the inductive hypothesis may be applied to R Thus Istl ~ 191 n stl 3 = 1(fj'13, so Ii!I ~ 19111(fj'12. But

1£: (fj'1 = IQ1 (fj/(fj')1 = I (fj/(fj' : (fj/(fj')21 = l(fj: 4>(fj)I·

Hence

Suppose then that (fj' = 91; thus 4>(fj) = 91 also. Since (fj is of class 2, 91 ~ Z(fj). Suppose next that 91 < Z(fj). Choose z E Z(fj), z ~ 91. Let 9Jl be a maximal subgroup of (fj such that z ~ 9Jl. Thus

(fj = 9Jl<z), 9Jl n <z) = 3,

where 3 = <Z2) "# 1. We apply the inductive hypothesis to 9Jl/3. If x E 9Jl and x2 E 3, we have x2 = z2m for some m and xz-m E 91. Hence zm E 9Jl n <z) = 3 and x3 E 91/3. Thus

and

Thus we may suppose that G>' = 4>(G» = Z(G» = 91. Write j(X) for the number of elements of order at most 2 in the 2-group X. Thenj(G» =

1911 = 2n, and j(G>/91) = 1G>/911 = 2d, say. Let Jt denote the set of maximal subgroups of 91. Thus IJtI = 2n - 1. Hence by 5.1,

(1) I j(G>/9Jl) = (I.ifl - 1)j(G>/91) + j(G»

= (2n - 2)2d + 2n.

By 5.3,j(G>/9Jl) is an integer of the form 2d, 2d + 2d- r or 2d - 2d- r, where r is a positive integer such that 2r ~ d. Substituting in (1), we obtain an equation of the form

2n+d _ 2d+1 + 2n = I (±2S,), i

where Si ~ td. Suppose now that n < td. Then Si > n for all i, so the right-hand side is divisible by 2n+1. But also, the terms 2n+d, 2d+1 of the left-hand side are divisible by 2n+1, whereas 2" is not. This is a contradiction. Hence n ~ td and IG;I = 2d+" :::; 23" = 19lj3. q.e.d.

5.5 Theorem. Suppose that G; is a non-identity 2-group and that Aut G; permutes the set 1 of involutions in G; transitively. Let IQl(Z(G;))1 = q.

a) Ql(Z(G;)) = I u {l}. b) (GRoss[5]) IG;I = qm for some m. c) IfG; is non-Abelian and of exponent 4, then G;' = Z(G;) = <P(G;)

and IG;I is q2 or q3.

Proof a) Since G; # 1, Ql(Z(G;)) contains an element tEl. Thus 1 = {tcxlo: E AutG;} ~ Ql(Z(G;)) and 1 u {l} = Ql(Z(G;)).

b) Given tEl, write mk = I{YIY E G;, l'-1 = t}1 (k ~ 1). Since Aut G; permutes 1 transitively, mk is independent of t, and the number of elements of order 2k in G; is (q - l)mk (k ~ 1). Hence

IG;I - 1 = (q - 1)(ml + m2 + ... ),

and q - 1 divides IG;I - 1. Since q and IG;I are both powers of 2, it follows that IG;I is a power of q.

c) By a), 1 ~ Ql (Z(G;)), so G;IQl (Z(G;)) is of exponent 2 and hence elementary Abelian. Thus G;' :::; Ql(Z(G;)). Since G; is non-Abelian, 1 # G;'. But then, if G;' < Ql(Z(G;)), there are involutions both inside and outside G;', which is not possible on account of the transitivity of Aut G; on I. Hence G;' = Ql (Z(G;)). Therefore G;/G;' is elementary Abelian and <P(G;) = G;'. If G;' < Z(G;), Z(G;) contains an element z of order 4. Now if Y E G; - Z(G;), y2 E I. Since also Z2 E I, y2 = (Z2)0: for some 0: E Aut G;. But then y2 = (Zo:)2 and y(zo:t l E 1 u {1} ~ Z(G;). Thus y E Z(G;), a contradiction. Hence G;' = Z(G;) = Ql(Z(G;)).

By b), IG;I = qm for some m. By 5.4, m :::; 3. Since G; is non-Abelian, m > 1. q.e.d.

5.6 Definition. An Abelian p-group ~ is called homocyclic if ~ is the direct product of cyclic groups of the same order.

5.7 Lemma. Let ~ be a homocyclic p-group of exponent pn. a) Suppose that ~ = <al' ... , ad>' where pd = I~: <P(~)I. Given

elements Xl' ... , Xd in any Abelian p-group X of exponent at most pn, there exists a unique homomorphism 0: of ~ into X for which aio: = Xi

(i = 1, ... , d).

b) Suppose that P is a homomorphism of 21 into !B/G:, where !B is an Abelian p-group of exponent at most pn and G: ~ !B. Then there exists a homomorphism ~ of 21 into !B such that P = ~v, where v is the natural epimorphism of!B onto !B/G:.

Proof a) This is obvious since

b) IfaiP = biG:,thehomorphismUorwhichai~ = bihastherequired property. q.e.d.

5.8 Theorem. a) If 21 is an Abelian p-group and Aut 21 induces an irreducible group on 0 1(21), 21 is homocyclic.

b) If'll is an Abelian p-group and Aut 21 permutes the set of subgroups of 21 of order p transitively, then 21 is homocyclic.

c) A homo cyclic p-group 21 possesses an automorphism a such that <a) permutes the set of elements of'll of order p transitively.

Proof a) Suppose that 21 is not homo cyclic and let pn be the exponent of 21. Then 1 < 21p ·-1 < 0 1(21), so Aut 21 induces a reducible group on 0 1 (21).

b) This follows at once from a). c) 21/<1>(21) is isomorphic to the additive group of some Galois field

K = GF(pk). If KX = <w), the mapping A --+ AW (), E K) is an additive automorphism y of K such that ypk-1 = 1 and y permutes transitively the set of elements of order p. Thus 21/<1>(21) possesses such an automorphism p. Let v be the natural epimorphism of 21 onto 21/<1>(21). Then by 5.7b), vp = av for some endomorphism rx of 21. Then <a) induces on 21/<1>(21) a group of automorphisms which permutes the non-identity elements transitively. If the exponent of 21 is pn, x<1>(21) --+ XP'-l is an (End 21)-isomorphism of 21/<1>(21) onto 0 1 (21), so (a) induces on 0 1 (21) a group of automorphisms which permutes the non-identity elements transitively. Thus a is an automorphism of 21. q.e.d.

5.9 Theorem. Suppose that 21 is an Abelian p-group and that X is a p'group of automorphisms of'll· Then 21 is the direct product of X-invariant homocyclic subgroups 21 1 , ..• , 21" and 21i/<1>(21i) is an irreducible lgroup.

Proof This is proved by induction on 1211. Let pn be the exponent of 21.

Suppose first that Q 1 (m:) is an irreducible X-group. Then Q 1 (m:) is irreducible under Aut m:, so by 5.8a), m: is homo cyclic. Hence m:jtP(m:) is X-isomorphic to Q 1 (m:) and is therefore an irreducible X-group. The theorem is thus proved in this case.

Now suppose that Q 1 (m:) is a reducible X-group. Since m:p'-' =I 1, there is a minimal non-identity X-invariant subgroup m of m:p'-'. Since Q 1 (m:) is reducible, m < Q 1 (m:). By the Maschke-Schur theorem (I, 17.7), Q 1 (m:) = m x U for some X-invariant subgroup U of Q 1 (m:). Thus U =I 1. If v E m - {1}, there exists x E m: such that x p '-' = v. Then the order of xU is pn, since v ¢ U. Thus the exponent of m:jU is pn.

Since U =I 1, it follows from the inductive hypothesis that m:jU is the direct product of X-invariant homocyclic subgroups with X-irreducible Frattini factor groups. Since the exponent of m:jU is pn, at least one of these factors, say f)jU, is of exponent pn. We write

m:jU = (f)jU) x (mjU),

where f), m are X-invariant subgroups of m:. Also f)jU is homocyclic of exponent pn, and its Frattini factor group is X-irreducible.

We observe that f)P n U = 1. For if Z E f)P n U, there exists y E f) such that z = yp. Then yU is an element of NU of order at most p. Since f)jU is homocyclic of exponent pn,

and y E f)pHU. Hence z = yP E f)P'UP = 1. Thus f)P n U = 1. Now Uf)Pjf)P is an X-invariant subgroup of Nf)P. Hence by the

Maschke-Schur theorem,

for some X-invariant subgroup 'D of f). Thus f) = 'DUf)P, and f) = 'DU. Since m:jU = (f)jU)(mjU), it follows that

Thus 21 = :n x 58. Then ~ = :n x (58 n ~) = :n x u, so :n ~ ~jU is homocyclic and 'DjtP('D) is X-irreducible. Also 'D =I 1, so the inductive

hypothesis may be applied to 'B, and the assertion of the theorem follows at once. q.e.d.

5.10 Theorem. Suppose that m: is an Abelian p-group and that 1 is a p'group of automorphisms of m:. The following conditions are equivalent.

a) m: is I-indecomposable. b) m:;<P(m:) is I-irreducible. c) The only I-subgroups ofm: are Qi(m:) (i = 0, 1, ... ). d) Q 1 (m:) is I-irreducible.

Proof First we deduce b), c) and d) from a). Suppose that m: is X-indecomposable. By 5.9, m: is homocyclic and b) holds. Therefore Q 1 (m:) is I-isomorphic to m:;<P(m:). Hence d) holds. To prove c), let 'B be a nonidentity I-subgroup of m:. If the exponent of 'B is pm, 'Bpm- 1 is a nonidentity I-invariant subgroup of Q 1 (m:), so by d), Q 1 (m:) = 'B pm- 1 = Q 1 ('B). Thus 'B is homo cyclic and 'B = Qlm:) for some i.

The assertions c) => d), d) => a), b) => a) are all trivial. q.e.d.

5.11 Corollary. Suppose that m: is an Abelian p-group, that 1 is a p'group of automorphisms ofm: and that m: is I-indecomposable.

a) Every I-subgroup ofm: is I-indecomposable. b) Every 11actor group ofm: is I-indecomposable.


5.12 Examples. a) It is possible to have a homocyclic p-group m: with a p-group 1 of automorphisms for which m: is I-indecomposable but m:;<1>(m:) is reducible.

For let m: = <al) Ef> <az), where a1 , az are of order pZ, and let 1 be the group of all automorphisms !Y. which induce the identity automorphism of m:;<1>(m:). Then

so 1 is elementary Abelian of order p4. Also 1 induces the identity automorphism on Q1(m:). But m: is an indecomposable I-group. For otherwise m: = <bz) with bi!Y. E <b) for all !Y. E 1. But then

so /1/ :::;; pZ, a contradiction.


b) There exist non-homocyclic Abelian p-groups which are Xindecomposable for some automorphism group X.

For let m: = (al) EEl (a2)' where al is of order p2 and a2 is of order p. Let rJ. be the automorphism of m: for which

If m: is (rJ.)-decomposable, m: = (b l ) EEl (b2), where p2bl = pb2 = ° and birJ. E (b;) (i = 1, 2). Since rJ. induces the identity mapping on m:jcP(m:), b2rJ. - b2 E (b2) n (pb l ) = 0, so b2rJ. = b2. Since C~(rJ.) n .QI(m:) =

(pal) = pm:, b2 E pm:, a contradiction.

5.13 Theorem. Suppose that (fj is a 2-group, m: is an Abelian subgroup of (fj and x E (fj. If x2 E m:2 and Em:, x] ::;; m:4 , there exists a E m: such that (xa)2 = 1.

Proof If this is false, there exists a least positive integer I such that (xa)2 ¢: m:2' for all a E m:. Since x2 E m:2, I > 1. Thus there exists a E m: and b E m: such that (xaf = b2'-i. Let c = b21-2. Then

(xac- I )2 = (xaf[xa, c]c- 2 = [xa, c].

Now Em:, x] ::;; m:4 , so [xa, b] Em: and

[xa,c] = [xa,bY-2 = [x,b]2I-2Em:2'.

Thus (xac- I )2 E m: 2', contrary to the definition of l. q.e.d.

5.14 Lemma. Let ~l' ... , ~n be linear transformations of afinite-dimensional vector space v. Suppose that ~i~j = ~j~i (i =F j) and a = ~i (i = 1, ... , n). Then ~l' ... , ~n have a common eigen-vector.

Proof Let U be a non-zero subspace of minimal dimension for which U~i s;; U for all i = 1, ... , n. Suppose that dim U > 1. Let u be a nonzero element of U. Then U is not spanned by u, so by minimality of dim U, U~i does not lie in the space spanned by u for some i. Hence the restriction ~ of ~i to U is neither the zero nor the identity mapping. Since ~2 = ~, it follows that ker ~ is a proper non-zero subspace of U. Since ~i~j = ~j~i' (ker ~gj s;; ker ~ for allj = 1, ... , n. This contradicts the minimality of U. Hence dim U = 1, and the non-zero elements of U are eigen-vectors of all the ~i. q.e.d.

5.15 Theorem. Suppose that m is a normal homocyclic Abelian subgroup ofthe2-group<fJsuchthatm4 #- 1,[.Ql(m),<fJ] = 1 and [m,«p(<fJ)] = 1. Then m contains a cyclic normal subgroup of<fJ of order 4.

Proof We show first that given x E <fJ, there is an endomorphism I'/(x) of m such that

(2)

for all a E m. To do this, write m = <at) x ... x <ar ). Since m is homocyclic, m/m2 is <fJ-isomorphic to .Q 1 (m). Hence em, <fJ] :::;; m2 and [ai' x] = a? for some a; E m. By 5.7a), th~re is an endomorphism I'/(x) of m such that ail'/(x) = a;-l (i = 1, ... , r). Clearly I'/(x) satisfies (2).

If x, yare in <fJ and a E m,

a(1 - 21'/(xy» = aXY = (aj(1 - 21'/(Y» = a(1 - 21'/(x»)(1 - 21'/(Y».

Hence

(3)

for all x, y in <fJ. Of course if Z E Co;(m), then 21'/(z) = 0 by (2), and 21'/(xz) = 21'/(x) for all x E <fJ by (3). Thus for any x, y in <fJ, 21'/(xy) = 21'/(Yx), since [x, y] E «P(<fJ) :::;; Co; (m). Hence by (3),

(4)

Also, putting x = y in (3), we get

(5)

as x 2 E «P(<fJ) :::;; Co;(m). Let r;(x) be the restriction of I'/(x) to m = .Ql(m). Thus r;(x) is an

endomorphism of m. Since m4 #- 1 and m is homo cyclic, m :::;; m4. Thus (4) and (5) give

r;(x)r;(y) = r;(y)rt(x),

r;(xf = r;(x).

Since m may be regarded as a vector space over GF(2), it follows from 5.14 that m contains an element b such that brt(x) E <b) for all x E <fJ. Then b = a2 for some a and (a 2)r;(x) = a2j for some j. Thus

Hence <a) is a cyclic normal subgroup of (fj of order 4. q.e.d.

In the application of this, we shall need the following elementary fact.

5.16 Lemma. Let V be a vector space over a field K, let S be a set of linear transformations of V, and let ¢l' ... , ¢n be distinct mappings ofS into K. Let

Vi = {vlv E V, VX = ¢Jx)v for all XES} (i = 1, ... , n).

If U is the K-subspace of V spanned by V1 , ... , Vn,

Proof Vi is a K-subspace of V, so

Suppose that the sum is not direct. Let d he the smallest positive integer for which there exist non-zero elements v1 , ... , Vd such that

and Vl E Vi , ... , Vd E Vi with i 1 < i2 < ... < id. Then d > 1, so there 1 d

exists XES such that ¢i ·(x) =f. ¢i (x). But 1 2

° = V1 X + ... + VdX

Hence

(¢i (x) - ¢i (X))V2 + ... + (¢i (x) - ¢i (X))Vd = 0, 2 1 d 1

contrary to the definition of d. q.e.d.

Exercise

4) Let (fj be a special non-Abelian 2-group in which every involution lies in Z(fj). Suppose that the number of involutions in (fjjIDl is the same

for all maximal subgroups IDl of Z(~). Prove that I~I is IZ(~W or IZ(~W·

§ 6. Commutators and Bilinear Mappings

The general formulae

(1)

(2)

[xy, z] = [x, z]Y[y, z],

[x, yz] = [x, z ] [x, y y

closely resemble the conditions for a mapping to be bilinear over 7L. We make this more precise.

6.1 Lemma. Suppose that i)1' i)2' i)1' i)2'~' ~ are subgroups of a group ~. Suppose that ~ <J ~, i)i <J i)i and that i)/i)i is Abelian (i = 1, 2). Suppose also that

Then there exists a 7L-bilinear mapping y of(i)t!i)l) x (i)2/i)2) into ~/~ such that ifx E i)1 and y E i)2'

Proof Let g be the mapping of i)1 x i)2 into ~/ ~ for which

We verify that if u E i)1 and v E i)2' then

(xu, yv)g = (x, y)g.

Indeed

(xu, yv)g = [xu, yv]~

= [x, v] u [x, y] vu [u, v] [u, y] v ~,

by (1) and (2). From the hypotheses we see that all these terms lie in ~ except [x, y]vu. But also [x, y, vu] E ~ since [~,~] ::; ~, so

§ 6. Commutators and Bilinear Mappings 287

(xu, YV)g = [X, y] Sl = (x, y)g,

as asserted. It follows that there exists a mapping y of (f)df)l) x (f)z/f)z) into

Sl/ Sl such that if x E f)l' Y E f)z,

If also x' E f) 1 ,

(XX'f)l, y~z)y = [xx', y]Sl

= [x, yY[x', y]Sl

by (1). Since [x, y, x'] E [Sl, (tj] ::; Sl,

(XX'f)l'y~Z)Y = [x,y][x',y]Sl

= ((Xf)l' y~z)Y)((X'f)l' y~z)y)·

This and the similar equation in the second variable show that y is Z-bilinear. q.e.d.

In our first applications of 6.1, (tj has an automorphism ~ which induces irreducible automorphisms on the f);/f)i.1t follows from 11,3.10 that under these circumstances, there are isomorphisms of these groups onto the additive groups of fields. Thus we examine bilinear mappings on fields.

6.2 Lemma. Suppose that K, F are fields and that F is a Galois extension of K. Let (tj be the Galois group of F over K and let V be a finite dimensional vector space over F.

a) If IY. is a K-linear mapping of F into V, there exists a unique set of elements v~ E V (~ E (tj) such that

alY. = I (a~)v~ (a E F). ~E(!;

b) Suppose that K S; Li S; F, Li is a Galois extension of K and f)i is the Galois group of Li over K (i = 1, 2). If f3 is a K-bilinear mapping of Ll x L2 into V, there exists a unique set of elements V~q E V (~ E f)l' '1 E f)z) such that

(a, b)f3 = I (a~)(b'1)v~q (a E L1, b E Lz). ~.q


c) Let y be the K-bilinear mapping of F x F into V given by

(a, b)y = L (a~)(brOv~w ~.~

If (a, a)y = Ofor all a E F, then v~~ = v~~ + v~~ = Ofor all ~ E (f), 11 E (f).

Proof a) Let M be the K-space of all row vectors (v~) with v~ E V and let N be the K-space of all K-linear mappings of F into V. Thus

dimK M = I (f)ldim K V = [F: K] dimK V = dim K N.

Define a mapping J.l of Minto N by (v~)J.l= a, where

aa = L(a~)v~ (a E F). ~

Thus J.l is K-linear. Suppose that (v~)J.l = O. Then

forallaEF.lfu1' ... ,umisanF-basisofVandv~ = Li=1A~,iUi,(A~,iEF), then for any a E F,

Since distinct homomorphisms of a group into the multiplicative group of non-zero elements of a field are linearly independent (Dedekind's lemma), it follows that all A~,i = 0 and (v~) = O. Thus J.l is a monomorphism. Since dimK M = dimK N, it follows that J.l is an isomorphism. Thus a) is proved.

b) For each bE L2, the mapping a -+ (a, b)/3 is a K-linear mapping of L1 into V, so by a), there exists a unique set of elements v~(b) E V (~ E f,1) such that

(a, b)/3 = L (a~)v~(b) (a E L1, b E L2)' ~EDI

But then v~ is a K-linear mapping of L2 into V, so again by a), there is a unique set of elements v~~ E V (11 E f,2) such that

v~(b) = L (bl1)V~~ (b E L2)' ~ED,

The assertion now follows easily.

c) By b), the K-dimension of the space N' of mappings

(a, b) ~ L (a~)(brJ}v~~, ~,~Eo;

where v~~ = v~~ + v~~ = 0, is tj<DI(I(f)1 - 1)dimK V, which is also the dimension of the space M' of bilinear mappings y of F x F into V for which (a, a)y = O. Since N' !:;;; M', it follows that N' = M'. q.e.d.

6.3 Corollary. Let F be a Galois extension of K, Suppose that F = K(A.), fl E F and rx is a non-zero K-linear mapping of F into a vector space V over F such that (A.a)rx = fl(arx)for all a E F. Then there is afield automorphism () of F over K such that fl = A.() and arx = (a()) v for some v E V.

Proof Let (f) be the Galois group of F over K. By 6.2, there exist v~ E V (~ E (f)) such that

for all a E F. Thus

fl L (a~)v~ = fl(arx) = (A.a)rx = L (a~)(A.~)v~. ~ ~

By the uniqueness assertion of 6.2, (fl - A.~)v~ = 0 for all ~ E (f), Since rx =I 0, Vo =I 0 for some () E (f). Thus fl = A.(). If ~ =I (), then fl = A.() =I A.~,

since F = K(A.). Hence v~ = 0 for all ~ =I (), and arx = (a())vo. q.e.d.

6.4 Lemma. Let K = GF(2), F = GF(2n). Suppose that A. E FX and that the multiplicative order of A. is k, where 2n - 1 divides kn. Suppose that fl E F and that /3 is a K-bilinear mapping of F x F into F such that (a, a)/3 = 0 and (A.a, A.b)/3 = fl((a, b)/3)for all, a, b in F. Then there is a K-linear transformation rx of F and a field automorphism () such that

(a, b)/3 = (a (b()) + (a())b)rx.

Proof If n = 1, /3 is the zero mapping since /3 is symplectic. If /3 = 0, the assertion is trivial with () = 1. Suppose that /3 =I O. Thus n > 1.

F is a Galois extension of K and the Galois group is generated by the automorphism x ~ x2. Thus by 6.2,

n (a, b)/3 = L Gija2Hb2j-1

i,j=l

for suitable 8ij E F with 8ij + 8ji = 8ii = O. However, not all 8ij are zero; suppose 8h,h+r =f. 0, where 0 < r < n. Then

n L j.l8ija2i-lb2i-1 = j.l((a, b)P) = (A.a, A.b)p i,j=1

and by the uniqueness assertion of 6.2,

for all i, j. Since 8h•h+r =f. 0,

Thus

for all i, j. Hence if 8ij =1= 0,

and

1 + 2r == 2i - h+n + 2j - h+n (k).

But k, n satisfy the conditions (1) of § 4. Hence by 4.4c), j - i == ±r (n). Hence, since 8ij + 8ji = 0,

n (a, b)P = L 8i(ab2' - a2'b)2i-l

i=1

for some 8i • If !Y. is the K-linear transformation of F for which

the assertion is clear, () being the automorphism a -+ a2 '. q.e.d.

6.5 Theorem. Let (£) be a 2-group, c; an automorphism of (£), ~ a C;-invariant normal subgroup of (£). Suppose the following hold.

(i) I(£): ~I = I~I = 2". (ii) (£) is not elementary Abelian.

(iii) c; induces irreducible automorphisms on (£)/~ and ~, and if k is the order of that induced on (£)/~, 2" - 1 divides nk. Let K = GF(2), F = GF(2").

a) There exist isomorphisms p, 0" of (£)/~, ~ respectively onto the additive group ofF, and there exists afield automorphism 0 ofF such that if x E (£) and (x~)p = a, then (x 2)0" = a(aO). Also 0 is not of order 2, and () = 1 if and only if (£) is Abelian.

b) There exists A E F such that F = K(A(A()) = K(A) and.

«xc;)~)p = A«X~)p), ye;O" = A(A()(Yo")

for x E (£), y E ~.

Proof First suppose that (£) is Abelian. By (iii), ~ is elementary Abelian, so 01«£» ~ ~. Since 01«£» is C;-invariant and c; is irreducible on (£)/~, 0 1 «£» = ~ or (£). By (ii), 0 1 «£» = ~. By 5.8, (£) is homocyclic, so there exists an isomorphism X of (£)/~ onto ~ such that

By (iii) and II, 3.10, there is an isomorphism p of (£)/~ onto the additive group of F and there is an element A of F such that F = K(A) and

for all x E (£). For y E ~, let yO" = (YX -1 p)2; thus 0" is an isomorphism of ~ onto F. If (x~)p = a, then (x 2 )0" = a2 = a(aO), where () = IF. Finally, if y E ~, then y = x 2 for some x E (£) and

ye;O" = (yc;X- 1 p)2 = «Xc;)2X-1 p)2 = «(xc;)~)p)2

= A2«X~)p)2 = A2(yO").

Thus 6.5 is proved in this case, except for the assertion F = K(A(AO». Now suppose that (£) is non-Abelian. Thus n > 1. Since (£)', [(£), ~]

are c;-invariant and 1 =i: (£)' :::; ~, [(£),~] < ~, it follows from (iii) that (£)' = ~ and [G>,~] = 1. Hence by 6.1, there exists a Z-bilinear mapping y of «£)/~) x «£)/~) into ~ such that if x, yare in (£),


(3) (x~, y~)y = [x, y].

But by (i), (iii) and II, 3.10, there exist isomorphisms p, p of (fj/~, ~ onto the additive group of F, and there exist elements Ie, f.1 in F such that F = K(Ie) = K(f.1) and

(4) ((x~)~)p = ),((x~)p), y~p = f.1(yp)

for all x E (fj, Y E ~. For a, b in F, write

(5)

Then P is a K-bilinear mapping of F x F into F. If a = (x~)p, b = (y~)p,

(lea, leb)P = (((x~)~)p, ((y~)~)p)P (by (4))

= [x~,y~Jp

= [x, YJ~P

= f.1([x, yJp)

= f.1((a, b)P)

(by (5) and (3))

(by (4))

(by (5) and (3)).

Clearly, (a, a)p = 0 for all a E F, and k is the order of l Hence by 6.4, there exists a K-linear transformation rx of F and a field automorphism 8 of F such that

(6) (a, b)P = (a(b8) + (a8)b)rx.

Since ~ = (fj', ~ is generated by [x, y J with x, y in (fj. Hence F is the additive group generated by all [x, yJp; hence F is generated by imp. Thus rx is an epimorphism and F is generated by all a(b8) + (a8)b.

It follows that rx is non-singular. Let a = prx- 1. If x, yare in (fj and a = (x~)p, b = (y~)p, then by (3) and (5),

[x,yJa = ((x~,y~)y)a = (a,b)prx- 1•

Hence by (6),

(7) [x, y J a = a(b8) + (a8)b.

Since ((x~)~)p = Aa and ((y~)~)p = Ab, (7) gives

[x~, y~Ja = ),(le8)(a(b8) + (a8)b) = A(le8)([x, yJa).


Hence z~a = A(A8)(za) for all z E ~ = (fj'. Thus the equations of b) are proved. It follows also from (7) that 8 i= IF, since (fj is non-Abelian.

To prove a), observe that if a E F, then a = (x~)p for some x E (fj

and x 2 depends only on a. Write ar = (x 2)a. If b = (y~)p, then a + b = (xy~)p and by (7),

(a + b)r = ((xyf)a

= (x 2l[ x, y])a

= (x2)a + (l)a + [x, y] a

= ar + br + a(b8) + (a8)b.

Hence if ar' = ar + a(a8),

(a + b)r' = ar + br + a(a8) + b(b8) = ar' + br'.

Thus r' is a K-linear mapping of F into F. By 6.2,

for suitable ei, or

Thus

But

n

ar' = L eia2i-1 i=l

ar = a(a8) + t eia2'-I. i=l

n

(Aa)r = A(A8)a(a8) + L eiA 2i-la2'-I. i=l

n

= (A(A8))(x2)a = A(A8)(ar) = A(A8)a(a8) + L eiA(A8)a 2'-I. i=l

By the uniqueness assertion of 6.2,


for all i. But () #- 1; hence if a(} = a2', r $. 0 (n). Hence by 4.3c), 1 + 2r $. 2i-1 (k), and 2(20) #- 22'-' for all i. Thus 8i = 0 for all i, and (x2)11 = at = a(aO).

Whether or not 6) is Abelian, let F1 = K(2(20». If yl1 E F1 (y E ~),

then y~11 E Fl' Hence F111-1 is a ~-invariant subgroup of ~. Since 2 #- 0, it follows that F111-1 = ~ and F1 = ~11 = F. Thus F = K(2(20». It follows in particular that 0 is not of order 2, since otherwise 2(20) lies in the field of elements fixed by O. q.e.d.

We show next that the group 6) in 6.5 is determined to within isomorphism by O.

6.6 Theorem. Suppose that 6)1' 6)2 are 2-groups. Suppose that ~i is an elementary Abelian subgroup of the centre of6)i and that 6)i/~i is elementary Abelian (i = 1,2). Suppose that p is an isomorphism of6)tI~l onto 6)2/~2' that 11 is an isomorphism of ~1 onto ~2 and that if (X~l)P = y~2' then (x2)11 = y2 (x E 6)1' Y E 6)2)' Then there is an isomorphism of 6)1 onto ffi2 which carries ~1 onto ~2 and induces p, 11 on 6)tI~l' ~1 respectively.

Proof Let X1~1' ... , Xm~l be a basis of the elementary Abelian group 6)tI~l' For i = 1, ... ,m, choose Yi E 6)2 such that (Xi~l)P = Yi~2' Then (xr)11 = land ((xixl)11 = (Yiyl· Since (xixl = xrxJ[Xj' x;], it follows that [Xj' x;] 11 = [Yj' y;]. Now if x E 6) 1, X can be written uniquely as x = xi' ... x:';'y with 0 :::; Ii :::; 1 and Y E ~1' so we may put

xx = Yi' ... y:';'(YI1)·

It is easy to verify that X is a homomorphism of 6)1 into 6)2' Also X carries ~1 onto ~2' X induces 11 on ~1 and XiX = Yi (i = 1, ... , m). Since p, 11 are isomorphisms, so is X, and X induces p on 6)1/~1' q.e.d.

To establish the existence of groups satisfying the conclusion of 6.5, we give matrix representations of them.

6.7 Example. Suppose that F = GF(2n) and 0 is an automorphism of F. We denote by A(n, 0) the set of all matrices of the form

(1 °

u(a, b) = a 1

b aO


with a, b in F. Then

u(a, b)u(a', b') = u(a + a', b + b' + a'(a8)),

and

u(a, btl = u(a, b + a(a8)).

Then A(n, 8) is a group of order 22n with unit element u(O, 0). The mapping u(a, b) --+ a is a homomorphism of A(n, 8) onto the

additive group of F with kernel

ill = {u(O, b)lb E F}.

Thus there is an isomorphism p of A(n, 8)jill onto F such that

(u(a, b)ill)p = a.

Also, there is an isomorphism 1J of ill onto F given by

u(O, b)1J = b,

and

(u(a, b)2)1J = a(a8).

Note that ill - {I} is the set of involutions in A(n, 8), and that A(n, 8) is Abelian if and only if 8 = 1.

Note also that if A E F, there is an automorphism ~A of A(n, 8) given by

u(a, b)~A = u(Aa, A(A8)b).

Theorem 6.5 may now be restated as follows.

6.8 Theorem. Let m be a 2-group, ~ an automorphism ofm, f) a ~-invariant normal subgroup of m. Suppose that the following hold.

(i) 1m: f)1 = 1f)1 = 2n. (ii) m is not elementary Abelian. (iii) ~ induces irreducible automorphisms on m/f) and f), and if k is

the order of that induced on m/f), 2n - 1 divides nk. Then (f) ~ A(n,8), where 8 is some automorphism of F = GF(2n) not

of order 2. Also, if ~ is the automorphism of A(n, e) corresponding to ;, there exists A E F such that F = GF(2)(A(Ae)) and ~, (, induce the same automorphisms on 91 and on A(n, e)/91.

Proof An isomorphism IjJ of (f) onto A(n, e) is obtained by applying 6.6 to the isomorphisms pp-l, aa-1, where p, a are as in 6.5 and p, a are as in 6.7. The assertion about ~ is equivalent to 6.5b). q.e.d.

Not all the groups A(n, e) are non-isomorphic, and not all A can arise in 6.8.

6.9 Theorem. Let F = GF(2n), and let e be an automorphism of F. a) A(n, e) ~ A(n,e- l ).

b) The mapping a --+ a(ae) of F into F is injective if and only if e is of odd order.

c) Ife is of odd order, there exists A E F such that the set of involutions of A(n, e) is transitively permuted by < ;).).

d) If e IS of even order, the set of involutions of A(n, e) is intransitively permuted by the group of automorphisms of A(n, e).

Proof a) Let IDl be the set of elements of order at most 2 in A(n, e- l ),

and let p', a' be isomorphisms of A(n, e- l )/IDl, IDl respectively onto F such that

(see 6.7). The isomorphisms pp'-1, ae- l a'-l (where p, a are as in 6.7) satisfy the conditions of 6.6. Hence A(n, e) ~ A(n, e- l ).

b) Put ax = a(ae) (a E F X). Then X is an endomorphism of F x. If ker X =F 1, there exists a =F 1 such that ae = a-1, so the order of e is even. If the order of e is even, the subfield of elements fixed by e2 is different from the subfield Fl of elements fixed bye. Thus there exists a E F such that ae =F a, ae2 = a. Then a(ae) E Fr, so, since I Fr I is odd, there exists bE Fl such that b2 = a(ae). Let c = ab- l ; c =F 1 since a ¢ Fl. Thus ce = (ae)(bet l = (ae)b- l = ba-1, by definition of b. Hence ce = c -1 and ker X =F 1.

c) Suppose that e is of odd order. Let w be a generator of FX. By b), there exists A E F such that A(Ae) = w. Then u(O, b);). = U(AO, wb), so ;). has the stated property.

d) Suppose that e is of even order. By b), X is not injective and thus not surjective. Hence there exists a E F such that a is not of the form b(blJ) with b E F; thus aa- l is an involution in A(n, e) but is not a square.

Since A(n, 8) is of exponent 4, some involutions are squares. The assertion follows at once. q.e.d.

A(n, 8) was constructed above by a rather special representation. We give a more theoretical way of constructing it.

6.10 Theorem. Let K = GF(2) and suppose that U, V are finite-dimensional vector spaces over K. Suppose that r is a mapping of U into V such that the mapping f3 of U x U into V, defined by

(u, u')f3 = (u + u')r + ur + u'r (u E U, U' E U),

is bilinear. Then there exists a 2-group (fj, a subgroup i) of the centre of (fj and isomorphisms p, a of (fj/i), i) onto U, V respectively such that if (xi))p = u, then (x 2)a = ur.

Proof Let u1, ••• , Urn be a K-basis of U. Put

(i < j),

and extend ~ to a bilinear mapping of U x U into V. Then

(u', u")~ - (u + u', u")~ + (u, u' + u")e - (u, u')~ = ° and (u,O)e = (0, u)e = 0. Let (fj = {(u, v)lu E U, v E V}, and define multiplication in (fj by the rule

(u, v)(u', v') = (u + u', v + v' + (u, u')~).

By I, 14.2, (fj is a group with unit element (0, 0), and

(u, vtl = (u, v + (u, u)~).

Obviously, the mapping (u, v) ~ u is a homomorphism of (fj onto U with kernel i) = {(O, v)lv E V}. Thus there is an isomorphism p of (fj/i) onto U such that «u, v)i))p = u. i) is contained in the centre of (fj, and the mapping (0, v) ~ v is an isomorphism a of ~ onto V. It remains to verify that «u, v)2)a = ur, that is, (u, u)~ = ur. Suppose that

U = U,' + ... + Ui with i l < ... < ir ; we use induction on r. It is , ' clear for r = ° and r = 1. If r > 1, the inductive hypothesis gives (x, x)~ = x't, where x = ui , + ... + Ui,_i But by definition of /3,

(x, Ui)/3 = U't + x't + Ui, 'to

Since /3 is bilinear, it follows that

Thus

(x, Ui)/3 = "(Ui' Ui)f3 = "(Ui' uJ~ = (Ui , xK r ~ J r ~ r) r

j<r j<r

U't = (Ui" x)~ + (x, x)~ + (Ui" Ui)~

= (u, u)~ - (x, ui,K

By definition of ~, (x, u;)~ = 0, so u't = (u, uK q.e.d.

We remark that the groups A(n, () of 6.7 may be constructed from 6.10 by taking U = V = GF(2n) and m = u(u(), for if

13 is bilinear.

Exercises

(U, u')f3 = (u + u')'t + u't + u''t

= u(u'() + (u()u',

5) Show that for any group (fj and any m ~ 1, there exists a 7l. -multilinear mapping f of (fj/(fj' onto Ym(fj)/Ym+1 (fj) given by

6) Let F be a Galois extension of K with Galois group (fj. Let e~ (~ E (fj)

be elements of F. Prove that the K-linear mapping

a -+ L e~(a~) (a E F) ~E<ii

is non-singular if and only if the matrix (e~~-'1J) is non-singular.

§ 7. Suzuki 2-Groups 299

7) Formulate and prove the analogues of 6.6 and 6.10 for an arbitrary prime p.

8) (G. IDGMAN [2]). Prove that if A(n, ()) ~ A(n, cjJ), then cjJ is () or ()-l.

9) Let () be an automorphism of F = GF(2n). Suppose that () is not of order 2.

a) If w is a generator of FX, w(w()) generates F over GF(2). b) The only characteristic subgroup of A(n, ()) other than A(n, ())

and 1 is the group generated by the involutions. c) A(n, ()) has a cyclic group of automorphisms which permutes

transitively the set of all maximal subgroups of A(n, ()).

10) (ALPERIN and GORENSTEIN [1]). Suppose that (f) is a p-group of class at most p + 1, that (f)/(f)' is elementary Abelian of order q and that LX is an automorphism of (f) of order q - 1 which induces an irreducible linear transformation on (f)/(f)'. If C(»(LX) =I 1, q is 2,4, 8 or 9.

§ 7. Suzuki 2-Groups

Suppose that (f) is an Abelian p-group and that Aut (f) permutes transitively the set of subgroups of (f) of order p. Then by 5.8b), (f) is homocyclic. The corresponding question when (f) is possibly non-Abelian has been considered, and it has been proved by SHULT [2, 3] that if p is odd, (f) must be Abelian. This is not the case for p = 2, since by 6.9, if () is an automorphism of GF(2n) of odd order, A(n, ()) possesses an automorphism ~ such that < 0 permutes the set of subgroups of order 2 transitively. All 2-groups having such an automorphism were determined by G. HIGMAN [2], and his results were extended by Goldschmidt, Shaw and Gross. In this section, one of these results, on 2-groups having a soluble automorphism group which permutes the set of involutions transitively, will be proved (Theorem 7.9).

If the 2-group (f) has only one involution, the identity automorphism permutes the set of involutions in (f) transitively, but the only nonAbelian 2-groups with just one involution are the generalized quaternion groups (III, 8.2). We therefore exclude this case and make the following definition.

7.1 Definition. A Suzuki 2-group is a group (f) which has the following properties.

a) <» is a non-Abelian 2-group. b) <» has more than one involution. c) There exists a soluble group of automorphisms of <» which

permutes the set of involutions in <» transitively.

Throughout this section, <» denotes a Suzuki 2-group.

7.2 Lemma. a) If 3 = Q 1 (Z(<»)), 3 is the set of elements of <» of order at most 2. Let 131 = 2n.

b) There exists a soluble group ID of automorphisms of <» such that ID permutes the set of involutions in <» transitively, and any prime divisor of I ID I divides 2n - 1. In particular I ID I is odd.

c) If m is the kernel of the representation of ID on 3, ID/m has a cyclic normal subgroup 31m of order k, where k divides 2n - 1 and 2n - 1 divides nk. 3 permutes the set of involutions in <» in orbits of length k, and the representation of 3 on 3 is irreducible.

Proof a) This follows from 5.5a). b) By 7.1, there exists a soluble group X of automorphisms of <»

which permutes the set of involutions of <» transitively. Let n be the set of prime divisors of 2n - 1 and let ID be a Halln-subgroup of X. If 6 is the stabiliser in X of an involution t, IX: 61 = 2n - 1. Hence IX: 61 and IX: IDI are coprime. By I, 2.13, 6ID = X. Hence the orbit of t under ID is the same as that under X, that is, the set of all involutions of <». Hence ID permutes the set of involutions in <» transitively.

c) If ID/m is regarded as a group of automorphisms of 3, the con-ditions of 3.5 are satisfied, and the assertion is clear. q.e.d.

As far as Theorem 7.9, the symbols 3, n, ID, m, 3, k will have the same connotation as in 7.2. Write q = 2n. Note that n, k satisfy the conditions (1) of § 4. F will denote the field GF(2n). Since 31m is cyclic, there exists ~ E 3 such that 3 = m<o.

7.3 Lemma. a) ~ permutes the elements of 3 - {I} in orbits of length k and induces an irreducible automorphism on 3. Hence 3 is contained in every non-identity ~ -invariant subgroup of <».

b) Suppose that f, is a ~-invariant subgroup of<» such that f, > 3 and f,/3 is elementary Abelian. Then < 0 induces a group of permutations of the non-identity elements of f,/3 in which the length of every orbit is divisible by k. Further N3 is of order a power of q.

Proof a) Since 3/m is cyclic and m is represented trivially on 3, ~ has the same orbits as 3 on 3. Hence ~ permutes the elements of 3 - {I} in orbits of length k, by 7.2c). Similarly, since the representation of 3 on 3 is irreducible, ~ induces an irreducible automorphism on 3. Any nonidentity ~-invariant subgroup of (fj possesses an involution and therefore contains 3.

b) Suppose that kl is the length of the orbit of ~ on ~/3 containing x3 (x ¢ 3). Then X~kl = xy for some y E 3. Since i = 1 and y E Z(fj), (X2)~kl = x2 • But x ¢ 3, so by 7.2a), x2 =1= 1; thus x2 is an involution. By a), kl == 0 (k). Hence 1~/31 - 1 == 0 (k). By 4.3a), IN31 is a power of 2n = q. q.e.d.

7.4 Lemma. Suppose that m: is a normal Abelian ~-invariant subgroup of (fj. a) m: is homocyclic. The only ~-invariant subgroups of m: are the m:2',

and [m:2', (fj] ~ m:2'+1.

b) Ifi 2:: 0 and m: 2' =1= 1, m: 2'/m:2'+1 is ~-isomorphic to 3. c) If '!3 is a minimal normal ~-invariant subgroup of (fj such that

'!3 > m:, ~ induces an irreducible automorphism on '!3/m:.

Proof a) Let!) be a non-identity ~-invariant subgroup of m:. By 7.3a), 3 = Q1 (!), so ~ induces an irreducible automorphism on Q1 (!). By S.8a), !) is homocyclic. In particular, m: is homocyclic and, since!) 2:: 3, !) = m:2' for some i. Since [m:2', (fj] is a proper ~-invariant subgroup of m:2', [m:2', GJ] ::; m:2'+1.

b) If the exponent of m: is 2m, xm:2'+1 --+ X 2rn -'-1 is a ~-isomorphism of m:2'/m:2i+l onto 3.

c) Since '!3 > m:, '!3 > ['!3, (fj]m: 2:: m:. Since ['!3, (fj] is a normal ~-invariant subgroup of (fj, it follows that ['!3, (fj] ~ m:. Thus if '!3 2:: '!3o 2:: m:, '!3o <l (fj. Hence '!3 is a minimal ~-invariant subgroup of (fj for which '!3 > m:. q.e.d.

7.5 Lemma. Let m: be a normal Abelian ~-invariant subgroup of (fj, and let '!3 be a minimal normal ~-invariant subgroup of (fj for which '!3 > m:. If'!3 is non-Abelian and '!3' < m:, then <P('!3) = <P(m:).

Proof By 7.4a), '!3' = m:2'; by hypothesis, r > 0 and m:2' =1= 1. Hence Qr(m:) =1= m: and so Qr(m:) ~ m:2.

If x E '!3, put x¢ = X 2,+1 m:2,+1 . Evidently x2 E m: and X 2,+1 E m:2', so ¢ is a mapping of '!3 into m:2' /m:2,+1 . We show that ¢ is a ~ -homomorphism. Clearly ¢~ = ~¢. If x, yare in '!3, (xyf = x 2y2[y, x]y, and since x 2 , l and [y, x] all lie in m:,

But [y, x] E ~' :::;; ~e, so (xy)4> = (x4>)(y4>). Thus im4> is a ~-subgroup of 212'/212'+1. By 7.4b), 212'/212'+1 is ~

isomorphic to ,3, so ~ is irreducible on 212'/212'+1. Thus either im 4> = 1 or im 4> = 212'/212'+1. Suppose first that im 4> = 1. Then given x E ~, X2'+1 E 212'+1, so X2'+1 = y2'+l for some y E m. Thus (X 2y-2f' = 1 and x2y-2 E Qr(m). Thus x2 y-2 E 212 and x2 E 212. Since <P(~) = <x2lx E ~>, it follows that <P(~) :::;; 212 = <p(m) and <P(~) = <p(m), as asserted.

Suppose then that im 4> = 212'/212'+1. Thus ker 4> =I ~. But obviously 21 :::;; ker 4> and ker 4> is a ~-invariant subgroup; thus by 7.4c), ker 4> = m. Hence there is a ~-isomorphism ljJ of ~/m onto 212'/212'+1 such that

By 7.4, 21 is homocycIic and 21 ~ ,3. Thus 1m2': 212'+11 = 2n and by II, 3.10, there exist an isomorphism (J of 212'/212'+1 onto the additive group of F and an element A of F such that

for all y E 212'. Put P = ljJ(J; thus p is an isomorphism of ~/m onto F, and if x E~,

Suppose that [~, m] :::;; 212'+1. Then, since [212', (fj] :::;; 212'+1, we can apply 6.1. Thus there exists a bilinear mapping y of (~/m) x (~/m) into 212'/212'+1 such that if x E ~ and y E~,

A bilinear mapping f3 of F x F into F is thus defined by

Obviously (a, a)f3 = 0, so by 6.2c) there exist Gij E F with Gii = Gij + Gji = 0, such that

n

(a, b)f3 = L Gija 2i-!b2H. i,j;l

§ 7. Suzuki 2-Groups

Now if a = (xill)p, b = (yill)p,

(a, b)f3 = ((x~, yill)y)O" = ([x, y]~2'+1)0".

Thus, since).a = ((xe)~)p and )'b = ((ye)~)p,

Hence

().a, )'b)f3 = ([xe, ye]~2'+1)0" = (([x, y]e)~2'+1)0"

= ).([x, y]~2'+1)0" = ).(a, b)f3.

for all a, b in F. By the uniqueness assertion of 6.2b),

Gij().2'-1+2 i - 1 _ ).) = 0

303

for all i,j. Since ~2'/~2'+1 is e-isomorphic to 3, the order of). is k. Hence by 4.3c), F-1+2 i - 1 =I- Afor i =I- j. Hence Gij = 0 for all i,j. Thus (a, b)f3 = 0 for all a, b and [x, y] E ~2'+1 for all x, y in ~. However, this implies that ~' ~ ~2'+1, whereas ~' = ~2'. This is a contradiction.

Hence [~, ~] f;, ~2'+1. We show now that [~, <P(~)] ~ ~2'+1. To do this, it suffices to prove that [x, y2] E ~2'+1 for any x, y in ~. We have

[x, y2] = [x, y] [x, y]Y = [x, y]2[X, y, y].

But [x, y] E~' ~ ~2' and [~2', (fj] ~ ~2'+1, so [x, y, y] E ~2'+1. Since also [x, y]2 E ~2'+1, our assertion follows.

Thus <P(~) =I- ~. Hence <P(~) < ~, and since <P(~) is a e-invariant subgroup, <P(~) ~ ~2 = <P(~). Hence <P(~) = <P(~). q.e.d.

7.6 Lemma. Let ~ be a normal Abelian e-invariant subgroup of (fj, and let ~ be a minimal normal e-invariant subgroup of (fj for which ~ > ~. If~' = ~, then ~ is elementary Abelian.

Proof Suppose that this is false. By 7.4, ~ is homo cyclic, ~ 2::: 3 and [~2', ~] ~ ~2'+1 for all i 2::: O. On the other hand ~/~' = ~/~ is elementary Abelian, so by III, 2.13, y)~)jYj+1 (~) is elementary Abelian for allj 2::: 1. It follows by induction onj that Yj(~) = ~2i-2, for ifj > 2, Yj-l (~) = ~2i-3 by the inductive hypothesis and

Let ~ = ~1 = !s/ID4, ~2 = ID/ID4, ~3 = ID2/214. Thus the lower central series of ~ is ~ = ~l > ~2 > ~3 > 1. ~ is a ~-group, and ~d~2 is ~-isomorphic to !s/21. By 7.4c), ~ induces an irreducible automorphism on !S/ID and hence on ~1/~2' By hypothesis ID2 #- 1, so by 7.4b), ~3 and ~2/~3 are ~-isomorphic to 3. In particular ~k induces the identity automorphism on ~2/~3 and ~3' Thus if Y E ~2' y~k = yz for some z E ~3' and if x E~, [x, y] E ~3' so that

[x, y] = [x, y]~k = [x~k, yz] = [x~k, y].

Thus (X~k)X-l commutes with y. Hence (X~k)X-l E C~(~2)' But C~(~2) is ~-invariant and ~2::::; C~(~2) <~; hence C~(~2) = ~2' Thus X~k == x mod ~2 for all x E ~, and ~k induces the identity mapping on ~/~2' Hence the order kl of the automorphism induced on N~2 by ~ is a divisor of k. But if x, yare in ~,

since ~2/~3 is contained in the centre of N~3' Thus ~kl induces the identity mapping on ~2/~3' Hence kl = k. Thus N~2 has an irreducible automorphism of order k. By II, 3.10, IN~21 = 2n1 , where nl is the smallest positive integer such that 2n1 == 1 (k). By 4.3, nl = n; thus I~: ~21 = 2n.

~/~3 therefore satisfies the conditions of 6.5. It follows that there exist isomorphisms p, (J of ~/~2' ~2/~3 respectively onto the additive group of F, and there exists a field automorphism 8 of F such that if x E ~ and (X~2)P = a, then (X2~3)(J = a(a8). Also 8 is not of order 2; and 8 #- 1 since N~3 is non-Abelian. Further, there exists A E F such that F = K(A) and

for all x E ~, Y E ~2'

«X~)~2)P = A«X~2)P),

«Y~)~3)(J = A(A8)«Y~3)(J)

If Z E ~3' then z = t2 for some t E ~2 and t~3 depends only on z; thus we may put ZT = (t~3)(J. Then T is an isomorphism of ~3 onto the additive group of F and

Z~T = A(A8)(ZT)


By 6.1, commutation induces a bilinear mapping y of (N~2) x (~2/~3) into ~3' A bilinear mapping f3 of F x F into F is defined by putting

and

(A.a,2(2(})b)f3 = A.(2(})((a, b)f3).

By 6.2,

n

(a, b)f3 = L Gija 2 '-lb 2j- 1

i,j=l

for certain Gij E F, and by the uniqueness part of 6.2,

for all i,j. Suppose Gij =1= O. If, then, a() = a2 ' for all a E F, 2r =1= 0 (n) as (}2 =1= 1,

and

1 + 2r == 2i - 1 + 2j - 1 + 2j +r - 1 (k).

The solutions to this may be read off from 4.5; they are (i) i == j == r (n) if2r == 1 (n),(ii)i == j - r - 1 == 0(n)if2r == -1 (n),(iii)n = 6,k = 21, r == j == i - 3 == 2 (n) or (iv) n = 6, k = 21, r == j + 2 == i + 1 == 4 (n). Thus for any given values of n, r, there is at most one non-zero Gij' and we may write

Thus

But if a = (X~2)P, then a(a(}) = (x2 ~3)11 and

Thus G = 0 and (a, b){J = 0 for all a, b in F. Thus [~, ~2] = 1, a contradiction. q.e.d.

7.7 Lemma. Suppose that 21 is a maximal normal Abelian ~-invariant

subgroup of (fj. Then 214 = 1 and (fj/21 is elementary Abelian.

Proof. Suppose that 214 =F 1. Since (fj is non-Abelian, there exists a minimal normal ~-invariant subgroup ~ of (fj such that ~ > 21. Thus ~ is non-Abelian. By 7.6, ~' < 21, and by 7.5, ct>(~) = ct>(21). Since there are no involutions in ~ - 21, it follows from 5.13 that [21, ~J $ 214. Thus [Q 2 (21), ~J =F 1 and if (£: = C~(Q2(21)), 21 ::-::; (£: < ~. Since (£: is ¢-invariant, it follows from the minimality of ~ that (£: = 21.

Since ct>(~) = ct>(21), [21, ct>(~)] = 1. Hence by 5.15, 21 contains a cyclic normal subgroup <a) of ~ of order 4. Now if v E,3 and x E ~, write v = u2 with u E 21 and

VIJ(x) = [u, xl

Then lJ(x) is a mapping of,3 into itself, and indeed f/(x) is an endomorphism of ,3. Since [a, x] E (a) for all x E ~,

where g(x) E GF(2). Let !F be the set of mappings of ~ into GF(2), and iff E !F, define

,3f = {vlv E,3, vlJ(x) = vf(X) for all x E ~}.

f'(x) = f(x¢-1) (x E ~).

Hence DfE §',3f is a non-identity ¢-invariant subgroup of,3. By 7.3a),

By 5.16, ,3 is the direct product of the ,3f· By 3.6, ,3 = ,3g, so

for all u E Q2(21), and UX = U1+2g(x). Hence if x E ~ - (£:, UX = u- 1 for all u E Q 2 (21). Thus I~: (£:1 = 2.

It follows that I~: 211 = 2 and ~ = <21, x) for some x; also X( == x mod 21. Let

211 = {aXala E 21}, 212 = <y21 y E ~ - 21).

Thus 211, 212 are ~-invariant subgroups of 21. If y E m - 21, then y = xa for some a E 21 and / = xZaxa. Hence 211 ~ 21z and 21z = <211, XZ). By 7.4, 211 = 21z' and 21z = 21zi for some i, j, so 21zi /21 z' is cyclic. But 3 is not cyclic, so by 7.4b), i = j and 211 = 21z. Thus x- z E 211 and x- z = aXa for some a E 21. Therefore (xa)Z = 1 and xa E 3 ~ 21. This gives x E 21, a contradiction. Hence 214 = 1.

Now suppose that (b/21 is not elementary Abelian. Then 21 ~ cI>((b), so 21cI>((b) > 21. Let !'l be a minimal normal ~-invariant subgroup of (b such that 21cI>((b) ~ !'l > 21. Then !'l = 21(!'l n cI>((b)) and !'l n cI>((b) 1;, 21. By 7.4a), [21, (b] ~ 212 and [21Z, (b] = 1; thus [21, (b, (b] = 1. Hence if x E 21 and y E (b,

[x,/] = [x, y][x, yr = [x,y]Z = 1.

Thus [21, cI>((b)] = 1. If Z E !'l n cI>((b) and Z ~ 21, it follows from 5.13 that ZZ ~ 212. Hence cI>(!'l) :1; 21z. By 7.4a), cI>(!'l) = 21. By 7.5, !'l' = 21, and by 7.6, 21 is elementary Abelian. Hence 21 = 3. In particular, 21 ~ Z((b). Thus 21 = Z((b) on account of the maximality of 21.

Let 1 < Zl((b) < Zz((b) ~ ... be the upper central series of (b. Then Zz((b) n (b' is Abelian, since [Zz((b), (b'] = 1 (III, 2.11). Since 21 = !'l', 21 ~ Zz((b) n (b'. Again, it follows from the maximality of 21 that 21 = Zz((b) n (b'. Hence Zz((b) n (b' = Zl ((b) n (b'. By III, 2.6 applied to (b' Zl ((b)/Zl ((b), (b' ~ Zl ((b). Hence Zz((b) = (b. By 111,2.13, (b/Z1((b) = ZZ((b)/Zl((b) is elementary Abelian, since 21 = 3 = Z((b). Thus cI>((b) ~ Zl ((b) = 21, a contradiction. q.e.d.

7.8 Lemma. cI>((b) is elementary Abelian.

Proof Suppose that this is false. The long argument needed to obtain a contradiction will be divided into 5 steps.

Step 1. Preliminaries

Let 21 be a maximal normal ~-invariant Abelian subgroup of (b. By 7.7, 214 = 1 and 21 ~ cI>((b). Since cI>((b) is not elementary Abelian, cI>((b) :1; 21z. But by 7.4, the only ~-invariant subgroups of 21 are 21, 21z and 1. Hence cI>((b) = 21. Thus cI>((b) is homocyclic of order qZ and is a maximal ~-invariant Abelian subgroup of (b. Also [cI>((b), (b] ~ 3, by 7.4.

We show that if b is a minimal ~-invariant subgroup of (b such that b > cI>((b), then b/3 is elementary Abelian. Indeed, as 21z =f. 1, b' < cI>((b) by 7.6, so cI>(b) = cI>(21) = 3 by 7.5.

We deduce that C(fj(cI>((b)) = cI>((b). For otherwise, we can choose a minimal ~-invariant subgroup b of (b such that cI>((b) ((b)).

Then f>/'J is elementary Abelian. Hence if x E ~ - cP(ffi), x2 E'J = (cP(ffi))2, so x2 = i for some y E cP(ffi). But [x, y] = 1, so xy-l E'J and x E <'J, y) ~ cP(ffi), a contradiction.

We prove next that ~ is of order k. Since ~ is of odd order and the order of the restriction of ~ to 'J is k, it suffices to show that ~k induces the identity automorphism on ffi/<1>(ffi), <1>(ffi)/'J and 'J, by I, 4.4. This is true for <1>(ffi)j'J by 7.4b). Suppose that x E ffi. Then for any y E <1>(ffi), [y, x] E 'J and y~k == ymod'J, so

[y,x] = [y,X]~k = [Y~\X~k] = [y,x~k].

Hence yX = yX~k. Thus (X~k)X-l E C(!)(<1>(ffi)) = <1> (ffi), so ~k induces the identity automorphism on ffi/<1>(ffi).

Since ~ is of odd order, it follows from the Maschke-Schur theorem that ffi/<1>(ffi) is the direct product of groups ~d<1>(ffi), where ~/ is a minimal ~-invariant subgroup of ffi such that ~/ > <1>(ffi). From above, ~d'J is elementary Abelian. But ffi/'J is not elementary Abelian, so [~d'J, ~2/'J] =F 1 for appropriate ~l' ~2' Thus ~l =F ~2 and [~l' ~2] f, 'J. Since [~l' ~2] is a ~-invariant subgroup of <1> (ffi), [~l' ~2] = <1> (ffi). Note also that ~/ is non-Abelian on account of the maximality of cP(ffi). Hence~; = 'J. The remainder of the argument only concerns the groups ~l' ~2'

Step 2. Structure of <1> (ffi), Sll' Sl2

By the Maschke-Schur theorem, ~Jj = (<1>(G>)jj) x (Sldj) for some ~-invariant subgroup Sl/. By 7.3b), ~ permutes the non-identity elements of Sld.3 in orbits the lengths of which are divisible by k. Hence the order of the automorphism induced by ~ on Sld.3 is k. By II, 3.10 and 4.3, I Sld.31 = q. Thus I~/I = q3 and ~; = .3 (l = 1,2), since ~; is a nonidentity ~-invariant subgroup of.3.

We shall now apply 6.5 to each of the groups <1> (G», Sll' Sl2' First, consider <1> (ffi), which is Abelian. There exist isomorphisms n, (J of <1>(ffi)/'J, 'J respectively onto the additive group of F such that

(1)

further there exists 2 E F such that F = K(22), (where K = GF(2)), and

(2)

for x E <1>(G», Y E .3. (The relevant field automorphism is 1 in this case, since <1>(G» is Abelian).

Next, consider R, (l = 1, 2). There exist isomorphisms P;, (J, of R'/3, 3 respectively onto the additive group of F and a field automorphism ()" not of order 2, such that if x E R, and (x3)p; = a, then

(3)

Also there exists 1l; E F such that F = K(Il;(Il;(},)) and

for all x E R" y E 3. Note that replacing (J, by (J,(},-l is equivalent to replacing (), by (},l. Thus if a(), = a2 ", we may suppose that 0 ~ r, < in; (r, =/; in since (), is not of order 2).

We now have 3 mappings of3 onto F, namely (J, (Jl' (J2' Our next aim is to remove the necessity of considering the last two; thus we must derive equations corresponding to (3), (4) involving (J instead of (J" To do this, define (x, = (J,l(J. Then (x, is a K-linear mapping of F onto F, and by (2) and (4), A 2 (a(X,) = (1l;(Il;(}I)a)(X, for all a E F. By 6.3, there is a field automorphism 1/1, of F such that A 2 = 1l,(Il,(},), where III = 1l;1/I, and a(X, = v;(al/l,) for some v; E F. Since IFI = 2n, we can write v; = vf. Put P, = Pil/l" We prove that if x E R, and (X3)Pl = a, then

(5)

Indeed, (x:j)pf = al/l,- l , so by (3),

hence (x2)(J = (X2)(Jl(X1 = v?(a(a(},)), We observe also that if x E R" then

(6)

For by (4),

«X~)3)Pl = «x~)3)P;I/I, = (1l;(X3)Pf))I/I, = 1l,{(X3)P,),

Step 3. Commutation between 11J«fj) and R,

By 6.1, there is a ;I-bilinear mapping Y, of (11J«fj)/:j) x (R,/:j) into:j such that if x E 11J«fj) and y E R"

(x:j, y:j)Yl = [x, y].


A K-bilinear mapping PI of F x F into F is defined by putting

Since C(!j(4J(fj)) = 4J(fj), PI # O. Thus by 6.2, there exist e~j E F, not all zero, such that

(7) n

(b){3 "I 2'-lb2i-l a, I = L..., eija i,j=l

for all a, b in F. Put (vla)n-1 = x3, bpI-1 = y3, where x E 4J(fj), Y E ftl. Then by (2) and (6),

Thus, by definition of PI'

(2a, Illb)PI = vl- 2(((vI2a)n-1, (lllb)PI-1)Yla)

= VI- 2( ((x~)3, (y~)3)Yla).

Using the definition of Yl and (2),

Hence

(2a, 1l1b)PI = VI- 2([X, y]~a) = VI- 222([x, y]a)

= VI- 222((x3, y3)y1a)

= VI-222(((vla)n-l, bpl-1)y1a).

It follows in the usual way from 6.2 that

1(12 _ 12'-1 2i- l ) - 0 eijA A III -

for all i,j = 1, ... , n. Since 22 = Ill(1l1BI) = Illl+2",

Since k is the order of the automorphism induced on SlI/3 by ~, it follows from (6) that k is the multiplicative order of J11' Hence if elj '" 0,

1 + 2r, == 2i - 2 + 2i-2+r, + 2j - 1 (k).

By 4.5b), the only solutions of this satisfying ° :5: rl < !n are (i) rl = 0, i = j = 1, (ii) rl = !(n - 1), i = rl + 2, j = 0, and (iii) n = 6, k = 21, rl = 2, i = 3, j = 5. In case (iii), (a, b)PI = Ga4b16 for G '" 0. Now by 4.6, there exist a, b in F, not both zero, such that

Suppose that (vla)n- 1 = x3 and bpl- 1 = y3; thus x, yare not both in 3, and so xy ¢ 3. Then a2 = vl- 2((x2)a) by (1) and b5 = b(b(}l) = vl-2((y2)a) by (5). Thus

(x2y2)a = (x2)a + (y2)a = vl(a2 + b5) = vl2ea4 b16

= vl((a,b)PI) = [x,y]a.

Thus [x, y] = x2l and (xy)2 = 1. However xy ¢ 3, so this contradicts 7.2a), and (iii) cannot occur. Thus either r, = ° and (a, b)PI = Glab, or rl = !(n - 1) and (a, b)PI = GI((a2)(}I)(bel); in either case GI '" 0(1 = 1, 2).

Step 4. Commutation between Sll and Sl2

The subgroups f>1' f>2 were so chosen that [f>1' f>2] $. 3. But f>1 =

SllcP(fj) and cP(fj)/3 is a central factor of (fj. Hence [Sll' Sl2] $. 3. But [5\1,5\2] :5: cP(fj) and again [(fj, cP(fj)] :5: 3. Hence by 6.1, there is a non-trivial bilinear mapping y of (Sll/3) x (Sl2/3) into cP(fj)/3 such that if Xi E Sl;(i = 1, 2),

We derive the non-zero bilinear mapping P of F x F into F by putting

By 6.2, there exist eij E F, not all zero, such that

n

(a, b)P = L Gija2'-lb2i- 1

i,j=l

for all a, b in F. Put a = (X3)Pl' b = (y3)P2' so X E ftl' Y E ft 2. By (6), ((X03)Pl = J.ll a and ((y~)3)P2 = J.l2b, so by the definitions of f3 and y,

and

(a,b)f3 = (x3,y3)yn = ([x,y]3)n.

But by (2), (([x, y]~)3)n = A(([X, y]3)n), so

(J.l1a, J.l2b)f3 = A((a, b)f3).

Hence by 6.2,

for all i, j. If eij :/; 0, it follows that A = J.lr'J.lr'. Since A2 = J.l11+2 r,

(p. 309), we find by taking the (1 + 2'1)(1 + 2'2)-th power that

Step 5. The contradiction

We obtain a contradiction by using the Witt identity, which takes the form

[x, y,z][y, z,x][z, x,y] = 1,

since (fj is metabelian (III, Aufg. 1)). To express this in the notation we have developed, suppose that a, a', b, b' are elements of F. Write

Then (a, b)f3 = ([x, y]3)n, so

(vz1((a, b)f3), b')f32 = VZ2(([x, y]3, y'3)Y2<T)

= vz2([x, y, y']<T).

Since y, y' are in ft2, [y, y'] E 3, so the above form of the Witt identity reduces to [x, y, y'] [y', x, y] = 1, or [x, y, y'] = [x, y', y J. Hence

(9) (VZ1((a, b)f3), b')f32 = (vz1((a, b')f3), b)f32'

Similarly,

(10)

First suppose that rl = r2 = O. Then (8) and 4.3c) imply i = j = n; hence (a, b)P = ea2n-1b2"-' and (a, b)P2 = B2ab. Thus by (9) a2"-'b2"-'b' =

a2"-'b'2"-'b for all a, b, b' in F, which is absurd, as IFI > 2. If rl = 0 and r2 = !(n - 1), (8) and 4.5b) imply i = !(n + 1),

j = n - 1, so (a, b)P = Ba2!(·-I)b2"-2 and (a, b)PI = Blab. Again (10) yields an absurdity. The case r l = !(n - 1), r2 = 0 need not be considered on account of symmetry.

Finally, suppose rl = r2 = !(n - 1) = m, say. Since (2m + 1, k) = (2m + 1,2" - 1) = 1, (8) and 4.4c) imply that {i,j} = {O, m}. Thus

( b)P - - 2"-'b2m-' + - 2m-'b2"-' a, -Bla B2a ,

where81, 82 are not both o. Substituting in (9) yields 62 = 0; substitution in (10) then yields 61 = o. q.e.d.

7.9 Theorem. Let GJ be a Suzuki 2-group. a) GJ' = cP(GJ) = Z(GJ) = {xix E GJ, x2 = 1}. b) Either (i) GJ ~ A(n, 0) for some non-identity automorphism 0 of

GF(2") of odd order, or (ii) I GJI = IZ(GJ)j3.

Proof a) By 7.8, <P(GJ) ~ 3 ~ Z(GJ), so the exponent of GJ is 4. Hence by 5.5c), GJ' = Z(GJ) = <P(GJ) = 3.

b) By 5.5, IGJI is q2 or q3. If IGJI = q2, IGJ: 31 = q, so by 7.3b), e is irreducible on GJ/3, and GJ satisfies all the conditions of 6.8. Hence GJ ~ A(n, 0) for some automorphism 0 of GF(2"). Since Aut GJ permutes the set of involutions of GJ transitively, it follows from 6.9d) that 0 is of odd order. Since GJ is non-Abelian, 0 =1= 1, by 6.7. q.e.d.

7.10 Remark. G. IDGMAN [2] and SHAW [1] proved that if GJ is a Suzuki 2-group of order I Z(GJ)13, there exist isomorphisms r of GJ/3 onto F ffi F and (1 of:3 onto F such that one of the following holds.

a) If x E GJ and (x:3)r = (a, b), then

(x2 )(1 = a(aO) + ea(bO) + b(bO);

here 0 is an automorphism of F of odd order and B is a non-zero element of F such that B #- c- 1 + (cO) for all c E F.

b) n = 2m + 1 and if (x:3)r = (a, b), then


here () is the automorphism a --+ azm and e =1= c + c((c- Z)()) for all c E F.

c) n = 5m and if (x3)r = (a, b), then

here () is the automorphism a --+ azm and e =1= c- 1(1 + (C())(C()4)) for all c E F.

This is proved by showing that (!j/3 is the direct product of two groups N3, .R/3, where ~, .R both satisfy the conditions of 6.5. There thus arise two mappings of 3 onto F, and these may be compared by using 6.3. Methods similar to those used in this section then lead to the congruence

(1 + 2')(1 + 2S) == (1 + 2S)2 i- 1 + (1 + 2')2 j - 1 (k).

The solution of this by the methods of § 4 is a long, wearisome process, but it leads to the various possibilities for the "square mapping" x3 --+ x2 •

By 6.6, the group (!j is determined to within isomorphism by the conditions a), b), c). The existence of groups satisfying these conditions follows easily from 6.10; further, matrix representations in F are easily found. Each of these groups possesses an automorphism which permutes the set of involutions transitively.

Let q = 2". By II, 10.12b), the matrices

(1 x Y)

Q(x, y) = ° 1 x q ,

° ° 1

where x, yare elements of GF(q2) and y + yq + xxq = 0, form a Sylow 2-subgroup n of SU(3, qZ). Since

the involutions in n are the elements Q(O, y) with y E F = GF(q). Now for ° =1= J. E GF(qZ), put

("-' 'H ,). H(J.) = I\, I\,

Then

and .Q is a Suzuki 2-group of order q3. If I} is an element of GF(qz)' of order q + 1, {1, I}} is a basis of

GF(qZ) over F. Thus there is an isomorphism r of .QjZ(.Q) onto FEB F given by

Since

(Q(a + bl},O)Z(.Q))r = (a, b) (a E F, b E F).

Q(a + bl},O)Z = Q(O, (a + bl})(a + bl})q)

= Q(O, aZ + ab(1} + I}q) + bZl}q+l)

= Q(O, aZ + cab + bZ),

where e = I} + I}q = I} + I}-\ .Q occurs above in a). Of course e #- c + c-1 for c E F.

7.11 Remarks. a) It was proved by SHULT [2,3] that if p is odd, (f; is a p-group and Aut (f; permutes transitively the set of subgroups of order p, then (fj is Abelian. The proof is also by linear methods, and the main lemma is the following.

Let K = GF(q), let ffi be a finite group and let U, V be (finite) K(fjmodules. Suppose that ffi permutes the one-dimensional subspaces of V transitively. Let f.1 be a K(fj-homomorphism of U @K V into V and suppose that for each U E U, the linear transformation v --+ (u @ v)f.1 of V is nilpotent. Then f.1 = 0.

The proof of this makes use of the Feit-Thompson theorem for q = 2 and a deep theorem of BRAUER [3] for q > 2.

b) The remaining problem is thus the complete determination of the 2-groups ffi for which Aut ffi permutes the involutions transitively. Using a combination of the method of Shult with that of Higman presented in this section, GROSS [5] has proved that one of the following holds for such a group.

(i) ffi is homocyclic. (ii) ffi is special and Iffil is IZ(ffiW or IZ(ffiW.

(iii) (fj has lower central series ffi = ffi 1 > ffiz > (fj3 > 1, where ffiz = (fj' = <J>(fj) = C(f;(fj') and (fj3 = [(fjz, (fj] = ffi~ = Z(fj).

c) Another result related to the determination of the Suzuki 2-groups is the following theorem of LANDROCK [1].


Suppose that (f) is a non-Abelian 2-group, rx is an automorphism of (f) of order 2 and [ = CQj(rx) is elementary Abelian of order 2". Suppose that (f) has an automorphism p of order 2" - 1 which commutes with rx. If p permutes transitively the set of non-identity elements of [, then (f) is isomorphic to a Sylow 2-subgroup of PSU(3, 2") or of PSL(3, 2n).

§ 8. Lie Algebras

In § 6 we considered commutation as a bilinear form. But we wish to study a whole series of commutations together, such as those which arise from the lower central series. In order to do this; we need the notion of a Lie ring. One of the advantages obtained by doing this is the possibility of extending the ring of coefficients. We therefore define Lie algebras over any commutative ring. Throughout this section A denotes a fixed commutative ring with identity.

8.1 Definitions. a) Let 9 be an A-module, and suppose that a multiplication is defined in 9 in which the product of a and b is denoted by [a, b ]. 9 is called a Lie algebra over A if the following hold for any a, b, c in 9 and any A., Il in A:

(i) [A.a + Ilb,c] = A.[a,c] + Il[b,c], (ii) [a, A.b + Ilc] = A.[a, b] + Il[a, c],

(iii) [[a, b], c] + [[b, c], a] + [[c, a], b] = 0, (iv) [a, a] = O.

(iii) is called the Jacobi identity. A Lie algebra over 7L is called a Lie ring. If a, b are elements of a Lie algebra, then

[a + b,a + b] = [a, a] + [a,b] + [b,a] + [b,b]

by (i) and (ii). It follows from (iv) that

[a, b] + [b, a] = 0

and hence

(iv') [a, b] = - [b, a].

If a 1, ... , a" (n ~ 2) are elements of g, the n-fold product [a1' ... , an] is defined by induction on n; for n > 2,

§ 8. Lie Algebras 317

Thus the Jacobi identity may be written

(iii) [a, b, c] + [b, c, a] + [c, a, b] = O.

b) If 91,92 are Lie algebras over A, a (Lie) homomorphism of 91 into 92 is an A-module homomorphism p of 91 into 92 for which

[a, b]p = [ap, bp]

for all a, b in 91' The homomorphism p is called an epimorphism if it is surjective and a monomorphism if it is injective. An isomorphism is a homomorphism which is both injective and surjective. An automorphism of a Lie algebra is an isomorphism of it onto itself. The product of two homomorphisms is a homomorphism, and the automorphisms of a Lie algebra form a group.

c) If 9 is an A-module and U, V are A-submodules of 9, write

U + V = {a + bla E U, v E V}.

Thus U + V is an A-submodule of 9, and

U + V = V + U, (U + V) + W = U + (V + W).

If also 9 is a Lie algebra over A, denote by [U, V] the A-submodule spanned by all products [a, bJ with a E U, b E V.

By (iv'), [U, V] = [V, UJ. Clearly, if U s V, [U, W] s [V, W] and by (i) and (ii),

[U + V, W] = [U, W] + [V, W], [U, V + W] = [U, V] + [U, WJ.

d) An ideal of a Lie algebra 9 is an A-submodule j such that [j, 9] s j. By c), this is equivalent to the condition [9, j] s j. The kernel of a homomorphism is an ideal.

If j is an ideal of 9, the A-module 9/j is a Lie algebra in which

[a + j, b + j] = [a, b] + j,

as is easily verified. The mapping v of 9 onto 9/j defined by putting av = a + j (a E 9) is a Lie epimorphism, called the natural homomorphism v of 9 onto 9/j. The kernel of v is j.

The intersection of a set of ideals of 9 is again an ideal of 9. If X is a subset of 9, the intersection of the (non-empty) set of ideals which contain X is an ideal, called the ideal of 9 generated by X.


If i1' h are ideals of g, so is i1 + b, for by c),

e) A subalgebra of a Lie algebra 9 is an A-submodule 1) of 9 such that [1), 1)] ~ 1). The intersection of a set of subalgebras of 9 is again a subalgebra. If X is a subset of g, the intersection of the (non-empty) set of subalgebras which contain X is a subalgebra, called the subalgebra of 9 generated by X. An ideal of a Lie algebra is a subalgebra; hence the ideal of 9 generated by X contains the subalgebra of 9 generated by X.

If i is an ideal of 9 and 1) is a subalgebra, i + 1) is a subalgebra, for by c),

[i + 1), i + 1)] ~ [j, g] + [1), 1)] ~ i + 1).

8.2 Example. Let ~ be an associative algebra over A. (By this is meant an A-module in which multiplication is defined satisfying the associative and distributive laws; also ~ is to have an identity element. No assumption analogous to that of finite dimension in V, § 1 is made, but it will always be supposed that a homomorphism of associative algebras carries the identity element into the identity element.) Given elements a, b of~, write

[a,b] = ab - ba.

The axioms (i), (ii), (iv) of 8.1a) are obviously satisfied by this product. Further,

[[a,b],c] = [a,b]c - c[a,b]

= abc - bac - c[a,b]

= abc - bea + b[e, a] - e[a,bJ.

Hence (iii) also holds. Thus we may define a Lie algebra I(~) as follows: the underlying A-module of I(~) is ~, and the product of two elements a, b of I(~) is [a, b J.

Suppose that ~, ~ are associative algebras and p is an associative homomorphism of ~ into ~; thus (ab)p = (ap)(bp) and 1p = 1. Put I(p) = p; then I(p) is a Lie homomorphism of I(~) into I(~).

If (\: is a subalgebra of ~, 1«(\:) is a Lie subalgebra of I(~). If 3 is a (two-sided) ideal of~, the set 3 is an ideal of I(~) and 1(~/3) = 1(~)/3.

8.3 Lemma. Let 9 be a Lie algebra over A. a) For n > 1, let 6: be the set of permutations I:t. of {t, ... , n} for

which there exists an integer k (0 ::::;; k < n) such that

11:t. > ... > kl:t. > (k + 1)1:t. < (k + 2)1:t. < ... < nl:t.,

and put s(l:t.) = (_1)k. Then ifb, aI' ... , an are elements of g,

b) If X is a subset of g, the subalgebra of 9 generated by X is the Asubmodule spanned by all products [aI' ... , an] with n > ° and ai E X. (The product [a] of a single element a is understood to be a).

c) Suppose that X, Yare subsets of 9 and that 9 is the subalgebra generated by Y. The ideal of 9 generated by X is the A-submodule spanned by all products [b, aI' ... , an] with n ~ 0, b E X, ai E Y.

Proof a) is proved by induction on n. For n = 2, 6i = 6 z, and the assertion is

(1)

This is clear, however, since by (iv'), (i) and (iii),

[b,al,az] - [b,a2,al] = [b,a l ,a2] - [-[az,b],a l]

= [b,al,aZ] + [aZ,b,al]

= -[al,az,b]

= [b, [al' az]]'

For n > 2, observe that for I:t. E 6:, either nl:t. = nor 11:t. = n. Hence the right-hand side of the required equation is

Applying the inductive hypothesis to both sums, this becomes

[b, [aI' ... , an-l], an] - [b, an, [al' ... , an-l]].

By (1), this is [b, [aI' ... , an]].


b) It suffices to show that the A-submodule f) spanned by all products [a1' ... , an] (n > 0, ai E X) is a subalgebra. Thus we must show that if ai E X and aj E X, then

This is clear if n = 1 and follows at once from a) if n > 1. c) Let i be the A-module spanned by all products [b, a1' ... , an]

with n ~ 0, b E X, ai E Y. By a),

[[b, a1' ... , am], [a'l, ... , a~]] E i

if bE X, ai E Y, aj E Y. Since 9 is the Lie algebra generated by Y, it follows from b) that 9 is the A-module spanned by all [a'l, ... , a~] with aj E Y. Hence

[b,a1' ... ,am,c]Ei

if b E X, ai E Y, C E g. Thus [i, g] S; i and i is an ideal of g. Hence i is the ideal generated by X. q.e.d.

8.4 Definition. If U1, ... , Un are A-submodules of the Lie algebra g, define [U1, ... , Un] to be the A-submodule spanned by all [Xl' ... , xn] with Xi E Ui . By a trivial induction, [U 1, ... , Un] = [[U 1, ... , Un- 1], Un] if n > 1.

8.5 Lemma. Suppose that U, V, Ware A-submodules of a Lie algebra g. a) (cf I II, 1.10) [U, V, W] s; [V, W, U] + [W, U, V]. b) If P is a homomorphism of g, [Up, Vp] = [U, V]p for any A

submodules U, V. Iii is an ideal of g, [U + iii. V + iii] = ([U, V] + j)/i. c) Ifi1' iz are ideals ofg, so is [i1' iz]'

Proof a) follows immediately from 8.4 and the Jacobi identity. The first part of b) is trivial, and the second part is the application of the first to the natural homomorphism of 9 onto g/i· For c), we note that [iz, g, i1] s;

[iz, i1] = [i1' iz] and [g, i1' iz] s; [i1' iz]; thus [it> iz, g] s; [i1' iz] by a). q.e.d.

8.6 Definition. If 9 is a Lie algebra over A, write gn = [g, ... , g]. Thus n

gl = 9 and gn = [gn-1, g] for n > 1. By 8.4, gn is the A-module spanned by all [x 1, ... ,xn] with Xi E g. By 8.5c), gn is an ideal of g. Hence gn S; gn-1 (n > 1); the series


is called the lower central series of g.

8.7 Lemma. Let 9 be a Lie algebra over A. a) Ifi is an A-submodule of 9 and gn+1 sis gn, then i is an ideal of g. b) If 9 is the subalgebra generated by X, gn is the A-module spanned

by all [Xl' ... , Xk] with Xi E X and k ~ n. c) If i is an ideal of g, (ginn = (gn + j)/i. d) If 9 = il ~ i2 ~ ... is a series of ideals of 9 and [in' g] S in+1

for all n ~ 1, then in ~ gn. e) [gm, gn] S gm+n for all m ~ 1, n ~ 1.

Proof a) [i, g] s [gn, g] = gn+l S i. b) is proved by induction on n. For n = 1, the assertion is the same

as that of 8.3b). For n > 1, gn-l is the A-module generated by all [Xl' ... , xk] with Xi E X and k ~ n - 1, by the inductive hypothesis. Hence gn is the A-module generated by all

with Xi E X, I > k ~ n - 1. The result follows by applying 8.3a). c), d), e) are all proved by induction 01). n and are all trivial for n = 1.

In c) for n > 1, we have

(g/it = [(g/irl, g/i] = [(gn-l + j)/i, g/i]

= ([gn-l, g] + j)/i = (gn + D/i

by 8.6, the inductive hypothesis and 8.5b). In d) for n > 1,

• :::> [. ] :::> [n-l ] _ n In - In-I> 9 - 9 ,g - 9 .

Finally, to prove e), observe that for n > 1,

and

by the inductive hypothesis and 8.6. By 8.5a), [gn-l, g, gm] S gm+n, whence

q.e.d.


8.8 Theorem. Let 9 be a Lie algebra over A and let ~ be a Lie subalgebra such that ~i + gi+1 = gi for some i ~ 1. Then ~n + gm = gn whenever m > n ~ i.

Proof This is proved by induction on m - i. If m - i = 1, the assertion is the hypothesis. Suppose that m - i > 1. We have

gi+1 = [gi, g]

= [l)i + gi+1, g]

= [l)i, g] + gi+2.

Now [l)i, g] = [g, ~i] is the A-module spanned by all [b, [a1' ... , a;]] with bEg, aj E l) (1 :::;; j :::;; i). Hence by 8.3a),

[~i, g] £ [g,~, . '.' , ~]. I

But now

= [l)i + gi+1, l)]

= [~i, ~] + [gi+1, 1)]

Thus

gi+1 = [1)i, g] + gi+2

£ [g,1), . '.' ,1)] + gi+2 I

The assertion in the case when n > i now follows at once from the inductive hypothesis; in particular

Hence

which is the assertion in the case n = i. q.e.d.


8.9 Definition. The Lie algebra 9 is called Abelian if g2 = 0, that is, if [a, b] = 0 for all a E g, bEg. Any A-module can be regarded as an Abelian Lie algebra by simply defining the product of two elements to be zero.

9 is called nilpotent if there exists an integer n such that gn+l = O. The smallest such integer n is called the class of g. Thus 9 is Abelian if and only if 9 is nilpotent of class at most 1.

8.10 Theorem. Let 9 be a Lie algebra over A. a) If i is an ideal of g, g/i is nilpotent of class at most n if and only if

i ~ gn+1. In particular, g/i is Abelian if and only if g2 s;;; i. b) 9 is nilpotent of class at most n if and only if there exists a series

9 = i1 ~ i2 ~ ... ~ in+1 = 0

of ideals of 9 such that [ji> g] s;;; ii+l (i = 1, ... , n). c) If 9 is nilpotent of class at most n, so is any subalgebra or quotient

algebra of g. d) If 9 is nilpotent and 1) is a subalgebra such that gi = 1)i + gi+1 for

some i ~ 1, then gj = 1)j for all j ~ i.

Proof a) This follows from 8.7c). b) This follows from 8.7d). c) This follows from a) and the fact that l)m £; gm for any subalgebra

1) of g. d) If m is the class of g, gj = 1)j + gm+l = l)j, by 8.8. q.e.d.

8.11 Definitions. For any Lie algebra g, the A-submodule g(n) is defined inductively by

g(O) = g, gIn) = [g(n-l), g(n-l)] (n ~ 1).

In particular, we write g' = g(1) = [g, g] = g2, g" = g(2), etc. By 8.5c), gIn) is an ideal of g. By 8.l0a), g(n-l)/g(n) is Abelian for n ~ 1. The senes

9 ~ g' ~ g" ::> ...

is called the derived series of g. By 8.7e), g(n) s;;; g2' (cf. III, 2.12). 9 is called soluble if there is an integer d such that g(d) = 0, and the smallest such integer d is called the derived length of g. Thus 9 is Abelian if and only if 9 is soluble of derived length 1. A nilpotent Lie algebra is soluble.

8.12 Theorem. Let 9 be a Lie algebra over A. a) If i is an ideal of g, (g!i)<n) = (g(n) + Wi. Thus g/i is soluble of

derived length at most d if and only if i ~ g(d).

b) If 9 = io ~ i1 ;2 iz ~ ... is a series of subalgebras of 9 such that in is an ideal Ofin-1 and in-din is Abelian for each n > 0, then in ;2 g(n). Thus 9 is soluble of derived length at most d if and only if there exists such a series with id = 0.

c) If 9 is soluble, so is any subalgebra or quotient algebra of g. d) If i is an ideal of 9 and i, g/i are both soluble, then 9 is soluble.

Proof a) follows from 8.5b). b) is proved by induction on n. For n > 0, in-1/in is Abelian, so in ~ i~-1 by 8.10a). Thus

I· => [I' I' ] => [g(n-1) g(n-1)] = g(n) n - n-1, "-1 -, .

c) follows from a) and the obvious fact that l)(m) s; g(m) for any subalgebra l) of g. To prove d), observe that g(d) s; i for some integer d, by a). Hence g(d+n) s; i(n) for all n. Since there exists an integer n for which i(n) = 0, 9 is soluble. q.e.d.

These results are obvious analogues of the elementary theories of nilpotent and soluble groups. But in Lie algebras, one can extend the ground-ring; there is no analogue of this for groups. In the following, A is a subring of a commutative ring A * containing the identity element of A*.

8.13 Theorem (cf. V, 11.1). Suppose that 9 is a Lie algebra over A and let g* = 9 ®AA*.

a) g* has the structure of a Lie algebra over A * with

(2) A(a ® A') = a ® AX,

(3) [a ® A, a' ® X] = [a, a'] ® AX

for all A, A' in A* and a, a' in g. b) If r:t. is an automorphism of g, r:t.* = r:t. @ 1 is an automorphism of g*. c) Suppose that A*, regarded as an A-module, is free. If U, V are A

submodules of g, U ® A* and V ® A* can be regarded as A*-submodules of g*. Then U @ A* s; V @ A* if and only if U s; V. Also

[U, V] ® A* = [U ® A*, V @ A*],

and ifi is an ideal of g, i ® A* is an ideal of g*.


d) Suppose that A* is a free A-module, that a is an automorphism of 9 and a* = a @ 1. If U is an A-submodule of 9 which contains all elements of 9 invariant under a, then U @ A* contains all elements of g* invariant under a*.

Proof a) Certainly g* is an A*-module in which (2) holds. Multiplication in 9 is an A-bilinear mapping of 9 x 9 into g; this can therefore be extended to an A*-bilinear mapping of g* x g* into g* satisfying (3). The verification of the Jacobi identity and the skew-symmetry of the multiplication in g* are tedious but routine matters.

b) If a* = a @ 1 and f3* = a- 1 @ 1, then a* and f3* are A*-endomorphisms of g* and a* f3* = f3* a* = 1. Since also a* preserves Lie products, IX* is an automorphism of g*.

c) Since A* is a free A-module, it follows from V, 9.5 that U @ A*, V @ A* can be monomorphically embedded as A*-modules in 9 @ A* and may therefore be regarded as A*-submodules of g*. It is clear that U @ A* £; V @ A* if and only if U £; V. By 8.1c), [U @ A*, V @ A*] is spanned over A* by all [u @ A, v ® X] = [u, v] @ AX =

AA'([u, v] @ 1) with A, X in A* and u E U, V E V, so

[U @ A*, V ® A*] = [U, V] ® A*.

If j is an ideal of g,

[j @ A*, g*] = [j ® A*, 9 ® A*] = [j, g] ® A* S i @ A*.

Thus i ® A* is an ideal of g*. d) Suppose that b is an element of g* invariant under a*. There are

distinct elements Al' ... , Am of an A-basis of A* such that

m

b = L ai ® Ai, i=1

where ai E g. Since blX* = b, L (ai - ailX) ® Ai = O. By V, 9.5, ai = aia, i

so ai E U. Thus b E U ® A*. q.e.d.

Exercises

11) Let 9 be the Lie algebra generated by X and let j be the ideal of 9 generated by Y. Show that [j, g] is the ideal of 9 generated by {[a, b] I a E X, bEY}. Show that 9n is the ideal of 9 generated by {[a l , ... , an] I ai EX}.

12) Prove that a proper subalgebra of a nilpotent Lie algebra is an ideal of a strictly larger subalgebra.

13) Suppose that n > 1 and 9 is the Lie algebra of all n x n matrices (Ai) with Aij E A and Aij = 0 if i > j. Prove the following.

a) ~ = g2 = {(Ai)IAij = 0 ifi ;;::j}. b) g2 = g3. c) ~m = {(Aij)IAij = 0 ifi > j - m} (1 ~ m ~ n). d) ~ is nilpotent. 9 is soluble but not nilpotent.

14) Formulate and prove the analogue of Theorem 8.8 for groups.

§ 9. The Lie Ring Method and an Application

In order to use Lie algebras for the study of nilpotent groups, we make a construction which enables one to associate a Lie ring with any strongly central series of a group.

9.1 Definition. A strongly central series of a group ffi is a series

(S)

of subgroups of ffi for which [ffii, ffiJ ~ ffii+j for all i,j. This implies that [ffii, ffi] = [ffi;, ffi 1] ~ ffi i+1 for all i ;;:: 1, so (S) is a central series in the sense of III, 2.1. Hence ffii <J ffi and ffi;/ffii+1 is contained in the centre of ffi/ffii+1' In particular ffi;/ffii+1 is Abelian.

For example, the lower central series of any group is strongly central (III, 2.11 b ».

With a strongly central series we obtain something resembling the Jacobi identity.

9.2 Lemma. Suppose that

(S)

is a strongly central series of a group ffi. If x E ffii, Y E ffij , z E ffi k ,

[x, y, z] [y, z, x] [z, x, y] E ffii+j+k+1'


[x,y,z] = [[y,x-IY,(XZX-Iy]

= [y, X-I, (XZX-IZ-I)ZY

= [y, X-I, ZY[y, x-I, [X-I, Z-I]]"X.

Since (S) is strongly central,

Since also (f;i+j+J(f;i+j+k+1 is Abelian, it follows that

[x, y, z] [y, z, x] [z, x, y] (f;i+j+k+1

= [y, x-I, zY[x, Z-I, y]"[z, y-I, x]y(f;i+j+k+1 = 1,

by III, 1.4.

9.3 Theorem (cf Ill, Aufg. 8). Let

(S)

be a strongly central series of a group (f;.

327

q.e.d.

a) There exists a Lie ring g, unique to within isomorphism, having the following properties.

(i) For each i ~ 1, there exists a monomorphism (Ji of(f;;/(f;i+1 into the additive group of g, and if Gi = im ai'

(ii) If x E (f;i and y E (f;j'

b) Suppose that I) is a Lie ring and that for each i ~ 1, there is a homomorphism Pi of (f;;/(f;i+ I into I) such that if x E (f;i and y E (f;j'

Then there is a Lie ring homomorphism e of g into I) such that Pi = (Jie for each i ~ 1. If each Pi is a monomorphism and the sum of the im Pi is direct, e is a monomorphism. If I) is spanned by the impi, e is an epimorphism.

c) IfIY. is an automorphism of(f) and (f)iIY. = (f)i for all i, there exists an automorphism p of 9 such that if x E (f)i,

Proof a) Let ai be an isomorphism of (f);/(f)i+1 onto an additive group Gi

(i = 1, 2, ... ), and let 9 be the direct sum of the Gi :

Since (S) is strongly central, it follows from 6.1 that for i ~ l,j ~ 1, there exists a Z-bilinear mapping Yij of ((f);/(f)i+l) x ((f))(f)j+l) into (f)i+)(f)i+j+1 such that if x E (f)i and y E (f)j,

Thus there exists a Z-bilinear mapping {3ij of Gi x Gj into Gi +j such that if a E Gi and b E Gj ,

Since 9 is the direct sum of the Gi , there exists a bilinear mapping (a, b) -+ [a, b] of 9 x 9 into g, the restriction of which to Gi x Gj is {3ij. Thus if x E (f)i and y E (f)j,

[(x(f)i+l)ai, (y(f)j+l)aJ = ((x(f)i+l)ai, (y(f)j+1)a){3ij

= ((X(f)i+l, y(f)j+l)Yi)ai+j

= ( [x, y] (f)i+ j+1)ai+ j,

as asserted in (ii). We show that 9 is a Lie ring. The distributive laws are equivalent to the bilinearity already estab

lished. To verify the skew-symmetry, suppose that a E g. We can write

with ai E Gi and ai = (Xi(f)i+l)ai (i = 1, ... , k). Now

k

[a, a] = I [ai' aj] i,j=l

k

= L [a;, aJ + L ([ai' aj] + [aj' aJ). i=l i<j

§ 9. The Lie Ring Method and an Application 329

Thus, to prove [a, a] = 0, it suffices to prove that [ai' aJ = ° and [ai' aj] + [aj, aJ = 0. But by (ii)

and

[ai' aJ + [aj' aJ = ([Xi' Xj](fji+j+l)O"i+j + ([Xj' XJ(fji+j+l)O"i+j

= ([Xi' Xj] [Xj' xJ(fji+j+l)O"i+j = 10"i+j = 0.

As for the Jacobi identity, it is clear from the linearity that it is sufficient to prove that

[[a, b], c] + [[b, c], a] + [[c, a], b] = ° if a E Gi, b E Gj , c E Gk • Let a = (X(fji+l)O"i, b = (y(fjj+l)O"j, c = (Z(fjk+l)O" k'

Then, by two applications of (ii),

and

[[a, b], c] = ([x, y, Z](fji+j+k+l)O"i+j+k'

Thus

[[a,b],c] + [[b,c],a] + [[c, a], b]

= ([x, y, z][y, z, x][z, x, y](fji+j+k+l)O"i+j+k

= 10"i+ j+k = 0,

by 9.2. Thus, in a), it only remains to prove the uniqueness assertion; first we prove b).

b) It is clear that for each i ~ 1, there is an additive homomorphism Bi of Gi into l) such that O"iBi = Pi' Since 9 is the direct sum of the Gi ,

it follows that there is an additive homomorphism fJ of 9 into l) such that Bi is the restriction to Gi of B. To show that fJ is a Lie ring homomorphism, it suffices in view of the linearity to show that

[a,b]fJ = [aB,bB]

for a E G;, bE Gj . But if a = (X(fji+l)O"i and b = (y(fjj+l)O"j, then aB = afJi = (X(fji+l)O"iBi = (X(fji+l)Pi and similarly bB = (y(fjj+l)Pj, whence

[a, b]e = [(X(f)i+l)O"i, (y(f)j+l)O"Jei+j

= ([x, y](f)i+j+l)O"i+A+j

= ([x, y](f)i+j+l)Pi+j

= [(X(f)i+l)Pi' (y(f)j+l)Pj]

= Cae, be],

as stated. The assertions concerning monomorphisms and epimorphisms are easy to verify.

The uniqueness assertion of a) follows at once from this. c) Suppose that IY. is an automorphism of (f) and (f)ilY. = (f)i for all i.

Then IY. induces automorphisms lY. i of (f);/(f)i+l:

We may apply b) to the isomorphisms lY.iO"i of (f);/(f)i+l onto Gi, since if x E (f)i and y E (f)j'

[(X(f)i+l)lY.iO"i' (y(f)j+l)lY.pJ = [«XIY.)(f)i+l)O"i' «YIY.)(f)j+l)O"j]

= ([XIY., ylY. ] (f)i+j+l)O"i+j

= ([x, y](f)i+j+l)IY.i+Pi+j.

Thus there exists an automorphism P of 9 such that O"iP = lY.iO"i for all i, so that if x E (f)i'

9.4 Lemma. Suppose that (f) is a group and that 9 is a Lie ring. Suppose that for i ~ 1, O"i is a homomorphism of"Y;(f)/Yi+l «f) onto an additive subgroup Gi of 9 such that 9 = G 1 + G 2 + .... Suppose further that if x E Yi«f)), y E Yi(f),

(1) [(XYi+l«f))O"i, (YYj+l«f))O"j] = ([x, yhi+j+l«f))O"i+j.

a) If(f) = <X), 9 is the Lie algebra generated by

b) gn = G n + G n+ 1 + ... (n = 1,2, ... ). c) gn/gn+l is isomorphic to Gn/(Gn n if+1). d) g(d) s; Li;;'1 «(f)(d) n y;(f)Yi+1«f)/Yi+1«f)))O"Jor all d ~ o.


Proof a) If x E X, write x = (XY2(G»)al . By a simple induction on n, it follows from (1) that

By III, 1.11, Yn(G»/Yn+1(G» is generated by all [Xl' ... , Xn]Yn+l(G» with Xi E X. Since Gn = im an' Gn is generated by all [Xl' ... , xnJ. Hence 9 is generated as an additive group by all [Xl' ... , Xn] (Xi E Y, n ~ 1) and as a Lie algebra by Y.

b) By 8.7b), gn is the additive group generated by all [Xl' ... , Xk] with k ~ n. Hence gn is generated by all Gk with k ~ n.

c) By b), gn = Gn + gn+1 and gn/gn+l ~ Gn/(Gn n gn+l). d) This is proved by induction on d. For d = 0, it is clear. If d > 0,

put f)i = G>(d-l) n Yi(G». Then by the inductive hypothesis,

Hence

g(d-l) ~ L (f)iYi+l (G»/Yi+l (G»)ai. i:2:l

g(d) = [g(d-l), g(d-l)]

~ L [(f)iYi+l (G»/Yi+l (G»)ai, (f)jYj+l (G»/Yj+l (G>))ajJ. i,j

Using (1), it follows that

g(d) ~ L([f)i, f)j] Yi+j+1 (G>)/Yi+j+l(G>))ai+j. i.i

But from the definition of the f)i' it is clear that

so d) is proved. q.e.d.

Theorem 9.3 has a number of applications in group-theory; the use of it is known as the Lie ring methodfor nilpotent groups. The idea is to prove a theorem about Lie rings and then use Theorem 9.3 to derive information about groups.

Even in simple cases when the group-theoretical result can be proved directly, some simplification is often effected by using the Lie ring method, since calculation is usually easier in the Lie ring than in the group. For example, put

f(a,b,c) = [a,b,c] - [b,c,a]

for any elements a, b, c of a Lie ring. Then it is very easy to verify the following identities:

f(b,c,a) + 2f(c, a, b) = -3[a,b,c];

f([a, b], c, d) - f(a, [b, c], d) - f(a, b, [c, d])

= 2[a, b, c, d] - [J(a, b, c), d] + [J(b, c, d), a];

f(a,b,c) = [b,c,c] + [b,a,a] - [b,c + a,c + aJ.

Suppose then that 9 is a Lie ring in which [a, b, b] = ° for all a, b in g. It follows at once that 3 [a, b, c] = 2[ a, b, c, d] = 0, so [a, b, c, d] = 0. The proof of the corresponding result for groups (III, 6.5) was not as simple as this, but it can be deduced from these facts about Lie rings without much difficulty.

The first example of the Lie ring method that we give is a theorem of Vaughan-Lee on p-groups (9.12). We begin by proving the following simple lemma.

9.5 Lemma. Let t 1, ... ,tn be independent indeterminates over a com-mutative ring A. Suppose that 9 E A[tl' ... , tn], that g(O, ... ,0) = 1 and that g(A 1, ... ,An) = 0 whenever A l' ... , An are elements of {O, I} not all zero. Then the degree of 9 is at least n.

Proof We may write

g(t 1,···,tn)=I+ '\' ti···tih···;(ti,···,ti )· ~ 1 r 1 r 1 r il< ... < i,

It will be proved by induction on r (r = 1, ... , n) that h ... i (1, ... , 1) = (-1)'. To do this, substitute 1 for ti , ... ,ti and 0

1 , 1 r

for all the other t i; this gives

r-1

o = 1 + '\' (~)( - I)S + h ... ;( 1, ... , 1), L.., 1 , s=l

the inductive hypothesis being used here if r > 1. The assertion thus follows from the binomial theorem.

In particular, f12 ... n(I, ... , 1) = (-It Hencef12 ... n i= 0 and the degree of 9 is at least n. q.e.d.

9.6 Lemma. Let V, W be finite-dimensional vector spaces over a field K such that dimK V > 1, and let f be a symplectic bilinear mapping of V x V into W. If there exists a maximal subspace U of V such that f(U, U) = f(V, V), there exist at least two such subspaces.

(If X is a subspace ofV,f(X, X) denotes the subspace ofW spanned by allf(x, y) with x E X, Y EX.)

Proof Suppose that this is false. Then f"# 0, since dim K V > 1. Let a1' ... , an be a K-basis of U. Thenf(V, V) = f(U, U) is spanned by all f(ai' a) (1 :$; i < j :$; n). Let

S = {( i, j) 11 :$; i < j :$; n}.

Since f"# 0, n > 1 and S is non-empty. If s E S, write s = (is,js) and put Cs = f(ai" aJ. Since U is maximal, V = U + Ka for some a E V. For i = 1, ... , n, write

f(a, aJ = L J.lisCs (J.lis E K). seS

Suppose that A1 , ••• , An are elements of K not all zero. Put bi = Aia + ai (i = 1, ... , n), and let U 1 be the subspace spanned by b1, ... , bn• Since A l' ... , An are not all zero, U "# U l' Since the assertion of the theorem is false,J(U 1, U1) "# f(V, V). But f(U 1 , U1) is spanned by all

f(b i, b) = f(Aia + ai' Aja + a)

= L (AiJ.ljs - AjJ.liS)cs + f(a i, a) seS

with i < j. If r, s are elements of S, put

ars = Ai J.l1· s - AJ. J.li S' r, r r

and let A be the S x S matrix the (r, s)-coefficient of which is ars ' Then

f(bi" bj) = L (a rs + brs)cs , SES

where brs is the Kronecker b. These elements do not span f(V, V), so A + I is a singular matrix and

det(A + I) = 0.


Now let t l' ... , tn be independent indeterminates over K, and let B be the S x S matrix the (r, s )-coefficient of which is ti J.lj s - tj J.li s. Let

r r r r

g(tl, ... , tn) = det(B + f).

Thus g E K[tl' ... , tnJ and g(j'l' ... , An) = 0 whenever AI, ... , An are elements of K not all O. Also g(O, ... , 0) = 1. By 9.5, the degree of g is at least n.

Let Br, fr be the r-rows of B, f respectively (r E S). Since the determinant is multilinear in its rows, g(t 1, ... , tn) is the sum of the determinants of certain matrices the rows of which are all taken from the Br or the fr. Since the degree of g is at least n, it follows that there exists a non-singular S x S matrix C, n rows of which are taken from the Br. Thus there exist n of the Br which are linearly independent over the field of rational functions K(t I' ... , tn). But if Ci is the row-vector (J.lis) (i = 1, ... , n),

B(i,j) = tiCj - tjCi

= titj(tj-lCj - ti-ICJ,

so all the Br are linear combinations of the n - 1 row-vectors ti1Ci - t;;lCn (i = 1, ... ,n - 1). Hence any n of the Br are linearly dependent. This is a contradiction. q.e.d.

9.7 Definition. Let V, W be finite-dimensional vector spaces over a field K and let f be a symplectic bilinear mapping of V x V into W. If v E V, we put

v.L = {ulu E V, f(u, v) = O}.

Thus v.L is a K-subspace of V, and since f is symplectic, Kv ~ v.L.

9.8 Theorem. Let V, W be finite-dimensional vector spaces over a field K and let f be a symplectic bilinear mapping of V x V into W. Suppose that f(X, X) c f(V, V) for every proper subspace X of V. Then V is spanned by

{vlv E V, v.L = Kv}.

Proof Suppose that this is false and that we have a counterexample for which dimK V is minimal. Then f =I- 0; in particular, dimK V > 1.


(1) There exists a maximal subspace U of V such that if v E V and v ¢ U, then v.L n U "# o.

Since the conclusion of the theorem is false, there exists a maximal subspace U of V containing all v E V for which v.L = Kv. Thus if v ¢ U, v.L ::::> Kv and v.L n U "# O.

(2) There exists a maximal subspace A of U such that if v E V and v ¢ U, then v.L n A "# o.

Since U is a maximal subspace of V, V = U EEl Kg for some g E V. Thus

f(V, V) = f(U EEl Kg, U EEl Kg) = f(U, U) + f(U, Kg) = f(U, V).

Hence f(U, V) "# f(U, U). But if v E V and v ¢ U, then

f(v.L n U, V) = f(v.L n U, Kv EEl U)

= f(v.L n U, Kv) + f(v.L n U, U) s; f(U, U).

Hence U is not spanned by the v.L n U for all v E V, V ¢ U. Hence there exists a maximal subspace A of U such that v.L nUs; A whenever v E V, V ¢ U. By (1), v.L n A "# o.

(3) Suppose that Y is a proper subspace of V and X is a maximal subspace of Y. Suppose also that y.L n X "# 0 whenever y E Y, y ¢ X. Then there exists a maximal subspace Xl of Y such that Xl "# X and f(X l , Xl) = f(Y, V).

Let Yl be the subspace of Y spanned by all y E Y for which y.L n Y = Ky. Ify E Yandy ¢ X,y.L n X "# 0, whencey.L n Y "# Ky. Hence Yl "# Y. But then, by minimality of dim K V, there exists a maximal subspace Xl of Y such that f(X l , Xl) = f(Y, V). Since X "# 0, we have dimK Y > 1, so by 9.6, we may suppose that Xl "# X.

Choose g E V such that V = U EEl Kg, and let L = A EEl Kg. By (2), A and L satisfy the conditions of (3). Hence there exists a maximal subspace M of L such that M "# A, and

(4) f(M, M) = f(L, L). Let B = M n A and write A = B EEl Ka. Then a ¢ M, so

L = M EEl Ka and g = v + A.a for some v E M, A. E K. Since g ¢ A, v ¢ B and M = B EEl Kv. Write U = A EEl Kz. Thus V = U + L = L EEl Kz. Let P = B EEl K(v + z).

(5) If x E P and x ¢ B, x.L n B "# O. Since x ¢ B, x ¢ U, so by (2), Y = x.L n A "# O. Also x = /1Z + y,

where /1"# 0 and y E L, so f(Y, Kz) = f(Y, K(x - y)) = f(Y, Ky) s; f(L, L). Thus


f(M + Kz + Y, Y) = f(M, Y) + f(Kz, Y) + f(Y, Y)

£; f(L, L) = f(M, M),

by (4). Hence

f(M + Kz + Y, M + Kz + Y) £; f(M + Kz, M + Kz) c f(V, V),

so M + Kz + Y c V. Since M EB Kz is a maximal subspace of V, Y £; M EB Kz. Hence

Y £; (M EB Kz) n A = (M EB Kz) n L n A = M n A = B,

and Y = x-L n B =f. O. By (3) and (5), there is a maximal subspace N of P such that N =f. B

and (6) f(N, N) = f(P, P). Let C = B n N and write B = C EB Kb. Since P = N + B, there

exists b' E B such that v + z - b' EN. Put w = v - b', so that M = B EB Kw and w + zEN. Finally, let Q = N EB K(b + z).

(7) f(P, Kb) £; f(Q, Q). For f(P, Kb) £; f(P, P) = f(N, N) £; f(Q, Q). (8) f(Kw EB N, Kz) £; f(Q, Q). For

by (7).

f(Kw EB N, Kz) £; f(K(w + z) + Kz + N, Kz)

£; f(N + Kz, Kz)

£; f(N, Kz)

£; f(N, K(z + b)) + f(N, Kb)

£; f(Q, Q),

(9) f(M, C) £; f(Q, Q). For M = B EB Kw = C EB Kb EB Kw. Since C £; Q,

f(C, C) £; f(Q, Q). By (7), f(Kb, C) £; f(Q, Q). And

f(Kw, C) £; f(K(w + z), C) + f(Kz, C)

s; f(N, C) + f(Kz, C)

£; f(Q, Q),


by (8). Thusf(M, C) ~ f(O, 0).

u

Now M = C E9 Kb E9 Kw, so

f(M, M) = f(M, C) + Kf(b, w).

But by (4),f(M, M) = f(L, L), so

f(L, L) ~ f(O, 0) + Kf(b, w)

by (9). Hence there exist p, y in K such that

(10)

(11)

f(a, w) - Pf(b, w) Ef(O, 0),

f(a, b) - yf(b, w) Ef(Q, 0).

337

For tx E K, put R = 0 + K(a + txz). Then R + Kz contains z, a and O. Since b + Z E Q and C ~ Q,

R + Kz ;2 C + Kb + Kz + Ka = B + Ka + Kz = A + Kz = U.

But also w + zEN ~ 0, so W E R + Kz and R + Kz ;2 U + Kw = V. Hence dimK R = dimK V - 1, so f(R, R) i= f(V, V). Since R + Kz = V, then, f(R, Kz) ¢. f(R, R). Since R = N + K(b + z) + K(a + txz), it follows from (8) that

(12) f(Kb + Ka, Kz) ¢. f(R, R).

On the other hand,

f(b + z, w + z) = f(b, w) + f(b, z) - f(w, z)

f(a + txz, w + z) = f(a, w) + f(a, z) - txf(w, z)

f(a + rxz, b + z) = f(a, b) + f(a, z) - rxf(b, z)


all lie in f(R, R). By (8), f(w, z) Ef(R, R). Using also (10) and (11) to eliminate f(a, w),f(a, b), it follows that

f(b, w) + f(b, z), Pf(b, w) + f(a, z), yf(b, w) + f(a, z) - rxf(b, z)

all lie in f(R, R). Hence, eliminating f(b, w),

f(a, z) - Pf(b, z), f(a, z) - (rx + y)f(b, z)

also lie inf(R, R). But then, if rx is chosen to be different from P - y, it follows thatf(a, z),f(b, z) lie inf(R, R), contrary to (12). q.e.d.

The main theorem of this section concerns the number of elements in conjugacy classes, or equivalently, the index of centralizers. We shall need similar notions for Lie algebras.

9.9 Definition. Let 9 be a Lie algebra over A. If x E g, put

g(adx) = [g, x]

for all g E g; thus ad x is an A-homomorphism of 9 into g. Put

Cg(x) = ker(ad x);

thus Cg(x) is an A-submodule of 9 and g/Cg(x) is A-isomorphic to the A-submodule

{[g, x] Ig E g}.

In particular, if 9 is finite, I 9 : Cg(x)1 is the number of [g, x] (g E g).

We now apply 9.8 to Lie algebras.

9.10 Theorem. Let p be a prime and let 9 be a nilpotent Lie ring for which Igi is a power ofp and pg n g2 = pg2. Let I) be a minimal element of the set of subrin~ of 9 for which I)2 = 92. Put I I) : I)2 + pI) I = pd, i = pg + g3, 9 = g/i and I) = (I) + i)/i.

a) I) is a minimal element of the set of subrings of gfor which I)2 + i = g2 + i.

b) G is a minimal element of the set of subrings of 9 for which G2 = 92.

c) q is generated byd elements Xl' ••• , xdfor which II) : CIi(xi)1 = pd-t,

where Xi = Xi + i (i = 1, ... , d).


Proof a) Clearly i is an ideal of 9. Suppose that f is a subring of I) and F + i = 92 + i. Then 92 = F + (i II 92). But

so 92 = F + P9 2 + 93• Hence U = 92/(F + 93) is an additive p-group for which pU = U. Thus U = 0 and 92 = F + 93 . By 8.10d), 92 = F. By minimality of I), f = I).

b) We have

~2 = (I)2 + i)/i = (92 + i)/i = ff.

Suppose that 1 is a subring of ~ for which 12 = ~2. Then 1 = (f + i)/i for some subring f of I), and F + i = I)2 + i = 92 + i. By a), f = I). Thus 1 = ~.

c) ~/~2 and ~2 are vector spaces over K = GF(p), and since ~3 = 0, there is a symplectic K-bilinear mapping! of ~/~2 into ~2 given by

Any proper subspace of ~/~2 is of the form f/~2 for some proper ideal f of~, and by b),

Hence by 9.8, ~/~2 is spanned by the set X of those x + ~2 (x E ~) for which (x + ~2).l = K(x + ~2). But

(x + ~2).l = {u + ~21u E~, [x, u] = O}

= C~(X)/~2.

Thus, putting dimK(~/~2) = n, x + ~2 E X if and only if dimK(~/C~(x)) = n - 1. Also ~/~2 is spanned by linearly independent elements Xl + ~2, ... , xn + ~2 of X. Hence I~: C~(xi)1 = pn-l (i = 1, ... , n), and ~ is spanned by Xl' ... , xn and ~2.

Put Xi = Xi + i with Xi E I), and let I)l be the sub ring generated by Xl' ... , xn• Then (I)l + iii) + ~2 =~. Squaring, ~2 = (I)i + i)/i, so I)i + i = I)2 + i. By a), I)l = I), so I) is generated by Xl' ... , xn. Since I)2 + i :2 I)2 + pI), it follows that I)/(I)2 + pI)) is spanned over K by the linearly independent elements Xi + (I)2 + pI)) (i = 1, ... ,n), and n = d. q.e.d.

9.11 Lemma. Let I) be a nilpotent Lie ring for which I I) I is a power of the prime p. Suppose I) js generated by Yl' ... , Yn' where II): C~(Yi)1 ~ pb (i = 1, ... ,n). Let I) = I)/(pI) + I)3), Yi = Yi + (pI) + I)3), and suppose that I~: C~(Yi)1 ~ pc (i = 1, ... , n). Put i = pI)2 + I)3. Suppose also that I is a subring of I) and

a) 1I)2: rz + il ~ p!a(a-l)+a(n-a).

b) II2 + i: 121 ~ pa(b-c)+(n-a)f.

c) Ii I ~ pn(b-c).

Proof a) Put i = pI) + I)2. Then I)/i is a vector space over K = GF(p) spanned by Yl + i, ... , Yn + i. Renumbering the Yi if necessary, we may suppose that I) is spanned by I + i and Y 1, ... , Ya. Then I + i is spanned .by i and n - a elements Z 1, ... , Zn-a of 1. Hence I)2 is spanned by the ta(a - 1) elements [Yi' yJ (1 ~ i < j ~ a), the a(n - a) elements [Yi' Zj] and 12 + pI)2 + I)3 = 12 + i. Thus

b) If I)l is the subalgebra generated by X = {Yl' ... , Ya' Zl' ... , zn-a}, I) = I)l + i = I)l + pI) + I)2. Thus A = f)/(I)l + I)2) is an additive p-group for which pA = A and f) = f)l + f)2. By 8.10, f) = f)l. Thus f) is generated by X. By 8.7, i = pI)2 + f)3 is the additive group generated by all P[Xl' x 2] and all [Xl' ... , xm] for which m ~ 3 (Xi E X). Hence if

Ui = {[U,y;]IUEi} (i = 1, ... ,a),

Vj = {[u,Zj]IUEi} (j = 1, ... ,n - a),

then

Since II + i: II = pf, we see from the mapping u + I ~ [u, Zj] + 12 (u E i) that

(13)

To estimate IUd, let

Ti = {[v, Y;]lv E f)} (i = 1, ... , a).

Thus U; s;; T; (') i s;; T; (') (pI) + I)3). Since T; is an additive subgroup of I),

Now by 9.9, lTd = II): C~(Y;)I ~ pb. Similarly

Thus

and

(14)

l(Ti + pI) + I)3)/(pI) + I)3)1 = I{[v, Y;]lv E ~}I

= 16: c~cYi)1 ~ pC.

I U;j ~ pb-c (i = 1, ... , a).

By (13) and (14),

a n-Q

W + i:I21 ~ fllUJ fllVj + 12:121 ;=1 j=l

~ pa(b-cl+(n-alf•

c) This follows from b) by taking I = 0, in which case we may suppose that n = a. q.e.d.

9.12 Theorem (VAUGHAN-LEE [2]). Suppose that is a finite p-group and that every conjugacy class of contains at most pb elements. Then

More generally, if fJ ~ and I : fJl = pr, then

where v = max(O, b - r + 1).

Proof Let c be the class of and let ; = y;(<I» (i = 1, ... , c + 1). By 9.3 there exists a Lie ring 9 = G1 EB ... EB Gc and, for i = 1, ... , c, an isomorphism (J; of ;/;+1 onto G; such that for any x E ;, y E j,


gm = Gm EEl Gm+1 EEl ... EEl Gc (m = 1, ... , C + 1).

i = pG 1 EEl pG2 EEl G3 EEl ... EEl GC'

i = pG2 EEl G3 EEl ... EEl Gc•

Thus i, i are ideals of g, i = pg + g3 and i = pg2 + g3. Let f) be a minimal element of the set of subrings of 9 for which f)2 = g2. Put If): f)2 + pf)1 = pd and ~ = (f) + i)/i. We show that f) satisfies the conditions of 9.11.

a) f) is generated by d elements Yl' ... , Yd for which If): Cbi)1 ::; pb (i = 1, ... , d). Also I~ : C~CyJI = pd-l, where Yi = Yi + i, so d - 1 ::; b.

Since pg n g2 = pg2, it follows from 9.10c) that f) is generated by d elements XI' ... ,Xd for which I~: C~(.xJI = pd-I, where Xi = Xi + i. Write Xi = Yi + Zi' where Yi E G1, Zi E g2 = f)2. Thus Yi E f) and if f)1 is the subring generated by Y 1, ... , Y d, f) I + f)2 = f). By 8.10, f) = f) 1·

Also Xi - Yi E ~2, so C~(yJ = q(.xJ and I~: q(yJI = pd-l. Since Yi E GI , Yi = (gi(fJ2)0"1 for some gi E (fJ. Since [G 1, Gj] S; Gj+1,

But for j = 1, ... , c,

Gj n C9(Yi) = {(g(fJj+l)O"jlg E (fJj' [gi' g] E (fJj+2}

2 {(g(fJj+l)O"jlg E (fJj n C(!j(gJ}.

Since O"j is an isomorphism, it follows that

IGj n Cg(Yi)1 ~ 1((fJj n C(!j(gJ)(fJj+t!(fJj+11

= l(fJj n C(!j(gi): (fJj+1 n C(!j(gi)l.

Thus

c

ICg(yJI = n IGj n Cg(YJI ~ IC(!j(gi)l, j=1

and since Igl = Iml,

Ig:Cg(Yi)1 ~ 1m: C(jj(g;) I ~ pb.

Hence 11): C~(Yi)1 ~ pb (i = 1, ... , d).

Let 6 = 1)/(p1) + 1)3), Yi = Yi + (p1) + 1)3). b) 16 :'C~(ji;)1 2 pd-l (i = 1, ... , d). For, using 9.9,

16: QCYi) I = I{[z,y;] + (p1) + 1)3)lzE1)}1

2 I{[z, y;] + ilz E 1)}1

= I~: c~cY;)1

by a). c) i = p1)2 + 1)3. By 8.10d), gi = 1)i for each i 2 2, so i = pg2 + g3 = p1)2 + 1)3. For i = 1, ... , c, put Li = ((~ n m;)(f)i+t!<f>i+l)O'i' Let

and I = 1* n 1). Put

d) I is a subring of 1) and 1m': ~'I ~ 11)2: 121· Also a + f ~ r. [Li' Lj ] is the additive group spanned by all

Thus 1* is a Lie subring of 9 and 1*2 ~ L~ EB ... EB L~, where

Hence I is a subring of 1). Now

343

Thus 11*1 = I~I and 11*21 :s; I~l Since

pa+ f = 11):11 = 11):1) n 1*1 = 11) + 1*:I*I:s; Ig:I*1 = I<»:~I = pr,

a + f :s; r. Also 1<»': ~'I :s; Ig2: 1*21 :s; 11)2: 121. Thus d) is proved. By 9.11c), Iii :s; pd(b-d+l), so

W + i: 121 :s; Iii :s; pd(b-d+1).

Combining with 9.11a), we find that 11)2 : 121 :s; pX, where

x = d(b - d + 1) + ta(a - 1) + a(d - a).

Now

x = tb(b + 1) - tv(v - 1) - x',

where

x' = t(b - d)(b - d + 1) + t(d - a)(d - a - 1) - tv(v - 1).

Thus if x' 2:: 0, we have

as required. Suppose then that x' < 0. Then v > d - a. Since d - a 2:: 0, it follows that v > 0, so v = b - r + 1 and b - r + 1 > d - a. Hence b - d - r + a 2:: 0. By a), d - 1 :s; b, so

2b - 2d - r + a + 1 2:: 0.

In this case we use 9.11a) and b), from which it follows that 11)2: 121 :s; pY, where

y = a(b - d + 1) + (d - a)f + ta(a - 1) + a(d - a).

Now

tb(b + 1) - tv(v - 1) - y

= t(r - a)(2b - 2d - r + a + 1) + (d - a)(r - a - f) 2:: 0,

since by d), r - a 2 r - a - f 2 O. Hence

Taking ~ = 1, we obtain 1(f;'1 ~ p1b(b+1). q.e.d.

9.l3 Remarks. a) The inequality for I (f;'1 in 9.12 is best possible, as is shown by the group generated by Xo, ... , Xb with defining relations

b) VAUGHAN-LEE [1] proved the analogue of Theorem 9.12 for Lie rings.

c) It is not difficult to prove by slightly strengthening IV, 2.3 that there is a function f(n) such that whenever every conjugacy class of (f; contains at most n elements, 1(f;'1 ~ f(n). It has been proved by P. M.

NEUMANN and M. R. VAUGHAN-LEE [1] that we may take

f(n) ~ nl:(3 + Slog.) ,

where the logarithm is taken to base 2. In the same paper it is shown that for soluble groups

f(n) ~ nl:(5 + logn) •

The proof utilizes Theorem 9.12 in the case when I(f;: ~I < pb.

It was conjectured for some years that if every conjugacy class of the p-group (f; contains at most pb elements, the class c of (f; is at most b + 1 (see I. D. MACDONALD [1]), but a counterexample to this was given recentlyl. However, it is known that c ~ 2b, and this will now be proved.

9.14 Lemma. Suppose that

is a central series of the p-group (f;. If x E (f; and there exist s suffices i for which 1 ~ i ~ k - 1 and x ¢ C(!j((f;;!(f;i+2)' then the number of conjugates of x in (f; is at least pS.

1 See FELSCH, NEUBUSER and PLESKEN [1].

Proof This is proved by induction on k and is trivial for k = 1. If k > 1 and x E C(fj«f)k-1), it follows from the inductive hypothesis that X(f)k has at least pS conjugates in (f)/(f)k and the assertion follows trivially. If k > 1 and x ~ C(fj«f)k-1)' the inductive hypothesis shows that X(f)k has at least ps-1 conjugates in (f)/(f)k, whence I(f): [I ;::: pS-1, where [/(f)k = Cffi/ffi,(X(f)k). Now C(fj(x) ~ [. But (f)k-1 ~ [ since the given series is central, and (f)k-1 $, Cffi(X). Hence C(fj(x) < [and I(f): Cffi(X) 1 ;::: pS.

q.e.d.

9.15 Theorem. (LEEDHAM-GREEN, NEUMANN and WIEGOLD [1]). If(f) is a p-group and the number of elements in any conjugacy class of (f) is at

most pb, then the class c of (f) is less than ~1 + 1 and is thus at most p-

2b.

Proof For each element x E (f) and for 1 ~ i ~ c - 1, write

Thus for i = 1, ... , c - 1,

since Yi«f))/Yi+2«f)) is not a central factor. Hence

c-1 1 L L ti(x) ~ ~1(f)1· i=l XE(fj p

But Ix:t ti(x) = c - 1 - s(x), where s(x) denotes the number of suffices i for which 1 ~ i ~ c - 1 and x ~ C(fj(Yi«f))/Yi+2«f))). Thus

c - 1 L (c - 1 - s(x)) ~ -1(f)1· XE(fj p

But s(1) = 0, so

c-1 I c-l I L (c - 1 - s(x)) ~ --I(f) - (c - 1) < --(\(f) - 1). l*xE(fj P P

Hence there exists x E (f) such that

By 9.14, s(x) ::; b, so

c - 1 c - 1 - s(x) < --.

p

b > (c _ 1) _ c - 1 = (c - 1)(p - 1). p p

This gives the stated inequalities.

347

q.e.d.

Under special conditions, there are other ways for forming a Lie ring from a group. The following is a very simple example, which will be used in Chapter X.

9.16 Lemma. Suppose that (fj is a (not necessarily finite) nilpotent group of class at most 2 in which every element x has a unique square root x!. F or any elements x, y of (fj, put

x + y = xy[y, x]t.

Then (fj is a Lie ring with respect to addition and commutation. Every element of (fj has the same order with respect to the two group operations on (fj, and the automorphisms of the multiplicative group (fj are the same as the automorphisms of the Lie ring (fj.

Proof First observe that since (fj is of class at most 2,

[x!, y]2 = [x, y1]2 = [x, y],

so [x, y]t = ext, y] = [x, yl:] E Z((fj). Thus

[x, y]l:[y, x]l: = [x!,y][y, x!] = 1.

Hence

xy[y, x]! = yx[x,y][y,x]l: = yx[x,y]l:

and x + y = y + x. Again,


(x + y) + z = (x + y)z[z, x + yJ!

= xy[y,xJ!z[z,xyJ!

= xyz[y,xJ![z,yJ![z,xJ!

= x(y + z)[yz, xJ!

= x + (y + z).

Thus (f) is an additive group with zero element 1, and - x = X-I.

We have

[x + y, zJ = [xy, zJ = [x, zJ [y, zJ = [x, zJ + [y, z].

Also [x, x J = 1 and [x, y, z J = 1. Thus (f) is a Lie ring. If n E 7L, nx = xn. Thus x has the same order with respect to the

two group operations on (f). Also!x = xt. Finally, note that

xy = x + y + !Ex, yJ,

so the automorphisms of (f), regarded as a multiplicative group or as a Lie ring, are the same. q.e.d.

9.17 Remark. The procedure in 9.16 is the first step in the inversion of the Baker-Hausdoiff formula. The Baker-Hausdorff formula (see, for example, JACOBSON [1J, page 173) is

( 1)m-1 [P, q, Pz q", J X*y = I I - X, ... ,x,y, ... ,y,x, ... ,X, ... ,y, ... ,y m~l Pi+qi>O mL.(pi + qi)PI! ql! ... Pm! qm!

= X + y + !Ex, y J + /2 [X, y, y J - /2 [X, y, x J + .. '.

If 9 is a Lie ring and the right-hand side of this formula has a meaning for all x, y in g, then 9 becomes a group with respect to *, the identity element being 0 and the inverse of x being - x. The commutator [x, y J* of x and y is

[x, yJ* = [x, yJ + !Ex, y, xJ + ![x, y, yJ + .. '.

Conversely, the formula can be inverted, so that x + y and [x, y J are expressible in terms of the *-operation. Thus, under appropriate con-

§ 10. Regular Automorphisms 349

ditions, a group can be regarded as a Lie ring (see, for example, AMAYO

and STEWART [IJ, Chapter 5). The above special case arises in a quite different way in a paper of

c. HOPKINS [1].

§ 10. Regular Automorphisms

Our next aim is to use Theorem 9.3 to prove some theorems about groups having fixed point free automorphisms. To do this, it is necessary to obtain first the corresponding results for Lie rings. We therefore wish to study Lie rings having an automorphism a which leaves only the zero element fixed, that is, for which ker(lI. - 1) = O. The results obtained are more general, and the hypothesis on a can be replaced by assertions about the ideal generated by the elements fixed by a. Consider, for example, the following theorem about automorphisms of order 2.

10.1 Theorem. Suppose that 9 is a Lie ring, that a is an automorphism of 9 of order 2 and that j is an ideal of 9 which contains all elements of 9 left fixed by a.

a) all. == -amodj,jor all a E g. b) If Gl = {ala E g, all. = -a}, G1 is an additive subgroup of g,

[G l , GlJ s; j and 2g S; j + Gl . c) a induces the identity automorphism on (g2 + j)/j. d) 4g2 S; j.

Proof a) If a E g, (all. + a)lI. = all.2 + all. = all. + a. Hence all. + a is fixed by a and lies in j.

b) Gl is clearly an additive subgroup. If a E G1 and b E G1 ,

[a,b]lI. = [all.,bll.] = [-a, -b] = [a,b],

so [a, b] E j. Thus [G 1, Gl] S; j. If a E g, 2a = (a + all.) + (a - all.). By a), a + all.Ej.Anda - all.EG1,since

(a - all.)lI. = all. - all. 2 = all. - a = -(a - all.).

c) Ifa,b are in g, [all.,bll.] == [-a, -b]modj,bya).Hence[a,b]lI. == [a, b] mod j. Thus a induces the identity automorphism on (g2 + i)/j.

d) 4g2 = [2g,29J S; [j + G1, j + GlJ by b). Using b) again,

q.e.d.


Suppose now that 9 has an automorphism of order 2 which leaves only the zero element fixed; then 9 is Abelian (cf. V, 8.18). This can be deduced from 10.1 in two ways. In both, we take j = O. On the one hand, it follows from c) that every element of g2 is fixed by a, so g2 = O. On the other hand, it follows from d) that 4g2 = 0; hence g2 = 0 since an automorphism of order p of a non-zero Abelian p-group always has a non-zero fixed point (see 10.6 below). The first of these procedures seems difficult to generalize to automorphisms of other orders, though this can be done for automorphisms of order 3 (see Exercise 16). The second procedure can, however, be generalized. Before doing so, we shall consider it for automorphisms of order 3.

10.2 Theorem. Suppose that 9 is a Lie ring, that a is an automorphism of 9 of order 3 and that i is an ideal of 9 which contains all elements of 9 left fixed by a. Then 27g 3 £; j.

Proof. In this case we are unable to make use immediately of the decomposition corresponding to that of 2a in lO.1b); first we must extend the ground-ring. Thus let OJ be a complex cube root of unity and let A be the ring of all numbers x + yOJ with x, y in Z. Thus {I, OJ} is a Z -basis of A. Let g* = 9 ®z A, a* = a ® 1, j* = j ®z A. By 8.13, g* is a Lie algebra over A, a* is an automorphism of g*, g3 ®z A = (g*)\ and j* is an ideal of g* which contains all elements of g* invariant under a*. Let

G1 = {ala E g*, ar:x* = wa}, G-1 = {ala E g*, aa* = OJ-la}.

Thus G1 , G-1 are A-submodules of g*. We have

since if a E Gl and b E G- l ,

[a,b]a* = [aa*,ba*] = [OJa,OJ-lb] = [a,bJ.

Also

For if a E g*, write

al = a + aa* + aa*2

a2 = a + OJaa* + OJ2ar:x*2

a3 = a + OJ2aa* + waa*2.

§ 10. Regular Automorphisms

Then clearly, a1 E j*, az E G- 1, a3 E G1 and a1 + az + a3 = 3a. Now

Hence

(3g*f s; [j* + G1 + G- 1 , j* + G1 + G- 1]

s; [j*, g*] + [G p G1] + [G 1 , G- 1] + [G_ 1 , G- 1]

s; j* + [G 1 , G1] + [G- 1 , G_ 1].

(3g*)3 s; [j* + [G 1 , G1] + [G-1 , G- 1], j* + G1 + G- 1]

s; [j*, g*] + [G 1 , G1 , G1] + [G 1 , G1 , G- 1]

+ [G-1 , G- 1 , G1] + [G-1 , G- 1 , G- 1].

351

But all these summands are contained in j*. For by 8.4, [G 1 , G1 , G1]

is spanned over A by all [a, b, c] with a, b, c in G 1, and

[a, b, c] a* = [aa*, ba*, ca*] = [wa, wb, we] = [a, b, c].

A similar reason holds for [G-1 , G- 1 , G- 1]. Finally, by 8.5a),

since [G 1 , G_ 1] s; j*, and a similar argument holds for [G- 1 , G_i , Gil Thus 27(g*)3 S; i*. Hence (27g3) ® A S; i ® A. By 8.13, 27g3 S; j.

q.e.d.

10.3 Lemma. Suppose that A is a commutative ring which contains a primitive n-th root of unity w. Suppose that 9 is a Lie algebra over A and that a is an automorphism of 9 of order n. For each integer j, write

a) Gj is an A-submodule of g, and Gi = Gj if i == j (n). b) [G i, Gj ] S; Gi+j for any integers i, j. c) If 1) = Go + ... + Gn _ p then 1) is a subalgebra of 9 invariant

under a and ng £ 1). Also if bi E Gi (i = 0, ... , n - 1) and bo + ... + bn - 1 = 0, then nbi = O.

d) If d is a non-negative integer,

n-l l)(d) = L (I) (d) n Gk),

k~O

and if d > 0,

n-1 n(£)(d) n Gk} = n I [£)(d-1) n Gi , £)(d-1) n Gk-;]

i=O

for any integer k.

Proof a} This is trivial. b} If u E Gi and v E Gj ,

[u, v]a = [ua, va] = [uiu, wjv] = wi+j[u, v J.

Thus [u, v] E Gi+ j' as required. c} £) is a subalgebra on account of b}. Since Gia = Gi, £) is invariant

under a. If a E g, write

for i = 0, ... , n - 1. Then aia = wiai, so ai E Gi . Since w is a primitive n-th root of unity, ao + a1 + ... + an- 1 = na. Hence na E £). Thus ng <;: £). Apply w-ija j to the equation bo + ... + bn- 1 = O. We find that for i,j = 0, ... , n - 1,

Summing over j = 0, 1, ... , n - 1, nbi = O. d) The first assertion is proved by induction on d. It is trivial for

d = 0.1f d > 0,

£)(d) = [£)(d-1), £)(d-1)] = ~]£)(d-1) n Gi , £)(d-1) n GJ, i,j

by the inductive hypothesis. By b} [T.(d-1) n G. T.(d-1) n G.] c T.(d) n G .. so the assertion is clear , 4, " J.., J - 4} ,+ J' .

Suppose that a E £)(d) n Gk (d > O). By definition of £)(d) and the previous assertion, a is a sum of elements of the form [x, y], with XE£)(d-1) n Gi , YE£)(d-1) n Gj. By b}, [x, y] EGi+j. Hence by c}, n(a - s} = 0, where s is the sum of those [x, y] for which i + j == k (n). The assertion follows at once from this. q.e.d.

10.4 Theorem. Suppose that A is a commutative ring which contains a primitive n-th root of unity w. Suppose that 9 is a Lie algebra over A and

that !Y. is an automorphism of 9 of order n. For each integer j, write

Let 9 = Go + ... + Gn- 1, let j be the ideal of9 generated by Go and let 9kbethesubalgebraofggeneratedbyGk+l' ... , Gn- 1 (k = 0, ... ,n - 1); in particular 9n-l = O.

a) For k = 1, ... ,n - 1,

b) For k = 1, ... , n - 1 and i = 1, ... , k,

) n-lh(2n-'-I) en-I· C n ') - n 1.

Proof a) and b) are proved together by induction on k. For k = 1, both state that n(9' n G1) s;;; n(91 + j). By 10. 3d), it suffices to prove that n[Gj , G1 - j ] s;;; n(91 + j) for j = 0, ... , n - 1. For j = 0 or j = 1, [Gj , G1- j ] s;;; j since Go s;;; j. If 2 ~ j ~ n - 1, then Gj s;;; 91 and G1- j = Gn+ 1 - j s;;; 91' since 2 ~ n + 1 - j ~ n - 1. Thus the result is clear.

Now suppose that 1 < k ~ n - 1. Suppose that a) is false. By 10.3d), there exist integers i, j such that 0 ~ i < n, 0 ~ j < n, i + j == k (n) and

Since Go s;;; j, i > 0 and j > o. Also either i ~ k or j ~ k; otherwise [G;, Gj ] would be contained in the algebra generated by Gk+l' ... , Gn- 1,

which is 9k. Hence i + j < k + n, but also i + j == k (n), so i + j = k. Hence 0 < i < k and 0 < j < k. Thus by the inductive hypothesis of b),

Now there exist u E 9(2k-l- 1) n Gi and v E 9(2'-1- 1) n Gj such that nk[u, v] ¢ nk(9k + j). But nk- 1v E nk- 1(9k_l + j), so there exist VI E 9k-l and V2 E j such that nk- 1(v - VI - V2) = O. By 8.3b), VI is a sum of products [Ul' ... , Uh], where h > 0, Ur E Gq, and k ~ qr < n; note that this product lies in Gq1 + ... +q. by lO.3b). Similarly by 8.3c), v2 is a sum of

products [WI' ... , WI]' where f > 0, Wr E Gp, and PI = 0; this product lies in Gl n i for some l. By substituting these sums in nk-I(v - VI - v2) = 0 and applying 10.3c), we find that nk(v - V'I - v~) = 0, where V'I is the sum of those [u I , ... , uh] for which qi + ... + qh == j (n) and v~ E Gj n j. Now

But nk[ U, v] ¢ nk(1)k + j) and nk[ u, v~] E nkj. Hence there is a product [Ul, ... , Uh] for which ql + ... + qh == j (n) and

By 8.3a), there is a permutation n of {1, ... , h} such that

Write a = [u, Ul 1t ' ... ,U(h-l)1t]' b = Uh1t' q = qh1t' Since

i + ql1t + ... + q(h-l)1t = i + (ql + ... + qh) - q

== i + j - q == k - q (n),

it follows from 1O.3b) that a E Gk - q . Also bE Gq , k :::;; q < n, and [a, b] ¢ 1)k + i· Thus a ¢ j, so k =1= q. Hence k < q < n, so a E GnH- q

and k < n + k - q < n. Thus a E 1)k and b E 1)k' a contradiction. Thus a) is proved.

Since 2k- l :::;; 2k - 1, it follows from a) that

which is the case i = k of b). Suppose now that 1 :::;; i < k. We apply

the inductive hypothesis to 1)(2k-i). To do this, note that

and, by 1O.3d),

n-l 1)(2k-1) = L (1)(2k-1) n GJ

j=O


where f)* is the subalgebra generated by Gk II f)(2k-l), ... , Gn - 1 II f)(2k-l).

All these subspaces are contained in f)k + j, except possibly Gk II f)(2k-l).

But by a), Gk II f)(2k-l) ~ f)k + j + t, where t = {bib E g, nkb = O}. Since t is an ideal of g, f)k + j + t is a subalgebra. Hence f)* + j ~ f)k + j + t and nk(f)* + D ~ nk(f)k + D. Combined with the above, this gives b).

To prove c), put k = n - 1 in b). This gives

since f)n-l = O. This is also true for i = 0, since Go ~ j. By lO.3d),

n-l n- 1h(2"-Ll) = n-l" (h(2"-1_1) G.) c n-l· n 'J n 1... 'J II I - n ). q.e.d.

i=O

10.5 Theorem. Suppose that 9 is a Lie ring and that rx is an automorphism of 9 of finite order n. If j is an ideal of 9 containing all elements of 9 which are invariant under rx,

Proof Let w be a primitive n-th root of unity in a field of characteristic o and let A = Z [ w] be the set of polynomials

where the Ci are rational integers. Thus A is a ring and A is a free Abelian group. Let g* be the Lie algebra 9 ®z A. By 8.13b), rx* = rx ® 1 is an automorphism of g*; clearly the order of rx* is n. Let j* be the ideal of g* generated by all elements of g* which are invariant under rx*. By 10.3c) and lO.4c), there exists a sub algebra f) of g* such that f) 2 ng* and

But by 8.13d), j* ~ j ® A. Thus


It follows from 8.13c) that

and hence the assertion. q.e.d.

In order to deal with the case i = 0, we need the following elementary lemma.

10.6 Lemma. Suppose that V is an additive Abelian group, that rx is an automorphism of V of finite order n and that the only element of V left fixed by rx is O.

a) rx n - 1 + ... + rx + 1 = O(cf V,8.9d)). b) Ifn is prime, a E V and na = 0, then a = O. c) If the additive order of every element of V is a power of the prime

p, then the only element of V left fixed by rx P is O.

Proof a) If a E V and b = a(rxn- 1 + ... + rx + 1), then

brx - b = a(rxn - 1) = o.

Thus b = 0, so rx n - 1 + ... + rx + 1 = O. b) Let B be the group generated by a, arx, ... , arxn- 1• Then B is a

finite Abelian group since na = O. Further rx induces an automorphism a: on B, and a: leaves fixed only the zero element of B. Decomposition of B into orbits under (a:) gives the congruence I B I == 1 (n), since n is a prime. Hence B = 0 and a = O.

c) Let C = {ala E V, arx P = a}. Then C is a subgroup of V and rx induces an automorphism IX on C of order a divisor of p. Also IX leaves fixed no non-zero elements of C. It follows from b) and the fact that the order of every element of C is a power of p that C = O. q.e.d.

10.7 Theorem (KREKNIN [1]). Suppose that the Lie ring 9 possesses an automorphism rx of order n, which leaves only the zero element of gfixed. Then 9 is soluble and g(2'-2) = O.

More precisely, let

t = {ala E g, nka = 0 for some integer k ;;::: O}.

Then t is an ideal of g, and a) g{2.-1-1) £: t; b) t(2.-J- 1) = o.

Proof It is clear that t is an ideal of g. a) We may apply 10.5 with i = O. Thus

and g(2.-1- 1) s; t. b) Let PI' ... , Pr be the distinct prime divisors of n, and write

ti = {ala E g, pfa = 0 for some k ~ O}.

By 1,13.9,

t = t1 EEl ... EEl tr •

Also each ti is an ideal of 9 and tilX = ti. Write n = Pikmi' where (Pi' mi) = 1. If lXi is the restriction of IX to ti' write f3i = IXft. By 1O.6c), the only element of ti fixed by f3i is O. Since the order of f3i divides mi, a) gives

Wmi -1

- 1) s; {ala E ti' m!a = 0 for some integer I ~ O}.

Since the order of any element of ti is a power of Pi and (Pi' mi ) = 1, m!a = 0 implies a = 0, for any a E t i . Hence Wm,-l-l) = O. Since n ~ mi, W·- 1

_ 1 ) = O. But since each ti is an ideal of g,

for any integer d; hence t(2·- 1 -1) = O. Thus g(2'-2) = (g(2'-'-1))<2'-'-1) s; t(2.-1-1) = O. q.e.d.

10.8 Lemma. Suppose that P is a prime. a) Suppose that 1 ~ r < P and that Xl' ... , Xr are not necessarily dis

tinct non-zero elements of an additive group of order p. Then I X I ~ r + 1, where

X = {x. + ... + x·IO < s < r i < ... < /' } 11 Is - - 'Is·

b) Suppose that n l' ... , np -1 are not necessarily distinct integers for which (ni' p) = 1. Given any integer n, there exist iI, ... , is (s ~ 0) such that i1 < ... < is and n == nil + ... + ni, (p).

Proof a) This is proved by induction on r. For r = 1, it is clear, since X = {a, Xl} and Xl #- 0. For r > 1, let

Y = {x· +... + x· 10 < s < r - 1 i < ... O. Since x, 1= 0, this implies that I YI = p, so Ixl = p > r. Thus IXI ~ r + 1.

b) This follows at once from the case r = p - 1 of a), applied to the additive group of integral residue classes modulo p. q.e.d.

10.9 Lemma. Suppose that i is an ideal of a Lie algebra E). Suppose that u E i and VI' ... , Vn are elements of E). lfn is a permutation of {l, ... , n},

is independent of n.

Proof It is clearly sufficient to prove this in the case when n is the transposition (i, i + 1). Put x = [u, VI' ... , Vi-I]' y = Vi' Z = Vi+I; we must show that

[x, y, Z, ... ] - [x, z, y, ... ] E [i, l)2J.

But by the Jacobi identity,

[x, y, z] - [x, z, y] = -[y, z, x] = [x, [y, z]] E [i, E)2J.

The assertion follows because [i, E)2] is an ideal of E) (8.5c)). q.e.d.

10.10 Theorem. Suppose that p is a prime and that A is a commutative ring which contains a primitive p-th root of unity w. Suppose that 9 is a Lie algebra over A and that (l is an automorphism of 9 of order p. For each integer j, write

and let E) = Go + ... + Gp _ I ' Let i be the ideal ofE) generated by Go. a) [(E)2)n, E), ... ,E)] s:::: (E)2t+1 + j for all n :? 1.

p-I

b) E)(p-I)n+2 s:::: (E)2t+t + i for all n ~ O.

c) ~~(n)+l S; ~(n) + j for all n ~ 1, where

~(n) = 1 + (p - 1) + ... + (p - 1rl.

Proof a) By 10.3d), 1)2 = LG6 (1)2 " Gd. It follows that

By W.3b), the summand displayed here is contained in (1)2t " Gil + ... +i,; hence

p-I

(1) 2t = L ((1)2t " GJ i=O

To prove a), it must therefore be shown that

where 0 ::; i < p and 0 ::; jk < p. This is clear if any jk is 0, since Go S; j. Suppose, then, that 0 < jk < p for k = 1, ... , p - 1. We show that if u E (1)2)n " Gi and VI E Gjl (l = 1, ... , p - 1), then

[u, VI' ••• , Vp - I ] E (1)2t+1 + j.

We observe first that (1)2t is an ideal of I) by 8.5c); it therefore follows from 10.9 that if n is any permutation of {1, ... , p - 1},

It therefore suffices to prove that

for some permutation n. But by 10.8b), there exist kl , ... , ks (s ::; p - 1) such that ki < ... < ks and -i == jk + ... + jk (p). Let n be a per-

l ,

mutation of {1, ... , p} such that In = kl for I = 1, ... , s. Then

i + jl" + ... + js" == 0 (p);

by W.3b),


Thus [U, VI", ... , V(P-l)"J E j, as required. b) This follows easily from a) by induction on n: for n > 0,

1)(p-l)n+2 = [1)(p-l)(n-l)+2,1), ... , 1)J ~ [(1)2)n + j,1), ... , 1)J, p-l p-l

by the inductive hypothesis. The result follows by applying a). c) This is also proved by induction on n. The case n = 1 is trivial

since ~(1) = 1. For n > 1, we apply the inductive hypothesis to 1)2. Since 1)2 = Lf:6 (1)2 (l Gk), this gives

But

1)~(n)+l = 1)2+(p-l)~(n-l) ~ (1)2)~(n-l)+l + j

by b), whence the assertion. q.e.d.

10.11 Theorem. Suppose that 9 is a Lie ring and that rx is an automorphism of 9 of prime order p. Let j be an ideal containing all elements of 9 which are invariant under rx. Then

a) (pg)~(n)+l ~ gin) + j for all n ~ 1; b) pP-l(pg)~(2P-l-l)+1 ~ pp-lj,

where ~(n) = 1 + (p - 1) + ... + (p - 1)n-l.

Proof As in lO.5, it may be assumed that g is a Lie algebra over A = Z [ w J, where w is a primitive p-th root of unity. Let 1) be the subalgebra of 9 defined in 10.3; thus 1) ;;;2 pg. By lO.lOc),

1)~(n)+ 1 ~ 1)(n) + j

for all n ~ 1, from which a) follows at once. It also follows from this that

But by lO.4c),

p-lJ..(2 P- 1-l) C p-l' P 'J - P ).

q.e.d.

§ lO. Regular Automorphisms 361

10.12 Theorem (G. HIGMAN [1], KREKNIN and KOSTRIKIN [1]). Suppose that the Lie ring 9 possesses an automorphism rt. of prime order p, which leaves only the zero element of 9 fixed. Then 9 is nilpotent of class at most ~(2p-1 - 1). If g(d) = 0, the class of 9 is at most ~(d). Here ~(n) = 1 + (p - 1) + ... + (p - 1)"-1.

Proof By 10.11 b),

It follows from 10.6b) that

so 9 is nilpotent of class at most ~(2P-1 - 1). If g(d) = 0, then by 10.11a),

(pg)~(d)+ 1 = O.

Again the result follows from 10.6b). q.e.d.

10.13 Remarks. a) MEIXNER [1] has shown that in the previous theorems, ~(n) can be replaced by (p - 1)"-1. To do this, the essential point is to replace 10.10b) by £)(p-1)n+1 S (£)2)"+1 + j.

b) As we see from the case p = 3, the bound on the class given in 10.12 is far from being sharp. HIGMAN [1] showed that the bound is at least i(p2 - 1) for all p ~ 5; this has been shown to be sharp for p = 5 by Higman and for p = 7 by Scimemi.

We shall now use 9.3 and 9.4 to apply 10.11 and 10.12 to groups. For this the following lemma is needed.

10.14 Lemma. a) Suppose that rt. is an automorphism of a finite group (fi and that 91 is a normal subgroup of (fi for which 91rt. = 91. Let a be the automorphism of (fi/91 induced by rt.. If [ = {xix E (fi, xrt. = x} and [ = {xix E (fi/91, xiX = x}, then 1[1 ~ I [ I·

b) (V, 8.10) Suppose that rt. is an automorphism of a group (fi, which leaves no non-identity element of (fi fixed. If 91 is a normal subgroup of (fi for which 91rt. = 91, the automorphism of (fi/91 induced by rt. leaves fixed no non-identity element of (fi /91.

Proof a) Suppose that (£: = Xl/ill If x E Xl, put xp = x- 1(xrt.); thus P is a mapping of Xl into in. Now p carries every element of the coset


(£:x into x-1(xC(). Conversely, suppose that xfJ = yfJ. Then x-1(xC() = y-1(yC(), so yx-1 = (yx- 1 )C(, yx-1 E (£: and y E (£:x. Hence lim fJl = l'!l: (£:1. SinceimfJ s; ~,I'!l:(£:1 ~ 1~I,andl(£:1 = 1'!l:~1 ~ 1(£:1.

b) This follows at once from a). q.e.d.

10.15 Theorem. Suppose that GJ is afinite group and that GJ has an automorphism C( of prime order p, which leaves no non-identity element of m fixed. Then GJ is nilpotent. Let c be the class of GJ.

a) If p = 2, c ~ 1. If P = 3, c ~ 2. In general, c ~ ~(2p-1 - 1), where ~(n) = 1 + (p - 1) + ... + (p - 1r1.

b) IfGJ(d) = 1, c ~ ~(d).

Proof By Y, 8.14, GJ is nilpotent. Let c be the class of GJ. By 9.3, there exists a Lie ring 9 = EEl i"l Gi and isomorphisms (1i of

yJGJ)/Yi+l(GJ) onto Gi (i ~ 1) such that

for all x E Yi(GJ), y E YiGJ); further 9 possesses an automorphism p for which

By 1O.14b), C( induces an automorphism on GJ/Yi+l(GJ) which leaves no non-identity element fixed. Thus p leaves no non-zero element of Gi

fixed. Hence p leaves no non-zero element of 9 fixed. Also the order of pIS p.

By 9.4, 9 is nilpotent of class c. The assertions in a) about c now follow from 10.1, 10.2, 10.6 and 10.12. If GJ(d) = 1, then g(d) = 0 by 9.4d), so by 10.12, c ~ ~(d). q.e.d.

10.16 Remarks. a) The assertion of 10.15 for p = 2 can, of course, be proved much more easily (see Y, 8.18). But it is not true for infinite groups: if!j is a free group with group-basis {x, y} and C( is the automorphism of !j such that xC( = y, YC( = x, then C( leaves only the identity element fixed.

b) A direct proof of the assertions for p = 3 is given in BURNSIDE

([1], p. 90). The analogue of this proof for Lie rings is given in Exercises 15, 16.

c) Theorem 10.15 remains valid if the word "finite" is replaced by "nilpotent" ( G. HIGMAN [1]).

d) In analogy with 10.7, it is conjectured that if the finite group GJ

possesses an automorphism a which leaves only the identity element fixed, then (fj is soluble. This is true if the order n of a is a prime (by V, 8.14) or a power of 2 (by the Feit-Thompson theorem). An alternative proof in the case n = 4 was given by GORENSTEIN and HERSTEIN [1] (see GORENSTEIN [1],10.4; also IX, 6.8).

It is also conjectured that if the finite group (fj possesses a group m: of automorphisms such that (I (fjl, 1m:1) = 1 and only the identity element of (fj is left fixed by every element of m:, then (fj is soluble. This will be discussed in Chapter X.

An application will now be given in which the number of fixed points is greater than 1. For this we need the following results.

10.17 Lemma. Let p be a prime. Suppose that V is a finite Abelian pgroup, that a is an automorphism of V of order p, and that the group of elements of V left fixed by a is of order pn. Then V can be generated by pn elements.

Proof. We write V in the additive notation. If B is a subgroup of V invariant under a, then B(a - 1) is contained in B and is isomorphic to B/(B n ker(a - 1)), so I B : B(a - 1)1 ::; Iker(a - 1)1 = pn. It follows that IV: V(a - l)pl ::; pnp. But

(a - l)P = aP - pa P- 1 + ... + (-1)P.

Hence

V(a - l)P <;; V(a P - 1) + pV = pV,

since p is the order of a. Thus IV: pVI ::; pnp. By III, 3.14, pV is the Frattini subgroup of V. Hence by III, 3.15, V can be generated by pn elements. q.e.d.

The following result may be compared with III, 7.10.

10.18 Theorem (P. HALL [1]). Suppose that 91 is a normal subgroup of the p-group (fj and that 91 ::; Yn(fj). If i > 1 and Yi+1 (91) oF 1, then IYi(91): Yi+1 (91)1 ~ pn.

Proof. Since Yi+1(91) oF 1, there exists a normal subgroup 9Jl of (}j such that 9Jl < Yi+1 (91) and IYi+1 (91): 9)11 = p. Thus [Yi(91),91] $. 9)1. It


follows since 9l ::;; Yn«fj) that [Yi(9l), Yn«fj)J f,. 9Jl. Let 3/9Jl be the n-th term of the upper central series of (fj/9Jl; thus by III, 2.11c), [3, Yn«fj)] ::;; 9Jl. Hence Yi(9l) 1, 3. By III,7.2b), IYi(9l): 9Jl1 ~ pn+1. Thus IYi(9l): Yi+1(9l)1 ~ pn. q.e.d.

10.19 Theorem (ALPERIN [IJ). For each prime p and each non-negative integer n, there exists an integer k(p, n) having the following property. If (fj is a finite p-group, IY. is an automorphism of (fj of order p and the group of elements of (fj left fixed by IY. is of order pn, then (fj(k(p,n)) = 1.

Proof. Let q = ~(2p-1 - 1), where ~(n) = 1 + (p - 1) + ... + (p - 1)n-1. Let d be the least integer such that 2d > (n + p + q)pn and put k(p, n) = d + q + 2.

Suppose that IY. is an automorphism of order p of the finite p-group (fj and that the group of elements of (fj left fixed by IY. is of order pn. Let f) = (fj(d). By 9.3, there exist a Lie ring I) = EBi~1 Hi and isomorphisms (Ii of Yi(f)/Yi+1 (f) onto Hi; also the restriction of IY. to f) induces an automorphism P of f) such that (JiP(Ji-: 1 is the automorphism of Yi(f)/Yi+1 (f) induced by IY.. By 10.14a), the group of elements of Hi left fixed by P is of order at most pn. It follows that if a E f) and ap = a, then pna = O. Let i = {b E I), pnb = O}. Then i is an ideal containing all elements of l) which are invariant under p. Hence by 10.llb),

Hence pn+ p-1+(q+1)l)q+1 S; pn+ p-1j = O. But by 9.4,

I)q+1 = Hq+1 + Hq+2 + ... , so pn+ p+qHq+1 = O. By 10.17, Hq+1 can be generated by np elements. Hence

But by II1,2.12a), f) = (fj(d) ::;; Y2,«fj). Hence by 10.18, Yq+2(f) = 1. Hence (fj(k(P.n)) = f)(q+2) ::;; Yq+2(f) = 1. q.e.d.

It is clear that the estimate given in the proof of 10.19 is in general a very crude one.


We conclude this section by showing that 10.19 is valid also for soluble groups.

10.20 Theorem (ALPERIN [1]). For each prime p and each non-negative integer n, there exists an integer l(p, n) having the following property. If (f) is a finite soluble group, r:t. is an automorphism of (f) of order p, and the group of elements of (f) left fixed by r:t. is of order pn, then (f)(I(P.n)) = 1.

Proof Let IDl = Op,(f)), 91 = 0p',p(f)). Thus IDlr:t. = IDl, 91r:t. = 91, and if 6 = 91/IDl, r:t. induces an automorphism a on the p-group 6. By 10.14 a), the automorphism of 6/c1>(6) induced by a leaves at most pn elements fixed. Thus if O:/IDl = c1>(6), 91/0: is an elementary Abelian p-group of order at most ppn, by 10.17. By VI, 6.5, 91/0: is self-centralizing; hence there exists a monomorphism of (f)/91 into the group of automorphisms of 91/0:. Thus (f)/91 is a soluble subgroup of GL(r, p), where r ::; pn. There is therefore a bound on the derived length of (f)/91, which depends only on p and n. But by 10.19, the same is true of the p-group 91/IDl, for r:t. induces on 91/IDl an automorphism of order a divisor of p and this automorphism leaves at most pn elements fixed, by 10. 14a). Finally, since the group of elements left fixed by r:t. is a p-group and IDl is a pi-group, r:t. leaves no non-identity element of IDl fixed. Hence by 10.15, there is a bound on the derived length of IDl which depends only on p. Combining these statements, there is a bound on the derived length of (fj which depends only on p and n. q.e.d.

Exercises

15) Suppose that r:t. is an automorphism of a Lie ring g. For any elements a, b of g, write

f(a) = [a,ar:t.], g(a,b) = [ar:t.,b] + [br:t., a].

Verify the identity

g(a, [br:t. - b, a]) - [J(a), br:t. - b] - 3[g(a, br:t.), a]

= [b(r:t. 2 + r:t. + l),ar:t.,a] - 3[br:t.2 ,a,a].

16) Suppose that the Lie ring 9 possesses an automorphism r:t. of order 3, which leaves only the zero element fixed. Prove (using 1O.6a)) that f(a) = g(a, b) = 0 for all a, b in g. Deduce that 9 is nilpotent of class at most 2.


17) (G. HIGMAN) If the (associative) ring 'H has an automorphism a of prime order p and a leaves no non-zero element of the ideal :5 fixed, then :5P = 0. (Extend the ground-ring to Z[ w J, where w is a primitive p-th root of unity. Show as in the Lie algebra case that if:5P t= 0, there is a product Xl ... xp t= ° with Xia = Wk'Xi' There exist r, s such that kr + ... + ks == ° (p).)

18) Let (j be a linear transformation of a vector space V over a field K. Show that V EB K is a Lie algebra, if the product is defined by

19) (KREKNIN and KOSTRIKIN [1J) Let n be a composite integer. There exists a Lie algebra 9 which is not nilpotent, such that 9 has an automorphism a of order n, which leaves no non-zero element fixed. (If n = 1m, I > 1, m > 1 and V is a vector space with basis {YI' ... , Ym} over the algebraically closed field K of characteristic 0, construct 9 = V EB K as in 18) with Yi(j = Yi+l' Ym(j = YI' Let (Yi + A)a = WI+(i-I)IYi + AWl, where w is a primitive n-th root of unity.)

20) (KREKNIN and KOSTRIKIN [lJ) Let n be a composite integer. Given any integer c, there exists a nilpotent Lie algebra 9 of class c and an automorphism a of 9 of order n, such that a leaves no non-zero element of 9 fixed. (Proceed as in 19), with V having a basis {yll1 :::;; j :::;; c, 1 :::;; i :::;; m} and Y/ (j = y/:l, y~(j = yr 1, Yic (j = 0. Then gr is spanned by all Y/ withj Z r. Let (y/ + A)a = W(i-l)l+! Y/ + ).w l.)

§ 11. The Lower Central Series of Free Groups

The Lie ring method was originally developed for its use in a theory, due mainly to Magnus and Witt, which gives some insight into the structure of the lower central series of a free group. This theory will be described in this section. It uses the fact that a free associative algebra can be embedded in the group-ring of a free group, and so it is necessary to study the Lie structure of the free associative algebra. The basic result used to do this is the Birkhoff-Witt theorem (11.2).

In order to unify the terminology, we shall use the word "basis" in connection with free groups and algebras. Thus a group-basis of a free group IY is a subset X of IY such that X generates IY and any mapping of X into a group is the restriction of some homomorphism of IY.

§ 11. The Lower Central Series of Free Groups 367

We recall that with each associative algebra 21 there is associated a Lie algebra 1(21) (see 8.2).

11.1 Lemma. Let 9 be a Lie algebra over a commutative ring A and let 21 be an associative A-algebra. Suppose that a is a Lie homomorphism of 9 into 1(21) and that 21 is the associative algebra generated by im a. For g E g, write g = ga. Let B be a fully ordered set such that 9 is the Amodule spanned by B.

a) For each integer n 2 0, let Vn be the A-module spanned by all products gl ... gm with gi E 9 and m ~ n. (In particular, Vo = A1.) Given gi E 9 (i = 1, ... , n; n > 0), then

for any permutation n of {I, ... , n}. Also for n > 0, Vn is the A-module spanned by Vn - 1 and all products b1 ••• bn , where bi E Band b1 ~ ••• ~ bn •

b) 21 is the A-module spanned by all products b1 ... bn, where n 2 0, bi E Band b1 ~ .•• ~ bn •

c) If 9 is the Lie algebra generated by X, 21 is the associative algebra generated by X a.

Proof a) To prove the first assertion, it is clearly sufficient to show that

Butgi+lgi - gigi+l = [gi+l a,gia] = [gi+1,g;]aEV1,sothisisciear. Since every element of 9 is an A-linear combination of elements of

B, Vn is spanned by Vn- 1 and all products b1 ... bn with bi E B. By the previous statement, we may add the condition b1 ~ b2 ~ ..• ~ bn-

b) Since 21 is the associative algebra generated by all g (g E g), 21 is the A.module spanned by all Vn (n 2 0). The assertion thus follows from a).

c) If 21* is the associative algebra generated by X a, then im a <;; 21*, so all b1 ... bn E 21*. Thus 21* = 21 by b). q.e.d.

11.2 Theorem (BIRKHOFF-WITT). Let 9 be a Lie algebra over a commutative ring A. Suppose that 9 has an A-basis B and that B is fully ordered. Then there exists an associative algebra 21 over A and a monomorphism e of 9 into 1(21) such that

B* = {(b 1 e) ... (bne) I n ~ 0, bi E B, b1 ::; ... ::; bn}

is an A-basis of m.

Proof Let ~ be the A-algebra of polynomials in a set B of independent indeterminates for which there exists a bijective mapping b ...... b of B onto B. We assert that it is sufficient to define an A-bilinear mapping ( , ) on ~ x 9 into ~ such that

a) (b1 ... bn, b) = b1 ... bnb if b1 ::; ... ::; bn ::; b (b i E B, b E B), and b) (J, [b, b']) = ((J, b), b') - ((J, b'), b) (f E ~, b E B, b' E B). For suppose that such a mapping exists. Given g E g, f ...... (J, g) is

an element ge of HomA(~' ~); thus f(ge) = (J, g). Now e is a Lie homomorphism of 9 into I(HomA(~' ~», for

f([gl' g2]e) = (J, [gl, g2])

= ((J, gl)' g2) - ((J, g2)' gl) (by b»

= f((gl e)(g2e) - (g2 e)(gl e»

= f[gl e, g2e].

By 11.1, the associative algebra m generated by ime is spanned over A by all (b 1 e) ... (bne) with bi E Band b1 ::; ... ::; bn. In fact, these elements form an A-basis of m, for the elements b1 ... bn with b1 ::; ... ::; bn are linearly independent and

1(b e)· .. (b e) = b ... b 1 n 1 n'

as is easily seen from a) by induction on n. Finally, e is a monomorphism, since the l(be) = b are certainly linearly independent.

For each n ~ 0, let On denote the A-module spanned by all b1 ... bm

with rn ::; n. We shall define a bilinear mapping ( , )n on On X 9 into ~ by induction on n, and it will satisfy the following.

a) (b ... b b) = b ... b b if rn < nand b < ... < b < b. n 1 m' n 1 m - 1 - - m-bn) Hrn ::; n andfE am, (J, b)n - fb E am; hence (J, b)n E 0m+1' cn) HfE 0n-1' (J, [b, b'])n = ((J, b)n' b')n - ((J, b')n' b)n· dn) On 0n-1 X g, ( , )n coincides with ( , )n-1 .

It is clearly sufficient to do this, for then by dn), a form may be defined on ~ x 9 which satisfies a) and b) on account of an) and cn).

For n = 0, put (J., b)o = J.b (J. E A). Then ao), bo) are satisfied trivially, co), do) vacuously.

For n > 0, we define (J, g)n for f E 0n-1 and f = b 1 •.. bn; the definition is then completed by linearity. Forf E 0n-1, put (J, g)n = (J, g)n-1'

If b <... < b < b put (h ... h b) = h ... h h. If b <... < b 1 - - n -'In' n 1 n 1 - - n and b < bn, then by bn- l ), U = (hI' .. hn- l , b)n-l - hI ... hn- lh EOn-I,

and we may put

Then ( , )n is bilinear and an), bn), dn) are satisfied trivially. It only remains to verify cn) with I = hI ... bn- l . We may suppose that b < b' andb l ~ '" ~ bn - l ·

We prove it first when bn - l ~ b'. The definition then gives

((hI'" hn- l , b')n, b)n = (hI'" hn- lh', b)n

= hI ... bn- lbb' + (hI'" hn- 1 , [b', b])n-l

where Ul = (hI' .. hn- l , b)n EOn' U2 = hI ... hn- lh E On and Ul - U2 E On-I' Hence (U2' b')n = hI ... hn_1hb' and

((hI'" hn- l , b')n, b)n = hI ... hn_1hh' + (hI'" hn- l , [b', b])n-l

+ (ul,b')n - (u2,b')n

= ((hI' .. bn- 1 , b)n, b')n - (hI' .. bn- 1, [b, b'])n-1,

as required. Now suppose that b < b' < bn- l . Then n > 1. We put

I = bl ... bn- 2, a = bn- l and c = b'. We then have IE 0n-2, b < a and c < a, and we must prove that

(fa, b)n = ((f, a)n-l' b)n-l

= ((f, b)n-l' a)n-1 + (f, [a, b ])n-1

= (v + Ib,a)n-l + (f,[a,b])n-1'

where v = (f, b)n-l - Ih E ~-2' Hence


We apply Cn-t) to the first and third terms, and we apply the case of cn) already proved to the second. Thus

((fa, b)n' c)n = ((v, c)n, a)n + (v, [a, c])n + ((fb, c)n, a)n

+ (fb, [a, c])n + ((J, c)n' [a, b])n + (J, [a, b, c])n"

Since v + /5 = (J, b)n-t,

((fa, b)n, c)n = (((J, b)n' c)n, a)n + ((J, b)n, [a, c])n

+ ((J, c)n, [a, b])n + (J, [a, b, c])n"

Next we interchange band C and subtract. This gives

((fa, b)n, c)n - ((fa, c)n, b)n

= (((J, b)n' c)n, a)n - (((J, c)n, b)n' a)n + (J, [a, b, c] - [a, c, b])n·

But by two applications of Cn-t),

(((J, b)n, c)n, a)n - (((J, c)n, b)n, a)n = ((J, [b, C ])n, a)n-t

= ((J, a)n, [b, c])n + (J, [b, c, a])n·

Hence

((fa, b)n' c)n - ((fa, c)n, b)n

= ((J, a)n, [b, c])n + (J, [b, c, a] + [c, a, b] + [a, b, c])n

= ((J, a)n, [b, c])n = (fa, [b, c])n,

by the Jacobi identity. q.e.d.

Theorem 11.2 is not true if it is not assumed that 9 is a free A-module. A counterexample was given by CARTIER [1].

11.3 Definitions. a) Let m: be an associative algebra over A. The subset X of m: is called an associative basis of m: if (i) m: is the associative algebra generated by X, and (ii) any mapping of X into any associative algebra over A is the restriction of a homomorphism of m:. An associative algebra which possesses an associative basis is called a free associative algebra.

b) Let 9 be a Lie algebra over A. The subset X of 9 is called a Lie basis of 9 if (i) 9 is the Lie algebra generated by X, and (ii) any mapping


of X into any Lie algebra over A is the restriction of a homomorphism of g. A Lie algebra which possesses a Lie basis is called a free Lie algebra.

11.4 Lemma. a) Let ~, Wl be Lie or associative algebras over A and let p be a homomorphism of ~ into Wl. Suppose that X generates ~, that p is injective on X and that X p is a Lie or associative basis of Wl. Then p is an isomorphism of ~ onto Wl.

b) Let ~, Wl be free Lie or associative algebras over A. Suppose that X is a basis of~, Y is a basis ofWl and that a is an injective mapping of X onto Y. Then a is the restriction of an isomorphism of ~ onto Wl.

Proof a) Since X p is a basis of Wl, X p generates Wl. Hence p is an epimorphism. Since X p is a basis of Wl and p is injective on X, there exists a homomorphism a of Wl into ~ such that pa is the identity mapping on X. Since X generates ~, it follows that pa = 1. Hence p is also a monomorphism.

b) Since X is a basis of ~, a is the restriction of a homomorphism p of ~ into Wl. By a), p is an isomorphism. q.e.d.

11.5 Remarks. a) Given a set X, let ~ be a free A-module with basis {(Xl' ... , xn)ln ~ 0, Xi E X}. It is easy to see that ~ has the structure of an associative algebra in which (Xl"'" Xm)(Xm+l, ... , Xn) = (Xl' ... , xn)and that ~ is a free associative algebra with basis {(x)lx EX}. Thus there exists a free associative algebra having a basis with given cardinal number. By 11.4b), it is unique to within isomorphism.

b) The proof of the existence offree Lie algebras is not quite so simple. In fact, however, if m is a free associative algebra over A with basis X, the Lie subalgebra f of I(m) generated by X is a free Lie algebra with basis X. This is proved in 11.10 for the case A = lL. In the case when A is a field K, it can be deduced from the Birkhoff-Witt theorem as follows.

Suppose that a is a mapping of X into a Lie algebra g. Since K is a field 9 has a K-basis. Hence by 11.2, there is a monomorphism f. of 9 into I(~) for some associative algebra ~. By l1.3a), af. is the restriction to X of an associative homomorphism p of m into ~.

)7L J : L

Now im t: is a Lie subalgebra of I(~) containing X at: = X p. Since f is the Lie algebra generated by X, it follows that imt: :2 fp. Since t: is a monomorphism, there exists a mapping /3 of f into 9 such that /3t: = p. /3 is a Lie homomorphism, since if a, b are in f,

([a,b]/3 - [a/3,b/3])t: = [a,b]p - [ap,bp] = O.

And the restriction of /3 to X is a, since if x E X, x/3t: = xp = xat:. c) Let 9 be a Lie algebra over A. By a), there exists a free associative

algebra ~ having a basis X for which there exists an injective mapping e of 9 onto X. Let 3 be the ideal of ~ generated by all [g, h]~ - [g~, h~], as g, h run through g. The algebra <f(g) = ~/3 is called the universal enveloping algebra of g. Let a be the mapping of 9 into <f(g) given by ga = g~ + 3; thus a is a Lie homomorphism of 9 into I(<f(g)). <f(g) is characterized by the following properties.

(i) <f(g) is generated by im a. (ii) If m is an associative algebra and /3 is a Lie homomorphism of

9 into I(m), /3 = a'1 for some associative homomorphism '1 of <f(g) into m.

The Birkhoff-Witt theorem asserts that if B is a fully ordered A-basis of g,

is an A-basis of <f(g).

11.6 Definition. Let m be a free associative algebra over A with basis X. Let Ho = AI, and for each integer n > 0, let Hn be the A-module generated by all products Xl ... Xn with Xi E X. Hn is called the homogeneous component of m with respect to X of degree n. The Hn have the following properties.

(i) Hn is a free A-module with basis {Xl' .. XnlXi EX}. (ii) m = HO Ee Hl Ee H2 Ee .. '.

(iii) If a E Hm and b E Hn, then ab E Hm+n. If further A is an integral domain and a '# 0, b '# 0, then ab '# O.

Of these, (iii) is an immediate consequence of (i) and (ii). To prove these, observe that they hold trivially in the algebra ~ constructed in 11.5a) and that. by 11.4b), there is an isomorphism between ~ and m in which (x) and X correspond (x EX).

11.7 Theorem. a) Given a set X, there exists a free Lie algebra over A with a Lie basis in (1, 1) correspondence with X.


b) Let 9 be afree Lie algebra with Lie basis X. For each n ~ 1, let Un be the A-module spanned by all products [a1' ... , an] with ai E X. Then gn = Un EEl Un+1 EEl ... and nn;;'l gn = O.

Proof By 11.4b), it is sufficient to prove b) for a free Lie algebra with a basis in (1, 1) correspondence with X. Both parts of the theorem will therefore be proved if we (i) establish the existence of a free Lie algebra f having a Lie basis X in (1,1) correspondence with X, and (ii) prove that f and X satisfy the assertion of b).

First we construct the non-associative words in X. For n ~ 1, we define a set X n by induction on n, with Xl = X and

n-1 Xn = U Xi X Xn- i

i=l

for n > 1. Thus Xm n Xn = 0 for m # n. Let Mn be the free A-module with basis Xn and let

Wl is thus the free A-module with basis X = Un;;'l Xn, and if u E X, V E X, the ordered pair (u, v) also lies in X. We define multiplication in Wl by putting

Obviously the distribution laws hold. We define a set!/' of A-submodules ofWl as follows: the A-submodule

N lies in !/' if (iii) 17 (a) = a'aand((a,b,c) = ((a'b)'c) + ((b·c)·a) + ((c·a)·b) lie

in N for all a, b, c in Wl; and (iv) a . bEN and b . a E N if a E Nand b E Wl. !/' is non-empty since Wl E !/'. Let 3 = nNe9' N. Then 3 also satisfies

conditions (iii) and (iv). From (iv) it follows that there exists a multiplication in the module Wl/3 in which

[a + 3, b + 3] = a' b + 3.

The distributive laws hold, and by (iii), Wl/3 is a Lie algebra f. Define a mapping ~ of X into f by putting x~ = x + 3 (x E X). We

show that X ~ generates f. Certainly the subalgebra of f generated by X ~ is an A-submodule of Wl/3 and is thus of the form U/3 for some A-


submodule U of9Jl.1f U f= 9Jl, there exists an integer n such that Xn ¢. U. Let n be the smallest such integer. Then n f= 1, for if x E X, then x~ E U 13 and x E U. Suppose that wE X n , W ¢ U. Since n > 1, W = (u, v) for u E Xi' V E Xn- i (1 ~ i ~ n - 1). Then u E U, V E U, so u + 3 E U/3, v + 3 E U 13, and since U 13 is a subalgebra, [u + 3, v + 3] E U 13. Thus U'V E U, and since U' v = (u, v) = w, we have a contradiction. Hence U = 9Jl and f is generated by X~.

We now show that given a mapping f3 of X into a Lie algebra 9 over A, there exists a homomorphism (J of f into 9 such that f3 = ~(J.

To do this we first define a mapping f3n of Xn into 9 by induction on n, with f3I = f3 and (u, v)f3n = [Uf3i' vf3n-J if U E Xi' V E Xn- i. There exists an A-module homomorphism p of 9Jl into 9 such that up = uf3n for all U E Xn' Thus (u' v)p = [up, vp] if U E Xm, V E Xn, and, indeed, if u, v are any elements of 9Jl. It follows easily that ker p E f/. Thus ker p :2 3 and there exists an A-homomorphism (J of 9Jl/3 = f into 9 such that (a + 3)(J = ap for all a E 9Jl. It is easy to see that (J is a Lie algebra homomorphism and that x~(J = xp = xf3 for all x E X.

By choosing f3 to be injective, we see that ~ is necessarily injective. Thus if X = X~, X is a Lie basis off in (1, 1) correspondence with X. Hence (i) is proved.

To prove (ii) let I n = 3 n Mn and let

Thus J is an A-submodule of 3, and J satisfies (iv), for if a E Jm and b E Mn, then a' b E 3 n Mm+n = Jm+n and b· a E Jm+n. Also J satisfies (iii). Indeed ((a, b, c) is a linear combination of elements ((a', b', c') with a' E MI, b' E Mm, c' E Mn, and ((a', b', c') E MI+m+n n 3. As for ,,(a), ,,(a) is a linear combination of elements ,,(a') and 1f(a', b') = [a', b'] + [b', a'] with a' E MI, b' E Mm. Now ,,(a') E 3 n M21 . Also 1f(a', b') E Ml+m' and

1f(a', b') = " (a' + b') - ,,(a') - ,,(b') E 3,

so 1f(a', b') E 3 n Ml+m' Thus J satisfies (iii). Hence J :2 3. Since obviously J ~ 3, we have J = 3 and

It follows at once that

f = 9Jl/3 = (Ml + 3/3) Ef> (M2 + 3/3) Ef> ....

Now let Vn be the A-module spanned by all products [bl, ... , bn] with bi E X. By a trivial induction, [bl' ... , bn] = u + 3 for some u E Xn , so Vn ~ (Mn + 3)/3. By 8.3b), f = VI + V2 + .... Hence f = VI EB V2 EB .... By 8.7b), fn = Vn EB Vn+1 EB .... It follows at once that nn;>:1 fn = 0, so (ii) is proved. q.e.d.

11.8 Theorem. Let 21 be afree associative algebra over Z with associative basis X and let 9 be the Lie subalgebra of1(m) generated by X. For n ~ 0, let Hn be the homogeneous component of 21 with respect to X of degree n (11.6), and let Gn = 9 n H ..

a) 9 = GI EB G2 EB ... , and Gn is the additive group generated by all products [x I' ... , xn] with Xi E X.

b) Gn and Hn/Gn are free Abelian groups. c) If X is finite, Gn is a free Abelian group of rank dn(IXI). Here

the integer dn(q) is defined for each positive integer q by induction on n by Witt's formula

Proof By 8.3b), 9 is the additive group generated by all [x I, ... , xnJ with n ~ 1 and Xi E X. Since [XI' ... , xnJ E Hn n 9 = Gn,

9 = GI + G2 + .... Since Gn ~ Hn and 21 = Ho EB HI EB H2 EB ... , a) follows at once.

By 11.6, Hn is a free Abelian group. Since every subgroup of a free Abelian group is free (cf. I, 13.4), Gn is a free Abelian group. Let Bn be a Z-basis of Gn and let B = Un>O Bn. By a), B is a Z-basis of g. We suppose B to be fully ordered in such a way that b l < b2 whenever bl E Hm, b2 E Hn and m < n. Since 21 is the associative algebra generated by g, it follows from 11.1 that 21 is the additive group generated by

However by 11.2 there is an associative algebra (£: and a monomorphism c; of 9 into 1((£:) such that

is linearly independent over Z. Since X is an associative basis of 21, there is a homomorphism (1 of 21 into (£: such that X(1 = xc; for all X E X. Hence if Xi E X (i = 1, ... , n),


It follows from 8.3b) that aa = at; for all a E g. Thus (b 1 ... bn)a = (b 1 t;) ... (bnt;), so B* is linearly independent over 7L. Hence B* is a 7Lbasis ofm.

Now each element of B, and hence each element of B*, lies in some Hn' It follows, since m = Ho E9 H1 EB ... , that B: = B* n Hn is a 7Lbasis of Hn (n ~ 0). Since Bn ~ B:, Hn = Gn EB Kn, where Kn is the additive group generated by B: - Bn. Thus HnlGn is a free Abelian group, and b) is proved.

Suppose that Ixl = q. Thus by 11.6, Hn is a free Abelian group of rank qn. Hence I B: I = qn. But

so I B: I is the coefficient of t n in the formal power series

= [1 (1 + tr + t2r + ... )d" r2:1

where dr = I BrI- Hence

[1(1 + tr + t2r + ···t = 2:qntn, rZl n

and

1 1 Il (1 - tr)d, 1 - qt'

Taking logarithms,

00 00 d trs 00 qntn 2:2:-r =2:-. r=l s=l S n=l n

Hence

d qn = n 2: ~ = 2:rdr . rs=n S rln

Hence dr = dr(q). Since dr = I Brl is the rank of G" c) is proved. q.e.d.


11.9 Lemma. Suppose that U is a finitely generated additive (Abelian) group but that U is not free. Then there exist primes p, q such that IU/pUI = pm, IU/qUI = qn and m =I n.

Proof By I, 13.12, U is the direct sum of a finite number of cyclic groups Ul , ... , Ur. Since U is not free, at least one of these cyclic groups is finite, say Ul . If P is a prime divisor of I Ull and q is a prime for which U has no element of order q, m - n is the number of Ui for which p divides I Uil, so m - n > O. q.e.d.

11.10 Theorem. Let m be afree associative algebra over Z with associative basis X, and let 9 be the Lie subalgebra of I(m) generated by X. Then 9 is afree Lie algebra with Lie basis X.

Proof Suppose this is false. By 11. 7a), there exists a free Lie algebra f 1

having a Lie basis Yl in (1, 1) correspondence with X. Hence there is a Lie epimorphism p of f 1 onto 9 such that the restriction of p to Yl is a bijection onto X. Since the theorem is false, ker p =I o. Suppose u E f l' U =I 0, up = O. By 8.3, u lies in the Lie algebra f generated by a finite subset Y of Yl . It is clear from the definition (l1.3b)) that f is a free Lie algebra with basis Y. Let T = Yp. If m is the associative subalgebra of m generated by T, m is free with basis T.

Suppose that f has a Z-basis. By 11.2, there exists an associative algebra \i: and a monomorphism e of f into I (\i:). If x E T, x = yp for a unique Y E Y. Thus there is an associative homomorphism rJ. of minto \i: such that XrJ. = ye and prJ. = e on f. But then Ue = uprJ. = O. Since e is a monomorphism, this implies u = 0, a contradiction. Hence f has no Z-basis.

For each n ~ 1, let Un be the additive group spanned by all products [Yl' ... , Yn] with Yi E Y. By 11.7b),

Hence Un has no Z-basis for some n. By 11.9, there exist primes Pl' P2 such that IUn/PiUnl = P;"; with ml =I m2 • We shall obtain a contradiction by proving that for any prime p, I Un/pUn I = pd,(q), where q = I YI·


Let K = GF(p). Then f/pf can be regarded as a Lie algebra over K. Now m/pm is a free associative algebra over K with basis Xp = {x + pmlx E T}. Hence by l1.5b), the Lie subalgebra fp of I(m/pm) generated by Xp is a free Lie algebra with basis Xp. But since f is a free Lie ring, there is a homomorphism a of f into fp such that ya = yp + pm for all y E Y. Clearly pf s;;; ker a, so a induces a Khomomorphism (f of f/pf into fp such that (y + pf)(f = yp + pm. By 11.4a), (f is an isomorphism off / pf onto f 11' Thus (Un + pf)/ pf is isomorphic to the additive subgroup generated by all [x I' ... ,xn] + pm with Xi E T. Hence if Gn is the additive group generated by all [x 1, ... , Xn], Un/pUn is isomorphic to (Gn + pm)/pm. By 11.8a), Gn is a direct summand of m, so Gn n pm = pG n• Hence Un/pUn ~ GjpG n• But by 11.8, Gn is a free Abelian group of rank dn(q), so I Un/pUn I = pd,(q). q.e.d.

Next we consider the augmentation ideal of the group-ring of a free group.

11.11 Theorem. Let tY be a free group with basis X and let 3 be the augmentation ideal (see 2.2) of AtY.

a) For each integer n > 0, 3n is a free AtY-module with basis the set Bn of all products (Xl - 1)· .. (xn - 1) with Xi E X.

b) nn>o 3n = 0.

Proof a) This is proved by induction on n. The case n = 1 is III, 18.5. For n > 1, observe that 3 n - l is the AtY-module generated by all (Xl - 1)· .. (xn- l - 1), by the inductive hypothesis. Hence 3 n- l is the A-module generated by all (Xl - 1)··· (xn- l - l)a with a E AtY. By definition (2.2), 3 n = 3 n - 13 is the additive group generated by all (Xl - 1)·· . (xn- l - l)b with b E 3. From the case n = 1,3 is the AtYmodule generated by all x-I with X E X. Hence 3 n is the AtY-module generated by Bn' If

"(X - 1)··· (x - l)b = ° L... 1 n Xt, ••• ,x" Xi

with bxl, ... ,x, E AtY, it follows from the fact that Bl is an AtY-basis of 3 that for each Xl EX,

Hence by the inductive hypothesis, all bx x are zero. 1,···, "

b) (R. H. FOX [1]). For f E (j, let l(f) denote the smallest integer I for which there exists a sequence (x11, ... , x;') with Xj EX, Sj = ± 1 and f = X11 ... xt'· For u E A(j, u =F o and u = LJelj AJ f (AJ E A), let I(u) denote the greatest integer I for which there exists f E (j such that AJ =F 0 and l(f) = I. Finally, put 1(0) = O.

We prove that every non-zero element u of 3 is expressible in the form

(1) u = L (x - 1)(x- I ux + vx ), xeX

where ux, Vx are elements of A(j for which I(ux) < I(u) and I(vx) < I(u). Since 3 is the additive group generated by all f - 1 with f E (j, it suffices to establish this when u = f - 1, where f E (j and f =F 1. We use induction on l(f). Since l(f) > 0, we may write f = x'g, where x E X, S = ± 1, g E (j, leg) < l(f). If S = 1, we have f - 1 = (x - 1)g + (g - 1); the result follows from this, trivially if g = 1, and by application of the inductive hypothesis to g - 1 if g =F 1. Ifs = -1,1- 1 = -(x - 1)x- l g + (g - 1), and the assertion follows in the same way.

Now let .3 = nn>O 3n and suppose that .3 =F O. We observe that if a = Lxex(x - 1)ax E.3 with ax E A(j, then ax E.3. Indeed for any positive integer n, a E 3n+1, so by a), a can be written as an A-linear combination of elements of the form (x - 1)(XI - 1)· .. (xn - l)b, with x, Xl' ... , Xn in X and b in Atj. Since the x - 1 form an A(j-basis of 3, it follows that ax is an A-linear combination of products of the form (Xl - 1)· .. (xn - 1)b. Thus ax E 3n for all n > 0 and ax E.3.

Now choose a non-zero element u of.3 for which leu) is minimal. Then leu) =F O. By (1), we can write

u = L (x - 1)(x- I ux + vJ, xeX

where I(ux) < leu), I(vx) < leu). By the above remark, X-lux + Vx E.3. Thus if Ax is the sum of the coefficients of ux, Ux - Ax1 E 3 and

By (1), we may write

Ux - Ax l = L (y - l)ux.y, Vx + Ax l = L (y - l)vx,y, yeX yeX

where l(ux) ::;; l(ux - A.x1) < l(u) and l(vx.y) ::;; l(vx + A.xl) < l(u). Now Ux + xVx = x(x-1ux + vx) E 3, and

Ux + xVx = {L (y - 1)(ux.y + Vx)} + (x - 1)(vx + Ux,x + vx,x)' y",x

Since Ux + xVx E 3, it follows from the above remark that Ux,y + Vx,y E 3 (y =F x) and Ux x + Vx x + Vx E 3. By minimality of l(u), it follows that Ux,y + Vx,y = ~x,x + ~x,x + Vx = O. Hence Ux + xVx = 0 for all x, and u = 0, a contradiction. q.e.d.

11.12 Theorem. Let fJ be afree group. a) If p is any prime and the subgroups "nm) are.defined as in 1.10,

then

b) (MAGNUS) n n~l Yn(m = 1.

Proof. Let K = GF(p), and let 3 be the augmentation ideal of the group-ring KfJ. If f E "n(m, then f - 1 E 3n by 2.7. It follows that if fEn n~l "n(m, then f - 1 E n n~13n; hence f - 1 = 0 by 11.11b). This gives a), and b) follows at once, as "n(ffi) ~ Yn(ffi). q.e.d.

We now describe the embedding of the free associative algebra in the group-ring of a free group, which was mentioned earlier.

11.13 Theorem. Let fJ be afree group with basis X. a) The subalgebra m of AfJ generated by the x - 1 (x E X) is afree

associative algebra with basis B = {x - 1/x EX}. b) If Hn is the homogeneous component ofm of degree n with respect

to B,

where 3 is the augmentation ideal of AfJ.

Proof. For each n ~ 0, let Bn be the set of all products (Xl - 1)·· . (xn - 1) with Xi E X. Let H~ be the A-module spanned by Bn' Thus

m = H~ + H~ + ....


By 11.11a), Bn is an A~-basis of :r. Afortiori, Bn is an A-basis of H~. Also 3n = H~ + 3n+1, for if bE Bn and u E~,

bu = b + b(u - 1) E H~ + 3n+1•

v = L Abb = L b(x - 1)Ub,x, bEB, bEB"XEX

with Ab E A and Ub,x E A~. Hence

Ab = L (x - 1)Ub,x, XEX

Ab = 0 and v = O. Hence 3n = H~ EB 3n+ 1 and

m- = H~ EB H~ EB ....

Since Bn is an A-basis of H~, it follows at once that m- is a free associative algebra and that Hn = H~. q.e.d.

By 11.13b),

Thus if U E 3, u has an expansion of the form

where hi E Hi and r m E 3m+1. We examine the first non-zero term of this expansion in the case when A = 7L and u = f - 1 with f E ~ - {1}. We shall see that if f E 1'n(m - 1'n+1 (m (n :s;; m), the first non-zero term is hn and it lies in Gn , (defined as in 11.8). Further, every non-zero element of Gn arises in this way.

11.14 Theorem (MAGNUS, WITT). Let ~ be afree group with basis X, let 3 be the augmentation ideal of the integral group-ring 7L ~ and, for each n ~ 1, let Gn be the additive group generated by all Lie products [Xl - 1, ... , Xn - 1] with Xi E X. There exists an isomorphism Pn of 1'n(~)jyn+1(m onto Gn such that iffE 1'n(m,

f - 1 - (/1'n+1 m))Pn E 3n+1.

ProofLet~betheassociativealgebrageneratedbyB = {x - llxEX}. By 11.13a), ~ is a free associative algebra with basis B. For each n ~ 0, let Hn be the homogeneous component of ~ with respect to B; thus ,3n = Hn ffi ,3n+1 by 11.13b). By 11.10, the Lie algebra 9 generated by B is free with basis B, and by 11.8a), Gn = 9 n Hn.

For n ~ 1, let

tJn = {fifE tJ, J - 1 E ,3n}.

By 2.3a), tJn <J tJ, and by 2.3b),

Finally, by 2.3c), if J E tJm and g E tJn, then

(2) [f, g] - 1 == [J - 1, g - 1] mod,3m+n+1.

Hence tJ = tJ1 ~ tJ2 ~ tJ3 ~ ... is a strongly central series of tJ. Thus tJn ~ Yn(tJ). ThusifJE Yn(tJ),J - 1 E,3n = Hn ffi ,3n+1. Hence there exists a unique element u E H n such that J - 1 - U E ,3n+ 1. If also g E Y n(tJ), v E Hn and g - 1 - V E ,3n+1, then u + v E Hn and

Jg - 1 - (u + v) = J(g - 1 - v) + (f - 1 - u)(l + v) + uv E ,3n+1,

since uv E H2N s;; ,32n s;; ,3n+1. Hence the mapping J -+ u is a homomorphism of Yn(tJ) into Hn. Yn+1 (tJ) is contained in the kernel, since Yn+1 (tJ) ::; tJn+1· Hence there exists a homomorphism Pn of Yn (tJ)/Yn+1 (tJ) into Hn such that for any J E Yn(tJ),

(3)

Note in particular that if x E X, (XY2(tJ»P1 = x - 1. It follows from (2) and (3) that i! J E Ym(tJ) and g E Yn(tJ),

[f, g] - 1 == [(hm+1(tJ»Pm' (gYn+1(tJ»Pn] mod,3m+n+l.

Since the term on the right-hand side lies in Hm+n and [J, g] E Ym+n(tJ), it follows that


By 9.3, there exists a Lie ring f = Fl E9 F2 E9 ... and, for each n ~ 1, an isomorphism G'n of YnmVYn+l (IJ) onto Fn such that

for f E Ymm), g E Ynm). Also, by 9.3b), there exists a Lie homomorphism () of f into I(m) such that G'n(} = Pn for all n ~ 1.

By 9.4, f is generated as a Lie ring by {(XY2 (fj»G' 11 x EX}. Now () carries this set bijectively onto B, since

Hence im () is contained in g. Since 9 is a free Lie ring with basis B, it follows from 11.4a) that () is an isomorphism of f onto g.

Since G'n, () are both monomorphisms, so is Pn. Since x - 1 E im () for x E X, it follows from the definition of Gn that Gn s im (). But

so Fn(} S Hn n 9 = Gn. Since 9 = Gl EEl G2 EB ... , it follows that Gn = Fn(}. Hence Pn is an epimorphism of Yn (IJ)IYn + 1 (IJ) onto Gn. Hence Pn is an isomorphism. q.e.d.

11.15 Theorem. Let fj be afree group. a) For each n ~ 1, Yn(IJ)/Yn+l(1J) is a free Abelian group, and iffj

has afinite basis with q elements, the rank ofYn(fj)/Yn+ 1 (IJ) is dn(q). b) Let,3 be the augmentation ideal of7Lfj and suppose thatfE fj. Then

f- 1 E,3nifandonlyiffEYn(lJ)(n ~ 1).

Proof LetmbetheassociativealgebrageneratedbyB = {x - 1IxEX}, where X is a basis of fj. By 11.13a), m is a free associative algebra with basis B. If Gn is the additive group generated by all Lie products [Xl - 1, .. "xn - 1] with X;EX, then by 11.8, Gn is a free Abelian group, and if IXI = q, the rank of Gn is dn(q). By 11.14, there exists an isomorphism Pn of Yn(IJ)IYn+1 (IJ) onto Gn and

for all f E Yn(lJ). Thus a) follows at once. If f E Yn(1J) - Yn+1(IJ), then fYn+1(1J) oj; 1 and u = (fYn+l(IJ)Pn oj; 0,

since Pn is an isomorphism. Thus u E ~n - ~"+1. Since f - 1 - U E ~n+t,

f - 1 E,3n - ,3n+1. This implies b). q.e.d.


11.16 Remarks. a) For any group (f), the dimension subgroup Yn(f)) is defined by

where:5 is the augmentation ideal ofZ(f). By 2.3, Yn(f)) S; Yn(f)). Theorem 1U5b) asserts that Yn(m = Yn(m for a free group ~, and it was conjectured that this is the case for all groups. This is true for n = 1, 2, 3, but RIPS [lJ gave a counterexample for n = 4. SJOGREN [lJ proved that there is a bound on the exponent of Yn(f))IYn(f)) independent of (f).

b) Let 91 be a normal subgroup of the free group ~, let 3 be the kernel of the natural homomorphism of Z~ onto Zm/91) and let

where :5 is the augmentation ideal of Z~. The question of the characterization of 91n has been raised by FOX [1]. In the case when 91 = ~, the answer is given by 11.15, and the answer is trivial if n = O. It was proved by Schumann that 91 1 = Y2(91) (see FOX [lJ) and by HURLEY [lJ that

Exercises

21) Show that the function dn(q) defined in 11.8 is given by

here J1 is the Mobius function, defined by J1(1) = 1, J1(r) = 0 if r is divisible by the square of a prime and J1(PI ... Pm) = (- l)m if PI' ... , Pm are distinct primes. (First prove this for q = 1.)

(It is remarkable that this formula coincides with that of Gauss for the number of irreducible polynomials of degree n over GF(q).)

22) If A is a commutative ring with identity and f is a free Lie algebra over Z, f ®1' A is a free Lie algebra over A.

Deduce 11.8 and 11.10 for algebras over A.

23) Suppose that ~ is a free group and that P is a prime. Let :5 be the augmentation ideal of Z~. Suppose that! E Ynm) but that !Yn+l (m is

§ 12. Remarks on the Burnside Problem 385

not the p-th power of any element of Ynm)/Yn+1 (~). Prove that f - 1 + 3n+1 ~ p(3n/3n+I ).

24) Let 9Jl be the Magnus algebra on a set X over A, that is, the set of all (possibly infinite) formal sums

with X as a set of non-commuting indeterminates. Show that if x E X, 1 + x is a unit of 9Jl. Let ~ be the subgroup of the group of units of 9Jl generated by all 1 + x (x E X). Prove that ~ is a free group and that the subalgebra of 9Jl generated by ~ is the group-ring A~.

(Hint: This follows from 11.11 and 11.13.)

25) Let ~ be a free group and let 'l) be the ideal of £:~ consisting of all elements LIE'ija If for which LIE'ija I is divisible by the prime p. Show that

§ 12. Remarks on the Burnside Problem

It has been proved by Adyan and Novikov (see, for example, ADYAN

[1]) that the answer to the Burnside problem is negative. In this book we shall be more concerned with the restricted Burnside problem (III, 6.7). In the case of prime-power exponent, linear methods have been used in investigations of this, and we shall now sketch some of the ways in which this has been done. We begin by interpreting 2.16 in the Lie ring of a group of exponent p. The following lemmas are needed (cf. III, 1.11).

12.1 Lemma. Suppose that GJ = <X). a) If a E GJ and b E GJ, then ab = ac I ... cn , where each Ci is a com

mutator [a, YI' ... , Yr]for which r ;::: 1 and Yi E X U X-I. b) Yn(GJ) is generated by all commutators [YI' .. " Yr] for which

r ;::: nand Yi E X U X-I.

Proof a) Suppose that b = YI ... Ym' where Yi E X U X-I and m ;::: O. We use induction on m. If m = 0, the assertion is trivial with n = O. Ifm > 0, then

The inductive hypothesis then gives the result at once. b) By III, U1a), Yn(<») is generated by all [Yl' ... , Yn]b, where Yi E X.

The assertion follows at once by applying a). . q.e.d.

12.2 Lemma. Suppose that <» is a group of prime exponent p. If u E Yn(<») and Ui E Yn,(<») (i = 1, ... , p - 1), then

where m = n + n1 + ... + np- 1 '

Proof This is obvious for p = 2 on account of the identity

[ ] -2( -1)2 2 U,U1 = U UU 1 U1•

Henceforth we suppose p odd. Let !j be a free group with basis x, Y l' ... , Yp-l' If Y = Yl ... Yp-l' by 2.16,

Hence

[x, y, ... , y] E !jPYp+l(m. p-l

By a trivial induction based on III, 1.2,

[x,y, ... ,y] == Il[x'Yi' ···,Yi] modYn+2(m, n 1"

where the product is taken with any order of the factors over all suffixes ij with 1 ~ ij ~ P - 1. Taking n = p - 1, we find that

(1)

Since x, Y 1, ... ,Yp-l is a basis of !j, there is a homomorphism of !j into !j which carries each of x, Y 1, ... , Y p-l into a preassigned element of !j. Such a homomorphism always carries !jPYP+1 (m into itself, so (1)


remains true if x, Yl, ... , Yp-l are replaced by other elements of ty. Thus, putting Yl = 1, we see that if Z is the product of those factors of the left-hand side of (1) for which no ik is equal to 1, then Z E tyPYp+I (ty). Removing all such factors from (1), we see that (1) remains true if the product is taken over those factors for which at least one ik is equal to 1. We now repeat this argument on the formula thus obtained for the suffix 2, then for 3, and so on. The conclusion is thereby reached that (1) is valid if the product is restricted to those suffixes for which (iI' ... , ip-I) is a permutation of(1, 2, ... , p - 1). Thus

By 12.1b), there exist commutators gi = [ZI' ... , Zk] (i = 1, ... , r) such that k ~ p + 1, Zi = X±I or yfl and

(2) n [x, YI,,, ... , Y(P-I)U] == g~l ... g:' mod tyP, UE6p _ 1

where Ci = ± 1. Putting x = 1, we see that if q is the product of those gfi for which no Zj is X±I, taken in the correct order, then q E tyPo We may therefore replace the right-hand side of (2) by q-I gfl ... g:'. But it is easy to see that q-Ig~l ... g:' is a product of factors (gn\ where only those gi occur for which some z)s X±I. By 12.1a),gibi is a product of commutators of the same form as g;, but with the additional condition that some Zj is X±I. Thus in (2) we may assume that all gi satisfy this condition. Indeed, by repeating this argument, we find that it may be assumed that in each

II f ±I ±I ±I Ad' k gi, a 0 x , Yl , ... , Yp_1 occur among Zl, ... , Zk' n SInce ~

p + 1, at least one of x, Y 1, ... , Yp-l occurs twice. Let P be the homomorphism of ~ into (f) for which XP = u and

YiP = Ui (i = 1, ... , p - 1). By III, 2.11b), it follows that giP E Ym+l(f)), since Ui E Yn(f)) and U E Yn(f)). Since (f) is of exponent p, ~p P = 1. The assertion th~refore follows by applying P to (2). q.e.d.

12.3 Theorem. Suppose that (f) is a group of prime exponent p, and that 9 is a Lie ring. Suppose that for each n ~ 1, a group homomorphism an of Yn(f))/Yn+l (f)) into 9 is defined, and that the following conditions are satisfied:

a) 9 = Ln~l Gn, where Gn = im an; b) [(XYm+l(f)))am, (YYn+l(f)))an] = ([x, Y]Ym+n+l(f)))am+n, whenever

x E Ym(f)), Y E Yn(f)). Then (ad a)p-l = Of or any a E g.

Proof Suppose that an E Gn (i = 1, ... , p - 1) and that bE Gn. There exist u E Yn((fj), Ui E Yn,(<») su~h that

By induction on k (k 2: 0), it follows from b) that

b(adan )··· (ad an ) = ([u, u1 , ••. , Uk]Ym +1((fj))O'm' 1 k k k

where mk = n + n1 + ... + nk' Taking k = p - 1,

where m = n + nl + ... + np-l' It follows from 12.2 that

" b(ad a ) ... (ad a ) = O. L.., "It n(p-l)t tE6p_1

Since 9 = Ln~l Gn and the left-hand side is linear in b and in each an" it follows that

I b(adal t }'" (ada(p_ljt) = 0 tE6p_1

for any elements ai in g. Putting a1 = ... = ap- 1 = a, we find that (p - I)! (ad a)P-l = O. Since the additive group of 9 is an elementary Abelian p-group, it follows that (ad a)P-l = O. q.e.d.

A condition on Lie algebras about some power of ad a being zero is called an Engel condition; such a condition sometimes implies that the algebra is nilpotent. The deepest known result of this nature is the following theorem.

12.4 Theorem (KOSTRIKIN [1]). a) Suppose that 9 is a finitely generated Lie algebra over afield of characteristic p. If (ada)P-l = Ofor all a E g, then 9 is nilpotent.

b) Suppose that p is a prime and d is a positive integer. Then there exists a finite group (fj generated by d elements and of exponent p such that every finite group generated by d elements and of exponent p is an epimorphic image of (fj.

Kostrikin's proof of a) is an extremely difficult calculation which cannot be reproduced here. We shall however show how b) may be deduced from it.

Proof b) Suppose that this is false. Let Ij be a free group of rank d. Let Ijn = Yn(!j)/jP (n = 1, 2, ... ). By 2.1, Ij/ljn is finite. Since Ij/ljn does not have the property described in the assertion of the theorem, there exists a finite group f) generated by d elements and of exponent p, such that f) is not an epimorphic image of Ijj!jn. But since Ij is free of rank d, there is an epimorphism of Ij onto f); thus if IDl is the kernel, IDl ? Ijn. Since Ij/IDl ~ f), Ij/IDl is nilpotent and IDl ~ IjP. Hence Ijn > IDl (\ Ijn ~ IjP and 1j/(IDl (\ Ijn) is nilpotent (III, 2.5c)). By III, 2.6 applied to 1j/(IDl (\ Ijn)' mn' Ij] IjP < Ijn" Thus

Ijn+tiljP = [Yn(lj), lj]ljP/ljP = mn/ljP, lj/ljP] < Ijn/ljP,

so Ijn+1 < Ijn· Let m = nn;;'l Ijn" Then m ~ IjP, but Ij/m is infinite. The lower central series of Ij/m is

By 9.3, there exists a Lie ring 9 = G1 EB G2 EB G3 EB ... and, for each n ~ 1, an isomorphism (Tn of Ijn/ljn+1 onto Gn such that the conditions of 12.3 are satisfied. By 12.3, (aday-1 = ° for any a E g. By a), 9 is nilpotent. By 9.4, gn = Gn EB Gn+1 EB .. '. Thus there exists n such that Gn = 0. Hence Ijn = Ijn+1' a contradiction. q.e.d.

12.5 Remark. Let Ij be a free group with n generators. The restricted Burnside problem (III, 6.7) may be stated as finding whether or not Ij/!Jm has a maximal finite homomorphic image. Theorem 12.4b) states that if m is a prime, the answer is affirmative.

Let (fj be the maximal finite homomorphic image of lj/ljP. Let f be a free Lie algebra over K = GF(p) with n generators, and let i be the ideal of f generated by all b(ad a)P-1 (a E f, bE f). Then there is an epimorphism of f/i onto the Lie ring defined by the lower central series of (fj (in the sense of 9.3). WALL [1] has investigated the kernel of this epimorphism and, in particular, he has shown that for p = 5, 7 or 11 and n ~ 3, it is non-zero.

In the negative direction, we now show that there do exist nonnilpotent finitely generated groups in which the order of every element is a power of p. To construct these, the following lemma will be,used.

12.6 Lemma. Let K be a field. Let ~ be a free associative algebra over K with associative basis Xl, ... , Xd' and for each n ~ 0, let Hn be the homogeneous component of~ of degree n with respect to this basis. For each n > 1, let Rn be a finite subset of Hn, let rn = I Rnl and let R = Un >1 Rn·

Let 3 be the ideal of m: generated by R, let ~ = m:/3, let B" = (H" + 3)/3 and let b" = dimKB".

a) 3 = (3 () H2) EB (3 () H3) EB ... and ~ = Bo EB B1 EB B2 EB .... b) b" ~ db,,-1 - Li'=2 rib,,-i (n ~ 2). c) Ifs" ~ r" and the coefficient oft" in the formal power-series

is non-negative for all n ~ 2, then ~ is not of finite dimension over K. d) If there exists a real number e such that 0 < e < fd and

rll ::=;; e2(d - 2er-2 for all n ~ 2, then ~ is not offinite dimension over K.

Proof. Let J" = 3 () H". Suppose that u E J". Since 3 is the ideal generated by R, 3 is spanned over K by all products P = Xi ... Xi ZXJ• • •• XJ. ,

1 r 1 .i

where r ~ 0, s ~ 0 and Z E R. Thus u is a linear combination of such elements P. But Z E Rm for some m and then Xi ... Xi ZXJ• • •• XJ. E Hr+s+m•

1 , 1 •

Since u E H" and m: is the direct sum of its homogeneous components, it follows by comparing terms in H" that u is a linear combination ~f those P for which Z E RII - r- s• Thus

(1) J" is spanned over K by all products Xi ... Xi ZXJ· ... xJ·, where 1 r 1 •

r ~ 0, s ~ 0 and Z E R,,-r-s. In particular, since R" is defined only for n > 1, (2) J o = J 1 = o. Observe also that since 3 is spanned by all products P, 3 is the sum

of the J". Since also J" s;;; H", (3) 3 = J 2 EB J 3 EB .... a) It is clear that ~ is the sum of the B", since m: is the sum of the H".

Suppose that L,,~o v" = 0, where v" E B". Then v" = u" + 3 for some u" E H", and L,,~o u" E 3. It follows at once from (3) that u" E J", and so v" = O. Thus ~ is the direct sum of the B".

b) Let Lm be a subspace of Hm such that Hm = ~ EB Jm (m ~ 0). For n ~ 2, let V be the subspace spanned by all elements UXj' where U E J"-1 and 1 ::=;; j :$ d. Clearly

(4) dimK V ::=;; d(dimKJ"_1). By (1), J" = V + V*, where V* is the vector space spanned over K by all products P = Xi, ... xi,z, where Z E R,,-r (O::=;; r ::=;; n - 2). Write Xi ... Xi = Y1 + Y2' where Y1 E Lr, Y2 E Jr· Thus P = Y1 Z + Y2Z. But , , it is clear from the definition of V that Y2Z E V. Thus V* s;;; U + V, where U is the vector space over K spanned by all products yz with Y E L" Z E R,,-r (0 ::=;; r ::=;; n - 2). Since U s;;; J", we have J" = V + U.


Clearly dimKU ~ Ii'=2IR;j(dimKLn-i). Since Lm ~ H"JJm ~ H"J(~ n HJ ~ (Hm + ~)/~ = 8m, dimK Lm = bm• Hence

(5) diml( U ~ Ii=2 ribn-i' Since I n = V + U, it follows from (4) and (5) that

(6) dimKJn ~ dimK V + dimK U ~ d(dimKJn- t ) + Ii=2 ribn- i. But dimKJm = dimK Hm - dimK Lm = dm - bm• Substitution in (6) now yields the desired inequality.

c) For n ~ 1, let

n

an = bn - dbn- l + I ribn-i' i=2

Thus al = 0 and, for n > 1, an ~ 0 by b). Let

B(t) = 1 + I bntn, n~l

D(t) = 1 - dt + I rntn, n~2

A(t) = 1 + I antn. n~l

Then B(t)D(t) = A(t) since bo = 1, bl = d. Write

D(trl = 1 + I cntn. n~l

Since B(t) = A(t)D(trl,

(7) bn = cn + Ii::! cn-iai + an (n ~ 1). Similarly write

D*(t) = 1 - dt + I Sntn, n~2

U(t) = I (Sn - rn)tn. n~2

Thus U(t) = D*(t) - D(t). By hypothesis the coefficients of tn in D*(trl and in U(t) are non-negative. But

00

D(tfl = D*(tfl(1 - D*(tfl U(t)tl = I D*(tfn-l U(t)"; n=O


thus the coefficient of tn in D(tt1 is non-negative, that is, Cn ;::: O. Since also an ;::: 0, it follows from (7) that bn ;::: Cn'

Suppose there exists no such that bn = 0 for all n > no. Then Cn = 0 for all n > no, and

no

D(tt1 = 1 + L cntn. n=1

On account of the definition of D(t),

Since Cn ;::: 0 and rn ;::: 0 for all n ;::: 1, this is obviously impossible. Hence bn =1= 0 for infinitely many n. By a), !8 is not of finite dimension

over K. d) Let Sn = e2(d - 2e)n-2 (n ;::: 2). Then

( 00 )-1 { 1 }-1

1 - dt + n~/ntn = 1 - dt + t2e2 1 _ (d _ 2e)t

1 - (d - 2e)t (1 - (d - e)t)2

oc

= 1 + L tn{(n + 1)(d - e)n

- (d - 2e)n(d - e)n-1} 00

= 1 + L tn(d - e)n-1(d + (n - 1)e). n=1

The coefficient of tn in this is positive. Since also rn ~ Sn for n ;::: 2, the result follows at once from c). q.e.d.

12.7 Theorem (GOLOD and SAFAREVIC). Let K be a countable field and let d be an integer greater than 1. There exists an associative algebra !8 over K which has the following properties.

a) !8 is generated by d elements. b) !8 is not offinite dimension over K. c) There exist subspaces Bn of!8 (n ;::: 0) such that Bo = {,U!!ll A E K},

and BmBn S; Bm+n for any m, n. In particular .3 = B1 EEl Bz EEl ... is an ideal of lB.

d) Let V be a finite subset of .3 having fewer than d elements. Then there exists an integer r such that the product of any r elements of V is O.

Proof Let m be a free associative algebra over K with associative basis Xl' .•. , Xd' For each n ;::: 0, let Hn be the homogeneous component of m of degree n with respect to this basis. Then m = Ho EEl H1 EEl .... If a E m and a = a1 + az + ... , where an E Hn , we refer to an as the homogeneous component of a of degree n. Let.3 be the ideal H 1 EEl Hz EEl ... of m. Let .xl be the set of all finite subsets U of.3 for which lUi = d - 1. Then .xl is countable; we write

( 1 )q-Z

Let q be a fixed integer such that q ;::: 4d and qZ:s;; 1 + 2d ;

clearly such an integer exists. We construct certain finite non-empty subsets Sm of Ur>q Hr for each m ;::: ° satisfying the following conditions.

(i) If Sm n Hr "# 0 and Sm+1 n Hs "# 0, then r < s. (ii) For each m > ° there exists an integer qm > q such that Sm is

the set of all homogeneous components of all products of qm elements of Um.

To do this we use induction on m. Put So = 0. For m > 0, let qm be the smallest integer greater than q such that every element of Sm-1 lies in some Hr with r < qm; this is possible since Sm-1 is finite. We then define Sm to be the set of all homogeneous components of all products of qm elements of Um' Since Um s; .3, it is clear that if Hr n Sm "# 0, then r ;::: qrn' Thus the conditions (i), (ii) are satisfied.

Let R = U m>oSrn' By (i), R n Hn is finite; let rn = IR n Hnl. Thus rn = 0 if n :s;; q. If rn "# 0, there exists a unique integer m such that R n Hn s; Sm by (i). Then each element of R n Hn is the homogeneous component of degree n of a product U1 ... uq .. , where Ui E Urn. Since I Urn I = d - 1, it follows that

Since rn "# 0, n ;::: qm' so rn :s;; (d - It Thus

Since q ~ 4d and n > q, it follows that

( 2)n-2( 1 )-(q-2) rn < dn 1 - q 1 + 2d .

On account of the choice of q this implies that

( 2)n-2 1 (d)2( d)n-2 rn < dn 1 - q q2 = q d - 2q .

Let 3 be the ideal of m: generated by R, let '13 = m:/3 and let Bn = (Hn + 3)/3 (n ~ 0). By 12.6d) with e = d/q, '13 is not of finite dimension over K. By 12.6a),

clearly BmBn £; Bm+n' Let V be a finite subset of ~ = BlEB B2 EB ... for which I Vi < d.

Then there exists Um such that V £; {u + 31u E Um}. If z is a product of any qm elements of Um' then by (ii) every homogeneous component of z lies in Sm. Hence z E 3. Thus the product of any qm elements of V is O.

q.e.d.

12.8 Theorem. Let p be a prime and let d be an integer greater than 1. There exists an infinite group (fj which has the following properties.

a) (fj is generated by d elements. b) The order of any element of (fj is a power of p. c) nn;el Yn(fj) = 1. d) (fj is not nilpotent, but any subgroup of (fj which can be generated

by fewer than d elements is nilpotent.

Proof Let K = GF(p) and let '13 be the associative algebra constructed in 12.7. Let

Now if v E ~, then by 12.7d), there exists an integer r such that vr = o. If pS is the smallest power of p greater than r, (1 + v)P' = 1 since K is of characteristic p. Thus every element of f) has an inverse in f). Hence f) is a group and the order of any element of f) is a power of p. By 12.7c), ~n = Bn EB Bn+l EB ... is an ideal of 'B. Let

By 2.3, ~n <l ~ and [~n'~] ~ ~n+1' Thus

"'='" >'" > ... -!) -!)1 - -!)2 -

is a central series of~. Hence ~n ~ Yn(~)' By 12.7a), 113 is generated by d elements Yl' ... , Yd; by 12.7c), we may

suppose that Yi E 3. Let (fj bethesubgroupof~generated by 1 + Yl' ... , 1 + Yd' Then y.(fj) ~ Y.(~) ~ ~n' By 12.7c), nn;;'l 3n = 0, so nn;;'l Yn(fj) = 1. Thus (fj satisfies a), b), c) of 12.8.

We prove next that (fj is infinite. There exists an associative homomorphism p of K(fj into 113 induced by the embedding of (fj in 113. Now im p contains 1 + Yi and 1; hence im p contains Yi and p is an epimorphism. Since 113 is not of finite dimension over K, neither is K(fj. Thus (fj is infinite.

It follows from 2.1 that (fj is not nilpotent. Let ft be a subgroup of (fj generated by fewer than d elements. Then there is a subset V of 3 having fewer than d elements, such that ft = (1 + vlv E V). By 12.7d), there exists an integer r such that the product of any r elements of V is O. If (J is the homomorphism of Kft into 113 induced by the embedding of ft in 113 and 3 is the augmentation ideal of Kft, then 3(J consists of sums of products of elements of V. Hence 3' (J = O. But it follows from 2.3 that if x E ys(ft), then x - 1 E 3 s• Since (J is injective on ft, it follows that y,(ft) = 1 and ft is nilpotent. Thus (fj also satisfies d). q.e.d.

12.9 Theorem. Let K be a countable field and let d be an integer greater than 1. There exists a Lie algebra g over K which has the following properties.

a) 9 is generated by d elements. b) 9 is not of finite dimension over K. c) nn;;,l gn = O. d) 9 is not nilpotent, but any subalgebra of 9 which can be generated

by fewer than d elements is nilpotent.

Proof Let 113 be the associative algebra constructed in 12.7. Suppose that 113 is generated by Yl' ... , Yd and let g be the Lie subalgebra of

1(113) generated by Yl' ... , Yd' Since gn ~ Bn EB Bn+1 EB .. " nn;;,l gn = O. It follows at once from 12.7d) that any subalgebra of g which can be generated by fewer than d elements is nilpotent.

Suppose that 9 is of finite dimension over K; let gl' ... , gn be a basis of g. Clearly the associative subalgebra of 113 generated by 9 is 113, since 9 contains Y 1, ... , Yd' By 11.1, it follows that 113 is spanned by the set of elements g~1 ... g~ •. But by 12.7d), there is an integer r i such that

g[; = 0 (i = 1, 2, ... , n). Hence !B is spanned by a finite set, contrary to 12.7b). Hence 9 is not of finite dimension over K. By 8.7, gmjgm+1 is finitely generated and hence of finite dimension for all m. Thus 9 is not nilpotent. q.e.d.

12.10 Remarks. a) Suppose that d > 2 in 12.9. If x, yare elements of g, x and y generate a nilpotent subalgebra, so there is an integer n such that x(ad y)n = O. Similarly, if d > 2 in 12.8 and x, yare elements of <fi, there is an integer n such that [x, y, ... , y] = 1. rhus there exist non-

n

nilpotent groups and Lie rings in which the Engel condition [x, y, ... , y] = 1 (or 0) holds.

b) Another construction of finite p-groups from the free associative algebra was given by WALL [2] to prove that for 2 :::; d :::; pm, there is a finite p-group <fi with the following properties.

(i) I <fi : </>(<fi) I = pd. (ii) The exponent of <fi is pm+1.

(iii) The centre 3 of <fi is cyclic, and if D1(3) = (z), z is a power of every element of <fi - </>(<fi).

§ 13. Automorphisms of p-Groups

If <fi is a finite p-group, <fij</>(<fi) is an elementary Abelian p-group and can therefore be regarded as a vector space V over GF(p). Thus the group of automorphisms of <fi induces a subgroup f) of GL(n, p) on V. In this section we shall prove a theorem of Bryant and Kovacs which asserts that any subgroup f) of GL(n, p) can arise in this way. We begin with an elementary lemma.

13.1 Lemma. Let V be a vector space over an infinite field K and let S be a finite set of linear transformations of V containing no scalar multiplications. Then there exists v E V such that v, vs are linearly independent for all s E S.

Proof Suppose that this is false. Then given v E V, there exists s E S such that vs = AV for some A E K. Hence V is the union of the eigen-spaces of the elements of S. Since no element of S is a scalar multiplication, no eigen-space of an element of S is the whole of V. Since S is finite, it follows that V is the union of a finite number of proper subspaces. But this is impossible, since K is infinite. q.e.d.

§ 13. Automorphisms of p-Groups 397

The proof of the main theorem is by Lie methods; the theorem about Lie algebras that we need is the following.

13.2 Theorem (BRYANT and KovAcs [1]). Let 9 be a free Lie algebra over a field K with Lie basis X. For each n ~ 1, let Gn be the K-space spanned by all products [Xl' ... , xn] with Xi E X. Let G} be a group for which GI is afaithful KG}-module.

a) 9 has the structure of a KG}-module in which GI is a KG}-submodule and given g E G}, the mapping a -+ ag (a E g) is an automorphism of the Lie algebra g. Each Gn is a KG}-submodule of g.

b) Suppose that 1 Xl> 1 and G} is finite. Let 3 be the set of elements of G} which induce scalar multiplications on GI , and put q = 131. Then there exists no such that for all n > no, Gn EEl Gn+l EEl ... EEl Gn+q- l has a KG}-submodule for the regular representation of G}.

Proof By 11.5b), 9 is the Lie subalgebra of l(~) generated by X, where ~ is the free associative algebra with basis X. Let Hn be the homogeneous component of ~ of degree n (n ~ 0). Then

~ = Ho EEl H I EEl ....

Since Gn s; Hn, we have

9 = GI EEl G2 EEl ...

and Gn = Hn II g. Suppose that g E G}. Since ~ is free, there is a unique endomorphism rxg of ~ such that xrxg = xg for each X E X. Then rxg1 rx g2 =

rxg1g2 and rx l = 1, so rxg is an automorphism of~. Also yrxg = yg for all y E HI = GI . We put yg = yrxg for all y E~. Then ~ is a KG}-module and GI is a KG}-submodule of~. Also Hn , g, Gn are all KG}-submodules, and rxg induces a Lie automorphism on g. Thus a) is proved.

We observe next that if b) holds for free Lie algebras over infinite fields, then it holds in general. For if K is any field, there exists an infinite field L containing K. Then ~ ®K L is the free associative algebra over L with basis X and 9 ®K L is the Lie subalgebra of l(~ ®K L) generated by X. Also Gn ®K L is the L-space spanned by all products [Xl' ... , xn] with Xi E X, and GI ®K L is an LG}-module. Finally, 3 is the set of elements of G} which induce scalar multiplications on G I ®K L. Assuming, then, that the theorem holds for Lie algebras over L, there exists no such that for all n > no, (Gn EEl ... EEl Gn+q- l ) Q9K L contains an LG}submodule for the regular representation of G}. By VII, 7.23, (or, if X is infinite, by its (immediate) extension to the infinite-dimensional case), Gn EEl ... EEl Gn+q - l contains a KG}-submodule for the regular representation of G}.

We shall suppose, then, that K is infinite. This will enable us to use 13.1.

Let T be a transversal of 3 in (f) containing 1. We prove the following. (*) Suppose that k > 3(1(f): 31 - 1). Then there exists a E Gk such

that {atlt E T} is linearly independent. If 3 = (f), there is nothing to prove, since IXI > 1, so we suppose

that 3 < (f). By 13.1, there exists b E H I such that b, bg are linearly independent for all g E (f) - 3. Choose c E HI such that b, c are linearly independent. Then there exist x, y in X and a non-singular linear transformation IX of H I such that XIX = b, ylX = c. But

k-I y(adx)k-I = L (ki l )(_I)ix iyxk-i-1 1= o.

i=O

By a), IX can be extended to an automorphism of g. Hence if

k-I (1) a = c(adb)k-I = L (ki1)(_I)ib icbk- i- 1,

i=O

a is a non-zero element of Gk •

We shall define two subspaces B, C of Hk such that Hk = B EEl C, aEB and atEC for all tET - {I}. To do this, let t1, ... ,tm be the non-identity elements of T. Thus m + 1 = I(f): 31 and k > 3m. Since b, btj are linearly independent (j = 1, ... , m), there exist subspaces Xj'

Yj of H 1 such that b E Xj' btj E Yj and H 1 = Xj EEl Yj . Then Hk is the direct sum of all its subspaces of the form

where Z3j-2, Z3j-1, Z3j are each either Xj or Yj (j = 1, ... , m) and Z1 Z2 ... Z3m H 1 ... H 1 is the space spanned by

Let B be the sum of all these subs paces for which at most one of the Zi is a Yj and let C be the sum of the remaining ones. Thus Hk = B EEl C and, since bE Xj' bicbk- i- 1 E B (i = 0, ... , k - 1). Thus a E B. But btj E Yj , so at least two of the terms in the (3j - 2)-th, (3j - l)-th or (3j)-th places of (bt)i(ct)(bt/- i- 1 lie in ~. Hence (b icbk-i-1)tj E C and, by (1), atj E C.

Since a 1= 0, it follows that a ¢ Lh <f3 Kah. Hence for any t E T, at ¢ Lh¢3t Kah, and {at I t E T} is linearly independent. Thus (*) is proved.

Now GI is a faithful KG)-module. Thus 3, being isomorphic to a finite subgroup of KX , is cyclic. Suppose that 3 = <z) and that az = Aa for all a E HI' Thus az = Aka for all a E Hk.

We take no = 3(/G): 3/ - 1). If n > no, then for k = n, ... , n + q - 1, (*) shows that Gk contains an element ak such that {akt/t E T} is linearly independent. Let Uk = Kak and

M = Vn EB ... EB Vn+q - I •

Then Uk is a K3-module for the representation z -+ Ak of 3, so N is a K3-module for the regular representation of 3. Also, Vk is a KG)-module isomorphic to the induced module U~, so M is a KG)-module isomorphic to N(lj. Hence M is a KG)-submodule of Gn EB ... EB Gn+q- I for the regular representation of G). q.e.d.

We shall give an application of 13.2 which involves the subgroups An(G)) defined in 1.4. For this the following lemma is needed.

13.3 Lemma. Suppose that fJ is afree group. a) If'R <l fJ and rx is an endomorphism of fJ/'R, there exists an endo

morphism f3 of fJ which carries 'R into itself and induces rx on fJ/'R. b) Suppose that p is a prime and f3 is an endomorphism of fJ· If f3

induces an automorphism on fJ/A2(fJ), then f3 induces an automorphism on fJ/An+l(fJ)for all n ~ 1.

Proof a) Let X be a basis of fJ. For each x E X, choose x' E fJ such that (x'R)rx = x''R. There exists an endomorphism f3 of fJ such that x' = xf3. Thus (x'R)rx = (xf3)'R for all x E X. Hence for any f E fJ, (f'R)rx = (f f3)'R, so 13 has the required properties.

b) Let rx be the automorphism of fJ/A2(fJ) induced by f3. By a), there is an endomorphism f3' of fJ which induces rx- I on fJ/A2(fJ)· Thus f3f3' and 13'13 induce the identity automorphism on fJ/A 2(fJ). By 1.7a), IW and 13' 13 induce the identity automorphism on A;(fJ)/ Ai+1 (fJ) for all i ~ 1. Hence f3 induces an automorphism on Ai(fJ)/ Ai+ I (fJ). It follows by a simple induction on n that f3 induces an automorphism on fJ/ An+1 (fJ). q.e.d.

We shall now use the main theorem of § 11 to deduce the following from 13.2.


13.4 Theorem (BRYANT and KOVACS [1]). Let 11 be a free group, let p be a prime and let K = GF(p). Let (fj be a group for which I1/A2(11) is a K(fj-module.

a) For each n ~ 1, )'n(I1)/An+1(11) has the structure of a K(fj-module such that whenever 13 is an endomorphism of 11, 9 E (fj and (f A2 (l1))g = (ff3)A2(l1)for all f E 11, then (fAn + 1 (l1))g = (ff3)An+1(l1)for all f E An(l1)·

b) Suppose that 11 is non-cyclic, (fj is finite and that I1/A2(11) is a faithful K(fj-module. Then given a positive integer r, there exists no such that for all n > no, An(I1)/An+1 (11) contains afree K(fj-submodule of rank r.

Proof a) Suppose that 9 E (fj. By 13.3a), there exists an endomorphism 13 of 11 such that (ff3)A2(11) = (fA2m))g for all f E 11· Then Ai(l1)f3 £ Ai (l1), so we may put

for all f E An(l1). It follows from 1.7a) that if 13' is any endomorphism of 11 such that (ff3')A2(11) = (fA2(I1))g, then

It follows at once that An(I1)/An+1(11) is a K(fj-module. b) Let X be a basis of 11. By 11.14, there exists an isomorphism Pi

of "Yi(I1)/"Yi+1 (11) onto the additive subgroup generated by all Lie products [x 1 - 1, ... ,Xi - 1] (Xj E X) in the group-ring £:11 of 11, and Pi is determined by the fact that if f E "Yi(I1),

where ~ is the augmentation ideal of £:11. Let

Since KI1 ~ £:11/P£:I1, it follows that there exists an isomorphism ai of (£i onto the additive subgroup Gi of KI1 generated by all elements [Xl - 1, ... , Xi - 1] (Xj E X), determined by the fact that if f E "Yi(I1),

where .3 is the augmentation ideal of KI1. In particular, a1 is an isomorphism of (£1 = I1/A2(11) onto G1 and if X E X, (xA 2 (I1))a1 = X - 1. If f1 E "Yi(l1), f2 E "Yil1),


[II - 1,f2 - 1]

== [(fl Yi(IJY'Yi+1 (fY)) tTi , (f2 Yj(fYY'Yj+1 (fY))tTJ mod 3i+ j+ I,

and by 2.3c),

[f 1 f 1] - [f f] 1 mod '(H j+l. I - , 2 - = I' 2 - v

Hence since C'fH j+1 11 G.+. = ° " I J '

(1) ([Ip f2] YH ifYY'YH j+1 (IJ))tTH j

= [(fl Yi(fYY'YHI (fY))tTi, (f2 Yj(fYY'Yj+1 (fY))tTJ

By 11.13, the associative subalgebra mof KfY generated by B = {x - 11x E X} is free with basis B, and, by 11.10, the Lie subalgebra 9 of I(m) generated by B is a free Lie algebra with basis B. We make GI into a Km-module in such a way that tTl is a Km-isomorphism. Thus GI is a faithful Km-module. By 13.2, 9 has the structure of a Km-module such that GI is a Km-submodule of 9 and, given gEm, the mapping a -+ ag (a E g) is an automorphism of the Lie algebra g. Each Gi is a Km-submodule of g, so [i can be made into a Km-module in such a way that tTi is a Km-isomorphism. We prove by induction on i that if /3 is an endomorphism of fY, gEm and (XA2(IJ))g = (X/3)A2(fY) for all x E fY, then

(2)

for all y E Yi(fY). This is trivial for i = 1, and for i > 1, it is sufficient to prove it for y = [I1,f2] with!t E Yi-t(fY),f2 E fY. Then by (1),

Since a -+ ag is a Lie automorphism,

Since the tTj are Km-isomorphisms,

Using the inductive hypothesis and (1), this gives


«YI'i(!j)PYi+1(!j))g)t1i = [«(flfJ)Yi-l(!'WYi(tJ))t1i- l , «(f2{3mpY2(!j))t1l ]

= ([fl{3, f2{3]y;(Ij)PYi+l (1j))t1i

= «[fl' f2]{3)Yi(!j)PYi+1(tJ))t1i

= «y{3)Yi(tJ)PYi+ 1 (tJ))t1i·

This implies (2). By 13.2, there exist integers no, q such that for all n > no,

Gn- rq+l EEl ... EEl Gn-(r-l)q has a KG>-submodule for the regular representation of G>. Hence Gn- rq +1 EEl ... EEl Gn has a free KG>-submodule of rank r. Thus the direct product [n-rq+1 x ... x [n is a KG>-module and has a free KG>-submodule of rank r. But since Yi(tJ)IYi+l (Ij) is torsion-free, it follows from 1.9 that, provided n - rq + 1 > 1, there is a monomorphism r:t. of [n-rq+ 1 X . .. x [n into An(lj)! An+ 1 (tJ) given by

(- - ) p,,-l 1 «(1:) an-rq+l, ... , an r:t. = an- rq+l ... anAn+l 0,

where ai E Yi(tJ) and ai = aiy;(Ij)PYi+1(tJ). To prove the theorem, it is thus sufficient to show that r:t. is a KG>-homomorphism.

Suppose that g E G>. By 13.3a), there is an endomorphism {3 of Ij such that

for allfE Ij. It follows from (2) that

Hence

«an-rq+l' ... , an)g)r:t. = (an- rQ+1{3Y',-l . .. (an{3)An+1(Ij)

= «a%~';:+l ... an){3)An+l (tJ).

By a), the right-hand side is

Thus b) is proved. q.e.d.

The main result of this section will be deduced from 13.4.

13.5 Theorem (BRYANT and KOVACS [1]). Let V be a vector space of dimension greater than lover GF(p), and let f) be a subgroup of the group of non-singular linear transformations of V. Then there exists a finite p-group 'l3 such that 'l3/([>('l3) is isomorphic to V and the group of automorphisms of'l3/([>('l3) induced by all the automorphisms of'l3 corresponds to f).

Proof We may suppose that V = (1/A.2(1Y) for some non-cyclic finitely generated free group (1. Let (f; be the group of all automorphisms of (1/A. 2(1Y). Thus f) ~ (f; and (1/A. 2(1Y) can be regarded as a faithful K(f;module, where K = GF(p). By 13.4, each A.n(IY)/A.n+1 ((1) has the structure of a K(f;-module such that whenever g E (f; and 13 is an endomorphism of (1, the equation

holds for all n provided that it holds for n = 1. Further, there exists n > 1 such that A.n(lY)/ A.n+! (IY) has a K(f;-submodule for the regular representation of (f;. Thus there is a K(f;-isomorphism () of K(f; into A.n (IY)/A.n + 1 (IY). Let 91/A.n+1((1) = (Kf))(). Thus 91/A.n+!(1Y) is a Kf)-submodule of A.n(lY)/ A.n+! (IY) and an element g of (f; lies in f) if and only if (91/A.n+l((1))g ~ 91/A.n+l((1). Since A.n+!(1Y) ~ 91 ~ A.n((1), 91 <J ~ and '.J3 = 3"/91 is a finite p-group, by 2.1.

Let IY. be an automorphism of 'l3 and let g denote the automorphism of (1/A.2(1Y) induced by IY.. By 13.3a), there exists an endomorphism 13 of (1 which carries 91 into itself and induces IY. on (1/91. Also (!A.2((1))g = (ff3)A.2(1Y) for allfE (1, so

for all f E A.n((1). Since 9113 ~ 91,

and g E f). Conversely, suppose that h E f). By 13.3a), there is an endomorphism

13 of (1 which induces h on (1/A. 2 (1Y). Then if f E 91,

since 91/An+10J) is a Kf)-module. Thusff3 E 91 and f3 carries 91 into itself. By 13.3b), f3 induces an automorphism on !j/An+1 (/J) and hence on !j/91 = ~.

Hence the group of automorphisms of !j/A2(!J) induced by all the automorphisms of ~ is precisely f). q.e.d.

13.6 Remarks. a) Theorem 13.5 shows that any subgroup of GL(n, p) is the linear group induced on ~/<P(~) by Aut ~ for some p-group ~. Since any finite group is isomorphic to a subgroup of GL(n, p) for some n, every finite group is isomorphic to the group induced on W<P(~) by Aut ~ for some p-group ~. It has been proved by HEINEKEN and LIEBECK

[1] that for p odd, any finite group is isomorphic to the group induced on (fj/Z(fj) by Aut (fj for some p-group (fj of class 2 and exponent p2.

b) HARTLEY and ROBINSON [1] have used Theorem 13.4 to prove that given a non-identity p' -group f), there exists a group (fj such that F(fj) is a p-group, (fj/F(fj) ~ f), Z(fj) = 1 and every automorphism of (fj is inner.

c) Taking f) = 1 in Theorem 13.5, we obtain a p-group in which every maximal subgroup is characteristic. However Heineken has constructed a p-group in which every normal subgroup is characteristic.

Notes on Chapter VIII

§ 1: The two central series discussed here are among many which were given by LAZARD

[1]. The K-series had been introduced earlier by 1ENNINGS [1] and ZASSENHAUS [4]. The treatment of the A-series follows BLACKBURN and EVENS [1]. § 3: The proof of Theorem 3.5 is that given in PASSMAN [3]. § 4: The method for solving the congruences in 4.1-4.5 is taken from SHAW [1]. § 5: Theorems 5.\-5.5 follow BLACKBURN [1]. Theorems 5.13 and 5.15 originated in G.

HIGMAN [2]. § 7: The results in this section are based on G. HIGMAN [2]. § 9: Theorem 9.12 was first conjectured by 1. WIEGOLD [I]. It was proved in various special cases prior to Vaughan-Lee's proof. § 10: The fact that 9 is nilpotent in 10.12 was proved by G. HIGMAN [1]. The bound on the class given here was obtained by Kreknin and Kostrikin for Lie algebras over a field. § 11: An alternative treatment of free Lie algebras is given by BOURBAKI [3]. The approach to the lower central series offree groups was given by MAGNUS [I, 2]. The treatment given here is derived fro,!11 WITT [1]. § 12: The Golod-Safarevic construction follows FISCHER and STRUlK [I]. § 13: We are indebted to R. M. Bryant for some helpful comments on this section.

Chapter IX

Linear Methods and Soluble Groups

Linear methods have been used extensively for quantitative investigations of soluble groups; for soluble linear groups, bounds in terms of the degree are known for such invariants as the order, derived length, etc. In 1956, P. Hall and G. Higman developed these ideas to obtain upper bounds for the p-Iength of a p-soluble group (fj in terms of various invariants of the Sylow p-subgroup of (fj. There emerged from this a body of techniques which have come to be known as Hall-Higman methods. The present chapter is an introduction to these methods. They are given in an elementary form in § 1, which is already sufficient to solve the restricted Burnside problem of exponent 6.

Let (fj be a p-soluble group and let m, !!3 be normal subgroups of (fj such that m ~ !!3 and m/!!3 is an elementary Abelian p-group. Then m/!B can be regarded as a vector space V over GF(p) and there is a natural representation of (fj on V. Representations over larger fields of characteristic p can be obtained by extension of the ground-field. Such representations were studied in general terms in Chapter VII, but here we are concerned with specific problems involving above all the minimum polynomial of the p-elements. These problems can be reduced to questions about faithful irreducible representations of groups of the form '13.0, where '13 is nilpotent and .0 is normal, in algebraically closed fields. Further conditions are usually imposed; in almost all such situations that have been studied, (1'131, 1.0 I) = 1 and the characteristic of the ground -field does not divide 1.0 I. Also, an extraspecial group often occurs in .0. The question asked is whether or not the representation module has a direct summand isomorphic to the group-ring of '13. If, for instance, this is the case and g is an element of '13 of order k, the minimum polynomial of g is t k - 1.

Hall and Higman's Theorem B (Theorem 2.9) is the classical case of such a theorem; it states that in a p-soluble group (fj of linear transformations of a vector space over a field of characteristic p satisfying the condition Op(fj) = 1, the minimum polynomial of an element of

406 IX. Linear Methods and Soluble Groups

order p" is in general (t - 1)p". But there are exceptional cases and these are studied in § 3, again with specific applications in mind. One of these is a bound on the p-Iength of a p-soluble group in terms of the exponent of a Sylow p-subgroup, and this can be used to reduce the restricted Burnside problem for soluble groups to the case of prime-power exponent. The effectiveness of this reduction is evident in view of Kostrikin's theorem on groups of prime exponent (VIII, 12.4).

Further applications of Theorem B are given in § 5. Among these are the theorem that for p odd, the p-Iength of a p-soluble group is at most the derived length of a Sylow p-subgroup, and similar bounds for the p-Iength in terms of the class and order of the Sylow p-subgroup are given. It is a consequence that the derived length of a soluble group is bounded in terms of the derived lengths of the Sylow subgroups. Another application is the proof that a simple group in which every proper subgroup is p-soluble for some odd prime divisor p of the order can always be generated by 3 elements.

The global invariant of a soluble group ffi corresponding to the p-Iength is the Fitting height h(ffi), and the Hall-Higman methods have been used to give upper bounds for this. For instance, Dade found such a bound for h(ffi) in terms of the composition length of a Carter subgroup. And if m is a soluble group of automorphisms of ffi for which (Iml, Iffi!) = 1 and Iml = PlP2 ... Pn' where Pl' ... , Pn are not necessarily distinct primes, Thompson and Kurzweil found bounds for h(ffi) in terms of h(C(!j(m)) and n. If, further, c(!j(m) = 1, this reduces to a bound on h(ffi) in terms of n alone, and it is conjectured that h(ffi) ::; n. Substantial progress towards the proof of this has been made by Berger. Unfortunately, simple proofs are known for none of these results. But the analogue of Theorem B, in which the characteristic of the field does not divide the order of the p-soluble group, can be used to prove some special cases of this conjecture (§ 6).

Further applications of the Hall-Higman theorems will be found in the next chapter, where they are used to obtained local information about possibly insoluble groups. Many applications of this nature require conditions under which [x, y, y] = 1 implies [x, y] = 1 for elements x, y of a group. In order to formalize this, the technical notion of p-stability is introduced. For p > 3, p-soluble groups are p-stable. This follows from Theorem B, but a more general result is that for p odd, all sections of a group ffi are p-stable if and only if ffi has no section isomorphic to the special affine group SA(2, pl. The proof of this makes use of the Dickson list of subgroups of SL(2, p) rather than the representation theory techniques of the earlier part of the chapter.

Finally, a further criterion for p-length 1 is given in § 8.

§ 1. Introduction 407

§ 1. Introduction

The main aim in this chapter is the derivation of bounds for the p-Iength of a p-soluble group in terms of invariants of the Sylow p-subgroup. Some preliminary remarks about this theory were made in VI, § 6. We begin by enlarging on these.

1.1 Notation. If re is a set of primes, the product of all the normal resubgroups of a finite group <D is a characteristic re-subgroup of <D denoted by O,,(<D). Thus O,,(<D) is the maximal normal re-subgroup of <D, and O,,(<DjO,,(<D)) = 1.

More generally, if reI' re 2 , ... are sets of primes, we define a normal subgroup 0"" ... , ".(<D) of <D by induction on i: for i > 1,

Thus O"" ... ,,,,(<D) is a characteristic subgroup of <D. For example, if p is a prime, the upper p-series of <D is

(cf, VI, 6.1). Observe also that any normal p-nilpotent subgroup 91 of <D is

contained in 0p',p(<D), for the normal p-complement of 91 is contained in Op,(<D). Thus 0p',p(f)) is the maximal normal p-nilpotent subgroup of <D. The Fitting subgroup of (f) is TIp Op (<D).

Dually, O"(<D) is defined to be the intersection of all normal subgroups 91 of <D for which <Dj91 is a re-group. Thus <DjO"(<D) is the maximal re-factor group of <D, and O"(<D) is a characteristic subgroup of <D.

For example, OP(<D) is the normal p-complement of a p-nilpotent group <D.

The basic lemma in VI, § 6 is Lemma 6.5 of Hall and Higman. We shall need slightly different versions of this.

1.2 Lemma. Suppose that re is a set of primes and ~ = O,,(<D). If Sl <J <D, Sl ~ C(!;(~) and ~Slj~ is a rei -group, then there exists a rei -subgroup (£: ofC(!;(~) such that (£: <J <D and Sl = (£: x (Sl (') ~),

Proof ~Slj~ ~ Slj(Sl (') ~) is a rei-group and Sl (') ~ is a re-group. It follows from the Schur-Zassenhaus theorem (I, 18.1) that there exists a subgroup (£: of Sl such that Sl = (£:(Sl (') ~) and (£: (') ~ = 1. Evidently,

(£: is a n'-subgroup of C(fj(~). Thus [(£:, ~] = 1 and S1 = (£: x (S1 n ~). Hence (£: is a normal Hall n'-subgroup of S1, whence (£: is a characteristic subgroup of S1 and (£: <J GJ. q.e.d.

1.3 Lemma. Suppose that GJ is a finite group and that n is a set of primes. Suppose that every chief factor of GJ is either a n-group or ani_group. Then

Proof Let ~ = GJ/O,,(GJ), so that O,,(~) = 1. The assertion is that (£: ~ !l, where (£: = CD(O",(~)) and !l = O",(~). Suppose that (£: f, !l. Then (£: > (£: n !l. Let ~/(£: n !l) be a minimal normal subgroup of ~/(£: n !l) for which ~ ~ (£:. Then ~/(£: n !l) is a chief factor of GJ and is thus either a n-group or a n'-group. If ~/(£: n !l) is a n' -group, then ~!l/!l and ~!l are normal n'-groups, whence ~ ~ O",(~) = !l and ~ ~ (£: n !l, a contradiction. Hence ~/(£: n !l) is a n-group. Thus ~!l/!l ~ ~/(~ n !l) = ~/(£: n !l) is a n-group. By 1.2, there exists a n-subgroup ~ of (£: such that ~ <J ~ and ~ = ~ x (~ n !l). Since 01[(~) = 1, ~ = 1 and ~ ~ !l, a contradiction. q.e.d.

We recall (VII, 13.3) that a group GJ is called p-constrained if C(fj(Op',p(GJ)/Op,(GJ)) ~ 0p',p(GJ).

1.4 Corollary. A p-soluble group is p-constrained.


1.5 Lemma. Suppose that G> is p-constrained. a) IfOp,(G» = 1 and Op(G» ~ ~ ~ G>, then Op,(~) = 1. b) If e E Sp(G», S1 is a pi-subgroup of G> and e ~ N(fj(S1), then

Sl ~ Op,(G».

Proof a) Since 0iG» and Op'(~) are normal subgroups of ~ of coprime orders, Op,(~) ~ C(fj(OiGJ)). Now C(fj(Op(GJ)) ~ Op(GJ), since Op,(GJ) = 1. Thus Op,(~) is a pi-subgroup of Op(GJ) and so Op,(~) = 1.

b) Let m = G>/Op,(G», 6 = eOp,(GJ)/Op,(GJ),3t = S10p' (GJ)jOp' (GJ). Thus Op,(m) = 1 and G> is p-constrained. Since 6 E Sp(GJ), Op(m) ~ 6. Also 6 ~ N~(~), so 6~ is a subgroup of ~ containing Op(~). By a), Op,(6ft) = 1. Since S1 is a pi-group, Op,(6ft) = 3t. Hence ft = 1 and Sl ~ Op,(G». q.e.d.

Lemma VI, 6.5 follows from 1.4 and the following.

1.6 Lemma. Suppose that G} is a p-constrained group and that the subgroup U of G} is defined by

Proof We may assume that Op,(G}) = 1. The~ U = cP(Op(G})), so Op(G})/U is elementary Abelian. Thus C(!j(Op(G})jU) ~ Op(G}). Suppose that <r = C(!j(Op(G})/U) > Op(G}). Since <r <J G}, <r/Op(G}) is not a p-group. Let a be an element of <r - Op(G}) of order prime to p, and let IX be the automorphism of 0iG}) given by XIX = a- l xa (x E 0p(G})). Since a E C(!j(Op(Gl)/cP(Op(G}))), IX induces the identity automorphism on 0p(G})/cP(Op(G})). By III, 3.18, IX = 1', since the order of IX is prime to p. Thus a E C(!j(Op(G})). Since G} is p-constrained and Op,(G}) = 1, C(jj(Op(G})) :::; Op(G}). Thus a E O/G}), a contradiction. Hence <r = Op(Gl). q.e.d.

1.7 Notation. Suppose that m is a group of operators on a group G}, that IB <J m and that gb = g for all b E lB. Given a E m, the mapping g -+ ga is an' automorphism of G} which depends only on alB: we denote this automorphism by p(alB). Then p is a homomorphism ofm/IB into the group of automorphisms of G}. We write (m/IB on G}) for p.

This notation is particularly useful in the following context. Suppose that f) :::; G}, ft <J f), m :::; N(!j(f)), m :::; N(!j(ft), IB <J m and [f), IB] :::; R Then m acts as a group of operators on f)/ft, with (hft)a = haft (h E f), a Em), and (hft)b = hft for all b E lB. We may thus speak of (mjlB on f)/ft). Note that

ker(mjlB on f)/ft) = C~/!lI(f)/ft).

The conclusion of 1.6 is that (G}/Op'.p(G}) on Op',p(Gl)/U) is faithful. Now 0p',p(Gl)/U is an elementary Abelian p-group and can therefore be regarded as a vector space over GF(p). The following lemma is an interpretation of commutators and powers in linear terms.

1.8 Lemma. Suppose that U <J G}, m <J G}, m :::; U and u/m is an elementary Abelian p-group. We regard U = wm as a vector space over GF{p) and write p = (G) on U).

a) Ifu E U and Xl' ... , Xn are elements ofG},


b) If U E U, X E (fj and n is a positive integer,

Proof a) We have

and the assertion follows by induction on n.

q.e.d.

From 1.8, we see that u'B is carried by (p(x) - l)n into [u, X, ... , x] 'B and by 1 + p(x) + ... + p(xr 1 into x-n(xut'B. Thus whether or not certain powers and commutators are equal to 1 depends on the vanishing of certain polynomials in p(x), and it is therefore necessary to study the minimum polynomial of p(x). To do this we prove the following lemma.

1.9 Lemma. Let V be a vector space of finite dimension over a field K, and let (fj be a group of non-singular linear transformations of V. Suppose that 91 is a normal Abelian subgroup of (fj, that char K does not divide 1911 and that K is a splittingfieldfor 91. Supposefurther that ~ :s; (fj and C\jl91) = 1. Then there exist K91-submodules V1, ... , Vm of V such that

and there exists a faithful permutation representation (J of ~ on {I, ... , m} such that Vig = Vi(f(gJor all g E ~ (i = 1, ... , m).

Proof Since K is a splitting field for 91, any irreducible representation of 91 in K is of degree 1 (e.g. V, 11.4). Let Pl' ... ,Pm be the distinct irreducible representations of 91 in K, and let

Vi = {vlv E V, VX = Pi(X)V for all x Em}.

By the Maschke-Schur theorem, V is the direct sum of irreducible K91-submodules. Since each such submodule is contained in some Vi'

By VIII, 5.16,

Now if g E~, X --+ Pi(gXg- l ) is an irreducible representation of in in K, so there exists j such that Pi(gXg- l ) = pix) for all x E in. We put j = iO'(g), so

Pi(gXg- l ) = Pit1(g)(X) (x E in).

If also h E~,

whence O'(gh) = O'(g)O'(h). Also 0'(1) = 1, so a is a permutation representation of ~ on {1, ... , m}.1f v E Vi and g E~,

for all x E in, so vg E Vit1(g). Thus Vig = Vit1(g) for all g E ~. It remains to show that a is faithful.

Suppose that g E ~ and g =1= 1. By hypothesis, g ¢ C\Il(in), so there exists y E in such that z = [y, g-I] =1= 1. Now zEin and there is some j such that z does not induce the identity mapping on Vj. Thus piz) =1= 1, and since yz = gyg-l,

Hence jO'(g) =1= j and O'(g) =1= 1. Hence a is faithful. q.e.d.

1.10Theorem (G. HIGMAN). Let V be a vector space offinite dimension over a field K, and let m be a group of non-singular linear transformations ofV. Let in be a normal, Abelian subgroup ofm such that char K does not divide linl. Suppose that a is an element of m of order pn and that aP·-1 ¢ C(!;(in). Then the minimum polynomial of a is tP' - 1.

Proof Since the minimum polynomial of a linear transformation is unaltered by an extension of the ground-field, we may suppose that K is a splitting field for in. We apply 1.9 with ~ = <a). Thus there exist Kin-submodules VI' ... , Vm of V such that

and there exists a permutation oe of {I, ... , m} such that oe is of order pR and Via = Via. (i = 1, ... , m). Now oe has a cycle (i, ioe, ... , ioe pn- 1) of length pR. If v is a non-zero element of Vi' vai E Via.i , so v, va, ... , vaP'-l are linearly independent. Hence 1, a, ... ,apn - 1 are linearly independent, and the degree of the minimum polynomial of a is at least pR. Since, however, apn = 1, the minimum polynomial of a is a divisor of t P' - 1. The minimum polynomial of a is thus precisely t P' - 1. q.e.d.

Theorem 1.10 can be applied, for example, to replace step r) in the proof of IV, 6.2. In 1.13, it will be seen how results of this kind may be used to obtain bounds on the p-length of"the p-soluble group (fj. We need the following, which will be much used in t.he sequel.

1.11 Theorem. Suppose that n is a set of primes, (fj is a n-group and that ~ is a nf -group of operators on (fj. Suppose ·that a s;;; n. Suppose further that either (1) (fj is soluble, or (2) a consists of just one prime p and either (fj or ~ is soluble.

a) There exists an ~-invariant Hall a-subgroup of(fj. b) If ~l' ~2 are ~-invariant Hall a-subgroups of (fj, there exists

x E C(!)(~) such that ~~ = ~2' c) Any ~-invariant a-subgroup of (fj is contained in an ~-invariant

Hall a-subgroup of (fj, provided that, in case (1), ~ is soluble.

Proof Let (fj be the semi direct product of (fj and ~; thus (fj = (fj~,

(fj <l (fj and (fj n ~ = 1. a) By VI, 1.8 in case (1), or Sylow's theorem in case (2), (fj possesses

a Hall a-subgroup .0 and all Hall a-subgroups of (fj are conjugate. By the Frattini argument, (fj = ft(fj, where ft = Nffi(.Q). Thus

and ftj(ft n (fj) is a nf_group. Since ft n (fj is a n-group, it follows from the Schur-Zassenhaus theorem (1,18.1) that there exists £ ~ ft such that ft = £ (ft n (fj) and (fj n £ = 1. Thus (fj = ft(fj = £(ft n (fj)(fj = £(fj, so £, ~ are both complements of (fj in (fj. Since either (fj or ~ is soluble, it follows from I, 18.2 that ~ = £X for some x E (fj. Since £ ~ ft = Nffi(.Q), ~ = £x ~ Nffi(.QX). Also .ox ~ (fj since (fj <l (fj, so .ox is an ~-invariant Hall a-subgroup of (fj.

b) Let 91 = Nffi(~l)' 91 = N(jj(~l)' Since ~ ~ 91, 91 = «fj n 91)~ = 91~. Now by VI, 1.8 or Sylow's theorem, ~l = ~~ for some g E (fj.

Hence ~g ~ Nffi(~~) = 91 and 1911 = 19111~1 = 191~gl, so 91 = 91~g also. Thus ~, ~g are complements of the normal Hall n-subgroup 91 of

91, and by I, 18.2, 21g = 21u for some u E 91. Write u = av for a E 21, v E 91; then 21V = 21av = 21u = 21g• Thus v = xg for some x E No;(21). Hence [21, x] ~ 21 n m = 1, so X E Co;(21). Also ~~ = ~~g-l = ~( = ~2'

c) We give different proofs in the two cases. Let 1 be an 21-invariant a-subgroup of m.

In case (1), m is soluble. Thus 121 is contained in some Hall (0' u n')subgroup 5l of m, by VI, 1.8. Thus 1 ~ 5l n m and 5l n m is an 21-invariant Hall a-subgroup of m.

In case (2), we proceed by induction on 1m: 11. If 1m: 11 is prime to p, 1 is an 21-invariant Sylow p-subgroup of m and there is nothing more to prove. Otherwise there exists 6 E Sp«fj) such that 1 < 6. By I, 8.8, 1 < Ns (1), so the order of a Sylow p-subgroup of No; (1) is greater than Ill. But No;(l) is 21-invariant, so by a), No; (1) has an 21-invariant Sylow p-subgroup :t. Thus :t ~ 1 and l:tl > Ill, so :t > 1. By the inductive hypothesis, :t is contained in an 21-invariant Sylow p-subgroup of G).

Hence, so is 1. q.e.d.

1.12 Corollary. Let n be a set of primes, let ~ be a n'-subgroup of a group G) and let 5l be a subgroup of~. Suppose that

(i) 91 is a n-subgroup of G);

(ii) ~ ~ No;(91) but 5l $ Co;(91); (iii) either ~ or 91 is soluble.

Then there exists a Sylow subgroup ~ of 91 such that ~ ~ N\lj(~) but 5l $. C\lj(~).

Proof For any pEn, there exists ~ E Sp(91) such that ~ ~ No;(~), by 1.11. But since 1911 = TIpe"I~I, 91 = <~Ip En). Since 5l f, Co;(91), 5l $ Co;(~) for at least one ~. q.e.d.

1.13 Theorem. Suppose that G) is a p-soluble group and thatfor each prime q #- p, the Sylow q-subgroups of G) are Abelian. If the exponent of the Sylow p-subgroups is pn, the p-Iength ofG) is at most n.

Proof This is proved by induction on the p-Iength I of G). If I = 1, G) is not a pi_group and there is nothing further to prove. Suppose that I > 1; let 91 = 0p'.p(G)) and ~ = G)/91. Then Ip(~) = I - 1. By the inductiye hypothesis, ~ possesses an element a of order pH. Let b = aP'-2.

Since 91 = 0p'.p(G)), Op(~) = 1, so by 1.3, C!;(Op'(~)) ~ Op'(~)' Thus b does not centralize Op'(~)' Let .Q be a minimal subgroup of Op'(~) which is normalised by a but not centralized by b. By 1.12, .Q is

a q-group for some prime q =f. p. By hypothesis, .0 is Abelian. Let Sl = <a, .0).

By 1.4, GJ is p-constrained, so by 1.6, C(!i(91jU) = 91, where UjOp,(GJ) = <l>(Op',p(GJ)/Op,(GJ)). Hence, if p = (~ on 91jU), p is faithful. We may apply 1.10 to p(Sl), since pt 1.01. Thus the minimum polynomial of p(a) is tp'-l - 1. Hence 1 + p(a) + ... + p(ay'-'-1 =f. 0, and there exists x E 91/U such that x(1 + p(a) + ... + p(ay'-1-1) =f. 1. We may choose a p-element g E GJ such that a = g91. If 6 E Sp(GJ) and g E 6, then (6 n 91)Op,(GJ) = 91 and we may choose x E 6 n 91 such that xU = x. By 1.8b), g-pl-1(gx)pl-l fj U. Since g and gx are both p-elements, it follows that either g or gx is of order at least pl. Thus the exponent of the Sylow p-subgroups of GJ is at least pl. q.e.d.

We can use this to obtain the best possible bound for the order of a finite group of exponent 6 with d generators. For this, we also need I, 19.11, which will now be proved.

1.14 Theorem (SCHREIER). Let ts: be a free group with group-basis X and suppose ~ :$ ts:. Then ~ is free. More precisely, there exists a transversal S of ~ in ts: with the following properties.

a) 1 E S. b) If s E S and x E X, SXS,-1 E ~ for a unique s' E S; put P(s, x) =

SXS' -1. There exists an injective mapping n of S - {1} into the Cartesian product S x X such that if sn = (Sl' Xl) with Sl E S, Xl E X, then P(S1' Xl) = 1 and either S1 = S or SlX1 = S.

c) Let Y = {P(s, x)ls E s, X E X, (s, x) =f. s'n for all s' E S}. Then Y is a group-basis of ~.

In particular, if ts: is of rank d and Its:: ~I = j, the rank of ~ is (d - l)j + 1.

Proof Since ts: = <X), every element f of ts: is expressible in the form

f = x" ... x'· 1 n ,

where n ~ 0, Xi E X and Gi = ± 1. We choose such an expression with n minimal and write n = l(f). Thus l(f) = ° if and only if f = 1. For any non-empty subset U of ts:' let

I(U) = min {1(f)lf E U}.

We construct a transversal S of ~ in ts: with the following properties. (i) 1 E S; more generally, l(s) = l(~s) for all s E S.

(ii) There exists a mapping A of S - {1} into X u X-I such that if S E Sand S =j:. 1, then SA(Sr l E Sand l(SA(Srl) < l(s).

We define the representative of the coset Sjf(f E m by induction on l(Sjf). If l(Sjf) = 0, let 1 be the representative. If l(Sjf) = n > 0, Sjf contains an element l' for which 1(1') = n. Hence l' = fl x', where l(fl) = n - 1, x E X and B = ± 1. By the inductive hypothesis, a representative Sl of Sjfl has already been chosen and l(sl) ~ n - 1. Since Sjf = Sj1' = Sjfl x' = Sjs 1 x', we may choose S = SIX' for our representative of Sjf and put A(S) = x'. Since l(sl) ~ n - 1, l(s) ~ n, but since S E Sjf, l(s) ~ n; thus l(s) = n. Also SA(Sr l = Sl E Sand l(SA(Srl) = l(sl) < n. Thus if S is a transversal constructed in this way, S has the properties (i) and (ii).

Suppose that S E S and x E X. Then sx E Sjs' for a unique s' E 5. Put f3(s, x) = SXS,-l; thus f3(s, x) E Sj. We define the mapping n of 5 - {1} into 5 x X as follows. If S E 5 - {l} and A(S) = X-I for some x EX, put sn = (S, x); since sx = SA(Sr l E 5, Sjsx n 5 = {sx} and f3(s, x) = 1. If S E 5 - {1} and A(S) = Y E X, put sn = (SA(srl, y); since SA(srly = S E 5, f3(SA(Sr l, y) = 1. We prove that n is injective. Suppose, then, that SIn = s2n, where Sl, S2 are in 5 - {l}. If A(SJ = xiI with XiEX (i = 1,2), then (Sl' Xl) = sIn = S2n = (S2' x2) and Sl = S2. If A(Si) = Yi E X (i = 1,2), (StYlI, Yl) = sIn = S2 n = (S2Y2I, Y2) and again Sl = S2. We must therefore show that it is not possible to have A(Sl) = x-I,

A(S2) = Y with x, Y in X. If this is the case, (Sl, x) = SIn = S2n = (S2y- l, y), so Y = x and S2 = StX. But then by (ii),

and

a contradiction. Thus a) and b) are proved. By I, 19.2, there exists a free group <D having a group-basis consisting

of elements g(s, x), where S E 5, x E X and (s, x) =j:. s'n for all s' E S. Thus there is a homomorphism rJ of <D into Sj such that g(s, x)rJ = f3(s, x). If (s, x) = s'n for some s' E 5, put g(s, x) = 10;; thus g(s, x)rJ = f3(s, x) for all S E S, X E X.

Our aim is to prove that rJ is an isomorphism. The fact that it is an epimorphism is equivalent to the assertion

Sj = <P(s, x)ls E 5, x EX).

To prove this, take h E ~ and write

h = X1' ... X~" (Xi E X, t:i = ± 1).

For i = 0, ... , n, there exists Si E S such that

~X1' ... xi' = ~Si'

Inparticular,so = Sn = 1,sincehE~.Thenifl ~ i ~ n,~si = ~si-1xi', ,. -1 "Th ,. -1 fJ( ) 'f 1 d ,. -1 SOSi-1Xi'Si E<!}. USSi- 1Xi 'Si = Si-1,Xi 1 t:i = + an Si-1Xi'Si =

fJ(Si' X;)-l if t:i = -1. Since

hE <fJ(s, x)ls E S, x E X). Hence 1] is an epimorphism. Now let Q be the Cartesian product (f) x S. For each x E X, we

define mappings x~, x~' of Q into Q as follows. If u E (f), S E Sand sx E ~s' (s' E S), put

(u, s)(x~) = (ug(s, x), s').

If u E (f), S E Sand sx-1 E ~s" (s" E S), put

(u, s)(x~') = (ug(s", x)-l, s").

Thus

(u, s)(x~)(xO = (ug(s, x)g(t, xtl, t),

where t E Sand s'x-1 E ~t; but then ~t = ~S'X-1 = ~s, so t = sand (x~)(xO = 1. Similarly (x~')(x~) = 1, so x~ is a permutation of Q. Since X is a group-basis of (y, ~ can be extended to a homomorphism of (Y into the group of all permutations of Q.

We prove by induction on I(s) that if u E (f) and S E S,

(1) (u, 1)(s~) = (u, s).

This is trivial if I(s) = 0. If I(s) > 0, we have s = SlX', where x E X, x' = A(S), Sl E Sand l(sl) < I(s) by (ii). By the inductive hypothesis,


Thus

If e = 1, then sn = (S1> x) and g(Sto x) = 1; thus

(u, 1)(s~) = (Ug(Sl' x), s) = (u, s).

If e = -1, then sn = (s, x), so g(s, x) = 1; thus

(u, 1)(s~) = (u, Sl)(X~rl = (u, Sl)(X~') = (ug(s, xr 1, s) = (u, s).

Thus (1) is proved. Next, we show that for any u E m, gem,

(2) (u, 1)(gl1e) = (ug, 1).

First note that the set of gem for which (2) is true is a subgroup of m. Thus it is sufficient to prove (2) for g = g(s, x). Suppose that sx E ~s' (s' E S). Then

g(s, x)11 = P(s, x) = SXS'-l,

and

(u, 1 )(g(s, x )l1e) = (u, 1 )(se)(xe)(s' er 1

= (u, s)(xe)(s'er1 by (1)

= (ug(s, x), s')(s' ~)-1

= (ug(s, x), 1) by (1).

Thus (2) is proved. It follows at once from (2) that 11 is a monomorphism. Hence 11 is an

isomorphism of m onto ~. This implies c). Iftjisofrankdandltj:~1 =j,IXI = dandlSI =j.Hencethetotal

number of pairs (s, x) is jd. Since n is injective, precisely j - 1 of these are of the form s'n for some s' E S. There remain jd - j + 1, so I YI = jd - j + 1 = (d - 1)j + 1. q.e.d.

1.15 Theorem (HALL and IDGMAN [IJ). Suppose that OJ is ajinite group of exponent 6 with d generators. Then Iml is a divisor of

where

m = 1 + (d - 1)' 3d+(~)+(~), n = 1 + (d - 1)' 2d.

Proof This is proved along the same lines as VI, 6.7. Since (f) is of exponent 6, I (f) I = 2a3b for suitable a, b. By V, 7.3, (f) is soluble. The Sylow 2-subgroups of (f) are of exponent 2 and are therefore Abelian. Hence by VI, 6.6a), 12(f)) ::; 1. Thus (f)/03, 2 (f)) is a grou~ of exponent 3 with d generators. Hence by III, 6.6, I (f)/03, 2(f))1 ::; 3d+(z)+(3). By 1.14, 03,2(f)) can be generated by m elements. Now if 6 E S2(f)), 0 3, 2 (f)) = 03(f))6, since 12(f)) ::; 1. Thus 6 is generated by m elements. Since 6 is Abelian and of exponent 2, 161 = 2a with a ::; m.

Since 6 is Abelian, 13(f)) ::; 1 by 1.13. Thus (f)/02 3(f)) is an elementary Abelian 2-group with d generators. Hence I (f)io2,3 (f)) I ::; 2d. By 1.14, 02,3(f)) can be generated by n elements. If l: E S3(f)), 02,3(f)) = O2 (f))l:, so l: is generated by n elements. Since l: is of exponent 3, Il:l = 3b with b::; n + (2) + (3). Thus 1(f)1 = 2a3b divides 2m'3n+(~)+m

q.e.d.

1.16 Remark. The bound in 1.15 is best possible. To see this, we need the fact that the bound for the order of a group with d generators and exponent 3 given in III, 6.6 is best possible. We shall not prove this, but we interpret it in the following way. Let fJ be a free group of rank d and let

Then any group of exponent 3 with d generators is a homomorphic image of fJ/fJ3, so I fJ/fJ3 I is the best bound for the order of a group with d generators and exponent 3. Thus

I fJ : fJ31 = 3d+(~)+(~).

Hence by 1.14, fJ3 is a free group of rank m, where

m = 1 + (d - 1)' 3d+(~)+(~).

Now if9Jl = <y21Y E fJ3), fJ3/9Jl is the largest group of exponent 2 with m generators. Hence 1 fJ3/9Jl1 = 2m. Note that 9Jl is a characteristic subgroup of fJ3 and hence of fJ. Also fJ/9Jl is of exponent 6.

§ 2. Hall and Higman's Theorem B 419

Similarly, if

tj/!J2 is of order 2d and tj2 is a free group of rank n, where

Finally, if 91 = <i Iy E tj2), 1 tj2/91 1 = 3n+(~)+m, 91 is a characteristic subgroup of tj and tj/91 is of exponent 6. Let ft = 9Jl (') 91. Then tj/ft is a finite group of exponent 6, and 1 tj/ftl is divisible by 1 tj3/9Jl1 = 2m

and by 1 tj2/91 1 = 3n+(~)+m. Using 1.15, it follows that

1 tj/ftl = 2m . 3n+(2)+m.

In fact, ft = <z6lz E tj). For there is no infinite group of exponent 6 with d generators (M. HALL [1]), and no finite one is larger than tj/R

§ 2. Hall and Higman's Theorem B

In order to apply results of the form of Theorem 1.10 more generally, it is necessary to remove the hypothesis that 91 is Abelian. Theorem 1.10 does not remain true if this is done, but it is clear from 1.12 and III, 13.5 that in a minimal counterexample, 91 is a non-Abelian special q-group for some prime q (that is, 4>(91) = 91' = Z(91) (III, 13.1)). By further reduction, additional restrictions are obtained, and we are led to consider the groups described in the following definition.

2.1 Defmition. A group (fj is called critical if it satisfies the following conditions.

a) (fj has an extraspecial normal q-subgroup .0 for some prime q. b) There exists a cyclic non-identity q'-subgroup 6 of (fj such that

(fj = 6.0 and 6 (') .0 = 1. c) [.0', 6] = 1 and each non-identity element y E 6 induces a fixed

point free automorphism on .0/.0'. Thus if 1 "# Y E 6 and x E .0 - .0', xy,o' "# x,o'.

(For the structure of extraspecial groups, see III, 13.7 and 13.8.)

2.2 Remarks. Suppose (fj = .06 is a critical group and 1.01 = q2m+l.


a) By 2.1c), (fj/.Q' is a Frobenius group (V, 8.1) with Frobenius kernel .0/.0' and Frobenius partition

(.0/.0') u (U ~},O,/,o,). \tET

where T is a transversal of .0' in .0. b) Since .0 is extraspecial, 1,0'1 = q; thus there is an isomorphism

r:t. of,O' onto the additive group of GF(q). We may write

(x,Q',y,Q') = [x,y]r:t. (XE'o,YE,Q);

then .0/.0' becomes a non-singular symplectic vector space over GF(q), by III, 13.7b). Since [.0', 6] = 1,

(xs'o',ys.Q') = [XS,yS]r:t. = [x,yJ'r:t.

= [x, y]r:t. = (x'o', y'o')

for all x E .0, y E ,0, S E 6. Thus 6 induces a subgroup of the symplectic group Sp(2m, q) on .0/,0'.

In V, 17.13, the complex characters of (fj were determined, and it was a side-result that qm == ± 1 (161). We give here a more direct proof of this, using the following lemma.

2.3 Lemma. Suppose that V is a non-singular symplectic vector space over K, (fj is a subgroup of the symplectic group on V and char K does not divide l(fjl. If every irreducible K(fj-submodule of V is isotropic, then V is the direct sum of an even number of irreducible K(fj-submodules.

Proof This is proved by induction on dimK V. Since char K does not divide I (fj I, we can write

for irreducible K(fj-modules Vi' By hypothesis, each Vi is isotropic, so r > 1. If r = 2, there is nothing to prove, so we suppose that r > 2. Suppose that 0 #- U E V1. Then (u, v) #- 0 for some v E V. Write v = V 1 + ... + Vn with Vi E Vi' Since (u, v1) = 0, there exists i > 1 such that (u, v;) "# O. Let U = V1 EB Vi' Then U is not isotropic and, since r > 2, U "# V. Let T be the radical of U. Since U is a K(!)-submodule, so is T.


Hence U = T EB W for some Km-submodule W. If 0 =1= w E W, then w ¢ T, so (w, u') =1= 0 for some u' E U. Write u' = t' + Wi with t' E T. Wi E W. Then (w, Wi) =1= O. Hence W is a non-singular proper Km-submodule ofV. Since U is not isotropic, W =1= O. By II, 9.4d), V = W EB W~. Since W~ is also non-singular, the inductive hypothesis may be applied to W and W~; the assertion then follows at once. q.e.d.

2.4 Lemma. Suppose that m = .06 is a critical group, where .0 is a normal extraspecial q-subgroup of m and 6 is a cyclic q'-subgroup. If 1.01 = q2m+1, qm == ± 1(1 6 1).

Proof Let V be an additive group isomorphic to .0/.0'. By 2.2b), V can be regarded as a non-singular symplectic space over K = GF(q). Also, V is a K6-module and 6 induces a subgroup of the symplectic group on V. And by 2.1c), vy =1= v if v E V - {O} and y E 6 - {l}. Thus 6 is represented faithfully on any non-zero K6-submodule of V.

Let 6 = (s) and let I be the smallest positive integer such that ql == 1 (161). Then by II, 3.10, any irreducible K6-submodule of V is of dimension I, since 6 is represented faithfully on it. By the MaschkeSchur theorem,

where the Ui are irreducible KG-submodules. Thus dimK Ui = I and Ir = dimK V = 2m.

If r is even, say r = 2r', then Ir' = m and qm = (ql)" == 1 (161), as required. If r is odd, then by 2.3, there is an irreducible K6-submodule U of V which is not isotropic. Thus U ~ U~ and U (') U~ c U. But U~ is a K6-submodule and U is irreducible. Thus U (') U~ = 0 and U is non-singular. Since dimK U = I, I is even, by II, 9.6b). If I = 21', m = l'r. By II, 9.23,161 divides ql' + 1. Since r is odd, { + 1 divides ql" + 1 = qm + 1. Thus qm == -1 (161). q.e.d.

2.5 Lemma. Suppose that m = .06, where .0 is a normal extraspecial q-subgroup of order q2m+l, 6 = (s) is cyclic and [6, Z(.o)] = 1. Let K be an algebraically closed field with char K =1= q, and let V be an irreducible Km-module on which .0 is represented faithfully. Then V is an irreducible K.o-module and dimK V = qm.

Proof Since Z(.o) :s; Z(f)), V is irreducible and K is algebraically closed, Z(.o) is represented on V by scalar multiples of the unit matrix (V, 4.3).


Thus there exists a non-trivial homomorphism a of Z(Q) into K x such that

vx = a(x)v (v E V, X E Z(.Q».

Hence if U is any irreducible K.Q-submodule of V,

ux = a(x)u (u E U, X E Z(.Q».

It follows from V, 16.14 that all irreducible K.Q-submodules of V are isomorphic. By V, 17.3, V is the direct sum of k isomorphic irreducible K.Q-submodules, and by V, 17.5, 6 ~ (fj/.Q has an irreducible projective representation p of degree k. If n is the order of sand p(s) = A, then A" = AI, where 0 =1= .Ie E K. But since K is algebraically closed, .Ie = fl" for some fl E K. Hence 6 has an ordinary irreducible representation s ~ fl- 1 A of degree k. By V, 6.1, k = 1. Thus V is a faithful irreducible K.Q-module. By V, 16.14, dimK V = qm. q.e.d.

Next we obtain the analogue of V, 17.13 in the case when 6 is a p-group and char K = p.

2.6 Theorem (HALL and IDGMAN). Suppose that (fj = .06 is a critical group,.Q is a normal extraspecial q-subgroup of order q2m+1 and 6 = (s) is a cyclic group of order pa. Let K be an algebraically closed field of characteristic p and let V be an irreducible K(fj-module on which .0 is represented faithfully. Then V is an irreducible K.Q-module and dimK V = qm. As a K6-module, V = FEEl W, where F is afree K6-module, W is indecomposable and dimK W is either 1 or pa - 1. (It is possible that F = 0.)

In particular, the minimum polynomial of the linear transformation v --+ VS of V is (t - 1)', where either r = pa or r = pa - 1 = qm.

Proof By 2.5, V is an irreducible K.Q-module and dimK V = qm. Let 3 = Z(.Q), ~ = 63. Thus l(fj: ~I = q2m. Denote V, regarded

as a K~-module, by U. By VII, 4.3, V is an indecomposable direct summand of Uffi • But if

where the Ui are indecomposable K~-submodules of U,


It follows from the Krull-Schmidt theorem (I, 12.3) that V is a direct summand of UT for some i. Put W = Ui . Thus W is an indecomposable Kf)-submodule of U, and V is a direct summand of WffJ.

If x E (f) - f), the co sets f)x, f)xs, ... ,f)xspa - 1 are distinct. For x = ysr with y E.o. Thus if f)xs i = f)xs i, YSi-iy-l = XSi-iX- 1 E f), whence s-(i-i)ysi-iy - 1 E f) (l .0 = 3 and 3y = 3 ys;-i. By 2.1c), Si = si, since otherwise y E 3 and x E f).

Hence there exists a set T such that

is a transversal of f) in (f). Thus

where W1 = W Q9 1 and, for each t E T,

WI = (W ® t) EB (W ® ts) EB ... EB (W ® ts pa- 1).

Now it is clear that WI is a free KS-module, for a K-basis of W ® t is a KS-basis of WI' Hence

where each Fi is a KS-submodule of WffJ isomorphic to KS. By VII, 5.3, Fi is an indecomposable KS-module. Also, W1 is an indecomposable KS-module. For as a Kf)-module, W1 is indecomposable, being isomorphic to W. But since W is a submodule of U, 3 is represented on W by scalar multiplications, so any KS-submodule of W is also a Kf)-module.

Since V is a direct summand ofWffJ, it follows from the Krull-Schmidt theorem that if

is a direct decomposition of V into indecomposable KS-modules, at most one summand is isomorphic to Wand the rest are isomorphic to KS.1f all are isomorphic to KS,


which is impossible since q =P p. Thus we may suppose that VI ~ W, so

By 2.4, qm == ± 1 (pa), so dimK VI == ± 1 (pa). By VII, 5.3, either dimK VI = 1 or VI is the unique indecomposable K6-module of dimension pa _ 1. In the latter case, the minimum polynomial of the mapping u -t us (u E VI) on VI is (t - l)pa- I . And if dimK VI = 1, h > 1. Hence the minimum polynomial of the mapping v -t vs on V is (t - l)pa, except when h = 1, in which case it is (t - Wa- I and dimK V = qm = pa _ 1. q.e.d.

The equation qm = pa - 1, which arises in the exceptional case of 2.6, will now be analysed.

2.7 Lemma. Suppose that pa = qb + 1, where p, q are primes and a, b are positive integers. Then either

a) p = 2, b = 1, a is a prime and q = 2a - 1 is a Mersenne prime, or

b) q = 2, a = 1, b = 2m and p = 22m + 1 is a Fermat prime, or c) pa = 9, qb = 8.

Proof Obviously p = 2 or q = 2. a) Suppose that p = 2. Then a > 1. If b = 2e,

q2C _ 1 = pa - 2 =1= 0 (4).

This is a contradiction, since q is odd. Thus b is odd and

2a = (q + l)(qb-I _ qb-2 + ... + 1).

Hence q + 1 = 2d for some d, and

2a- d = qb-I _ qb-2 + ... + 1 == b == 1 (2).

Thus a = d and b = 1. If a = ef,

q = 2a - 1 = (2e - 1)(2e(f-I) + ... + 1),

so either 2e - 1 = 1 and e = 1, or 2e - 1 = q and e = a. Hence a is prime.

b) Suppose that q = 2. Thus pa = 2b + 1 and

2b = pa _ 1 = (p - 1 )(pa-l + . . . + 1).

Thus p - 1 = 2C for some c, and pa-l + ... + 1 = 2b- c•

If a > 1, 2b- c = pa-l + ... + 1 > 1, so a == 2b- c == 0 (2). If a = 2d,

so pd - 1 = 2e and pd + 1 = 2f for suitable e, f. Thus 2f - 2e = 2, and e = 1, f = 2. Thus pd = 3, pa = 9 and qb = 8.

If a = 1, p = 2b + 1. If b = 2cd, where d is odd,

p = (22' + 1)(2(d-l)2' - ... + 1).

Thus 22' + 1 = p and b = 2c• q.e.d.

In order to apply 2.6 to more general situations we need the following lemma.

2.8 Lemma. Let V be a finite-dimensional vector space over a field K, and let (f) be a p-soluble group of non-singular linear transformations of V for which Op(f)) = 1 and char K does not divide IOp,(f))I. Let '.p be a p-subgroup of (f) and suppose that 1 # y E '.p. Then there exists a section f) = (f)t/(f)2 of(f) with the following properties.

a) (f)l ~ '.p, and y 1: (f)2' Let y = y(f)2' b) f) = .06, where .0 is a normal, special q-subgroup of f) for some

prime q # p, and 6 = '.p(f)2/(f)2 E Sp(f)). c) .0/<1>(.0) is a minimal normal subgroup of f)/<1>(.Q), and y centralizes

<1>(.0) but not .0. d) f) has a faithful, irreducible representation on a section Vt/V2 of

V. Further, if the representation of.Q on V1jV2 is absolutely irreducible, .0 is extraspecial.

Proof. By 1.3, C(j)(Op,(f))) ~ Op,(f)). Hence y does not centralize Op,(f)). Thus, if fI' is the set of subgroups I of Op,(f)) for which '.p ~ No; (I) and y 1: C(j)(I), Op,(f)) E fI'. Let I be an element of fI' of minimal order, and choose x E I such that z = [x, y] # 1. Let (f) 1 = '.pI and let

V = Vo > V1 > ... > Vn = 0

be a K(f)ccomposition series of V. Since char K does not divide the order

of z, there exists a K <z )-submodule Ui of Vi- 1 such that Vi- 1 = Ui EEl Vi (i = 1, ... , n), by 1,17.7. Thus

Since z f:. 1, there exists some Ui on which z induces a non-identity linear transformation. Let (fj2 be the kernel of the representation of (fj1 induced on Vi-dVi. Thus ~ = (fjd(fj2 has a faithful irreducible representation on Vi-dVi' Since Ui is K<z)-isomorphic to Vi- 1/Vi, z f/: (fj2' Since (fj2 <J (fj1' it follows that y f/: (fj2' Thus a) is satisfied.

Let .0 = X(fj2/(fj2' 6 = ~(fj2/(fj2' Then .0 <J ~, ~ = .06 and, since ~ E Sp(fj1)' 6 E Sp(~). If ID < X and ~ ::; Nfj(ID), then ID f/: ff, since X is an element of [I' of minimal order, so y E Co; (ID). Hence by 1.12, X is a q-group for some prime q f:. p. Thus .0 is a q-group. Since z f/: (fj2 and X(fj2 E .0, Y does not centralize .0.

Let 6 be the normal closure of y in 6. We show that 6 centralizes any proper 6-invariant subgroup .00 of .0. We have .00 = XO(fj2/(fj2' where Xo < X and ~ ::; No;(Xo). Thus y E C'll(Xo) and y E Cs(..Qo). But since 6 ::; N\)(.Qo), Cs(..Qo) <J 6. Thus 6 ::; Cs(.Qo).

By III, 13.5, .0/4>(.0) is a minimal normal subgroup of ~/4>(..Q) and [4>(.0), 6] = 1. Thus c) is proved. Also .0 is a special q-group. Now if the representation of..Q on VdV2 is absolutely irreducible, then by Schur's lemma, Z(..Q) is represented on V1/V2 by scalar multiples of the identity linear transformation. Since the representation is faithful and y ¢ Cs(..Q), it follows that .0 is non-Abelian and that Z(..Q) is cyclic. Thus 4>(.0) =

.0' = Z(..Q) is of order q (III, 13.1), and .0 is extraspecial. q.e.d.

The celebrated Theorem B of Hall and Higman will now be proved.

2.9 Theorem (HALL and HIGMAN). Let K be afield of characteristic p and let V be a finite-dimensional vector space over K. Let (fj be a finite p-soluble group of non-singular linear transformations of V such that Op(fj) = 1. If g is an element of (fj of order pn, the minimum polynomial of g is (t - 1)" where r ::; pn. If r < pn, there exist integers no ::; n for which pno - 1 is a power of a prime q and the Sylow q-subgroups of (fj are non-Abelian; further, if no is the least such integer, r ;;::: pn-no (pno - 1).

Proof Since extension of the ground field does not affect the minimum polynomial of a linear transformation, we may suppose that K is algebraically closed. The theorem will be proved by induction on n. Since o = gP" - 1 = (g - 1)P", the minimum polynomial of g is (t - 1)" where r ::; pn. Suppose that r < pn.

We apply 2.8 with ~ = <g), y = gp.-I. Thus there exists a section f) = m1/m 2 of m such that g E m1 and, if s = gm2, Sp·-I :f. 1. Thus the order of s is pn. Also f) = .06, where .0 is a normal q-subgroup of f) for some prime q :f. p, 6 = <s), .0/4>(.0) is a minimal normal subgroup of f)/4>(.Q) and Sp·-I centralizes 4>(.0). Finally, f) has a faithful irreducible representation on a section U of V. Thus Op(f») = 1 by V, 5.17, and by 1.3, .0 ~ Cl}(.Q). Since r < pn, u(s - 1)p'-1 = 0 for all u E U. It suffices to prove that

(i) .0 is non-Abelian, (ii) qml = pnl - 1 forsomem1,n1 withn1:S; n,and

(iii) u(s - !)p'-" (p" -1)-1 :f. 0 for some u E U. For then, the Sylow q-subgroup of m is non-Abelian, so there certainly exist integers no :s; n for which pno - 1 is a power of a prime qo and the Sylow qo-subgroups of m are non-Abelian. Further, if no is the least such integer, no :s; n1, so p"-"o(p"o - 1) - 1 :s; pn-"I(p"1 - 1) - 1 and (g - l)P'-" (p" -1)-1 :f. O.

First suppose that U is an irreducible K.Q-module. We prove that f) is a critical group in the sense of 2.1. By 2.8, .0 is extra special. Now .0/4>(,0) may be regarded as a vector space over the field GF(q), and if p = (6 on .0/4>(,0)), p is irreducible, since .0/4>(.0) is a minimal normal subgroup and f) = .06. By II, 3.10, ,0/4>(.0) is isomorphic to the additive group of a certain field, and p(s) corresponds to multiplication by an element A of this field. By I, 4.4, p(SP"-l) :f. 1, so A pH :f. 1. It follows that p(Si) is fixed point free (i = 1, ... ,p" - 1). Thus ~ is a critical group (2.1).

Since u(s - 1)P"-1 = 0 for all u E U, it follows from 2.6 that p" - 1 is a power of q and u(s - 1)P"-2 :f. 0 for some u E U. Hence (i), (ii) and (iii) are proved.

Now suppose that U is a reducible K,Q-module. By V, 17.3b),

where Ui is the sum of all K.Q-submodules of U isomorphic to some irreducible K.Q-module. By V, 17.3d), there is a transitive permutation representation (J of f) on {U1, ... , U/} such that Ui(J(h) = Uih (h E f»). Let f)o = ker (J. Then.Q :s; f)o, so f)o = 6 0 .0, where 6 0 = 6 n f)o. The restriction (J' of (J to 6 is transitive, and since 6 is Abelian, 6 0 = ker (J' is the stabiliser of any Ui in (J'. Thus I = 16: 6 0 1 = If): f)ol. Since U is reducible, U possesses non-isomorphic irreducible K.Q-submodules, by VII, 9.19. Thus I > 1 and 6 0 < 6. We observe also that 6 0 :f. 1. For if 6 0 = 1, (J(S)p"-1 :f. 1, so (J(s) has a cycle (Ui , Uis, ... ) of length p". But then, if u is a non-zero element of Ui , U, US, ... , US p"-l lie in distinct

summands of the direct sum U = U1 EB ... EB U1• Thus these elements are linearly independent, contrary to the fact that u(s - l)p"-l = o. Hence 1 < 6 0 < 6. Thus SP"-l E f)o. Suppose that 6 0 = <sp), so that 1 ::; m < n.

Since .0 ~ Cs(,Q), there exists x E .0 such that z = [x, SP"-'] #: 1. Then z E .0 ::; f)o, but there exists Uj on which z induces a non-identity linear transformation. No~ Uj is a Kf)o-module; let W be an irreducible Kf)o-submodule of Uj . If W is an irreducible K,Q-sub_module of W, Uj

is the direct sum of K,Q-submodules isomorphic to W, so z induces a non-identity linear transformation on IN and on W. Thus, if 91 is the kernel of the representation of f)o on W, z 1= 91 and SP"-I 1= 91.

Since 6 0 = <spm) is the stabiliser of Uj , W, Ws, ... , Wspm_1 are contained in distinct summands of the direct sum U = U1 EB ... EB U1•

It follows that if wE Wand w #: 0, then w(s - l)pm-l #: O. Let f) = f)0/91, h = spm91. Then W is a faithful, irreducible Kf)

module, so Op( f)) = 1. We wish to apply the inductive hypothesis to f) and the element h, which is of order pn-m since Sp·-l 1= 91. If w E W,

(w(h - 1)P"-m_1)(s - l)pm-l = w(s - 1)P"_pm(s - 1)pm-1

= w(s - 1)P"-1 = O.

Thus by the above remark, w(h - 1)p·-m-1 = 0 for all wE W. By the inductive hypothesis, pn, - 1 = qm1 for some ml' n1 with nl ::; n - m; also .0/91 is non-Abelian, and there exists WE W such that

Again, we have W1 (s - l)pm-l #: O. But

w1(s - l)pm-l = w(spm _ l)p·-m_p.-m-·'- l(s _ l)pm-l

= w(s - l)P"-P"-"'-l,

so (i), (ii), (iii) are again proved. q.e.d.

2.10 Remarks. a) The result corresponding to the crucial theorem 2.6 in the case when char K does not divide Iml is V, 17.13. This will be generalized along the lines of 2.9 in 6.2.

b) Several other proofs of 2.6 are known. The original proof of HALL

and HIGMAN [1, Theorem 2.5.1] is not quite so general. Dade gave a proof of it along lines similar to his proof of V, 17.13. Another proof

§ 3. The Exceptional Case 429

for q odd was given by BERGER [1, IV]. THOMPSON [4] gave a proof using vertices and sources similar to the one presented here.

c) A number of situations similar to that of 2.6 have been studied. Examples may be found in DADE [5] or BERGER [1].

§ 3. The Exceptional Case

Let <g) be a cyclic group of non-singular linear transformations of a vector space V and suppose that g is of order n. The minimum polynomial f(t) of g is then a divisor of tn - 1. We have seen, for example in 1.13, that the case when f(t) = tn - 1 leads to the desired theorems. However, this does not always follow from the appropriate hypotheses. To study the exceptional case, it is necessary to make some rather detailed calculations.

3.1 Definitions. Let rx be an automorphism of an extraspecial q-group.o. We say that rx bisects .0 if there exist rx-invariant subgroups .01 , .02 of .0 with the following properties.

a) .0 = .01 .02 , [.01 , .02] = 1 and .01 n .02 = Z(.o). b) «rx) on .odZ(.o)) is irreducible, and xrx = x for all x E .02 ,

In this case, we also have the following. c) .01 is extraspecial and either .02 is extraspecial or.o2 = Z(.o). To see this, observe that since [.01 , .02] = 1, Z(.o1) ~ Z(.Q) and

Z(.o2) ~ Z(.o). Hence Z(.o1) = Z(.o2) = Z(.o). Since «rx) on .o1/Z(.o)) is irreducible, .01 i= Z(.o). Thus .01 is extraspecial. Similarly, if .02 i= Z(.o), .02 is extraspecial.

We also observe that .01 , .02 are uniquely determined by the conditions a), b). For rx i= 1, so .02 = Ctlrx) and .Q1/Z(.Q) = [.Q/Z(.o), <rx)].

3.2 Theorem. Suppose that <f> = .06, where .0 is a normal, extraspecial q-group, 6 = <s) is cyclic and [Z(.o), 6] = 1. Suppose that s is of order pn for some prime p i= q, and that Sp·-l ¢ Ca;(.o). Let V be a faithful, irreducible K<f>-module, where K is an algebraically closed field and char K i= q, and suppose that the degree m of the minimum polynomial of the linear transformation v -+ VS (v E V) is less than pn. Then thefollowing assertions hold.

a) pn - 1 = qd for some d. b) The automorphism of .0 induced by s bisects .0, and

1.0: Co(s)1 = q2d. c) m = pn - 1.

Proof Let F = GF(q). The elementary Abelian q-group .Q/Z(.Q) may be regarded as an F6-module, so by the Maschke-Schur theorem, .Q/Z(.Q) is the direct product of irreducible F6-submodules. If Sp·-l centralizes all these submodules, Sp·-l centralizes .Q/Z(.Q) = .Q/<1>(.Q), and by III, 3.18, Sp·-l E C(ij(.Q). This is contrary to the hypothesis, so there exists an irreducible F6-submodule .QdZ(.Q) which is not centralized by Sp·-l.

Thus.Q1 is invariant under sand «s) on .QdZ(.Q)) is irreducible. Hence the centre of .Q1 is either .Q1 or Z(.Q). If it is .Q1' .Q1 is Abelian and Sp·-I1= C(lj(.Q1). But then, by 1.10 applied to im(6.Q1 on V), the minimum polynomial of the linear transformation v -+ VS (v E V) is t P' - 1. This is not the case, so the centre of .Q1 is Z(.Q) and .Q1 is extraspecial. Let .Qz = CO(.Q1)· Then .Qz is 6-invariant, [.Q1, .QzJ = 1 and .Q1 (\ .Qz = Z(.Q). If .Q/Z(.Q) is regarded as a symplectic space, .QdZ(.Q) is a non-singular subspace, and .Qz/Z(.Q) is the orthogonal complement (.QdZ(.Q)).L of .QdZ(.Q). Thus by II, 9.4d), .Q = .Q1.QZ.

Since Z(.Q) ~ Z(6)) and 6) is irreducible, there exists a non-identity linear character X of Z(.Q) such that

vx = X(x)v (v E V, X E Z(.Q)).

By V, 17.3, V is the direct sum of irreducible K.Q1-submodules, since .Q1 <J 6). By V, 16.14, all of these irreducible K.Qrsubmodules are isomorphic. Hence by the proof of V, 17.5, we may suppose that V = VI (8\ V2 , where Vl is an irreducible K.Ql-module, and that

where PI' pz are irreducible projective representations of 6) on VI' Vz respectively; further, V1P1(Y) = V1Y and pz(y) = Iv for all Y E .Q1.

2

If Z E .Qz, zy = yz for all Y E .Q1 and

V1P1(Y)P1(Z) ® vzpz(z) = (VI ® Vz)yz = (VI ® Vz)ZY

= VIPl(Z)Pl(Y) ® Vzpz(Z).

Thus PI (Y)Pl (z) = PI (Z)P1 (Y), and PI (Z) commutes with every element of the group 9t = <Pl(y)ly E .Ql)· Since V1Pl(Y) = V1Y for Y E .Q1' 9t is an irreducible group of linear transformations of VI. Thus PI (z) is a scalar multiplication.

Since

Pt(s)P' = ~1v and P2(S)P' = ~-tlv , where 0 =f:. ~ E K. Since K is alge-I 2

braically closed, ~ = 11 P' for some 11 E K. Thus, if Ut = 11- tpt(s), U 2 = 11P2(S),

Since VtPt(Y) = VtY and P2(Y) = Iv for all Y E Ot, 91 is a group 2

isomorphic to Ot. Also

so ul l Pt(y)ut = Pt(s-t ys) for all y E Ot. Let (fj = <UI> 91). Thus (fj is a group of non-singular linear transformations of Vt, 91 is a normal, extra-special q-subgroup of(fj, (fj = <ut)91and [Z(91), <ut) ] = 1. Since Sp'-I does not centralize OdZ(O) and «s) on OdZ(O)) is irreducible, Co,!Z(O)(SPH) = 1. Hence C9I/Z(91)(UC1

) = 1 and uC I induces a fixed point free automorphism on 9tjZ(9t). Thus (fj is a critical group. Also Vt is an irreducible K91-module, and the degree of the minimum polynomial of Ut is at most m. For if 1, Ut , ... , uf are linearly independent elements of HomK(Vt , Vt), then by V, 9.11,1, ut ® u2 , ••• , uf ® uf are linearly independent elements of HomK(Vt , Vt) ®K HomK(V2 , V2), and by V, 9.14, 1, u, ... , um are linearly independent elements of Homt«V, V), where vu = vs (v E V).

We deduce that if 1911 = q2d+1, then l = pn - 1 and Vt is K<ut )isomorphic to K<ul)jR, where R is a K<ut)-submodule of K<u t) and dim K R = 1. This follows from V, 17.13 if char K =f:. p and from 2.6 and VII, 5.3 in the case when char K = p. Now if rL is a linear transformation of a vector space X and Xo is an rL-invariant subspace of X, the minimum polynomial of rL divides the product of the minimum polynomials of the linear transformations of XjXo and Xo induced by rL. Thus the degree of the minimum polynomial of Ut is pn - 1; hence pn _ 1 :::;; m < pn and m = pn - 1. It only remains to show that O 2 = Cds). If O 2 f;, Co(s), there exists Z E O 2 and Wi E Vi (i = 1, 2) such that (Wt ® W2)SZ =f:. (Wt ® W2)ZS, or

Since Pt(z) is a scalar multiplication, it follows that U2P2(Z) =f:. P2(Z)U2, We now use different methods according to whether char K =f:. p or char K = p.

If char K =f:. p, K contains a primitive pn_th root of unity w, and U1 , (J'2 can be diagonalized. Since Vt is an epimorphic image of K<ut ),

the eigen-values of 0"1 are distinct. Thus the eigen-values of 0"1 are 1, h-1 h+ 1 p" -1 I' h N· ( )..../.. () w, ... , w ,w , ... , W lor some . ow sInce 0"2P2 Z -r P2 Z 0"2'

0"2 has at least two distinct eigen-values Wi, w j • Thus all wi+k and Wj+k

with k "# hare eigen-values of 0". Hence every power of w is an eigenvalue of 0". This shows that the minimum polynomial of 0" is divisible by t P" - 1, a contradiction.

If char K = p, we observe that 0"2 "# 1, since 0"2P2(Z) "# P2(Z)0"2. Thus there is an indecomposable K<0"2)-submodule of V2 of dimension greater than 1, and by VII, 5.3, there is an element U2 of this submodule such that U20"2 = U2 + u~, u~ "# 0, U~0"2 = u~. Since the degree of the minimum polynomial of 0"1 is pn - 1, there exists U1 E V1 such that U1(0"1 - 1)P"-2 "# 0. But for j ~ 1, by induction,

(U1 ® u2 )(s - l)j

= U1 (0"1 - l)j ® U2 + jU1(0"1 - l)j @ u~ + jU1(0"1 - 1)j-1 ® u~,

so

a contradiction. Hence.o2 ~ Co(s). Thus Co(s) = COJs).o2. Since Co';Z(O)(SP"-') = 1, Co, (s) = Z(.o), so

Co(s) = .02. q.e.d.

Suppose now that (}j = 6.0, where .0 is a normal, extraspecial q-group, 6 is a p-group (p "# q) and [.0', 6] = 1. As in 2.2, .o/Z(.o) is a non-singular symplectic space over GF(q) and 6 induces a p-subgroup of a symplectic group on .o/Z(.o). Accordingly, we need to describe certain Sylow p-subgroups of Sp(2m, q).

3.3 Lemma. Suppose that p, q are distinct primes and that m is the smallest positive integer such that q2m == l(p). Let F = GF(q2m). Let ( , ) be a non-singular, symplectic bilinear form on F over GF(q), and let .B be a Sylow p-subgroup of the corresponding symplectic group Sp(2m, q) on F.

Let X be a vector space over F with F-basis {Y1' ... , Yr}, and let 'l3 be the set of all mappings lI. of X onto X of the form


where (J 1, ... , (Jr run through £ and 1t runs through a fixed Sylow psubgroup of the symmetric group of degree r. Define a bilinear form ( , )' on X over GF(q) by putting

Then ( , )' is non-singular and symplectic, and ~ is a Sylow p-subgroup of the corresponding symplectic group Sp(2mr, q) on X.

Proof. It is clear that ( , )' is a non-singular, symplectic bilinear form on X over GF(q). ~ is a subgroup of the corresponding symplectic group on X, for if lI. E ~ and lI. has the form displayed in the statement of the lemma,

r

= L (Ai(Ji, J1.i(Ji) i=1

r

= L (Ai' J1.i) i=1

Thus it need only be shown that 1 ~ 1 is the order of a Sylow p-subgroup of Sp(2mr, q). By II, 9.13b),

For any positive integer x, let pV(X) be the highest power of p which divides x. Then by definition of m, V(q2i - 1) > 0 if and only if i is divisible by m. Thus

r

v(ISp(2mr, q)i) = L v(q2mi - 1). i=1

In particular, 1£1 = pI, where f = v(q2m - 1). Of course, f > 0, but note that if p = 2, f ~ 2. Now I~I = prf+v(r,), so it is sufficient to show that


V(q2mi - 1) = f + v(i) (i = 1, ... , r).

Suppose that q2m - 1 = pfj; thus j is not divisible by p. Let i = spk, where s is not divisible by p and k = v(i). Then

Using I, 13.18 and the fact that for p = 2, f ~ 2, it is easy to check that all the summands for h ~ 2 are divisible by pf+k+l. Thus

V(q2mi - 1) = f + k = f + v(i). q.e.d.

3.4 Definition. Let V be a vector space over a field K. We say that a nonsingular linear transformation a of V bisects V (over K) if V = Cv(a) Eel U for some irreducible K(a)-module U. If this is the case, U is unique, and U is a faithful K(a)-module.

3.5 Lemma. Let K = GF(q), F = GF(qm) (m > 1), and let X be a vector space over F with F-basis {Yl' ... , yJ Let £ be a group of K-linear transformations of F, let 5l be a subgroup of the symmetric group of degree r and let ~ be the set of all K-linear transformations ~ of X for which there exist (Ji E £ (i = 1, ... , r) and n = n(~) E 5l such that

a) If ~ is an element of ~ of prime-power order pk > 1 and ~ bisects X (over K), nPk-l = 1, where n = n(~).

b) Suppose that £ is Abelian. If ( is an element of~' of prime order p, ( does not bisect X.

c) Suppose that p is a prime and that any two elements of £ of order p commute. Suppose that ~, I] are elements of ~ which bisect X and that the orders of~, I] are pm, pn respectively. Then ~pm-l and I]p.-l commute.

d) Suppose that m = 2, q = 2n - 1 is a Mersenne prime and £ is a Sylow 2-subgroup of SL(2, q) (acting on F). Suppose that ~ is an element of~ of order 2n and that ~ bisects X. Then n(~) = 1 and

for all I] E ~.

Proof a) If nPk-l #- 1, n has a cycle (iI' ... , ip ') of length pk. Let

Y = Fy. q::> ... 1? Fy .. '1 \LI W 'p.

Thus Y is a K«O)-submodule. Now X = Cx(~) (f) Xl for some irreducible K( < 0 )-module Xl' Thus if Y n X I = 0, Y is K < O-isomorphic to a submodule of X/Xl and of Cx(~), which is impossible since pk > 1. Hence Y n Xl = Xl' Xl ::; Y and Y = Cy(~) EB Xl' Now it is easy to see that

Thus dimKCy(~) = dimK F = m. Since dimK Y = pkm, it follows that

Since Xl is a faithful, irreducible K<O-module, it follows from II, 3.10 that m(pk - 1) is the smallest of the set of integers I for which ql == 1 (pk). Hence pk-l(p - 1) is divisible by m(pk - 1) and pk-l(p - 1) ~ m(pk - 1). This is impossible since m > 1. Hence nPk-l = 1.

b) If ~ E ~ is as displayed in the statement of the lemma, write

since E is Abelian, the order of the factors in this product is immaterial. It is easy to see that b(~)b(1J) = b(~1J), so b is a homomorphism of ~ into the Abelian group E. Hence ~' ::; ker band b(O = 1. Now if , bisects X, it follows from a) that

for certain 'i E E, since, is of order p. Thus

but not all 'i are 1. Suppose X = Cx(O (f) Xl' where Xl is an irreducible K( < 0 )-module. If 'j #- 1, FYj n Xl #- 0, so FYj ~ Xl' Thus if i #- j, FYi n Xl = 0, FYi::; Cx(O and 'i = 1. Hence 1 = 'I ... 'r = 'j #- 1, a contradiction.

c) By a),

(I Aiy;)~pm-l = I (Aiai)Yi'

(I AiY;)1J p·- 1 = I(Air;)Yi

for elements ai' r i of £ such that af = rf = 1. It follows from the hypothesis that ai and ri commute. Hence ~pm-l and 1Jp.-l commute.

d) Note that n > 1; thus q ¢ 1 (2n) but q2 == 1 (2n). By II, 3.10, the dimension of any faithful, irreducible K < O-module is 2. Thus X = Cx(~) EEl Xl' where Xl is an irreducible K<O-module and dimKX l = 2. Let n = n(~).

First suppose that n2 "# 1. Then n has a cycle of length greater than 2, so n2 has two distinct cycles (iI' i2 , .•• ), (jl,j2' ... ) oflength greater than 1. Hence if Yl = FYi EEl FYi EEl··· and Y2 = FyJ. EEl FyJ• EEl···,

1 2 1 2

Yl and Y2 are disjoint K<~2>-modules on which e operates non-trivially. Thus Yi n Xl "# 0 (i = 1, 2). Since Yl n Y2 = 0 and dimK Xl = 2, Xl n Yl , Xl n Y2 are both of dimension 1, and

It follows that ~2 operates faithfully on either Xl n Yl or Xl n Y2 ,

whence q == 1 (2n- l ). Hence n = 2 and by a), n2 = 1, a contradiction. Thus n2 = 1. If n "# 1, n has a cycle (j, /) of length 2. Then

Xl :::;; FYi EEl FYI and FYi:::;; Cx(O if i "# / and i "# j. Suppose that

Then

Since ~2 "# 1, either aa' "# 1 or a' a "# 1. But in either case, a' "# a-I, so aa' "# 1 and a' a "# 1. Hence FYi n Xl "# 0 and FYI n Xl "# O. Thus FYi n Xl is a i-dimensional K<~2>-submodule of FYi' and ~2 operates trivially on Fy/(FYi n Xl). Since £ :::;; SL(2, q), det aa' = 1; thus ~2 operates trivially on FYi n Xl. Similarly, e operates trivially on FYI n Xl. Hence ~2 = 1, contrary to n > 1. Hence n = L

It follows that each FYi is a K( < 0 )-submodule of X, so either ~ operates trivially on FYi or FYi 2 Xl. Therefore

(IAiY;)~ = (Ap)Yj + I AiYi i,.j

for some j and some element a of 5! of order 2". Put p = a2"-2. Then p2 is an element of SL(2, q) of order 2, so p2 = -1. Write

We prove that

by direct calculation, distinguishing between the cases when jw =1= j and jw =j. Ifjw =l=j, one verifies that [e"-I,Il] and [e"-2,1l]2 both carry

LAiYi into

-AjYj - AjroYjro + L AiYi' i*j.i*jro

If jw = j, ~2"-1 and 11 commute, whereas

(LAiY;) [e"-2, 11] = (AJa2.-2, !j])Yj + L AiYi' i*j,i*jro

But by II, 8.1 0, 5! is a generalized quaternion group of order 2"+1, so

q.e.d.

3.6 DefInition. Let V be a K<g)-module, where <g) is a cyclic group of order n. We say that g is exceptional on V if the degree of the minimum polynomial of the linear transformation v -+ vg (v E V) is less than n.

3.7 Lemma. Suppose that (fj = 6.0, where .0 is a normal, extraspecial q-group and 6 E Sp(fj) (p =1= q). Let V be a Jaithful K(fj-module, where K is algebraically closed and char K =1= q. Suppose that all irreducible K.osubmodules oJV are isomorphic.

a) IJ p is odd, s is an element oJ6 oj order p, s ¢ Op(fj) and s is exceptiona I on V, then s ¢ 6'Op(fj).

b) Suppose that s, s' are elements oj 6, s is oj order pO, s' is oj order p"' and s, s' are both exceptional on V. Then Sp"-IOp(fj) and S,p"'-lOp(fj) commute.

c) Suppose that p = 2 and q = 2" - 1 is a Mersenne prime. IJ s is an element oj 6 oj order 2", S2H ¢ 02(fj) and s is exceptional on V, then

for all x E 6.

Proof Since all irreducible K.Q-submodules of V are isomorphic, Z(.Q) is faithfully represented on V by scalar multiples of the identity mapping. Thus Z(.Q) ~ Z(<fi). Hence, as in 2.2, .QjZ(.Q) is a symplectic space, and if p' = (6 on .QjZ(.Q)), im p' is a subgroup of the symplectic group on .QjZ(.Q). By III, 3.18, ker p' = C6 (.Q). Since C6 (.Q) <J .06 = <fi, C6 (.0) = Op(<fi).

In a), we have s ~ Op(<fi) and in c), S2"-1 ~ 02(<fi). In b), we may suppose that sP"-' ~ Op(<fi) and S,p"'-l ~ Op(<fi). In all cases, let U be an irreducible K<s).Q-submodule ofV. Then U and V are both direct sums of K.Q-submodules isomorphic to some irreducible K.Q-module, so ker(.Q on U) = ker(.Q on V) = 1. Now if X is the subgroup of <s) of order p, X n Op(<fi) = 1, so X i Co;(.Q) and [.0, X] "# 1. Thus ([.0, X] on U) is non-trivial and (X on U) is non-trivial. Therefore <s) is faithfully represented on U. Since ker(,Q on U) = 1, it follows that U is a faithful, irreducible K( <s).Q)-module. Also s is exceptional on U, so by 3.2, pi _ 1 = qi for certain positive integers i, j; also the automorphism of .0 induced by s bisects .0. Similarly, in b), the automorphism of,Q induced by s' bisects .0. Thus p'(s) and, in b), p'(s') bisect ,QjZ(,Q).

Let m be the smallest positive integer such that q2m == 1 (p) and let F = GF(q2m). Let ( , ) be a non-singular, symplectic bilinear form on F over GF(q), and let £ be a Sylow p-subgroup of the corresponding symplectic group Sp(2m, q). Let X be a vector space over F with F-basis {Yl' ... , Yr}, where r is an integer for which q2mr 2': 1.0: Z(.Q)I. Let ~ be the set of all mappings IY. of X into X of the form

where (J 1, ... ,(Jr run through £ and 11: runs through a fixed Sylow p-subgroup of the symmetric group of degree r. By 3.3, there exists a non-singular, symplectic bilinear form on X over L = GF(q) such that ~ is a Sylow p-subgroup of the corresponding symplectic group Sp(2mr, q) on X. By II, 9.6, X = Y EEl Z, where Y, Z are orthogonal, nonsingular L-subspaces of X and there exists a symplectic isomorphism () of Y onto .QjZ(.Q).

Given x E 6, there exists a L-linear transformation p"(x) of X such that zp"(x) = z for all z E Z and


yp"(x) = yep'(x)e-1

for all y E Y. Then p" is a monomorphism of 6 into Sp(2mr, q) and im p" is a p-subgroup of Sp(2mr, q). By Sylow's theorem, there exists {3 E Sp(2mr, q) such that im p" ::; {3-1~{3. For x E 6, write

p(x) = {3p"(X){3-1;

thus p is a homomorphism of 6 into ~. Note that if x E 6 and p'(x) bisects ,Q/Z(,Q), then p(x) bisects X, for if 0 is a p'(x)-invariant subspace of ,Q/Z(,Q) such that (p'(x» on 0) is irreducible, (p(x» on Oe-1p-l) is irreducible.

If p is odd, then by 2.7, p = 2e + 1 is a Fermat prime and q = 2. In this case m = e and £ is a Sylow p-subgroup of Sp(2e, 2). Since p2 does not divide 22e - 1,1£1 = p and £ is Abelian. On the other hand, if p = 2, then m = 1 and 51, being a Sylow 2-subgroup of Sp(2, q) = SL(2, q) (II, 9.12», is a generalized quaternion group.

a) In this case p(s) bisects X and £ is Abelian. By 3.5b), p(s) 1: ~', so s 1: 6'Op(Gl), since ker p = ker p' = Op(Gl).

b) Since £ has only one subgroup of order p, any two elements of order p commute. By 3.5c), p(s)P.-1 and p(s')P"'-' commute. Thus sP"-'Op(Gl) and s'P"'-'Op(Gl) commute.

c) This follows from 3.5d). q.e.d.

We now build these assertions into more general theorems.

3.8 Theorem (HALL and HIGMAN). Let K be a field of characteristic p and let V be a finite-dimensional vector space over K. Let Gl be a finite, p-soluble group of non-singular linear transformations of V such that 0iGl) = 1. Suppose that g, h are elements of a Sylow p-subgroup of Gl, and let c = [g, h]. If c =F 1 and (c - 1)P-l = 0, then either (g - 1)P-l(C - 1)P-2 =F ° or (h - l)P-l(C - 1)p-2 =F 0.

Proof. We may suppose that K is algebraically closed, and we apply 2.8 with ~ = (g, h), y = c. Thus there exists a section S = GldGl2 of Gl with the following properties. First, g E Gl 1, hE Gl 1 and c 1: Gl 2 . Put SI = hGl2, S2 = gGl2· S = cGl 2 . Then s = [S2' SI] and s =F 1. Also S = 6.0, where .0 is a special normal q-subgroup of S for some prime q =F p and 6 = (SI' S2)' Further, ,Q/Z(,Q) is a minimal normal subgroup of S;Z(,Q) and s centralizes Z(,Q) but not .0. Finally, S has a faithful irreducible representation on a section U of V, and .0 is extraspecial if U is an irreducible K,Q-module.

Thus Op(f» = 1 (V, 5.17), .0 2 Cii(.Q) (1.4) and u(s - 1)P-1 = 0 for all u E U. Since s "# 1, p is odd. Also u(sP - 1) = u(s - 1)P = 0 and sP = 1. If U is an irreducible K.Q-module, .0 is extraspecial; but then 3.7a) leads to a contradiction. Hence U is a reducible K.Q-module. By V,17.3b),

where U; is the sum of all K.Q-submodules of U isomorphic to some irreducible K.Q-module. By VII, 9.19, I > 1. By V, 17.3d), there is a transitive permutation representation a of f> on {U1, ... , U,} such that U;a(u) = U;u (u E f». Since .0 2 Cii(.Q), s ¢ Cii(.Q). Thus there exists x E .0 such that z = [s, x] "# 1. Thus the representation of z on one of the Uj is non-trivial. If f>o is the stabiliser of Uj, .0 ~ f>o < f>, since I > 1. Since f> = .06 and 6 = <S1' S2), it follows that S; ¢ f>o for i = 1 or i = 2. Hence the length of the cycle of a(s;) containing Uj is at least p, and Uj, Ujs;, ... , UjSr1 are distinct.

Since u(s - 1)P-1 = 0 for all u E U, the length of any cycle of a(s) is less than p and is therefore 1. Hence s E f>o and UjS;kS = UA (k = 0, ... ,p - 1). Let W be ~n irreducible Kf>o-submodule of ~ let in = ker(f>o on W) and let f> = f>o/in. Since (fj is p-soluble, f> is p-constrained, by 1.4. Since f> has a faithful, irreducible representation on W, Op( f» = 1. Also z is represented non-trivially on W, since Uj is the direct SUm of isomorphic K.Q-submodules. Thus in = sin, s is of order p. By 2.9, the minimum polynomial of the mapping w ~ ws (w E W) is (t - 1)" where p - 1 ~ r ~ p. Thus there exists WE W such that w(s - 1)P-2 "# O. Let u = W + wS; + ... + wsr1; thus

Since UjS~(s - ly-2 ~ UjS~, it follows that u(s - 1)p-2 "# O. But u = w(s; - 1)P-1, so w(s; - 1)P-1(s - 1)p-2 "# O. Hence either (g - 1)p-1(C - 1)P-2 # 0 or (h - 1)P-1(C - 1)P-2 "# O. q.e.d.

3.9 Theorem (HALL and HIGMAN). Let K be afield of characteristic p and let V be a finite-dimensional vector space over K. Let (fj be a finite, p-soluble group of non-singular linear transformations of V such that Op(fj) = 1. Suppose that g, h are elements of a Sylow p-subgroup of (fj and that gP"-\ hP'-! do not commute. Then either (g - 1)P"'-1 "# 0 or (h - 1)P"-1 "# o.

Proof We may suppose that K is algebraically closed. The theorem will be proved by induction on m + n. Suppose that (g _ 1)pm-1 = (h - 1)P'-1 = O. We apply 2.8 with '13 = <g, h), y = [gpm-', hP"-']. Thus there exists a section ~ = GltlGl 2 of Gl with the following properties. First, g E Gl 1, hE Gl 1 and y ¢ Gl 2 . Put Sl = hGl2 , S2 = gGl2 , S = yGl 2 •

Thus s = [sC', sC'] =F 1. Also ~= .06, where .0 is a normal q-subgroup of ~ for some prime q =F p and 6 = <Sl' S2)' Further ,QjZ(,Q) is a minimal normal subgroup of ~jZ(,Q) and s centralizes Z(,Q) but not .0. And ~ has a faithful, irreducible representation on a section U of V. Thus Op(~) = 1 (V, 5.17),.0 ~ Cij(,Q) (1.4) and U(S2 - 1)pm-1 = U(Sl - 1)P"-1 = 0 for all U E U. Then sf = sf" = 1. Since sf-' and sr-' do not commute, it follows that the orders of S2' Sl are pm, pn respectively. Finally, if U is an irreducible K,Q-module, .0 is extraspecial. But then 3.7b) leads to a contradiction. Hence U is a reducible K,Qmodule. By V, 17.3b),

where Ui is the sum of all K,Q-submodules of U isomorphic to some irreducible K,Q-module. By VII,9.19, I > 1. By V, 17.3d), there is a transitive permutation representation a of ~ on {U1, ... , U/} such that Uia(u) = Uiu (u E ~). Since .0 ~ Cij(,Q), s ¢ Cij(,Q). Thus there exists x E,Q such that z = [s, x] =F 1. Thus the representation of z on one of the Uj is non-trivial. Let ~o be the stabiliser of Uj . Since I > 1, .0 ~ ~o < ~.

Let pa, ph be the lengths of the cycles of a(sl), a(s2) containing Uj

respectively. Since U(S2 - l)Pm-1 = U(Sl - l)P"-1 = 0 for all U E U, a < nand b < m. Hence sf E ~o and S~b E ~o. Since .0 ~ ~o < ~ and 6 = <S1> S2), either a > 0 or b > O. Let W be an irreducible K~osubmodule of Uj , let 91 = ker(~o on W) and let ~ = ~o/91. Thus W is a faithful, irreducible K~-module. Hence OA)) = 1. Let g = s~b91, h = sf"91. Thus, g, h are elements of a Sylow p-subgroup of ~. Also, since a < nand b < m,s E ~o and

Since z is represented non-trivially on Uj , z E,Q and all irreducible K,Q-submodules of Uj are isomorphic, z is represented non-trivially on W. Thus z ¢ 91 and s ¢ 91.

It follows from the inductive hypothesis, applied to im( ~ on W), that either w(g - l)pm-'-l =1= 0 or w(h - 1)P"-'-1 #- 0, for some WE W.

Suppose that Wi = w(g - 1)pm-b -1 =f. O. By definition of b, the elements I I I pb-1 l' . d" U w, W S2' ••• , W S2 Ie m Istmct i' so

But

w'(l + S2 + ... + S~'-1) =f. O.

w'(1 + S2 + ... + sf-1) = w(sf - 1)pm-'-1(sz - 1)p'-1

= w(Sz - 1)pm_l,

so (g - lr-1 =f. O. Similarly, if w(lI - 1)P"-C1 =f. 0 for some wE W, (h - 1)P"-1 =f. O. q.e.d.

In order to generalize 3.7c), two lemmas are needed.

3.10 Lemma. Let (fj = 6,0, where ,0 is a normal q-subgroup of (fj for some odd prime q, 6 = <g) is cyclic of order 2n (n > 1) and gZ'-' ~ C(!;(,Q). Suppose that V is a faithful K(fj-module, where char K =f. q, and that the degree of the minimum polynomial of the linear transformation v -+ vg of V is at most 3. Then q = 3 and n = 2.

Proof Without loss of generality, we may suppose that K is algebraically closed. By 2.8, there exists a section f, = (fjt/(fj2 of (fj with the following properties. First, (fj1 ~ 6 and g2'-' ~ (fj2' Also f, = ,0161> where ,01 is a special q-group and 6 1 = <g), where 9 = g(fj2' Further, g2"-' does not centralize ,01' and f, has a faithful, irreducible representation on a section U of V. By V, 17.3,

where Ui is the sum of all the submodules of U which are K,Q1-isomorphic to an irreducible K,Q1-module, and there exists a transitive permutation representation a of 6 1 on {1, ... , l} such that Uix = Uia(x) for all x E 6 1, Now for any u E U, U, ug, ugZ, ug3 are linearly dependent. Thus i, ia(g), ia(?ff, ia(g? cannot be distinct, from which it follows that a(9)z = 1. If a(9) =f. 1, I = 2 since a is transitive. Now Uig2 = Ui , but since g2 ~ CS(,Q1)' g2 cannot be represented by scalar multiples of the identity mapping on both U 1 and U 2' Hence there exists u in either U 1 or U 2 such that u, ug2 are linearly independent. But then u, ug, ug2, ug3 are linearly independent, a contradiction. Therefore a(g) = 1.

Hence I = 1 and U is the direct sum of isomorphic, irreducible K,Q1-modules. ,01 is faithfully represented on anyone of these, so

[Z(.Ql)' 6 1] = 1, Z(.Q 1) is cyclic and .01 is extraspecial. By 3.2, 2n - 1 = qd for some d, and 2n - 1 S 3. Thus q = 3 and n = 2. q.e.d.

3.11 Lemma. Suppose that (fj is a soluble group and that the Fitting subgroup F(fj) of(fj is of odd order. If g E (fj and [g, x] = 1for every element x ofF(fj) of prime order, g E F(fj).

Proof Let 91 be the set of elements g E (fj such that [g, x] = 1 for every element x of F(fj) of prime order. Clearly 91 is a normal subgroup of (fj. If 91 -t F(fj), choose a minimal normal subgroup Wl/(91 n F(fj)) of (fj/(91 n F(fj)) contained in 91/(91 n F(fj)). Then Wl/(91 n F(fj)) is an elementary Abelian p-group for some prime p. For every prime q :f:. p, the Sylow q-subgroup .0 of Wl is also a Sylow subgroup of 91 n F(fj). Thus .0 <J Wl, since F(fj) is nilpotent. Also, if ~ is a Sylow p-subgroup of Wl, it follows from IV, 5.12 that ~ s CIDl(.Q), since ~ s 91 and IF(fj)1 is odd. Thus .0 s NIDl(~) for all q :f:. p. Since also ~ S NIDl(~)' ~ <J Wl. Thus all Sylow subgroups ofWl are normal and Wl is nilpotent. But Wl <J (fj, so Wl s F(fj) and Wl s 91 n F(fj), a contradiction. Hence 91 s F(fj). q.e.d.

3.12 Theorem (GROSS). Let V be afinite-dimensional vector space over a field K. Let (fj be a finite soluble group of non-singular linear transformations of V such that IF(fj)1 is odd and is not divisible by char K. Suppose that g is an element of (fj of order 2n and that g is exceptional on V. Then

Proof We may suppose without loss of generality that K is algebraically closed. If n = 1, the fact that g is exceptional on V means that g is a scalar multiple of the identity mapping; thus g E Z(fj) s F(fj), which is impossible since IF(fj)1 is odd. Hence n > 1.

Let i! be any normal, nilpotent subgroup of (fj. Thus i! s F(fj), so by hypothesis, char K does not divide ji!l. Hence V is a completely reducible Ki!-module. Write

(1 )

where Vi is the direct sum of isomorphic irreducible Ki!-submodules and, for i :f:. j, the summands of Vi are not isomorphic to those of Vj' Then it is easily deduced from the Jordan-Holder theorem that every irreducible Ki!-submodule of V is contained in some Vi' and Vi is the sum of all K£-submodules of V isomorphic to a given irreducible K£-module. We call (1) the Wedderburn decomposition of V for i!.

If X E G> and U, U' are K£l-isomorphic submodules of V, Ux and U'x are K£l-isomorphic. Thus Vix = Vj for some j. Put j = ia(x); then a(x) is a mapping of {1, ... , I} into itself. Since a(1) = 1 and a(xy) = a(x)a(y) (x, y E G», a is a permutation representation of G>. Let k(£l) = ker a.

We put

b = n k(£l), l!

where the intersection is taken over all normal, nilpotent subgroups £l of G>. Thus b <J G>.

) 2,-1 "'-a 9 E~.

For suppose that £l, a are as above. Then sillce 9 is exceptional on V, a(g) cannot have a cycle oflength 2n. Hence

and g2.-1 E k(£l). Thus g2.-1 E b. b) If m is a normal Abelian subgroup of G>, em, b] = 1. Suppose that (1) is the Wedderburn decomposition of V for m. Since

m is Abelian and K is algebraically closed, any irreducible representation of m in K is of degree 1; thus each element x of m is represented on each Vi by a scalar multiple of the identity mapping. Now if y E b, ViY = Vi' so [x, Y J is represented on each Vi by the identity mapping. Hence [x,yJ = 1. Thus [m,bJ = 1.

c) The class of lj = F(b) is 2. Let m = lj' n Z2(m. By III, 2.11, m is Abelian. By b), em, b] = 1.

Thus m ::;; Z(lj). By III, 2.6, lj' ::;; Z(m, so the class of lj is at most 2. If lj is Abelian, then lj ::;; Z(b) by b). Thus b ::;; Cs(m ::;; lj, since b is soluble (111,4.2). Hence lj = b. But by a), Ibl is even, whereas since lj ::;; F(G», Iljl is odd. Thus lj is non-Abelian, and the class of lj is precisely 2.

d) If p is odd, g2'-' centralizes 0p(b/m. Let Op(bjm = ljlj. Thus III is odd, so by 1.11, there exists 6 E Sil)

such that 6 9 = 6. Let 6 0 = [6, <g2'-1 > J. We show that if q is any prime divisor of Iljl other than p, [Oib), 6 0J = 1.

For suppose that this is false. Then by IV, 5.12, there exists an element of order q in 0ib) which is not centralized by 6 0 , By c) and III, 10.2, 0q(b) is a regular q-group; hence by III, 10.5, the elements of order at most q in Oq(~) form a characteristic subgroup .Q. We have [0,60J oF 1. Hence there is a summand U in the Wedderburn decom-

position of V for.Q such that ([.0, 6 0] on U) is non-trivial. Since g2·- 1 E ~ and 6 ~ ~, grl and 6 leave U fixed. It follows that g2.- 1 cannot be represented on U by a scalar multiple of the identity mapping.

Let 2m be the length of the orbit containing U in the permutation representation of <g) on {Uala E (fj}. Thus if h = g2m, h leaves U fixed, but U, Ug, ... , Ug2m - 1 form a direct sum. Thus m < n. Since

the linear dependence of 1, g, ... , g2'-1 implies that of 1, h, ... , hrL1

on U. Thus h is exceptional on U. Also, ifm = n - 1, then the restrictions to U of 1, h are linearly dependent, and h is represented on U by a scalar multiple of the identity mapping, a contradiction. Thus m < n - 1.

Let (fjo = <h,.Q) and let Sl = ker«fjo on U). Thus [.0, 6 0 ] $, Sl and g2.- 1 ~ R Since Sl <l (fjo and all the involutions in (fjo are conjugate, it follows that Sl contains no involutions. Hence ISlI is odd and Sl ~ .0. Thus Sl = ker(.Q on U). Since 6 leaves U fixed and 6 ~ N(!j(.Q), we have 6 ~ N(!j(Sl). We deduce that

For otherwise, we would have

and

But then it follows from III, 1.10 that

since Sl <l <h, 6, .0). This is a contradiction, so

(2)

Now U is the direct sum of isomorphic irreducible KO-modules, so O/Sl has a faithful, irreducible representation. Thus the centre 31ft of O/Sl is cyclic, and 3 is represented on U by scalar multiples of the

identity mapping. Thus [ <hSl), 3/SlJ = 1. By (2), 3 "# .0, so .o/Sl is nonAbelian. By c), the class of .o/Sl is 2. Also .0 is of exponent q, and the centre of .o/Sl is cyclic so .o/Sl is extraspecial.

Let U1 be an irreducible KGJo-submodule of U. Then U1 is the direct sum of faithful irreducible K(.o/Sl)-modules, so by (2), ([ <h2.-m

-,), .oJ on U 1) is non-trivial. Hence if Sl1 = ker(GJo on U 1),

h2·-m-

, ¢ Sl1 and Sl1 :::; .0. Thus Sl1 = ker(.o on U 1) = R Hence U 1 is a faithful K(GJo/Sl)-module. By 3.2, 2"-m - 1 = qd and

for some d. By 2.7, d = 1. Hence ifW is the elementary Abelian q-group .0/3 regarded as a vector space over GF(q),

Hence the degree of the minimum polynomial of the linear transformation w --+ wh ofWis at most 3. Also since 6 :::; No;(Sl), 6 is represented on W. Now (60 on W) is non-trivial. For otherwise, since 3 is represented on U by scalar multiples of the identity mapping, 6 0 operates trivially on .0/3 and on 3/R Thus by I, 4.4, (60 on .o/Sl) is trivial and [.0, 6 0J :::; Sl, a contradiction. It follows from 3.10 that p = 3 and n - m = 2. But then

q = qd = 2n- m - 1 = 3 = p,

a contradiction. Hence [Oi~), 6 0] = 1 for every prime divisor q of 13"1 other than

p. Hence 6 0 :::; Ce(Op'(3")), since 3" is nilpotent. Now if G: = Cx(Op'(3")), G: <J l, so 6 n G: E Sp(G:) and 6 n G: <J 6. Thus 6 n G: <J 60p'(3") = l. Hence the Sylow p-subgroup of G: is normal in G:. But if q "# p, the Sylow q-subgroup of l is normal; hence so is that of G:. Thus G: is a nilpotent normal subgroup of ~ and G: :::; F(~) = 3". Thus [6, <g2.- I

) J = 6 0 :::; G: :::; 3". Since l = 63", g2·- 1 centralizes l/3" = Op(~/3"). Thus d) is proved.

Since we cannot prove d) for p = 2, we introduce another normal subgroup ~1' For each prime divisor p of 13"1, Op(~) is regular by c) and III, 10.2; thus the elements of 0i~) of order at most p form a characteristic subgroup .Q1(Op(~)), by III, 10.5. Let

be the Wedderburn decomposition of V for QI(Op(~)). Let R)P) = ker(QI(Op(N) on VjP). Since ~ leaves VjP) invariant, RjP) <J ~. Let

so <£?) <J ~. Let ~l be the intersection of those (£:jP) which contain g2·- 1•

(If g2·- 1 lies in no (£:)p), put ~l = ~). In any event, ~l <J ~ and, by a), g2·- 1 E k Also V(P)g = ViP) for some i and -l)l· J' ,

g-IR\P)g = R(p) g-l(£:\p)g = (£:\P) } l , J l •

Thus g E N(!j(~l)' and ~l <J (g, ~). Note also that F(~l) = ~ n ~l' for F(~l) is a normal, nilpotent subgroup of ~.

e) If p is odd, g2·- 1 centralizes Op(~dF(~I)). If XdF(~I) = Op(~dF(~I»' Xl is a characteristic subgroup of ~l'

and Xl n ~ = ~ n ~l. Thus Xl <J ~ and Xl ~/~ is a normal p-subgroup of N~. By d), g2·- 1 centralizes Xl ~m. Hence g2·- 1 centralizes Xl/(X l n m = Op(~dF(~l))·

Now let ~/F(~l) = 02(~dF(~I». Let ~* be a Sylow 2-subgroup of (g, ~) containing g, and put ~ = ~* n ~. Thus ~ E S2(~) and g E N(!j(~).

[ 2·-1 ] [2.-2 ]2 f) If x E~, g , x = g , x . Let y = [g2·- 1

, x] [g2.- 2 , x] - 2, and suppose that y =1= 1. Since

g E N(!j(~), Y E ~ ~ ~l. Thus Y is a 2-element. Since I~I is odd, y rI~· By 3.11, there exists a prime q such that y rI C(!j(.Q), where.Q = QI (Oq(~». Thus ([..0, (y)] on Vr) is non-trivial for some j, and

Hence y rI (£:~q). But Y E ~ l, so it follows from the definition of ~ 1 that g2'-' rI (£:~q). Hence ([..0, (grl) ] on VJq) is non-trivial.

Write U = V?). Thus [.0, (y) ] and [.0, (gr') ] are both represented non-trivially on U. Let 2m be the length of the orbit containing U in the permutation representation of (g) on {Uala E (fj}, and let h = g2m. As in the proof of d), (p. 445), m < nand h is exceptional on U. Since ([.0, g2H

] on U) is non-trivial, g2·- 1 is not exceptional on U, and m < n - 1. Let (fjo = (h, ~,.o) and let R = ker(fjo on U). Then h2·-m - 1 = g2'- 1 rI R

Now U is the direct sum of isomorphic, irreducible K.o-modules, so .oR/R has a faithful irreducible representation. Thus the centre 3/R of .oR/R is cyclic, and 3 is represented on U by scalar multiples of the identity mapping. Thus [3, (fjo] ~ R Since [.0, (fjo] -.t R, 3 =1= ,OR

By c), the class of nft/ft is 2, and nft/ft is extraspecial, since .0 is of exponent q and 31ft is cyclic.

If U1 is an irreducible K( <h)n)-submodule of U, then U1 and U are direct sums of isomorphic irreducible Kn-submodules, so ker(O on Ud = ker(O on U). Hence ([.0, h2.-m

-1

] on U 1) is non-trivial. By 3.2, 2n- m - 1 = qd for some d. By 2.7, d = 1. By 3.7c), yft E 02(fjo/ft). Thus yft centralizes nft/ft and [.0, <y)] ::; R This is a contradiction, so f) is proved.

g) g2·-1 F(~l) E F(~dF(~l))' Let (fjF(~l) = F(~ljF(~l))' Now ~jF(~l) E S2(f/F(~1))' so

~ E S2(f). Let l) = <P(~)F(~l)' If x E~, [g2.- 2, x] E~, so by f),

[g2'-" x] E l). Thus g2.-1 centralizes ~F(~l)/l). But bye), g2·-1 centralizes all the other Sylow subgroups of (f/l). Thus g2·-1 centralizes (f/l).

Now (f/l) is nilpotent, so if (f*/l) = F(~dl)), (f ::; (f*. But l)jF(~l)::; 4>(f*jF(~l))' since ~F(~l)jF(~l) <J (f*jF(~l)' Thus by III, 3.7, (f*jF(~l) is nilpotent. Thus (f* = (f and

Hence g2·-1 l) E C);dll(F(~dl))). By III, 4.2b), l·-l E (f. h) g2·- 1 F(fj) E F(fj/F(fj)). By g), g2'- 1 E (f, where (f/F(~d = F(~ljF(~l))' Since (f is a charac

teristic subgroup of ~1' (f <J ~. Since (f n ~ = F(~l)' (f/(f n m is nilpotent. Thus (f~m is a normal nilpotent subgroup of ~/~. Hence if (f*jF(~) = F(~jF(~)), (f ::; (f* and g2·- 1 E (f*. But F(~) ::; F(fj), so (f*F(fj)/F(fj) is a normal nilpotent subgroup of (fjjF(fj). Thus

q.e.d.

3.13 Remark. HOARE [1] proved the following improvement of 3.9. Let (fj be a p-soluble group of non-singular linear transformations

of a vector space V over a field K of characteristic p. Suppose that Op(fj) = 1. If g, h are elements of a Sylow p-subgroup of (fj and gpm-" hP·- 1 do not commute, either (g - 1)pm-1 "# 0 or (h - 1)P'-1(g - 1ym-1 "# O.

Exercises

1) Let p = 2e + 1 be a Fermat prime and let F = GF(22e). For a E F, put

( ) 2 4 22.-1 tra = trF:GF(2) a = a + a + a + ... + a ,

§ 4. Reduction Theorems for Burnside's Problem 449

and for a, b in F, put (a, b) = tr(ab 2'). Verify that ( , ) is a bilinear form on F over GF(2), and that ( , ) is non-singular and symplectic. F has a primitive p-th root of unity 0); let 0) be the mapping a -4 aO) on F. Then (aO), bO)) = (a, b), so < 0) is a Sylow p-subgroup of the symplectic group Sp(2e, 2) on F.

2) Let q = 2f - 1 be a Mersenne prime and let F = GF(q2). Let 0) be a primitive 2f+l_th root of unity in F and define the mappings a, ' of F into F by

Then a, ' are linear over GF(q) and ,0 = <a, ,) is a generalized quaternion group of order 2f+1. If (a, b) = abq - aqb (a E F, b E F), ( , ) is a bilinear form on F over GF(q) and ( , ) is non-singular and symplectic. Also (aa, blX) = (a, b) for all a E ,0, so ,0 is a Sylow 2-subgroup of the symplectic group Sp(2, q) on F.

§ 4. Reduction Theorems for Burnside's Problem

The results of § 2 and § 3 will now be used to improve Theorem 1.13.

4.1 Defmition. Suppose that '-P E Sp(f)). a) The exponent of '-P will be denoted by pep«(f»).

b) For n ~ 0, let '2n('-P) = <xP"lx E '-P), and for n ~ 1, let

Then

Let e;(f)) be the smallest positive integer m for which 'm('-P) = 1. Hence

The invariant e;(f)) is introduced since it is not possible to prove that the p-Iength of any p-soluble group (f) is at most ~p(f)). We now investigate what happens to ep' e; on passage from (f) to (f)jOP',p(f)).


4.2 Lemma. Suppose that (f) is a p-soluble group and p divides 1(f)1.

a) ep(f)/Op',p(f))) :s; ep(f)) - 1, provided that one of the following three conditions is satisfied:

(i) p is odd and p is not a Fermat prime. (ii) p is a Fermat prime and the Sylow 2-subgroups of (f) are Abelian. (iii) p = 2 and the Sylow q-subgroups of (f) are Abelian for every

Mersenne prime q. b) lfp ~ 3, e;(f)/Op',p(f))) :s; e;(f)) - 1.

Proof Suppose that ~ E Sp(f)), and write

Thus V = 0p',p(f))/U may be regarded as a vector space over GF(p). Let p = (f) on V). By 1.4 and 1.6, ker p = Cm(Op',p(f))/U) = 0p',p(f)). Thus if (f) = im p, (f) is a group of non-singular linear transformations of V and Op(~) = 1. We shall apply 2.9 and 3.9 to (f).

First note the following consequences of 1.8. a) If y E ~ n 0p',p(f)), x E ~ and

(yU)(p(x) - 1)p'-1 "# 1,

then either x p' "# 1 or (xy)p' "# 1. For by 1.8b),

x-P'(xy)p'U = (yU)(1 + p(x) + ... + p(X)P'-l)

= (yUHp(x) - W'-l "# 1.

(yU)(p(x) - 1fP'-1 "# 1,

then [(xy)P', xP,] "# 1. For if v = (yU)(p(x) - 1)P'-1, then by 1.8b),

Hence by 1.8a),

v = (yUH1 + p(x) + ... + p(X)P'-l)

= x-P'(xy)P'U.


1 =1= V(p(X) - lyn = V(p(X pn ) - 1)

= [X-pn(xyyn, Xpn]U

= [(xyyn, Xpn]U.

a) Let ep«fj/Op',p«fj)) = n. Thus there exists x E 'lJ such that the order of xOp',p«fj) and p(x) is pn. By 2.7, there is no integer no such that pno - 1 is a power of a prime q and the Sylow q-subgroups of (fj are non-Abelian, ~ince one of (i), (ii) or (iii) holds. Hence, by 2.9, the minimum polynomial of p(x) is (t - lyn. Since 0p',p«fj) = U('lJ n 0p',p«fj)), it follows that there exists y E 'lJ n 0p',p(fj) such that

(yU)(p(x) - Wn- l =1= 1.

By 0:), either xpn =1= 1 or (xyyn =1= 1. Since x and xy are in 'lJ, it follows that ep(fj) > n.

b) Suppose first that e;(fj/Op',p(fj)) = 2n is even. If 2n = 0, there is nothing to prove, since 'lJ =1= 1. Thus we suppose that 2n > 0. Hence zpn E 0P' .p«fj) for all z E 'lJ, but there exist Xl E 'lJ, x 2 E 'lJ such that [xr-~ xr-'] ¢ 0p',p«fj). Thus p(xl)pn-, and p(X2yn-, do not commute. By 3.9, (p(x) - lyn- l =1= 0, where x = Xl or X = X2' Thus there exists y E 'lJ n 0p',p«fj) such that

(yU)(p(x) - ly"-l =1= 1.

By 0:), either xpn =1= 1 or (xyyn =1= 1. Thus e;«fj) > 2n. Next suppose that e;«fj/Op',p«fj)) = 2n + 1 is odd. Then there is

an element X in 'lJ such that the order of xOp',i(fj) and p(x) is pn+l. By 2.9, the minimum polynomial of p(x) is (t - 1)" where r ~ pn+1 _ pn. Since p ~ 3, r > 2pn - 1, so there exists y E 'lJ n 0p',p«fj) such that

(yU)(p(x) - 1)2pn-l =1= 1.

By (3), [(xyyn, xpn] =1= 1, so e;«fj) > 2n + 1. q.e.d.

This brings us to one of the main theorems on bounds for the p-length.

4.3 Theorem. Suppose that (fj is a p-soluble group of p-Iength I and that pe is the exponent of a Sylow p-subgroup of(fj.

a) I ~ e, provided that one of the following three conditions holds:

(i) p is odd and p is not a Fermat prime. (ii) p is a Fermat prime and the Sylow 2-subgroups of CfJ are Abelian.

(iii) p = 2 and the Sylow q-subgroups of CfJ are Abelian for every Mersenne prime q.

b) Ifp is a Fermat prime, I ~ 2e.

Proof a) This is proved by induction on l. If I = ° or I = 1, it is trivial. If I > 1, the p-Iength of CfJ = CfJ/Op' ,1'(CfJ) is I - 1. By the inductive hypothesis, I - 1 ~ ep(CfJ). By 4.2a), ep(CfJ) ~ ep(CfJ) - 1, so I ~ ep(CfJ) = e.

b) We prove by induction on I that I ~ e;(CfJ). This is clear for I ~ 1. If I > 1, I - 1 ~ e;(CfJ/Op',p(CfJ» ~ e;(CfJ) - 1 by the inductive hypothesis and 4.2b). Thus I ~ e;(CfJ). Hence I ~ 2e, since by 4.1, e;(CfJ) < 2e + 1. q.e.d.

4.4 Remark. HOARE [1J has introduced the invariant e~(CfJ) defined as follows. For 'lJ E Sp(CfJ), let O'o('lJ) = 'lJ, and for n ~ 1, let

0'3n-2('lJ) = <xp"lx E 'lJ),

0'3n-l('lJ) = <ExPo, ypHJ, zp"+1lx E 'lJ, y E 'lJ, Z E 'lJ),

0'3n('lJ) = < [x p", yp"J, zp"+'lx E 'lJ, y E 'lJ, Z E 'lJ).

Then e~(CfJ) is the smallest positive integer m for which O'm('lJ) = 1. It follows from 3.13 that if CfJ is a 2-soluble group of even order, e~(CfJ/02,,2(CfJ» ~ e~(CfJ) - 1. Hence the 2-length ofCfJ is at most e~(CfJ).

Since e2 (CfJ) = [H4 + e~(CfJ))], this implies that the 2-length of CfJ is at most 3e2(CfJ) - 2. However, for a soluble group CfJ of order divisible by 4, it has been proved by GROSS [1, 2J that the 2-length is at most 2e2(CfJ) - 2. A slightly weaker result will now be deduced from 3.12.

4.5 Theorem (GROSS). Let CfJ be a soluble group of 2-length I, and let 2e

be the exponent of a Sylow 2-subgroup of CfJ. a) If I > 0, e2(CfJ/02',2,2',2(CfJ» ~ e - 1. b) If I > 0, I ~ 2e - 1.

Proof a) This is trivial if I = 1 or I = 2, so we suppose that I > 2. Let

and let V = O2,, 2 (<»)/U, regarded as a vector space over GF(2), Let p = (<» on V) and let <» = imp, By 1.4 and 1.6, ker p = O2'.2(<>>)' so O2(<>>) = 1 and IF(<»)I is odd. Suppose that ~ E S2(<») and that x E~. For any y E ~ n O2'.2(<>>), (yU)(p(x) - 1)2'-1 = x- 2'(xy)2'U = 1, by 1.8b). Thus either p(X)2'-! = 1 or p(x) is exceptional. It follows from 3.12 that in either case, p(X)2.-! E £, where

But IF(<»)I is odd, so F(<») ::; O2,(<>>), Thus £02 , (<»)!02' (<») is nilpotent. Since p(X)2.-! is a 2-element, it follows that p(Xf'-l E O2',2(<>>)' Hence x2.-! E O2',2,2',2(<>>) for all x E~. Thus

b) This is proved by induction on I. If I = 1, it is trivial. If I = 2, it must be proved that e > 1. But if e = 1, the Sylow 2-subgroup of <» is Abelian and I = 1 by VI, 6.6a). If I > 2, the inductive hypothesis and a) give

so I ::; 2e - 1. q.e.d.

We now discuss the question of the extent to which these results are best possible.

4.6 Example (HALL and HIGMAN). A good deal of the proof of the theorems leading to 4.3a) was devoted to the investigation of a minimal counterexample to the proposition that I ::; e. To show, then, that Fermat primes really are exceptional in this regard, we simply have to verify that this minimal counterexample exists.

a) The first step is to construct a group of the kind discussed in 2.6 when p is a Fermat prime and q = 2. This could be done, for example, by observing that the order of an appropriate orthogonal group is divisible by p. A more concrete approach is as follows.

Let K = GF(2), L = GF(22m) (m ~ 1). Let 8 be a primitive (2m + l)-th root of unity in L. Then 82'-1 =1= 1 for i = 1, ... , 2m - 1, so L = K(8). Let IY. be the automorphism x --+ x2m of L; thus IY. is of order 2. Also, for x E L, write

2 4 22m - 1 tr x = trL : KX = x + x + x + ... + x .

Thus tr X E K, tr(x + y) = tr X + tr y and tr x2 = tr x. As is well-known, there exists c E L such that tr c = 1; (this can be seen from the properties of the Vandermonde determinant, for example). Put xr = tr(cx(xa)). Thus

(X + y)r + xr + yr = tr(c(x(ya) + y(xa))).

Hence by VIII, 6.10, there exists a 2-group .0, a subgroup 91 of the centre of.Q and isomorphisms p, a of .0/91,91 onto L, K respectively such that if (u91)p = x, then (u2)a = xr. Also (ex)r = tr(Ce1+2mX(Xa)) = xr, so by VIII, 6.6, there is an automorphism 13 of.Q such that ((uf3)91)p = e((u91)p) for all u E .0 and uf3 = u for all u E 91. Since L = K(e), 13 induces an irreducible automorphism on .0/91. Hence either Z(.Q) = .0 or Z(.Q) = 91. But in fact Z(.Q) = 91. For otherwise.Q is Abelian, and since 91 < Q I (.0) ~ .0, .0 is elementary Abelian. This, however, is not the case, since 1r = tr c = 1. Thus Z(.Q) = 91 and .0 is extraspecial. Also 132m +1 induces the identity automorphism on .0/91 and on 91. By I, 4.4, 132'" + I is of order 2a for some a. If e' = e2", e' is also a primitive (2m + 1 )-th root of unity and ((uf32")91)p = e'((u91)p) for all u E.Q. Thus we may assume that 132" +1 = 1. Let f) be the semidirect product .Q(s), where .0 <:J f) and US = uf3 for all u E .0. Then 1f)1 = 22m+1(2m + 1), Ct;(.Q/91) = .0 and 91 is the only minimal normal subgroup of f).

b) Next we construct a suitable representation of f). If F is a field and char F =j:. 2, consider the group-ring Ff) as an Ff)-module, and let

be an Ff)-composition series of Ff). Since char F does not divide 1911, Vi- I = Vi EB Wi for some F91-module Wi' and Vo = WI EB ... EB 'Nt. Now 91 is not represented trivially on Ff), so there exists some Wi on which the representation of 91 is non-trivial. Since 91 is the only minimal normal subgroup of f), f) is represented faithfully on Vi-dVi. Thus V = Vi-dVi is a faithful irreducible Kf)-module. If F is algebraically closed, it follows from 2.5 that V is an irreducible K.Q-module and dimFN = 2m.

If F is the algebraic closure of GF(p), f) is represented on V by a finite number of matrices, the coefficients of all of which lie in some finite field Fo = GF(pn). Thus there exists a faithful, irreducible Fof)-module U of degree 2m. Let (fj be the corresponding semidirect product f)U. Thus U is an elementary Abelian minimal normal p-subgroup of (fj and C(jj(U) = U. Thus 0p,(fj) = 1 and Op(fj) = U.

If p is a Fermat prime, we may take 2m = p - 1. Thus I~/.ol = p and 6> is of p-Iength 2. But if g is a p-element of 6>, g = hu, where h E ~,

U E U and hP = 1. Thus the minimum polynomial of the linear transformation ~ of U induced by h is (t - 1)' for some r. But r $ dimF U =

. 0

2m = p - 1, so (~ - l)p-l = O. Hence

gP = (hu)P

= hP(u(1 + ~ + ... + ~p-l)) = hP(u(~ - I)P-l)

= hP = 1.

Thus the exponent of the Sylow p-subgroup of 6> is p. c) We show that Fo may be chosen to be GF(p). First suppose p = 3. Then .0 is the quaternion group of order 8,

for the dihedral group of order 8 has no automorphism of order 3. Thus ~ ~ SL(2, 3), since SL(2, 3) is an extension of a quaternion group (consisting of ± 1 and the matrices of trace 0 and determinant 1) by a cyclic group of order 3. Thus ~ certainly has a faithful, irreducible representation of degree 2 in GF(3).

If p > 3, we show first that the matrices representing elements of .0 may be chosen with coefficients in GF(p). Since p - 1 = 2m > 2, GF(p) possesses a primitive 4-th root of unity i. By V, 16.14, the representation of.o on V is induced from the representation of degree 1 of a maximal normal Abelian subgroup m of .0. But since the exponent of m is at most 4, the matrices of such a representation are all of the form (in). Thus we may suppose that .0 is represented by matrices with coefficients in GF(p). Denote this matrix representation of ~ by p.

We know that the coefficients of all p(x) (x E~) lie in some finite field GF(pk). Let !X be an automorphism of GF(pk).1f x E.o,

p(x)P(S) = p(XS) = p(XS)!X = (p(x)!X)P(S)1Z = p(x)P(s)iZ.

Thus (p(s)!X)p(stl commutes with all p(x). Since the restriction of p to .0 is absolutely irreducible, it follows from Schur's lemma that p(s)!X =

lp(s) for some l E GF(pk). Taking p-th powers, lP = 1. Since the characteristic is p, l = 1. Thus p(s)!X = p(s) for every automorphism !X of GF(pk). Thus all the coefficients of p(s) lie in GF(p).

With this choice ofFo, 1<»1 = pP·22m+1•


4.7 Remarks. a) Let p, q be primes such that q divides p - 1 and let f) be the permutation group of order pq and degree p. Let

be the wreath product of n copies bf f) (I, § 15). Thus (})n is a permutation group of degree pn. By I, 15.5, if '13 E Sp(f)) and

'13n E Sp((})n). By III, 15.3b), the exponent of '13n is pn.

if But also, the p-Iength of (})n is n. To prove this, one observes that

(i) f)1' f)2 are p-soluble groups of p-Iengths 11, 12 respectively, (ii) f)2 is a transitive permutation group, and

(iii) either OP(f)l) = f)1 or Op(f)2) = 1, then f)1 ?, f)2 is p-soluble of p-Iength 11 + 12 •

Thus the inequality I ::; e in 4.3a) cannot be improved. b) Let p be a Fermat prime, and let (})1 be a group of p-Iength 2 in

which the exponent of the Sylow p-subgroup is p (cf. 4.6). We define (})n by induction on n: for n > 1,

(Here, the regular wreath product is meant, and 32 denotes the cyclic group of order 2.) Since Op(32 ?, (})1) = 1, it follows from the observation in a) that the p-Iength of (})n is 2n. But the exponent of a Sylow p-subgroup is at most pn, so the inequality I ::; 2e in 4.3b) cannot be improved.

c) For p = 2, it appears that no example is known for which I > e. As was remarked earlier (4.4), the inequality I ::; e has been proved by Gross if e = 2.

We use 4.3 to prove the following.

4.8 Theorem. Suppose that (}) is p-soluble of p-Iength I and that '13 E Sp((})). a) Suppose that p > 2 and that p is not a Fermat prime. If the wreath

product of two cyclic groups of order p does not occur among the sections of'13, 1 ::; 1.

b) Suppose that '13 is regular (in the sense of I II, § /0). If P is not a Fermat prime, 1 ::; 1; in any case, I ::; 2.

Proof Both assertions are proved by supposing that they are false and considering a counterexample of minimal order. Since the properties of 'P involved in the hypotheses are inherited by homomorphic images, the p-Iength of any proper homomorphic image of G> is less than I. It follows from VI,6.9 that 91 = 0p',p(G» is an elementary Abelian p-group, 91 is the unique minimal normal subgroup ofG>, C(fj(91) = 91 and there exists ~ ::;; G> such that G> = ~91, ~ n 91 = 1. Since 91 = Op(G», 91 ::;; 'P; thus 'P = m91, where m = 'P n ~.

a) Since 1 > 1, Op(G» "# G>; thus there exists a maximal normal. subgroup 9Jl of G> such that Op(G» ::;; 9Jl < G>. Since the property of 'P involved in the hypothesis is inherited by subgroups and G> is a counterexample of minimal order, the p-Iength of9Jl is 1. Since 9Jl <l G>, 0P' ,p(9Jl) ::;; 0p'.p(G» = 91, so 9Jl/91 is a pi_group. Thus 9Jl n 'P = 91, and since 1 > 1, G>/9Jl is of order p. Hence 1 m 1 = p.

By 4.3a), 'P possesses an element x of order p2. Write x = yz with y E m, Z E 91. Then yP = 1. Let 910 = <z, zY, ... , zyP-l>. Then y E N(fj(91o), so y induces an automorphism 11 on 910, By 1.8,

z(11 - 1)P-1 = z(I1P- 1 + ... + 11 + 1) = y-P(yzY = x P "# 1,

so (11 - 1y-1 "# O. But since yP = 1, I1 P = 1 and (11 - 1)P = O. Hence the degree of the minimum polynomial of 11 is p, so p ::;; dim GF(p)91o. Thus 19101 = pP and <y, 910> is the wreath product of two cyclic groups of order p. By hypothesis, this is impossible.

b) If p = 2, I ::;; 1 by VI, 6.6a), for a regular 2-group is Abelian (III, 10.3a)). If p is not a Fermat prime, I ::;; 1 by a), for regularity is inherited by subgroups and factor groups and the wreath product of two cyclic groups of order p is irregular (III, 10. 3d)). Suppose, then, that p is a Fermat prime. We have C(fj(91) = 91. But by III, 1O.8f),

Hence U1 ('P) ::;; C(fj(91) = 91, and

Thus m is of exponent p. But 'P = m91, so 'P is of exponent p, by III, 10.5. By 4.3b), 1 ::;; 2. q.e.d.

4.9 Remarks. a) The inequality 1 ::;; 2 for Fermat primes in 4.8b) is best possible, since, by 4.6, there exist groups of p-length 2 in which the Sylow p-subgroup is of exponent p and is therefore regular.

b) DADE [1] has given examples, for p odd, of p-soluble groups of p-Iength 3 in which no section is isomorphic to the Sylow p-subgroup of the symmetric group of degree p3.

We now prove the reduction theorem for Burnside's problem.

4.10 Theorem (HALL and IDGMAN [1 ]). For any positive integers m, d, let 6(m, d) denote the class of all finite soluble groups of exponent a divisor of m and with d generators. Also, let s(m, d) be the maximum of the orders of the groups in 6(m, d).

Suppose that p is a prime and p does not divide m. If s(pn, d) and s(m, d) are finite for all d, s(pnm, d) is finite for all d.

Proof Suppose that (fj E 6(pnm, d) and that ~ E Sp(fj). Then the exponent of ~ is at most pn. By 4.3 and 4.5, the p-Iength of (fj is at most 2n. Thus, it is sufficient to prove that there is a bound on the orders of the groups of p-Iength I in 6(pnm, d) for all I ~ 0, d ~ 1. This we do by induction on l. For I = 0, any p'-group in 6(pnm, d) lies in 6(m, d); the assertion thus follows since s(m, d) is finite. If I > 0, there exist terms ~, 91 of the upper p-series of (fj (VI, 6.1) such that ~ ::; 91 ::; (fj, the p-Iength of ~ is I - 1, mj~ is a p-group and (fjjm is a p' -group. Thus (fjjm E 6(m, d), so l(fj: 911 ::; s, where s = s(m, d). By 1.14, 91 is generated by

d' = 1 + (d - 1)s

elements. Thus mj~ E 6(pn, d') and 1m: ~I ::; s' = s(pn, d'). By 1.14, ~ E 6 (pnm, 1 + (d' - 1)s'). But ~ is of p-Iength I - 1. By the inductive hypothesis there is a bound r on the orders of the groups of p-Iength I - 1 in 6(pnm, 1 + (d' - 1)s/). Thus I~I ::; rand l(fjl ::; ss'r. q.e.d.

To apply this with m = 4, we need the solution of the Burnside problem for exponent 4.

4.11 Lemma. Suppose that (fj = <~, g), where ~ is finite and g2 E~. Suppose that given h E ~, there exist h', h" in ~ such that

(hgf = gh'gh".

Then (fj is finite.

Proof Since (fj = <f), g) and g2 E f), any element x of (fj is expressible in the form

We choose such an expression with n minimal. Thus hi "# 1 (2 :s; i :s; n - 1).

If 2 :s; j :s; i < n, define h(j, i) E f) by induction on i - j; we put h(i, i) = hi' and if j < i,

(1)

We now prove by induction on i - j that

for certain h: E f). This is obvious if j = i. If i - j > 0, we have

with h~ E f) and h = h(j + 1, i), by the inductive hypothesis. Now there exist h', h" in f) such that (hg)2 = gh'gh". Thus ghg = h-1gh'gh", so

x = h g' .. gh.h-1gh'gh"h'. g'" gh' 1 J J+ 2 n'

By (1), hjh-1 = h(j, i), so (2) holds. Consider the elements h(2, 2), h(2, 3), ... , h(2, n - 1).

If n - 2 > 1f)1, there exist k, I such that 2 :s; k < I < nand h(2, k) = h(2, I). By (1), h(j, k) = h(j, I) for j = 2, ... , k. Thus

hk = h(k, k) = h(k, I) = hkh(k + 1, Itl,

so h(k + 1, I) = 1. Hence by (2),

contrary to the definition of n. Thus n - 2 :s; 1f)1. Hence if 1f)1 = m, I (fj I :s; mm+ 2. q.e.d.

4.12 Theorem (SANOV [1]). A finitely generated group of exponent 4 is finite.

Proof Suppose that <» = <Xl, ... , xn) is of exponent 4. We use induction on n. Ifn = 1,1<»1::; 4. Ifn > 1, ~l = <Xl' ... , Xn-l) is finite by the inductive hypothesis. We apply 4.11 twice, first with (~, g) = (~l' x;), then with (~, g) = «~l' x;), xn). In both cases, since (hg)4 = 1,

and g-2h- l g- 2 E ~ since g2 E~. Thus <~l' x;) and <~l' x;, xn) = <» are finite. q.e.d.

We shall now apply 4.10 to the restricted Burnside problem (VIII, 12.5), which may be formulated as follows. Given positive integers m, d, let b(m, d) be the maximum of the orders of finite groups with d generators and exponent a divisor of m. We ask, what is b(m, d)? In particular, is b(m, d) finite? In the proof of the next theorem, we shall use the theorem of Kostrikin (VIII, 12.4b)), which asserts that b(p, d) is finite for any pnmep.

4.13 Theorem (HALL and HIGMAN [IJ). If p is an odd prime, b(2p, d) and b(4p, d) are finite for all d 2:: 1.

Proof By a theorem of Burnside (V, 7.3), any finite group of exponent a divisor of 4p is soluble. Thus b(4p, d) = s(4p, d), in the notation of 4.10. By 4.12, s(4, d) is finite, and by Kostrikin's theorem, s(p, d) is finite. The assertion thus follows from 4.10. q.e.d.

4.14 Remark. Taking p = 3, the value of b(6, d) was given in 1.16. It is not possible, however, to give a good estimate for b(12, d), since no good estimate for b(4, d) is known. The proof of Sanov's theorem (Lemma 4.11) gives a very poor estimate for b(4, 2) already; in fact, b(4, 2) = 212 (111,6.7).

The reduction theorem (4.10) effectively reduces the restricted Burnside problem to questions on p-groups and simple groups.

4.15 Theorem. Suppose that m = p~1 ... P;', where PI' ... , Pr are distinct primes. Suppose that

(1) for every positive integer d, b(pi\ d) is finite (i = 1, ... , r); and (2) the number of isomorphism types of finite simple groups of exponent

a divisor of m is finite. Then b(m, d) is finite for every d 2:: 1.

Proof. Let ~(m, d) be the class of finite groups with d generators and of

exponent a divisor of m. We prove by induction on m that 5B(m, d) contains only a finite number of non-isomorphic groups. If m is a prime-power, this follows at once from (1). Suppose that m is not a prime-power.

a) 5B(m, d) contains only a finite number of non-isomorphic groups (fj having an insoluble minimal normal subgroup 91 for which C(!j(91) = 1.

For such a group (fj, 91 = 5\1 X ... X 5\r' where 5\1' ... , 5\r are isomorphic simple groups and 5\j E {5\1' ... , 5\r} for all g E (fj, 1 ~ i ~ r (1,9.12). Thus a permutation representation p of (fj on {5\1' ... , 5\r} is defined by 5\iP(g) = 5\j. The stabiliser of 5\1 is N(!j(5\l)' Also P is transitive, since 91 is a minimal normal subgroup. Thus Iffi: N(!j(5\l)1 = r.

Letpbe a prime divisor ofl5\ll and write m = pnm',where(m',p) = 1. Suppose that g is an element of ffi of order pn. Then g E 6 for some 6 E Sp(ffi) and 6 n 5\1 E Sp(5\l)' Hence 6 n 5\1 "# 1; choose x E 6 n 5\1 with x "# 1. Then (gx)P' = 1, since the exponent of (fj divides m. But

( ) P' p' gP'-1 gP·-z ••. xgx. gx = g x x

Hence

But x gi E 5\1 p(g)i (i ~ 0). It follows that 5\1,5\1 p(g), ... , 5\lP(g)p'-l cannot be distinct factors of 91, since x "# 1. Thus the length of the orbit of p(g) containing ft1 is less than pn. Since g is a p-element, 5\1 p(g)p'-l = 5\1' Thus gPo-lE N(!j(5\l) for every p-element g in ffi.

Let 9Jl be the group generated by all gP'-" as g runs through the pelements of ffi. Thus 9Jl ~ N(!j(5\l) and 9Jl <l ffi. But ffi/9Jl E 5B(pn-1m', d), and by the inductive hypothesis, b(pn-1m', d) is finite. Thus

By (2), there is an integer no such that the order of every finite simple group of exponent a divisor of m is at most no. Then 15\11 ~ no, and

Since C(fj(91) = 1, (fj is isomorphic to a subgroup of the group of automorphisms of 91. The number of possibilities for (fj is therefore finite, and a) is proved.

b) 5B(m, d) contains only a finite number of non-isomorphic groups which have no non-identity soluble normal subgroup.

By a), there exists a finite number of groups ~1' .•. , ~k such that every group (f) E !8(m, d) having an insoluble minimal normal subgroup in for which C(!j(in) = 1 is isomorphic to one of the ~i' Let h be the maximum of the I ~J

Suppose that ~ is any group with d generators. If ~/9Jl ~ ~j' there is an epimorphism e of ~ onto ~j and 9Jl = ker e. Now if ~ = <Y1' ... , Yd)' there are at most h possibilities for Yie, and since e is determined by Y1 e, ... , Yde, there are at most hd possibilities for e. Hence there are at most hd possibilities for 9Jl. Thus there are at most khd

normal subgroups l! of ~ for which Nl! is isomorphic to one of the ~i' Now suppose that (f) E !8(m, d), (f) =f- 1 and (f) has no non-identity

soluble normal subgroup. Let in1, ... , inr be the distinct minimal normal subgroups of (f), and let (ti = C(!j(in;). Since ini is insoluble, {ti n ini = 1. Thus ini(£){ti is an insoluble minimal normal subgroup of (f)/{tj. If g{ti

commutes with X{ti for all x E inj, then [g, x] E ini n (ti = 1 for all x E ini, whence g E C(!j(ini) = (ti and g{ti = 1. Thus C(!j/It,(ini{t/{ti) = 1. Hence (f)/{ti is isomorphic to one of the ~j.

It follows that all the {ti (i = 1, ... , r) occur among at most khd

normal subgroups of (f) of index at most h. But ni <ti = 1, for otherwise, there would exist in j such that in j :::;; n i {ti :::;; (t j' Therefore (f) is isomorphic to a subgroup of the direct product of kh d subgroups all of order at most h. The number of possibilities for (f) is therefore finite.

c) !8(m, d) contains only a finite number of non-isomorphic groups. By b), there exists an integer n1 such that if (f) E !8(m, d) and (f) has

no non-identity soluble normal subgroup, I (f)1 :::;; n1 •

Now suppose (f) E !8(m, d). Let Sl be the maximal soluble normal subgroup of (f). Then I (f)/Sll :::;; n1' By 1.14, Sl can be generated by d' = 1 + (d - 1)n1 elements, so Sl E 6(m, d'), in the notation of 4.10. But by 4.10, s(m, d') is finite and 1(f)1 :::;; n1s(m, d'). Thus there is only a finite number of possibilities for (f). q.e.d.

The hypotheses (1) and (2) of 4.15 are satisfied for square-free exponents. To prove this, we need (i) the theorem of Kostrikin on the finiteness of b(p, d), (ii) the theorem of Feit and Thompson on the solubility of groups of odd order, and (iii) the following theorem of WALTER [2]; (see also BENDER [4]).

4.16 Theorem (WALTER). Suppose that the Sylow 2-subgroups of the finite group (f) are Abelian. Then 02'(f)/02,(f)) is the direct product of a 2 -group and some of the following simple groups.

(1) PSL(2, 2n), where n > 1. (2) PSL(2, pn), where pn == 3 or 5 (8) and pn > 3.

(3) Thefirst simple group 31 of Janko. (4) A simple group 91 of order (r 3 + 1)r3(r - 1), where r = 32n+ 1

(n > 0), in which the centralizer of any involution t is <t) x 5l, where 5l is isomorphic to PSL(2, 32n+1).

The groups arising in (3) and (4) will be discussed in XI, § 13.

4.17 Theorem. Ifm is square-free, b(m, d) is finite.

Proof b(p, d) is finite for every prime p, by Kostrikin's theorem; thus, by 4.15, it is sufficient to show that there is only a finite number of simple groups of exponent a divisor of m, to within isomorphism.

Let (f) be a simple group of exponent m. Since m is square-free, 4 does not divide m and the Sylow 2-subgroups of (f) are of exponent at most 2. Thus the Sylow 2-subgroups of (f) are Abelian. It follows from 4.16 that (f) is isomorphic to one of the simple groups mentioned in 4.16.

However, only a finite number of these groups are of exponent a divisor of m, for by II, 8.3, PSL(2, 2n) has an element of order 2n - 1 and PSL(2, pn) (p odd) has an element of order t(pn - 1). q.e.d.

We conclude by proving that the Burnside problem has an affirmative answer for groups of linear transformations of a finite-dimensional vector space.

4.18 Theorem (BURNSIDE). a) Let K be an algebraically closed field of arbitrary characteristic and let (f) be an irreducible subgroup of GL(n, K). If xm = 1 for all x E (f), 1(f)1 :::; mn3.

b) Let K be any field and let (f) be a finitely generated subgroup of GL(n, K). If xm = 1 for all x E (f), (f) is finite.

Proof a) By V,5.14, (f) contains n2 linearly independent elements g(1), ... ,g(n2 ). Write g(k) = (g!j»). Then if x E GL(n, K) and x = (xij),

n

t (k) - ,,(k) (k - 1 2) r g x - L... gij Xji -, ••• ,n . i,j=1

Now the coefficients (g&k») in these n2 equations form a non-singular matrix, since g(1), ... , g(n2

) are linearly independent. Hence there exists at most one x E GL(n, K) for which tr g(k)X takes a preassigned value for all k = 1, ... ,n2 . But if x E (f), (g(k)x)m = 1, the eigen-values of g(k)X are m-th roots of unity and there are at most mn possible values

for tr g(k)X (k = 1, ... ,n2). Hence there are at most mn3 possibilities for x.

b) We may suppose that K is algebraically closed. We proceed by induction on n. If (f) is irreducible, the assertion follows from a). If (f) is reducible, we may suppose that every element of (f) is of the form

= (gl 0) g - , g g2

where gl runs through a finitely generated subgroup (f)l of GL(r, K) (1 ~ r < n) and g2 runs through a finitely generated subgroup (f)2 of GL(n - r, K). By the inductive hypothesis, (f)l and (f)2 are finite. Now

is a homomorphism e of (f) into the direct product (f)l x (f)2' If 91 = ker e, (f)/91 is finite. By 1.14, 91 is finitely generated. But also 91 is Abelian. Since xm = 1 for all x E 91, 91 is finite. Thus (f) is finite. q.e.d.

§ 5. Other Consequences of Theorem B

5.1 Lemma. Suppose that (f) is p-soluble and ~ E Sp(f)). a) Suppose that ~ ~ ~ and ~ ~ Op,j(f)). If p > 3, there exist

x E ~ and y E ~ such that [y, x, x, x] =1= 1. If P = 3 and the Sylow 2-subgroups of (f) are Abelian, there exist x E ~ and y E ~ such that [y, x, x] =1= 1.

b) Suppose that 91 <l ~. If p > 3 and the class of 91 is at most 2, 91 ~ 0p',p(f)). If p = 3, 91 is Abelian and the Sylow 2-subgroups of (f) are Abelian, then 91 ~ 0p',p(f)).

Proof a) Write

and let p = (f) on 0p',p(f))/U). By 1.4 and 1.6, ker p = 0p',p(f)), and by 1.8,

(vU)(p(x) - It = [v,x, ... ,x]U (VEOp',p(f)),XE~). m

§ 5. Other Consequences of Theorem B 465

Choose x E ~,x rf 0p',p((f») and let (t - 1)' be the minimum polynomial of p(x). By 2.9, r 2:: p - 1 and, if p = 3 and the Sylow 2-subgroups of (f) are Abelian, r 2:: 3. Thus if p > 3, r 2:: 4, (p(x) - 1)3 "# 0 and there exists v E 0p',p((f») such that (vU)(p(x) - 1)3 "# O. If p = 3 and the Sylow 2-subgroups are Abelian, there exists v E 0p',p((f») such that (vU)(p(x) - 1)2 "# O. Since 0p',p((f») = U('l3 n 0p',p((f»)), vU = yU for some y E 'l3; thus [y, x, x, x] "# 1 in the first case and [y, x, x] "# 1 in the second.

b) Suppose that 91 1- 0p',p((f»). Since ['l3,91] ::; 91, it follows from a) that [91, 91, 91] "# 1 if p > 3 and [91, 91] "# 1 if p = 3 and the Sylow 2-subgroups are Abelian. This is contrary to the hypothesis. q.e.d.

In the semi-direct product (f) of SL(2, 3) with an elementary Abelian group <t of order 9, there are Abelian normal subgroups of a Sylow 3-subgroup which do not lie in 03',3((f») = <t. Thus the condition on the Sylow 2-subgroups in 5.1b) cannot be dropped. However, we have the following.

5.2 Theorem. Suppose that ~ is a p-soluble group, that 'l3 E Sp(~) and that ~ is a cyclic normal subgroup of'l3. Suppose further that either

(1) p is odd, or (2) no section of~ is isomorphic to the symmetric group 6 4 ,

Then ~ ::; 0p',p((f»).

Proof Suppose that ~ is a counterexample of minimal order. Let 9'l be the normal closure of ~ in (f). Then 0p',p(9'l) <l (f), so

0p',p(9'l) ::; 0p',p((f»). Hence ~ 1- 0p',p(9'l). But ~ <l 'l3 n 9'l E Sp(9'l), so by minimality of I ~ I, 9'l = ~.

Since ~ 1- 0p"p(~), 0p',p((f») "# ~ and ~ is not p-nilpotent. It follows that ~ is not of p-Iength 1. For otherwise, (f) would have a proper normal subgroup £ of index prime to p; but then ~ ::; £ <l ~

and so £ 2:: 9'l = ~. But if 1 < 91 <l ~, then ~91/91 ::; Op',p((f)/91) by minimality of 1(f)1. Hence ~/91 = 9'l91/91 = Op',p(~/91), so the p-Iength of (f)/91 is 1. By VI, 6.9, Op'(~) = 1 and (f) has just one minimal normal subgroup 9.n, which is elementary Abelian; further Op(~) = 9.n, C(f;(9.n) = 9.n and there exists a subgroup ~ of (f) such that ~ = 9.n~ and 9.n n ~ = 1. Also (f)/9.n is p-nilpotent. Let Sl/9.n = Op,((f)/9.n); thus ~Sl <l 'l3Sl = (f) and, since 9'l = (f), ~Sl = (f). Hence 'l3 = ~(Sl n 'l3) = ~9.n. Since ~ <l 'l3 and ~, 9.n are both Abelian, it follows that 'l3' ::; [~, 9.n] ::; ~ n 9.n. Since ~ is not of p-Iength 1, 'l3' '# 1. But I~ n9.n/ ::; p, since ~ is cyclic; thus 'l3' = ~ n 9.n is of order p. In particular, the class of'l3 is 2.

If p is odd, ~ is a regular p-group and the elements of ~ of order at most p form a characteristic subgroup ~l of~. Since 9Jl ~ ~l' ~l = (m: n ~l)ml But m: n ~l = m: n 9Jl, since 1m: n 9Jl1 = p. Thus ~l = 9Jl. Hence ~ 1 n i) = 9Jl n i) = 1. Since the elements of ~ n i) of order p lie in ~ 1 n i), it follows that ~ n i) = 1. Since (f) = 9Jli), ~ = 9Jl(~ n i) = 9Jl and m: ~ 9Jl, a contradiction.

Suppose then that p = 2. Let Slo/Sl be the subgroup of (f)/Sl of order 2. Then 02'.2(Slo) <:J (f) and 02'.2(Slo) ~ 02'.2«(f) = 9Jl. But Slo = (m: n Slo)Sl, so m: n Slo $. °2,. 2 (Slo)· Since m: n Slo <:J ~ n Slo E S2(Slo), it follows from the minimality of 1(f)1 that Slo = (f). Thus I(f): Sll =

1~:9Jl1 = 2. Of course, ~ is not normal in (f); let ~l be another Sylow p-subgroup

of (f) and let U = <~, ~l>' Since U ~ 9Jl = Cm(9Jl), 02'(U) = 1. Also 9Jl ~ 02(U) ~ ~ n ~l' and since I~: 9Jl1 = 2, 9Jl = 02(U), Thus m: $. 02'.2(U), so by minimality of 1(f)1, (f) = U = <~, ~l>' Thus Z(~) n Z(~l) ~ Z«(f). Since 9Jl is the only minimal normal subgroup of (f) and 9Jl = Cm(9Jl), Z«(f) = 1. Thus Z(~) n Z(~l) = 1. But since (f) = 9Jli), ~ = 9Jl1), where 1) = ~ n i) and Z('-l3) = C!l1/(1). Since 1) n 9Jl = ~ n i) n 9Jl = 1,11)1 = 2. Since also I~'I = 2,19Jl:C!l1/(1)1 = 2. Thus 19J1: Z(~)I = 2. Similarly 19J1: Z(~l)1 = 2, so 19J11 = 19J1: Z(~) n Z(~l)1 ~ 4. Since i) is isomorphic to a subgroup of Aut9Jl, i) ~ 6 3 and (f) ~ 6 4 , This is contrary to the hypothesis. q.e.d.

Our next result (Theorem 5.4) will be deduced from 5.1 for p > 3, but for p = 3, we need the following replacement of it.

5.3 Lemma. Suppose that (f) is 3-so1uble and ~ E S3«(f). If 9l <:J ~ and 9l' f,. 03'.3«(f), then [~, 9l, 9l, 9l'J =1= 1.

Proof Let U/03,«(f) = <P(03'.3«(f))!03,«(f))) and let P = «(f) on 03',3«(f)/U). By 1.4 and 1.6, ker P = 03'.3«(f), and by 1.8,

(VU)(P(Xl) - 1)· .. (p(xn) - 1)

= [v, Xl' ... , xnJU (v E °3"3 «(f), Xi E ~).

Since 9l' $ 0 3"3 «(f), there exist Yl' Y2 in 9l such that z = [Yl' Y2] ¢ °3"3 «(f). Thus [P(Yl), P(Y2)] = p(z) =1= 1. By 3.8, at least one of (p(z) - 1)2, (P(Yl) - 1)2(p(z) - 1) or (P(Y2) - 1)2(p(z) - 1) is non-zero.

If(p(z) - 1)2 =1= 0, there exists u E ~ n 03',3«(f) such that

[u,z,z]U = (uU)(p(z) - 1)2 =1= 1,


so [~, 91', 91'] =1= 1. But

[~, 91'] = [91', ~] = [91, 91, ~]

~ [91, ~, 91] [~, 91, 91] = [~, 91, 91],

by III, l.10b). Thus [~, 91', 91'] ~ [~, 91, 91, 91'] and [~, 91,91, 91'] =1= 1. Otherwise, there exists v E ~ II 03'.3(f) such that

[v, y;, y;, z]U = (vU)(p(y;) - 1)2(p(Z) - 1) =1= 1

for i = 1 or i = 2. Thus [v, y;, y;, z] =1= 1 and [~, 91, 91, 91'] =1= 1. q.e.d.

5.4 Theorem. Suppose that (f) is a p-soluble group, where p ~ 3, and that ~ E Sp(f).

a) If ~ =1= 1, the derived length of ~ exceeds that of ~Op,j(f)/Op,j(f),

b) The p-Iength of (f) is not greater than the derived length of ~.

Proof. a) Let k be the derived length of ~. Since ~ =1= 1, k > O. Suppose that p > 3. By 5.1b), ~(k-l) ~ Op',p(f), since ~(k-l) is a

normal Abelian subgroup of~, so (~Op',p(f)/Op',p(f))(k-l) = 1. Thus the derived length of ~Op',p(f)/Op',p(f) is less than k.

If p = 3, we observe that for k = 1, the assertion follows from VI, 6.6a). If k > 1, put 91 = ~(k-2). Then [~, 91] ~ 91, [~, 91, 91] ~ 91' and

[~, 91, 91, 91'] ~ 91" = ~(k) = 1.

By 5.3, 91' ~ 03',3(f); that is, ~(k-l) ~ 03'.3(f). Hence the derived length of ~03,,3(f)/03,,3(f) is less than k.

b) This is provea by induction on the p-Iength I of (f). If I ~ 1, it is clear. If I > 1, the p-Iength of (f)/Op',p(f) is I - 1. Thus, by the inductive hypothesis, the derived length of ~Op' ,p(f))jOp' j(f) is at least I - 1. Hence by a), the derived length of ~ is at least I. q.e.d.

The example given in 4.7a) shows that the inequality in 5.4b) is best possible, for the derived length of the wreath product of n cyclic groups of order p is n (III, 15.3d». BERGER and GROSS [1] have shown that if the derived length of the Sylow 2-subgroup of a soluble group (f) is d ~ 2, the 2-length of (f) is at most 2d - 2. It is conjectured that the 2-length is at most d, as in the case when p is odd.


Arguments along the same lines give inequalities involving other invariants of ~.

5.5 Theorem. Let Gj be a p-soluble group of p-Iength I, and suppose that 1 i= ~ E Sp(Gj). Let c be the class of~ and let I~I = pb.

a) If p i= 2 and p is not a Fermat prime, c ~ pl-l and b ~ (pi - l)!(p - 1).

b) Ifp is a Fermat prime and p > 3, c ~ ((p - 2)1 - l)/(p - 3) and b ~ ((p - 2)1+1 - I(p - 3) - p + 2)!(p - W.

c) If P = 3, c ~ 21- 1 and b ~ 21- 1 + I - 1.

Proof These assertions are proved by induction on 1 and are all trivial if 1 = 1. For I > 1, let UjOp,(Gj) = CP(Op',p(Gj)/Op,(Gj)) and let p = (Gj on 0p',p(Gj)/U). By 1.4 and 1.6, ker p = 0p',p(Gj).

a) Suppose that the exponent of ~Op',p(Gj)/Op',p(Gj) is pe. Thus there exists x E ~ such that the order of xOp',p(Gj) and p(x) is pe. By 2.9, the minimum polynomial of p(x) is of degree pe, so (p(x) - 1)p'-1 i= O. By 1.8, there is a commutator of weight pe which is not 1; hence c ~ pe. By 4.3a), e ~ 1 - 1, since I - 1 is the p-Iength of Gj/Op',p(Gj). Hence c ~ pl-1.

Also, since pe is the degree of the minimum polynomial of a linear transformation of 0P' jGj)/U,

Using the inductive hypothesis, it follows that if I~U/UI = pn,

n ~ (pl-l _ l)/(p - 1) + pl-l = (pi - l)/(p - 1).

Hence b ~ (pi - l)/(p - 1). b, c) Let c' be the class of ~Op',p(Gj)/Op',p(Gj). Since I > 1, c' ~ 1,

and if

ffi=ffi >ffi > ... 1-' 1-'0 - 1-' 1 -

is the lower central series of ~, ~c' $. 0p',p(Gj). Suppose x E ~c' and x ¢ 0p',p(Gj). Then the order of p(x) is at least p, so by 2.9, (p(x) - W- 2 i= O. By 1.8, Z = [y, x, ... , x] i= 1 for some y E ~.

p-2

But by III, 2.11, Z E ~1+(p-2)c" so C ~ 1 + (p - 2)c'. If p > 3, c' ~ ((p - 2)1-1 - 1)/(p - 3) by the inductive hypothesis, whence

c ~ ((p - 2)1 - 1)/(p - 3).

If p = 3, observe that by 5.4, ~(1-1) =I 1, so by III, 2.12, c ~ 21-1. For any p, we define a series

of ~-invariant subgroups of 0p',p(m) inductively; if n > 0, ~ operates on 0p',p(m)/un_1 and UjUn_1 is defined to be the set of elements of 0p',p(m)/un_1 left fixed by ~. By III, 2.8, [Un' ~;] ~ Un- i if i ~ n; thus ~c' centralizes UC'/U, U2c'/Uc" .. " Hence p(x) - 1 induces the zero linear transformation on these spaces, and (p(x) - W- 2 induces the zero linear transformation on U(p-2)c'/U, But (p(x) - W- 2 =I 0, so U(p-2)c' < 0p',p(m). Hence Ui - 1 < Ui for i = 1, ... , (p - 2)c', and if IOp',p(m)/ul = pm, m ~ (p - 2)c' + 1.

If p > 3, it follows by using the inductive hypothesis that

b ~ m + ((p - 2)1 - (I - 1)(p - 3) - p + 2)/(p - 3?

~ ((p - 3?(p - 2)c' + (p - 2)1 - I(p - 3) + p2 - 6p + 8)/(p - W.

But by the first part, (p - 3)c' ~ (p - 2)/-1 - 1, so

b ~ ((p - 2)1+1 - I(p - 3) - p + 2)j(p - W.

For p = 3, it follows by using the inductive hypothesis that

b ~ m + 21-2 + I - 2

~ c' + 1 + 21-2 + I - 2

~ 21-1 + I - 1. q.e.d.

The group constructed in 4.7 shows that the inequalities in 5.5a) are best possible.

5.6 Lemma. Let m be a soluble group of p-Iength l. Suppose that P1' ... , Pr are the prime divisors of I m I other than p and that Ii is the Pi-length of m. Then

r

I ~ TI (l + IJ i=l

Proof This is proved by induction on r and is trivial for r = O. If r > 0, let


be the upper P. -series of m. If kj is the p-Iength of 9l i~ j (j = 0, 1, ... , I.),

But the Pi-length of 9li~j is at most Ii (i = 1, ... , r - 1) and 9l/~j is a p~-group. Hence by the inductive hypothesis,

.-1

kj ~ n (1 + Ii), i=l

so

.-1 •

I ~ (1 + I.) n (1 + Ii) = n (1 + IJ q.e.d. i=l i=l

5.7 Theorem (HALL and HIGMAN). Let m be a soluble group of derived length d. For each prime divisor p of Iml, let dp be the derived length of a Sylow p-subgroup of m and let lp be the p-length of m.

a) d ~ Lp dplp-b) d ~ Lp>2 d; + d2 np>2 (1 + dp). c) If c is the class of the Sylow 2-subgroups of m,

d ~ c + L tdp(dp + 1). p>2

Proof. a) This is proved by induction on Iml. Suppose that m # 1. If 9l1, 912 are distinct minimal normal subgroups ofm and d; is the derived length of m/9li, d ~ max(d~, d;), since m is isomorphic to a subgroup ofthe direct product (m/9l1) x (m/9l2). But by the inductive hypothesis, d; ~ Lpdplp, so the assertion follows at once. We may suppose, then, that m has a unique minimal normal subgroup 9l. Since m is soluble, 1911 is a power of a prime p; also Op,(m) = 1. If ~ = 0iG» and d' is the derived length of m/~,

since the derived length of ~ is at most dp- Since ~ = Op,jm), the p-Iength of m/~ is lp - 1, so by the inductive hypothesis,

d' ~ dp(lp - 1) + L dqlq. q;&P

Thus

b) By Theorem 5.4, Ip :-::;; dp for p odd, so by a),

d:-::;; I d; + d2 12 • p>2

But by 5.6,

12 :-::;; n (1 + Ip) :-::;; n (1 + dp), p>2 p>2

so

d :-::;; I d; + d 2 n (1 + dp). p>2 p>2

c) This is proved by induction on J<f>J. As in a), we may suppose that <f> has a unique minimal normal subgroup 91 If J mJ is a power of p, Op,(<f» = 1; let ~ = 0i<f» and let 6 E Sp(<f».

Suppose that p = 2. Since ~:-::;; 6, Z(6):-::;; C<v(~). By 1.4, ~ ~ C<V(~), so Z(6) :-::;; ~ and Z(6) :-::;; Z(~). But Z(~) <J <f> and 6/Z(~) E Sp(<f>/Z(~)), so the class of the Sylow 2-subgroup of <f>/Z(~) is less than c. By the inductive hypothesis, the derived length of <f>/Z(~) is at most

c - 1 + I 1;didq + 1). q>2

Since Z(~) is Abelian, the assertion follows at once. If p > 2, then by 5.4a), the derived length of 6/~ is less than dp •

Hence by the inductive hypothesis, the derived length of <f>/~ is at most

c + L 1;didq + 1) + 1;dp(dp - 1). q>2 q",p

Since the derived length of ~ is at most dp ,

q>2

q.e.d.

In many ways, the most elementary way of using Theorem B is by means of the following lemma.

5.S Lemma. Suppose that a minimal normal subgroup 91 of the group (£) is an elementary Abelian p-group. Let [ = C(j;(91). Suppose that (£)/[ is p-soluble, and that either p > 3 or p = 3 and the Sylow 2-subgroups of (£)/[ are Abelian. If 9 E (£) and [~)(, y, y] = 1, then 9 E [.

Proof Let p = (£) on 91); thus ker p = [. Since 91 is a minimal normal subgroup of (£), p is irreducible. Thus by V, 5.17, op(£)/[) = 1 and 0p(imp) = 1. By 1.8a), (p(g) - 1)2 = 0, since [91, g, g] = 1. Thus

p(g)P - 1 = (p(g) - 1)P = 0,

and p(g)P = 1. If p(g) =1= 1, the order of p(g) is p and the minimum polynomial of p(g) is (t - If. By 2.9, p > 3 and 2 2 P - 1, a contradiction. Hence p(g) = 1 and 9 E [. q.e.d.

5.9 Theorem. Let (fj be a finite group. a) If 6 E Sp(fj) mid m: is a maximal normal Abelian subgroup of 6,

C(j;(m:) = m: x :n for some pi-subgroup :n. b) (THOMPSON) Suppose that (fj is soluble, n is a set of primes greater

than 3 and t) is a Hall n-subgroup of(£). Ifm: is a maximal normal Abelian subgroup of the Fitting subgroup F(t) of t), C(j;(m:) = m: x :n for some subgroup :n ofO",(£).

Proof a) Since m: <J 6, 6 E Sp(N(j;(m:)). Since C(j;(m:) <J N(j;(m:), 6 11 C(jj(m:) E Sp(C(j;(m:)). But 6 11 C(jj(m:) = Cs(m:) = m: by III, 7.3, so m: E SiC(j;(m:». By IV, 2.6, there exists :n::; C(j;(m:) such that C(jj(m:) = m::n and m: 11 :n = 1. Thus C(jj(m:) = m: x :n.

b) We use induction on 1(£)1. Let 91 = O",(fj). Then t)91/91 is a Hall n-subgroup of (fj/91 and m:91/91 is a maximal normal Abelian subgroup of F(t)91/91) = F(t)91/91. If 91 =1= 1, it follows from the inductive hypothesis that

since 0",(£)/91) = 1. Hence C(j;(m:) ::; m:91 and C(j;(m:) = m: x CiJl(m:). If 91 = 1, then Op(fj) = 1 for all p E n'. Hence the nilpotent normal

subgroup F(fj) is a n-group. Thus F(£) ::; t) (IV, 7.2), and F(fj) ::; F(t). Hence F(fj) ::; N(jj(m:). Since C(j;(m:) <J N(jj(m:), F(fj) ::; N (jj(C(jj(m:» and

Since F((\j) ::; F(b),

Now for each prime p, the Sylow p-subgroup mp of m is a maximal normal Abelian subgroup of the Sylow p-subgroup 6 p of F(b), so c~,(mp) = mp, by III, 7.3. Thus cF(f))(m) = m. Therefore

and

Now let

F((1)) = ~o > ~l > ... > ~m = 1

be the part below F((1)) of a chief series of (1), and let [i = C(!jmi-d~i)

(i = 1, ... , m). Hence F((1)) ::; [i' and for each i = 1, ... ,m, ~i-d~i is a p-group for some pEn, since F((1)) ::; b. Since 2 ~ n, 3 ~ nand

it follows from 5.8 that c(!j(m) ::; [i (i = 1, ... , m). Hence c(!j(m) ::; [, where [ = nT=l [i'

Also F((1)) ::; [. Now if [ # F((1)), there is a non-identity normal Abelian subgroup 9Jl/F((1)) contained in [jF((1)), since [ <:l (1) and (1) is soluble. But [ centralizes each ~i-l/~i' so

9Jl > ~o > ~l > ... > ~m = 1

is a central series of 9Jl, and 9Jl is nilpotent. Since 9Jl <:l (1), it follows that 9Jl::; F((1)), which is a contradiction. Hence [= F((1)) and c(!j(m) ::; F((1)) ::; F(b). Thus c(!j(m) = cF(f))(m) = m. q.e.d.

We shall prove next an application of 4.2 to the minimal number of generators of minimal simple groups. For this several lemmas are needed.

5.10 Lemma. Suppose that ffi = ~m, where m is a minimal normal elementary Abelian q-subgroup of ffi and ~ is a p-subgroup (p "# q). If ~ = <at> ... , an), where n > 1, then ffi = <aI' ... ,aiz, ... , an) for some Z Em - {1} and some i.

Proof Suppose that x E m - {1} and ~ = <a1 x, a2' ... , an). If ~ = ffi, there is nothing to prove, so we suppose that ~ "# ffi. Now ~m contains a1 = (aIx)x-t, a2' ... , an and m, so ~m = ffi. Since ~ "# ffi, ~ ';j. m. Thus ~ n m < m. But ~ n m <:J ~m = ffi, since m is Abelian. Since m is a minimal normal subgroup, ~ n m = 1. Hence I~I = Iffi: ml = I~I, and ~, ~ are Sylow p-subgroups of ffi. Thus ~ = ~y for some y E m, since ffi = ~m. Hence a~ E ~ and

Thus yanda2 commute. Since their orders are coprime, <a2Y) = <a2, y). HenceifSl = <a1,a2Y' ... ,an), ~ ::::;; Slandy E R HenceSl n m <:J ~m = ffi and either Sl n m = 1 or m ::::;; R If Sl n m = 1, then y = 1, ~ = ~and1 "# x = al1(alx)E ~ n m = 1,acontradictlon.SoSl ~ m and Sl ~ ~m = ffi. Thus ffi = <a1' a2Y, ... , an). q.e.d.

5.11 Lemma. Suppose that ffi is a group, 'It is a set of primes and O,,(ffi) = 1. Letmbeasubgroupofffisuchthatnils&<lJjO,,(m"# 1. Thenffi = <m,g) for some gEm.

Proof Suppose that u E n!!lS&<(JjO,,(~) and u "# 1. Since O,,(ffi) = 1, u ¢: O,,(ffi). Hence the group generated by the conjugates of u in ffi is not a 'It-group. Let k be the smallest integer for which there exist elements aI' ... ,ak of ffi such that y = ua, ••• ua• is not a 'It-element. Since u E O,,(m), u is a 'It-element. Thus k > 1. Let g = (u a2 ... ua.)a"j' . By definition of k, g is a 'It-element. Hence if ~ = <m, g), <g, O,,(~» is a 'It-group. But ug = ya,' is not a 'It-element, so ug ¢: <g, O,,(~». Hence u ¢: O,,(~). It follows from the definition of u that ~ = ffi. Hence ffi = <m, g). q.e.d.

5.12 Lemma. Suppose that ffi is a group of order divisible by the odd prime p, and suppose that every proper subgroup of ffi is p-soluble. Then there is a p-subgroup U of ffi such that U has 2 generators and

Proof Let x be a p-element of maximal order pe, and put U = <x). We may suppose that U does not have the stated property. Thus there is a subgroup f) of (fj such that U ::; f) < (fj and x pe-1 ~ 0P',p(f». Now f) is p-soluble and, in the notation of 4.1, e;(i>/0p',p(f»)) > 2e - 2. By 4.2b), e*(f» > 2e - 1. Hence there exist elements Xl' x 2 in a Sylow p-subgr~up of f) such that v = [xC" xr-1

] #- 1. Let 'D = <Xl' x2 ). If 'D ::; Sl < (fj, then Sl is p-soluble and e;(Sl) ::; 2e. By 4.2b), e;(Sl/Op',p(Sl)) ::; 2e - 1. Thus v E Op',p(Sl). Hence

V E n 0p',p(Sl) #- 1, '!l,;Sk(!;

and 'D has the required properties. q.e.d.

5.13 Lemma. Suppose that (fj is a group of order divisible by the odd prime p, and suppose that every proper subgroup of (fj is p-soluble. Then there exists a subgroup 'D generated by 2 elements, such that either

n Op(f» #- 1 or n Op,(f» #- 1. '!l,;v<Q; '!l,;v<(!;

Proof By 5.12, there is a p-subgroup U of (fj such that U has 2 generators and U* = nu,;v<Q; 0P',p(i» is a normal, non-identity subgroup of U. By III, 7.2a),Z(U) n U*contains an element u #- 1. Ifnu,;v<Q;Op(f» #- 1, there is nothing further to prove, so we suppose that this is not the case. Thus

is non-empty. Let 'D be an element of Y of minimal order, and let ,0 = 0p.('D). By 1.2 (with (70, (fj, ~, Sl) replaced by (pi, 'D, ,0, 0p',p('D) n C'!l(,o»,

0P',p('D) n C'!l('o) = <r x 1)

for some normal p-subgroup <r of 'D and some pi-subgroup 1). Since u ~ Op('D), u ~ <r; since u is a p-element, u ~ <r x 1). And since u E 0P',p('D), u ~ C'!l('o). In particular, u ~ O/,OU), so ,ou E Y. By minimality of I'DI, 'D = ,OU. But if ,00 is a U-invariant proper subgroup of ,0,

'ooU ~ Y, so U E Op('ooU) and u E C'!l('oo). By 1.12, ,0 is a q-group for some prime q. Since <u) <J U, it follows from III, 13.5 that'o is a special q-group and that (U on ,0/,0') is irreducible. Since U has 2 generators,

it follows from 5.10 that [;/0' = U,Q/O' has 2 generators. Since ,0' :::; cP(,Q) :::; cP([;), it follows that [; has 2 generators.

Since u ¢ CID(,Q), there exists y E ,0 such that z = [u, y] i= 1. Now if [; :::; f) < m, then U :::; f) < m and u E U* :::; Op,jf)). Thus Z E Op,jf)) n,Q, since ,0 <l [;. But since ,0 is a p'-group, Op',p(f)) n ,0 :::; Op,(f)). Hence Z E Op,(f)) for every proper subgroup f) of m containing [;. Thus

q.e.d.

5.14 Theorem (POWELL). Suppose that m is a simple non-Abelian group, p is an odd prime divisor of Iml and that every proper subgroup of m is p-soluble. Then m can be generated by 3 elements.

Proof By 5.13, there exists a subgroup [; generated by 2 elements such that n ID,;vdfjO,,(f)) i= 1, where n is either {p} or p'. But O,,(m) = 1, since m is simple and p divides Iml. By 5.11, m = <[;, g) for some gEm, so m has 3 generators. q.e.d.

Exercises

3) Show that the following are counterexamples to Theorem 5.9b) with the omission of the words 'greater than 3'.

(i) m is the semi direct product X~, where X is the quaternion group of order 8 and ~ is a group isomorphic to GL(2, 2) operating faithfully onX.

(ii) m is the semi direct product X~, where X is the non-Abelian group of order 27 and exponent 3 and ~ is a group isomorphic to SL(2, 3) operating faithfully on X.

§ 6. Fixed Point Free Automorphism Groups

The technique of Hall and Higman has been used in a number of other investigations. In this section, we use it to prove some theorems about fixed point free automorphism groups. First we need the following lemma.

6.1 Lemma. Suppose that p = 2S + 1 (s > 0) is a Fermat prime. If k > 0, 2" == 1 (pk) if and only if n == 0 (2Spk-l).

§ 6. Fixed Point Free Automorphism Groups 477

Proof Since (p - 1)2 == 1 (p), (p - 1)2p '-1 == 1 (pk). Thus 22sp '-1 == 1 (pk), and if n == 0 (2Spk-l), 2n == 1 (pk).

Conversely, suppose that 2n == 1 (pk). Write n = sq + r, where o::s; r < s. Then

Thus p divides 2r - (-1)q, so if r > 0,

2s - (- 1)q > 2r - (- 1)q :::: p = 2s + 1.

This is impossible, so r = 0, (-1)q == 1 (p) and, since p is odd, q is even. Hence n = 2ms for some integer m. Thus

Now if m = plm' and (m', p) = 1, it follows from I, 13.18a) that

(p - 1)2m _ 1 + 2mp == ° (pl+2).

Putting (p - 1)2m - 1 = pku, we obtain

Since (m', p) = 1, it follows that k ::s; I + 1. Thus m == 0 (pk-l) and n == o (2Spk-l). q.e.d.

We prove next an analogue of Theorem 2.9 in which the characteristic is prime to the order of the group.

6.2 Theorem. Let V be a non-zero, finite-dimensional vector space over a field K, and let (f) be a p-soluble group of linear transformations for which Op(f) = 1 and char Ktl(f)l. Let g be an element of (f) of order pn. Then one of the following occurs.

(1) The minimum polynomial of g is tP" - 1, and CAg) "# O. (2) The Sylow 2-subgroups of(f) are non-Abelian, p is a Fermat prime

and C(!;(g) n Op,(f) "# 1. (3) p = 2 and there exists a M ersenne prime q < 2n such that the

Sylow q-subgroups of(f) are non-Abelian.

Proof Since the conditions in (1) remain unchanged under an extension of the ground-field, we may suppose that K is algebraically closed. We

suppose that the theorem is false and choose a counterexample for which dimKV + l(fjl is minimal. We have 0p,(fj) =I- 1.

a) The minimum polynomial of g is not t P' - 1. For otherwise, v = w(1 + g + ... + gpn-i) =I- 0 for some W E V. But

v E Cv(g), so Cv(g) =I- 0 and (1) holds. b) (f) = 'l3~, where 'l3 = <g), ~ = 0p,(fj). For suppose that (fjo = <g)Op,(f)) and (fjo < (fj. Then Op(f)o)

::s; C(!j(Op,(fj)) ::s; 0p,(fj) by 1.3, since 0i(fj) = 1. Thus Op(f)o) = 1. By minimality of the counterexample, the theorem is valid for (fjo on V). But this implies that the theorem is true. Hence (f)o = (f).

c) V is an irreducible K(fj-module. If not, then by the Maschke-Schur theorem, V = V1 EB V2 for certain

non-zero K(fj-submodules V1 , V2 of V. If 91; = ker(fj on V;) (i = 1,2), 911 11 912 = 1. Let 6;/91; = Op(fj/91;) (i = 1, 2). Then 6 1 11 6 2

::s; Op(fj) = 1. Thus gP'-l rt 6; for either i = lor i = 2, and 'l3 11 6; = 1. Hence 91; is a p'-group, 6; = 91; and Op(fj/91;) = 1. By minimality of the counterexample, the theorem is valid for (fj/91; on V;). By a), (1) cannot hold in (fj/91;; also (3) cannot hold in (fj/91;, since otherwise it holds in (fj. Thus (2) holds in (fj/91;. Hence the Sylow 2-subgroups of (fj are non-Abelian, p is a Fermat prime and there exists x E ~ for which x991; = x91; =I- 1. By V, 8.10, Cs(g) =I- 1, and (2) holds in (fj, a contradiction.

d) Let b = gpn-i. ~ is a special q-group for some prime q =I- p, b centralizes ~', Cf)/f),(b) = 1 and ('l3 on ~/~') is irreducible.

Let ~o be a proper 'l3-invariant subgroup of ~. By minimality of the counterexample, the theorem holds in 'l3~o. But the validity of (1), (2) or (3) in 'l3~o would imply the validity of the same in (fj, so Op('l3~o) =I- 1. Since Op('l3~o) ::s; 'l3, b E Op('l3~o) and b E C(!j(~o). By 1.3, b if; C(!j(~)' Thus by 1.12, ~ is a q-group for some prime q =I- p. And by III, 13.5, ~ is a special q-group, b centralizes ~' and ('l3 on ~/~') is irreducible. Since Cf)/f),(b) is 'l3-invariant and Cf)(f),(b) =I- ~/~' by 1,4.4, Cf)/f),(b) = 1.

By V, 17.3,

where V; is the direct sum of isomorphic irreducible K~-submodules of V and V;, Vj have no isomorphic irreducible K~-submodules for i =I- j; also there is a permutation representation (J of 'l3 on {I, ... , I} such that V;g = V;a(g). Since V is an irreducible K(fj-module, (J is transitive. Thus if ker (J = <gpm), <gpm) is the stabiliser of each i and I = pm. Let c = gpm and Do = <c, D). Thus Do ...;::] m.

e) ~o = {xix E (fj, V;x = V;} and V; is an irreducible K~o-module.

Since i(1'(c) = i, V;c = Vi' Conversely, if V;ygj = Vi (y E ~), then Vigj = Vi' gj E (c) and ygj E f>o. Vi is an irreducible Kf>o-module by V,17.3e).

f) c # 1 and the minimum polynomial of the linear transformation u -+ uc of Vi is not t P'- no - 1. In particular, b E f>o.

Suppose that the minimum polynomial of the linear transformation u -+ uc of Vi is tP'- no - 1. Then, as is well-known, there exists u E Vi such that u, uc, ... , ucp·-no- 1 are linearly independent. But by a), u, ug, ... , ug P'- l are linearly dependent, so there exist Ajk E K (j = 0, ... , pn-m - 1; k = 0, ... ,pm - 1) such that

LAjkUgjpno+k = ° j,k

and Ajk # ° for some j, k. Since ugjpno+k = ucjgk E Vi (1'(gk), it follows that

and hence

p,,-rn-l

L AjkUgjpno+k = ° (k = 0, ... , pm - 1) j=O

p,,-m-l

L AjkUC j = 0, j=O

contrary to the linear independence of u, uc, ... , ucp·-no - 1.

In particular, c # 1 and b = cp ·-no-1 E f>o. Let Sl; = ker(f>o on Vi), and let G)i = f>O/Sli' Ci = CSli (i = 1, ... , pm).

Thus bye), Vi is a faithful irreducible KG)i-module. g) n; Sl; = 1, Sly = Sli<1(g) and Sli < f> (i = 1, ... , pm). For ni Sli ~ ker(f>o on V) = 1. Since Sli = ker(f>o on Vi), Sly =

ker(f>t on Vi g) = ker(f>o on Vi<1(g)) = Sli<1(g)' If p divides ISlll, (c) n Sll # 1, since (c) E Sp(f>o) and Sll <J f>o. Thus bE Sll' Since (1' is transitive and Sly = Sli<1(g), it follows that b E Sli for all i and that b = 1, a contradiction. Thus Sll is a pi_group and Sll ~ f>. If Sll = f>, then Sli = f> for all i = 1, ... , pm and f> = niSli = 1, a contradiction. Hence Sll < f> and Sli < f> for all i = 1, ... , pm.

Let 0; = f>/Sl; (i = 1, ... , pm). h) 0i is extraspecial, Z(O;) :=:;; Z(G)i), CoJo;(b) = 1 and Vi is a faithful

irreducible KOi-module.

Suppose that Vi is the direct sum of Si isomorphic irreducible K~modules. By V, 17.5, there exists an irreducible projective representation p of ~o/~ of degree Si' Then p(c~)p·-m = Al for some A E K, and since K is algebraically closed, A = JiP·-m for some Ji E K. Hence c ---+ Ji- 1 p(c~) is an ordinary irreducible representation of ~o/~ of degree Si' By V, 6.1, Si = 1. Thus Vi is an irreducible K~-module. Since S\i = ker(~ on V;) and .oi = NS\i' Vi is a faithful irreducible K.ocmodule.

By Schur's lemma, Z(.o;) is represented on Vi by scalar multiples of the unit mapping. Thus, since (ffii on Vi) is faithful, Z(.oi) ~ Z(ffi;) and since b E ~o, bS\i centralizes Z(.o;). But by d), CMy(b) = 1, so by V, 8.10, Ciijfj'.R.(b) = 1. Hence Z(.o;) ~ ~fS\;/S\i' Since ~ is special, ~fS\;/S\i ~ Z(~/S\;) = Z(.o;), and 4>(.0;) = 4>(~)S\;/S\i = ~fS\;/S\i = .0;. Hence Z(.oi) = n; = 4>(.0;). By g), .oi #- 1, so .oi is extraspecial. Since .0;/.0; ~ ~/~fS\i' CO.lo,(b) = 1.

Let I.o;j = q2d+1. By g), d is independent of i. i) p is a Fermat prime and q = 2. Also IN~fl = 22dpm and «Ci) on

.0;/.0;) is irreducible. By h), .0; is centralized by Ci and Co,/o,(b) = 1. We may therefore

apply V, 17.13 to (ffii on Vi), which is faithful and irreducible. Now by f), «c;) on Vi) has no component isomorphic to the regular representation «ci ) on K<c i »). Hence V, 17.13b) shows that qd + 1 = l<c;)1 = pn-m, In particular q < pn. Hence if p = 2 and q is a Mersenne prime, the conclusion (3) of the theorem holds. This is not the case, so by 2.7, q = 2 and p is a Fermat prime; also either n - m = lor n - m + 1 = p = d = 3. It follows from 6.1 that 2X == 1 (pn-m) if and only if x == 0 (2d) and 2Y == 1 (pn) if and only if y == 0 (2dpm).

It follows from II, 3.10 that 2d is the degree of any faithful irreducible representation of <c;) in GF(2). Since CoJo;(b) = 1, any irreducible component of «c;) on .0;/.0;) is faithful and hence of degree 2d. Thus «c;) on .0;/.0;) is irreducible.

Also by II, 3.10, 2dpm is the degree of any faithful irreducible representation of ~ = <g) in GF(2). By d), IN~fl = 22dpm.

Note that since (2) does not hold, Cfj(g) = 1. It follows that pm > 1, for otherwise S\1 = 1 by g),.o1 = ~,ffi1 = ffi and, by h), Z(~) ~ Z(ffi). Let

Ei = n S\j~f (i = 1, ... , pm). Ni

j) If i #- j, vx = V for all v E Vj' X E [Ei' ~]. For [Ei' ~J ~ [S\j~f, ~]. By d), ~f ~ Z(~), so

[£Ii' ~ ] :;; [S\j' ~J :;; S\j. Thus j) follows at once.

k) ~' = S!'l X ... X S!~m, and IS!;I = 2. Since I~: ftj~'1 = l,Qj: 'oj I = 22d, it follows from 1,2.13 that

I~: S!;/ ~ 22d(pm_l). From i), S!j i ~', so S!j > ~'. Thus

and S!l ... S!pm is ~-invariant. It follows from d) that

Hence

~' = S!'l ... S!~m 0 [S!j, S!J i".j

But by j), ([S!j, S!J on Vk) is trivial for all k (i -# j), so [Eil Ej ] = 1. Thus

Now by j), (E; on V) is trivial for allj =I- i. Hence (OJ,,.i 52] on V;) is trivial and ((E; n ONj S!j) on Vk) is trivial for all k. Thus S!; n OJ,,.i S!j = 1, and

If x E S!;, then x E ~', xftj E ,0; and xft j is represented on Vi by ± 1. Since x is represented on Vj by 1 for all j -# i, it follows that IE;I ~ 2. Since E? = S!;q(g), all the S!j are conjugate. Thus if E; = 1 for some i, S!; = 1 for all i and~' = 1, contrary to h). Therefore IE;I = 2 for all i.

By k),

I~'I = 2pm == 2 (p).

Thus the length of some orbit of «g > on ~') other than {1} is not divisible by p and Cs,(g) -# 1. Thus (2) holds in <fi. q.e.d.

In our application of this, we shall need the following elementary lemma.

6.3 Lemma. Suppose that G3 is a group, in <l G3 and IX is an automorphism of G3 which leaves fixed every element of in. Then IX leaves C(!;(in) fixed and IX induces the identity automorphism on G3jC(!;(in).

Proof If x E in and g E G3, then IX leaves fixed the elements x, x g of in. Hence x g = (xg)1X = (xlXr = x ga and x(ga)g-l = X. Thus, for any g E G3, (glX)g-l E C(jj(in). Hence IX leaves C(jj(in) fixed and induces the identity automorphism on G3/C(jj(in). q.e.d.

6.4 Theorem (SHULT [IJ, GROSS [3J). Let p be an odd prime. If G3 is a soluble group, IX is an automorphism of G3 of order pn and C(jj(IX) = 1, the Fitting height of G3 is at most n.

(By the Fitting height of X is meant the nilpotente Lange in the sense oflII, 4.7; it will be denoted by h(X).)

Proof This is proved by induction on I G31. It is trivial for G3 = 1. Suppose, then, that G3 =1= 1. Since C(jj(IX) = 1, the length of each orbit of (IX) on (3) other than {I} is divisible by p. Thus (p, 1(31) = 1.

Let in be a minimal normal subgroup of G3. By V, 8.10, C(jj/91(IX) = 1, so by the inductive hypothesis, h(G3jin) is at most n. As in III, 4.6, we define a series

of characteristic subgroups of G3 inductively by the rules that ~o = G3 and, for i > 0, ~i-d~i is the maximal nilpotent factor group of ~i-l' Since h(G3jin) ::;; n, ~n ::;; in. Hence either ~n = 1 or ~n = in. If ~n = 1, there is nothing to prove. We therefore assume that Sln = in. Thus in is the only minimal normal subgroup of G3. Since G3 is soluble, in is an elementary Abelian r-group for some prime r. It follows that Or,(G3) = 1 and 0r(G3) = F(G3). Let

U = cJ>(Or(G3»,

By 1.4 and 1.6, C(jj(Or(G3)/U) = 0r(G3), Hence Or' (G3jU) = 1 and 0r(G3)jU = F(G3/U). If U =1= 1, U ~ in and h(G3/U) is at most n. By III, 4.6, the Fitting height of (G3/U)/F(G3/U) ~ G3/0r(G3) is at most n - 1. Hence h(G3) ::;; n, and there is nothing further to prove. We therefore suppose that U = 1 and write m = 0r(G3). Thus m is an elementary Abelian r-group.

Let Wl = 0r,r,(G3). We show that f3 = IX P·- 1 induces the identity mapping on Wl/m. Suppose that this is not the case, and let ~ be the

semi-direct product 9Jl<ex). Since p # r,5ll = 0r(i;). If'l!/5ll = Op(N5ll), [9Jl, 'l!] ~ 5ll, since p does not divide 19J11· Hence fJ rt 'l! and <ex5ll) (") ('l!/5ll) = 1. Since <ex5ll) is a Sylow p-subgroup of N5ll and 'l!/5ll is a normal p-subgroup of i;/5ll, it follows that 'l!/5ll = 1. Thus Op(N5ll) = 1. Also, since 0r,(m) = 1, Or,(i;) (") 9Jl ~ Or,(9Jl) = 1 and 0r,(i;) is a p-group. Thus 0r,(i;)5ll/5ll ~ 0ii;/5ll) = 1, Or,(i;) ~ 5ll and, since 5ll is an r-group, Or,(i;) = 1. By 1.3, Cr,(5ll) = 5ll. Hence (i;/5ll on 5ll) is faithful. By 6.2, either (1) C\ll(ex) # 1, or (2) Cr,/\ll(ex) # 1, since p is odd. But C(!i(ex) = 1, so (1) is impossible. Also (2) is impossible, by V, 8.10. Thus fJ induces the identity mapping on 9Jl/5ll.

If [/5ll = C(!i/\ll(9Jl/5ll), it follows from 6.3 that fJ induces the identity mapping on m/[. But by 1.3, [ ~ 9Jl, so fJ induces the identity mapping on m/9Jl and on 9Jl/5ll. By I, 4.4, fJ induces the identity mapping on m/5ll. Thus ex induces an automorphism ex on m/5ll of order at most pn-l, and by V, 8.10, C(!i/\ll(ex) = 1. By the inductive hypothesis, h(m/5ll) ~ n - 1. Since 5ll is nilpotent, h(m) ~ n. q.e.d.

6.5 Remark. SHULT [1] has shown that the bound in 6.4 is best possible. More generally, GROSS [4] has shown that if 21 is a soluble group of order P1PZ' .. Pn' where Pl' Pz, ... , Pn are (not necessarily distinct) primes, there exists a finite soluble group m of Fitting height n for which 21 is a group of operators and C(!i(21) = 1.

To obtain results corresponding to 6.4 for P following lemmas.

2, we need the

6.6 Lemma. Suppose that m is of odd order and r is an automorphism of m such that rZ = 1.

a) If x E m, there exist unique elements y, z of m such that x = yz, yr = y and zr = z -1.

b) Suppose that x, yare elements ofm such that xr = x", (xY)r = (xY)" and (yr)y E C(!i(x), where B = ± 1. Then x and y commute.

Proof a) Since Iml is odd, there exists z E m such that Z2 = (xrflx. Then

Since Iml is odd, zr = Z-l. If y = xz- 1,

Clearly, y and z are unique.

b) We have

so x commutes with y2. Since I (fj I is odd, x commutes with y. q.e.d.

Note that by 6.6a), if Colr) = 1, then xr = x-1 for all x E (fj, whence (fj is Abelian (V, 8.18a».

6.7 Lemma. Suppose that (fj is soluble, ~ <J (fj, ~ is not nilpotent but ~5l/5l is nilpotent for every non-identity characteristic subgroup 5l of (fj. Then there exists a prime p such that 0p,«fj) = 1 and Op«fj) is elementary Abelian.

Proof Let p be the prime divisor of an elementary Abelian normal subgroup of (fj. Then ~ = Op«fj) :f:: 1. Suppose that 91 = 0p,«fj) :f:: 1. Then ~~/~ and ~91/91 are nilpotent, so £j(~ n ~) and £j(~ n 91) are nilpotent. By III, 2.5, £j(~ n ~ n 91) is nilpotent. But ~ n 91 = 1, so ~ is nilpotent, contrary to hypothesis. Thus Op,«fj) = 1.

If ~ = 0i(fj) is not elementary Abelian, cf>(~) :f:: 1. By III,3.3b), cf>(~) :::; cf>«fj), so cf>«fj) :f:: 1. Hence ~cf>«fj)/cf>«fj) is nilpotent. By III, 3.5, ~cf>«fj) is nilpotent. Thus ~ is nilpotent, a contradiction. Hence Op«fj) is elementary Abelian. q.e.d.

6.8 Theorem (GORENSTEIN and HERSTEIN [1 ]). Let (1 be an automorphism of the soluble group (fj. If (14 = 1 and C(!j«(1) = 1, (fj' is nilpotent.

Proof Suppose that this is false and let (fj be a counterexample of minimal order.

a) l(fjl is odd. If I is the length of any orbit of (1 on (fj - {1}, I divides 4 and I > 1,

so I is even. Hence I (fj I - 1 is even and I (fj I is odd. b) There exists a prime p such that Op,«fj) = 1 and ~ = Op«fj) is

elementary Abelian. Let X be any non-identity characteristic subgroup of (fj. By V, 8.10,

(1 induces a fixed point free automorphism on (fj/X. Since (fj is a counterexample of minimal order, (fj'X/X is nilpotent. The assertion ,b) thus follows from 6.7.

Let T = (12, (£ = C(!j(T). c) (£ is Abelian. For (£ is (1-invariant and the restriction (1' to (£ of (1 is an automor

phism of(£ of order 2 such that CI[«(1') = 1. Hence (£ is Abelian.

By 1.11 applied to the semi-direct product (f)<a), (f) has a a-invariant Hall p' -subgroup ~.

d) ~ i [. Let 9\ = Op,p,«f)). By 1.3, C(l;(9\/~) :$ 9\, Since 9\ <l (f), ~ n 9\ is

a Hall p'-subgroup of 9\ and 9\ = ~(~ n 9\). By c), [ is Abelian, so if ~ :$ [, [ centralizes ~ n 9\ and 9\/~. Thus [ :$ 9\, Hence ~[ :$ 9\ =

~m n 9\) :$ ~[ and 9\ = ~[. Thus 9\' :$ ~ and 9\' is nilpotent. Hence 9\ < (f). If y E (f) and y ~ 9\, y = UZ, where u E [ and zr = Z-I,

by 6.6a). Thus u E 9\ and z ~ 9\. Hence there exists x E 9\/~ such t~at x and z = z~ do not commute. But since 9\ = ~[, xr = x and (x7f = xZ, where r is the automorphism of (f)/~ induced by r. Since zr = Z-1, 6.6b) shows that x and z commute, a contradiction. Thus ~ f;, [,

By d), we may choose a minimal a-invariant subgroup .3 of ~ such that .3 i [. Let ft be a maximal normal a-invariant subgroup of .3. Thus .3/ft is an elementary Abelian group and «0') on .3/ft) is irreducible. Also ft :$ [, by definition of .3. It follows from c) that ft is Abelian.

By b) and the Maschke-Schur theorem, applied to <a).3,

~ = ~1 X .,. x ~"'

where <0').3 :$ N(l;(~;) and «a).3 on ~i) is irreducible. Also, since .3 f;, [, it follows by applying 6.6a) to .3 that there exists Z E.3 - {1} such that zr = Z-I. By b) and 1.3, ~ = C(l;(~), so z does not centralize some ~i' Denote such a ~i by m; thus «a).3 on m) is irreducible and z ¢ C(l;(m).

e) Eft, m] = 1. Ifnot, Cm(ft) is a proper <a) .3-invariant subgroup ofm, so Cm(ft) = 1.

But since ft :$ [ and [ is Abelian, [ n m :$ Cm(ft), so [ n m = 1. Hence Cm(r) = 1 and vr = V-I for all v E m. Hence (vZ)r = (v Zr1 and by 6.6b), z E C(l;(m), a contradiction. Hence Eft, m] = 1.

Let p = «a).3 on m). By e), ker p 2 R Since z ~ C(l;(m), .3 i ker p. Thus (ker p) n .3 is a proper, normal a-invariant subgroup of .3 containing R By definition of ft, ft = (ker p) n 3. But 0'2 = r does not centralize .3/R Also z ~ C(l;(m), so by 6.6b), r ~ ker p. Hence by 1.10 applied to im p, the minimum polynomial of the linear transformation of m induced by a is of degree 4. Thus there exists v E m such that u = v(va)(va 2 )(va3 ) =f 1. But ua = u, so C(l;(a) =f 1, a contradiction.

q.e.d.

6.9 Theorem (GROSS). Suppose that (f) is a soluble group and 0' is an automorphism of (f) such that Co;(a) = 1. If the order of a is 2" (n > 1), the Fitting height of (f) is at most 2n - 2,

Proof Suppose that this is false and let ffi be a counterexample of minimal order. By 6.8, n ~ 3.

a) Iffil is odd. As in 6.4, the length of any orbit of a on ffi - {1} is even, so 1 ffi 1 - 1

is even and 1 ffi 1 is odd. b) There exists a prime p such that Op,(ffi) = 1 and ~ = Op(ffi) is

elementary Abelian. Define the series

ffi = ffio ~ ffi1 ~ ...

of characteristic subgroups ffii of ffi by the fact that for i > 0, ffii-dffii is the maximal nilpotent factor group of ffi i- 1. Then ffi zn- z =/: 1, by III, 4.6, and ffi Zn - 3 is not nilpotent. But if X is any non-identity characteristic subgroup of ffi, C(l;/.I(a) = 1 by V, 8.10. Since ffi is a counterexample of minimal order, the Fitting height of ffi/X is at most 2n - 2, so ffi zn- 31/X is nilpotent. The assertion b) thus follows from 6.7.

Let r = a z·-1•

c) The restriction of r to ~ is not the identity mapping. Since r =/: 1, it follows from 6.6a) that there exists Z E ffi such that

zr = Z-1 =/: 1. But if c) is false, xr = x and (xZ)r = X Z for all x E ~, so by 6.6b), z E C(l;(~). But by b) and 1.3, C(l;(~) = ~. Thus Z E ~. Since zr = z-1, the restriction orr to ~ is not the identity mapping.

Let ffi be the semi-direct product (a>ffi, let IYd~ = F(ffi/~),

~2/tjl = F«(f)/~l)' d) r E ~z. If 91 <l (f) and 1911 is even, then 91 intersects every Sylow 2-subgroup

of (f) non-trivially, so r E 91. By c), r ¢ C(f;(~), so IC®(~)I is odd. Hence C®(~) = C(jj(~) = ~.

Thus if p = «(f)/~ on ~), p is faithful. If l~d~1 is even, r E ~1 and the assertion is clear; thus we may suppose that l~d~1 is odd. Clearly I ~d~1 is not divisible by p, since ~d~ is nilpotent and ~ = O/(f)). We also observe that p(a~) is exceptional on ~, in the sense of 3.6. For otherwise the degree of the minimum polynomial of p(a~) is 2n. Thus there exists v E ~ such that

u = v(va)· .. (va 2' -1) =/: 1.

But then ua = u, so C(l;(a) =/: 1, a contradiction. It follows from 3.12 that r = a2o-1 E IYz. e) r induces the identity automorphism on (f)/(~1 n (f)). By d), r E ~2; also, ~Z/~1 is nilpotent. Since Imz n (f)md~11 is odd,

it follows that r centralizes mz n (f)md~1' Thus r induces the identity

automorphism on 012 (\ (f))jm:l (\ (f)). But since (f) and (J2 are normal in (f), [(f), r] ~ (J2 (\ (f), so r induces the identity automorphism on (f)j((J2 (\ (f)). By I, 4.4, r induces the identity automorphism on (f)jm: 1 (\ (f)).

It follows from e) that if a' is the automorphism of (f)jm:l (\ (f)) induced by a, a,2 n

-1 = 1. By V, 8.10, a' is fixed point free. Since n ~ 3

and (f) is a counterexample of minimal order, it follows that the Fitting height of (f)j((Jl (\ (f)) is at most 2n - 4. Since m:l (\ (f))j'l3 and 'l3 are nilpotent, the Fitting height of ij) is at most 2n - 2. q.e.d.

6.10 Remark. The hypothesis of solubility in 6.8 and 6.9 can be omitted. F or the proof that I (f) I is odd in both theorems does not utilize the solubility of (f), and this then follows from the theorem of Feit and Thompson. A proof of the solubility of (f) in 6.8, independent of the theorem of Feit and Thompson, was given by Gorenstein and Herstein (see GORENSTEIN

[1], p. 342). It has been conjectured that if the group (f) has an automorphism

a such that Cm(a) = 1, then (f) is soluble.

We conclude this section with a theorem of J. N. WARD [1] on soluble groups having an elementary Abelian fixed point free automorphism group of order p2. For this we need the following lemmas.

6.11 Lemma. If 91 is a normal p'-subgroup and 'l3 is a p-subgroup of (f), Nm/91('l391j91) = Nm('l3)91j91 and Cm/91('l391j91) = Cm('l3) 91j91.

Proof Let SV91 = Nm/9l('l391j91). Then 'l391 <J ft and 'l3 E Sp('l391), so by the Frattini argument, ft = Nll('l3)('l391) ~ Nm('l3)91. Clearly Nm('l3) ~ ft, so ft = Nm ('l3)91.

If i?j91 = Cm/91('l391j91), then 91 ~ 5! ~ ft = Nm('l3)91. Hence 5! = rom, where IDl = 5! (\ Nm('l3). Thus ['l3, IDl] ~ 'l3. But also

['l3, IDl] ~ ['l3, 5!] ~ 91,

so ['l3, IDl] ~ 'l3 (\ 91 = 1 and IDl ~ Cm('l3). Thus 5! = Cm('l3)91. q.e.d.

6.12 Theorem. Suppose that (f) = m~, where m is elementary Abelian of order p2 and ~ is a soluble, normal p'-subgroup of (f). Let V be afaithful K(f)-module, where K is a field of finite characteristic q such that q =P p and Oq(~) = 1. Suppose that for any non-identity element a of m, Cf)(a) is nilpotent and any q'-element of Cf)(a) leaves fixed every element of Cv(a). Then ~ is nilpotent.


Proof If L is an extension of K and VL = V ®K L, CVl(g) = Cv(g) ®K L for all g E fjj. If a E m - {1}, Cy(a) $; Cv(g) for every q'-element g of Ci)(a), so CVl(a) $; CVl(g). We may therefore suppose without loss of generality that K is algebraically closed.

Suppose now that the theorem is false. Let fjj be a counterexample of minimal order, and let V be a faithful Kfjj-module of minimal dimension, such that Cv(a) $; Cv(g) whenever a E m - {1} and g is a q' -element of Ci)(a).

a) V is an irreducible Kfjj-module. To prove this, let

V = Vo > Vi > ... > Vr = 0

be a Kfjj-composition series ofV. If ~i = ker(fjj on Vi_dV;) (i = 1, ... , r) and ~ = ni ~i' ~ is a q-group (cf. I, 4.4). Since Oq(fjj) = 1, ~ = 1. Thus N(ni ~i) is not nilpotent. By III, 2.5, ~~;/~i is not nilpotent for some i. We prove that the conditions of the theorem are satisfied by the K(fjj/~;)-module Vi-dVi' First, 0ifjj/~i) = 1, since Vi-iNi is a faithful irreducible K(fjj/~;)-module. Next, suppose that a E m - {1}. By 6.11,

is nilpotent, and every q'-element of this group leaves fixed every element of Cv;_,/v; (a). By the Maschke-Schur theorem,

C V;-I!V; (a) = (C V;-l (a) + Vi)/Vi·

Finally !m~;/~i! = p2; otherwise, there exists bE (m n ~i) - {1}, so that ~~;/~i = Ci)51j51; (b) is nilpotent, contrary to the choice of i. It thus follows from the minimality of !fjj! and dimK V that ~i = 1, Vi- i = V and Vi = O. Thus V is an irreducible Kfjj-module.

b) V is the direct sum of isomorphic irreducible K~-modules. By Y, 17.3,

V = Wi EB ... EEl Ws '

where each Wi is the direct sum of isomorphic irreducible K~-modules, and there is a transitive permutation representation (J of m such that Wit1(a) = Wia (a Em). Let m = ker (J. Since m is Abelian, m is the stabiliser of i for all i = 1, ... ,s, so !m: m! = s. Thus it is to be proved that m= m.

Let (Y = F(~). By Y, 8.15, there exists a E m - {1} such that Clj(a) i= 1. On the other hand, Clj(a) i= (Y. For otherwise, a centralizes ~/Ci)(m,

by 6.3. But by III, 4.2, C!j(m :S ~,so a centralizes ~ and ~/~. By I, 4.4, a centralizes ~. Hence C!j(a) = ~, which is impossible since C!j(a) is nilpotent and ~ is not. Thus CIJ(a) #- ~,and CIJ(a) is a proper m-invariant subgroup of ~. Let ~ 0 be a minimal m-invariant subgroup such that CIJ(a) < ~o :S ~. Since ~ is nilpotent, CIJ(a) < NlJo(CIJ(a)), so by rninimality of ~o, CIJ(a) <J ~o· Let l: = ~o/CIJ(a). If CIJ(a) = 6/CIJ (a), then a centralizes 6 by I, 4.4 and 6 :s CIJ(a). Hence Cx(a) = 1. But again by V, 18.5, there exists b E m - {I} such that C;x(b) #- 1. Evidently, b ¢ (a), so m = (a, b). And since C;x(b) #- 1, CIJ(b) #- 1, by 6.11. Thus to prove that ~ = m, it suffices to show that if e E m and CIJ(e) #- 1, then e E~.

To do this, suppose that x E CIJ(e) and x #- 1. Then there exists an element w of some Wi such that wx #- w. Let

u = w + we + ... + weP- 1 ;

then U E Cv(c). Since 0i~) = 1, ~ is a q'-group, so by hypothesis, U is left fixed by all elements of CIJ(e). Hence ux = u. But since x E CIJ(e),

ux = wx + wxe + ... + wxcP-l,

and

o = u(x - 1) = w(x - 1) + w(x - 1)c + ... + w(x - l)cP- 1•

Now w(x - l)d E Wit1(c j ) and w(x - 1) #- 0; hence the sum

is not direct. Thus c lies in the stabiliser ~ of i. c) If i! is any non-identity Abelian normal subgroup of (fj and i! :S ~,

Ci!(m) #= 1. By V, 17.3,

where Vi is the direct sum of all Ki!-submodules of V isomorphic to some irreducible K2-module X;, and there exists a permutation representation p of (fj on {I, ... , n} such that Vjp(g) = Vjg for all g E (fj. Since 2 is Abelian and K is algebraically closed, dimKXi = 1. Thus there exist distinct homomorphisms Xl' ... ,Xn of i! into K x such that if x E 2, wx = xix)w for all w E Xj and hence for all wE Vj.

Let {jl' ... ,jk} be an orbit of the restriction of p to ~. Then V = W1 EB W2 , where W1, W2 are K~-submodules of V and W1 = EB~=l Vj ,

If W2 # 0, then the irreducible K~-submodules of W1 , W2 are isomorphic, by b), so W1 , W2 certainly have isomorphic irreducible KE-submodules. This, however, is impossible on account of the definition of the Vi' Hence W2 = 0 and the restriction of p to ~ is transitive.

It follows that I~I is divisible by n. Since ~ is a pi-group, we deduce that p does not divide n. Hence the restriction of p to m has a fixed point j and Vja = Vi for all a E m. Thus if a Em and x E E, then va-1 E Vi and

for all v E Vi' Hence v [a, x] = v. If Em, E] = E, it follows that vy = v for all v E Vi' Y E E. But then, if 1 ~ i ~ n, i = jp(g) for some g E ~, since the restriction of p to ~ is transitive. Thus if u E Vi' then ug-1 E Vi' ug-1y = ug-1 for all y E E, so ug-1yg = u and uy = u for all y E E. Hence E is represented trivially on all Vi' which is impossible since E # 1. Hence Em, E] < E.

Put Eo = Em, EJ. Then C'1lil/ilo (m) = mE/Eo, and by 6.11,

Hence E = Cil(m)Eo' Since Eo < E, Cil(m) # 1. d) Zm) = 1. Let 91 = CZ(t;) (m). Then 91 is contained in the centre of m~ = Qj. It

follows from a) that each element of 91 is represented on V by a scalar multiple of the identity mapping.

Now by Y, 8.12, there exists a E m - {1} such that Cv(a) # O. Since Oq(~) = 1, Zm) is a q'-group, so by hypothesis CZ(t;)(a) leaves fixed every element of Cy(a). But 91 ~ CZ(t;)(a). Thus if x E 91 and vx = AV for all v E V, then AV = v for all v E Cv(a), so A = 1, vx = v for all v E V and x = 1. Hence 91 = 1. By c), Z(~) = 1.

e) To obtain a contradiction, let !Y = F(~). Since Qj is a counterexample of minimal order, (m on ~/IY) is irreducible. Hence ~/!Y is an elementary Abelian r-group for some prime r. By 1.11, Qj possesses an m-invariant Sylow r-subgroup 9\. Thus ~ = 9\!y = 9\Or'(IY), since !Y is nilpotent. Hence Z(9\) n !Y centralizes ~ and Z(9\) n !Y ~ Z(~). By d), Z(9\) n !Y = 1. By III, 7.2, 9\ n !Y = 1. Thus 9\ is m-isomorphic to N!Y and (m on 9\) is irreducible. But by Y, 8.15, there exists a E m - {1} such that C!Il(a) # 1. Since C!Il(a) is m-invariant, C!Il(a) = 9\. But Ct;(a)


is nilpotent and C~(a) = 9lCIj(a), so 91 centralizes CIj(a). Thus CZ(Ij)(a) ::; Z(i») = 1. This contradicts c).

Hence i) is nilpotent. q.e.d.

6.13 Theorem (J. N. WARD). Let 0) be a soluble pi_group and let m be an elementary Abelian group of automorphisms of 0) of order p2. If C(p(a) is nilpotent for every non-identity element a of m, the Fitting height of 0) is at most 2.

Proof This is proved by induction on 10)1. Let 91 be the minimal normal subgroup of 0) for which 0)/91 is nilpotent. If .R is any non-identity characteristic subgroup of 0), C(p(,Il(a) is nilpotent for every non-identity element a ofm, since by 6.11, C(p(,Il(a) = C(p(a).R/R Thus by the inductive hypothesis, 91.R/.R is nilpotent. By 6.7, we may suppose that Oq.(O)) = 1 for some prime q and that Oq(O)) is elementary Abelian. Since 0) is a pi -group, q =P p.

Let i) denote the semi-direct product mO). Then Oq.(i») nO)::; Oq.(O)) = 1, so Oq.(i») is a p-group. Thus Oq.(i») ::; m. Hence, if Oq.(i») =P 1, C(p(Oq.(i»)) is nilpotent. But C(p(Oq.(i»)) = 0), so the proof is complete in this case. We may therefore suppose that Oq.(i») = 1. Thus if i) = i)j0iO)), (i) on 0iO))) is faithful, by 1.3. If a is a non-identity element of m, C(p(a) is nilpotent, so C(p(O,«(p)(a) is nilpotent and any q'-element ofC(p(o,«(p)(a) operates trivially on CO,«(p)(a). Hence O)/OiO)) is nilpotent, by 6.12. Thus the Fitting height of 0) is at most 2. q.e.d.

6.14 Theorem (J. N. WARD). Suppose that 0) is a soluble group, m is an elementary Abelian group of automorphisms of 0) of order p2 and Co;(m) = 1. Then the Fitting height of 0) is at most 2.

Proof Since c(p(m) = 1, the length of any orbit of m on 0) other than {l} is divisible by p. Hence 10)1 == 1 (p) and 0) is a pi_group.

If a Em - {l}, there exists bE m such that m = (a, b). If(£ = C(p(a), (£ is (b)-invariant and C(t(b) = C(p(a, b) = 1. By V, 8.14, (£ is nilpotent.

Thus by 6.13, the Fitting height of 0) is at most 2. q.e.d.

6.15 Remarks. a) For p = 2, it was proved by S. Bauman that under the conditions of 6.14, 0)' is nilpotent (see GORENSTEIN [1]). Shult has also proved that if m is a group of automorphisms of 0), m ~ (; 3' (10) I, 6) = 1 and c(p(m) = 1, then 0)' is nilpotent.

b) The hypothesis of solubility in 6.14 is not necessary. This will be proved in X, 11.18, and 6.14 will be used in the proof.

c) Theorems 6.4, 6.9 and 6.14 are all special cases of the following theorem of BERGER [2J.

Suppose that m is a soluble group, ~ is a group of automorphisms of m, (I~I, Iml) = 1 and C(!j(~) = 1. Suppose that ~ is nilpotent and that for every prime p, ~ has no section isomorphic to the wreath product of two cyclic groups of order p. Then, if I~I = PIP2 ... Pn' where PI' ... , Pn are (not necessarily distinct) primes, the Fitting height of m is at most n.

d) Let ~ be a soluble group of automorphisms of the soluble group m for which (I~I, Iml) = 1. Let I~I = PIP2 ... Pn for primes PI'· .. , Pn· We denote the Fitting height of the group X by h(X).

fHOMPSON [1J proved that

and later KURZWEIL [1 J improved this to

In particular, if C(!j(~) = 1, h(m) ::;; 4n + 1. e) DADE [4J proved that for any soluble group m,

h(m) ::;; 10(21 - 1) - 4/,

where <£: is a Carter subgroup of m and I<£:I = PI ... PI for primes PI'· .. , PI·

The proof of this is very lengthy. The same method was used by HARTLEY [1 J and RAE [1 J to prove that if P is odd and m is a p-soluble group of p-Iength at least 2k+2 - 1, then any subgroup of m of order pk is contained in more than one Sylow p-subgroup of m.

§ 7. p-Stability

In order to formalize the consequences of Lemma 5.8, we make the following definition.

7.1 Defmition. The finite group m is called p-stable if, whenever 'lJ is a p-subgroup of m, g E N(!j('lJ) and ['lJ, g, gJ = 1, then gC(!j('lJ) E Op(N(!j('lJ)jC(!j('lJ».

§ 7. p-Stability 493

7.2 Example. Any p-nilpotent group ijj is p-stable. For if'.p is a p-subgroup of ijj, No;('.p)/Co;('.p) is a p-group.

Lemma 5.8 implies the p-stability of a large class of p-soluble groups. To prove this, we need the following.

7.3 Lemma. Suppose that ijj is a finite p-group and

where 91 i <J ijj. Let ~ be a group of automorphisms rt of ijj for which 91i rt = 91 i (i = 0, 1, ... , k), and let

~ = {PIP E ~, (x91JP = x91i for all x E 91i-1 (i = 1, ... , k)}.

Then ~ is a normal p-subgroup of~. In particular, any non-identity p'-element of~ induces a non-identity

automorphism on 91 i -d91Jor at least one i.

Proof If rt E ~, then for each i = 1, ... , k, rt induces an automorphism rti on 91i-d91i• The mapping rt ~ (rt l , ... , rtk) is a homomorphism of ~ into the direct product ofthe groups of automorphisms ofthe 91i-d91i,

and ~ is the kernel. Thus ~ <J ~. By repeated application of I, 4.4, ~ IS a p-group. q.e.d.

7.4 Theorem. Let ijj be a p-soluble group. If p > 3, ijj is p-stable. If p = 3 and the Sylow 2-subgroups of ijj are Abelian, ijj is 3-stable.

Proof. Suppose that '.p is a p-subgroup of ijj, g E f) = No;('.p) and ['.p, g, g] = 1. Let

'.p = '.po > '.pI > ... > '.pm = 1

be the part below '.p of a chief series of f,. For 1 ~ i ~ m, let <£i = Cij('.pi-d'.p;). By 5.8, g E <£i' Thus if <£ = n~=l <£i' g E <£. Let <£ = Co;('.p), so <£ ~ <£. Now N<£ is isomorphic to a group ~ of automorphisms rt of '.p for which '.pirt = '.pi (i = 0, ... , m). In this isomorphism, <£/<£ corresponds to ~, where

~ = {PiP E~, (x'.p;)P = X '.pi for all x E '.pi-l (i = 1, ... , m)}.

By 7.3, !B is a normal p-subgroup of 21, so <£/<£ :s; 0p(f)/<£). Thus g<£ E 0p(f)/<£). Hence ijj is p-stable. q.e.d.

Our principal aim in this section is to obtain a generalization of 7.4 which gives a characterization of groups in which every section is p-stable, for p odd. First we give an example of a group which is not p-stable.

7.S Example. For any prime p, let (f) = SA(2, p) be the set of all matrices of the form

(a b t) e d u o 0 1

with coefficients in GF(p) and ad - be = 1. Let ~ be the set of all such matrices for which t = u = 0 and let 91 be the set of all such matrices for which a = d = 1, b = e = O. Then 91 <l (f), (f) = ~m and ~ n 91 = 1. Also Cu;(m) = 91 and ~ ~ SL(2, pl. Let 9 be a nonidentity p-element of~. Since 1911 = p2, [91, g, g] = 1. But 0i(f)/cijj(m» = Op(~m/m) ~ Op(~) = 1 and gm # 1, so (f) is not p-stable.

In particular, the soluble group SA(2, 3) is not 3-stable; its Sylow 2-subgroup is the quaternion group of order 8.

Note that SA(2, 2) ~ 6 4 •

7.6 Theorem. For p odd, let IDl be the subgroup of SL(2, pi) generated by

(~ ~) and (~ ~), where b # O. Let 3 = < -1) ~ SL(2, pi).

a) If p = 3 and b2 = -1, then f is even, IDl ~ 3, IDl ~ SL(2, 5) and IDl ~ SL(2, 3)X for some x E SL(2, 9).

b) Otherwise, IDl = SL(2, pm), where GF(pm) is the field generated by b over GF(p).

Proof We prove this in a number of steps. (1) 1f'l30 is a non-identity subgroup of SL(2, pi) consisting of elements

of the form (~ ~). the normaliser 910 of 'l30 in SL(2, pi) consists of

. (x y) matnces of the form 0 X-I ,and the only Sylow p-subgroup of 910

is {(~ ~)IY E GF(pf)}.

If xv - yu = 1,

Thus 910 consists of matrices of the stated form and the set of all matrices

of the form (~ n with Y E GF(pI) is a normal Sylow p-subgroup of

910 ,

(2) Let ~ be the set of all matrices (~ ~) which lie in 9Jl. Then

~ E Sp(9Jl), and ~ is the only Sylow p-subgroup of 9Jl which contains

(~ ~). Suppose that ~t E Si9Jl) and G ~) E ~t. Since the Sylow

p-subgroups of SL(2, pI) are Abelian, it follows from (1) that ~t consists

of elements of the form (~ x~t) and, further, that ~t ~ ~. Then

~t =~. (3) 9Jl ~ 3, and either 9Jl/3 ~ PSL(2, pn) for some n, or p = 3 and

9Jl/3 ~ Ws' It follows from (2) and the definition of 9Jl that 9Jl has more than

one Sylow p-subgroup, for G ~) ¢ ~. Also, if Sl <J 9Jl and 9Jl/Sl is a

p' -group, Sl = 9Jl, for 9Jl/Sl is generated by p-elements. It follows from II, 8.27 that 9Jl3/3 is isomorphic to W4 , Ws or PSL(2, pn) for some n. Thus the order of 9Jl is even. Since -1 is the only element of order 2 in SL(2, pI), 3 ~ 9Jl.

Suppose that 9Jl/3 is not isomorphic to PSL(2, pr) for any r. Then 9Jl/3 is isomorphic to W4 or Ws. Since p divides 19J11, p = 3 or p = 5. Since W4 ~ PSL(2, 3) and Ws ~ PSL(2, 5), the only possibility is p = 3 and 9Jl/3 ~ Ws.

(4) If p = 3 and 9Jl/3 ~ Ws, b2 = -1. Let rx be an epimorphism of 9Jl onto Ws such that 3 = ker rx. If

G ~)rx = (it, i2, i3) and G ~)rx = (jt, j2' j3)' then

Ws = «(it, i2, i3), (jt, j2' j3)' so {it, i2, i3, jt, j2' j3} = {1, 2,3,4, 5}. Hence we may suppose that jl = i3 and that i , . i2, i 3, j2' j3 are distinct. Thus

Hence

This gives b2 = -1. (5) We now prove a). Suppose that p = 3 and b2 = -1. Since GF(Y)

contains a square root of -1,f is even. Hence by 11,8.13, SL(2, 31)/3 has a subgroup £/3 isomorphic to 2ls. Since 2ls is generated by two elements of order 3, £ = 3<u, v) for elements u, v of SL(2, 31) of order 3. Now it follows from the double transitivity of PSL(2, 31) on the projective line that the equivalent representation of PSL(2, 31) by transformation of its Sylow 3-subgroups is doubly transitive. Hence there exists A E SL(2, 31) such that

for suitable x, yin GF(Yr. Let B = (~ ~), and replace £ by £.lAB.

Thus £.I = 3£.10 , where

By (3), £0 z 3, so £ = £0' By (4), (xy)2 = -1. Thus 9)( ~ £0 = £ and 9)(/3 is isomorphic to 2ls. Since 2ls is simple, it follows that 39)(' = 9)(.

But 3 ~ 9)(', since 9)( has only one element of order 2. Thus 9)( = 9)('.

By V, 25.7, SL(2, 5) is a representation group of PSL(2, 5) ~ 2ls, so the Schur multiplier of 2ls is of order 2. By V, 23.4, 9)( is a representation group of 2ls. Using V, 23.6, it follows that 9)( ~ SL(2,5).

Finally, 9)( contains

Thus if B = G -~), 9)(B contains (~ ~J = (~ ~) and


(1 - b 1)B = (1 1 + b2) = (1 1 l+b 1 1 1 ~). Since (~ ~) and G ~)

generate SL(2, 3), 9JlB contains SL(2, 3). (6) To prove b), suppose that either p > 3 or b2 =1= -1. By (4), either

p > 3 or 9Jl/3 ~ ~5' so by (3), 9Jl/3 ~ PSL(2, p") for some n. Thus pn is the order of a Sylow p-subgroup of 9Jl. Hence by (2), 'l3 E Sp(9Jl), where

for some additive subgroup A of GF(pf) of order p". If 91 = Nm('l3), 91/3 is the normaliser of a Sylow p-subgroup in PSL(2, pn), so 1911 =

pn(pn _ 1). By (1), 91 consists of matrices of the form (~ X~l). where

x runs through some multiplicative subgroup 1: of GF(pfr. Thus 11:1 = p" - 1 and 1: = GF(pnr. And if x E 1:, x2 E A, since

(x-l y)-l(1 1) (x-l y) = (x -y _ ) (x- l Y + x) = (1 X2). Ox 01 0 x 0 Xl 0 X 01

Thus A contains x2 + y2 for all x, y in GF(p"). Since IAI = p", it follows from II, 10.6 that A = GF(pn).

Now 9Jl/3 has a doubly transitive permutation representation on the co sets of 91/3, since 9Jl/3 ~ PSL(2, p"). Hence

But

(1 0) (1 1) ( 1 0) (1 - b 1) b 1 0 1 -b 1 = - b2 1 + b

does not lie in 91, so there exist x, y in 1: = GF(pnr such that

(1 - b _b2

1 ) (x x') (1 0) (y y,) 1 + b = ° X-l b 1 ° y-l

= (bx:ly :).

q.e.d.

We use this to prove the following.

7.7 Lemma. Suppose that (fj =f. 1 and that V is a faithful irreducible K(fj-module, where K = GF(p) for an odd prime p. Suppose that there exist elements a, b of(fj such that (fj = <a, ab ) and V(a - 1f = 0. Then there exists a K-subspace U of V such that dimK U = 2 and every linear transformation of U of determinant 1 is the restriction to U of some element of(fj·

Proof This is proved in several steps. (1) Put VI = V(a - 1), V2 = V(a b - 1). Then V2 = VI b, VI = Cv(a),

V2 = Cv(a b) and V = VI EB V2 ·

Clearly V2 = VI b. The mapping x --+ x (a - 1) is a K-linear mapping of V onto VI = V(a - 1) with kernel Cv(a). Thus if dimKVI = f and dimKCv(a) = 1', dimKV = f + f'. Since V(a - If = 0, VI ::;; Cv(a), so f ::;; f'. But Cv(a) n Cv(a b) ::;; Cy(fj) = 0, since (fj = <a, ab ) =f. 1 and V is a faithful irreducible K(fj-module. Thus dimK V ~ dimKCv(a) + dim KCy(a b). Since Cv(ab) = Cv(a)b, dimK Cv(ab) = 1'. It follows that f + I' ~ 21', or f ~ 1'. Hence I' = f, VI = Cv(a) and V2 = Cv(a b).

Thus VI n V2 = 0; also dimKVI + dim KV2 = dimKV, so V = VI EB V2 .

(2) If x E VI' put xoc = x(a b - 1), and if y E V2, put yf3 = y(a - 1). Then oc is a K-linear isomorphism of VI onto V2 and f3 is a K-linear isomorphism of V2 onto VI'

It is clear that oc, f3 are K-linear mappings of VI into V2 and V2 into VI respectively. If x E VI and xoc = 0, then x(a - 1) = x(ab - 1) = 0, since VI = Cv(a). Thus x E Cv(fj) = 0. Hence oc is a monomorphism. Since dimKVI = dim KV2 , oc is an isomorphism. Similarly, f3 is an isomorphism.

(3) Let I' = ocf3; thus I' is a non-singular K-linear transformation of VI' <I') is irreducible on VI' and the set F of polynomials in I' with coefficients in K is a field.

Weshowfirstthat <I') is irreducible on VI' Suppose that ° < Vo ::;; VI and Vol' = yo· Then (Vo EB Vooc)(a - 1) = ° + Voocf3 = Vol' = Yo, so (Vo EB Vooc)a = Vo EB Vooc. Again (Vo EB Vooc)(a b - 1) = Vooc + 0, so (Vo EB Vooc)a b =. Vo EB Vooc. Thus Vo EB Vooc is a K-subspace orv invariant under (fj. Since V is irreducible, V = Vo EB Vooc and VI = Yo.


By Schur's lemma (I, 10.5), any polynomial in I' is either zero or non-singular. Hence the minimum polynomial of I' is irreducible. Therefore any non-zero element of the set F of polynomials in I' has an inverse in F. Hence F is a field.

(4) Take a fixed, non-zero element Z in Vt. Given x E Vt, there exists a unique A E F such that x = ZA. In particular IFI = IVtl = pl.

For {zAIA E F} is a non-zero y-invariant K-subspace of Vp so {zAIA E F} = Vp since (I') is irreducible on Vt. And if ZA t = ZA2 with Ai E F (i = 1,2), then At - A2 is a singular linear transformation. By (3), At - A2 = O.

(5) V has the structure of a vector space over F, in which AX = XA for all x E Vt, A E F and AY = yf3Ap-t for all y E V2, A E F. {z, za} is an F-basis for V.

Given v E V, there exist a unique x E Vt and y E V2 such that v = x + y, since V = Vt EEl V2; if A E F, define AV = XA + yf3Af3-t. It is easy to check that the vector space axioms are satisfied. Z and za are Flinearly independent, since Z E Vt, za E V2 and Vt, V2 are F-subspaces. Since IVI = IFI2, dimFV = 2. Hence {z, za} is an F-basis ofV.

(Note that if k E K and z is the identity mapping on Vi' then (kz)v = kv for all v E V).

(6) If g E {f}, the mapping v -t vg (v E V) is F-linear. Thus V is an F{f}-module.

If x E Vt, then xa = x and x(ab - l)(a - l)a = x(ab - l)(a - 1). Thus (xy)a = XI' and (yx)a = (xy)a = XI' = yx = y(xa).

Also, if y E V2, then ya = yea - 1) + Y and yea - 1) E Vi' Y E V2; thus

y(ya) = yea - 1)1' + yf3yf3-i = yf3y + yf3a

= (yf3a)a = (yf3yf3-i)a = (yy)a.

Thus y(va) = (yv)a for all v E V. It follows that (Av)a = A(va) for all A E F.

If x E Vi' then xab = x + x(ab - 1) and x E Vi' x(ab - 1) E V2; hence

y(xab) = XI' + x(ab - 1)f3yf3-i = xaf3 + xaf3a = (xy)ab = (yx)a b.

If y E V2, then yab = y and yy E V2, so

Hence A(vab) = (Av)a b for all A E F, v E V. Since (fj = <a, ab ), it follows that A(Vg) = (AV)g for all g E (fj. Thus

the mapping v ~ vg is F-linear. Hence V is an F(fj-module. (7) With respect to the F-basis {z, za} of V, the matrices of the

F-linear transformations v ~ vab, v ~ va are respectively

since, for example,

(za)a = za + (zaHa - 1) = za + za/3 = zy + za = yz + za.

By 7.6, the group generated by these matrices contains a conjugate of SL(2, p) in SL(2, pI). Hence there exists an F-basis {u 1 , U2} of V such that given any matrix A of determinant 1 with coefficients in K, there exists g E (fj such that A is the matrix of the F-linear transformation v ~ vg with respect to the F-basis {u 1, u2} of V. If U is the K-space spanned by U1 and U2' dimKU = 2 and every K-linear transformation of U of determinant 1 is induced by some element of (fj. q.e.d.

In our application of 7.7, the stringent condition (fj = <a, ab) will be obtained by applying an important theorem of Baer (III, 6.15). We state the relevant consequence of that theorem.

7.8 Theorem. Suppose that a is an element of the finite group (fj and that <a, ab) is a p-group for every b E (fj. Then a E Op(fj).

Proof If bE (fj, [b, a] = (a bt 1a E <a, ab). By hypothesis, <a, ab ) is nilpotent. Hence [b, a, ... , a] = 1 for sufficiently large n. Thus a is a

n

right Engel element (III, 6.12). By III, 6.15, a E F(fj). But a is a p-element, so a E Op(fj). q.e.d.

7.9 Lemma. Let p be an odd prime, and let U, m be subgroups of a group (fj. Suppose that U <J m and m/U is elementary Abelian of order p2. If the group of automorphisms of m/U induced by N(!;(U) n N(!;(m) contains SL(2, p), then (fj has a section isomorphic to SA(2, p).

Proof. Let V = m/U, and let ~ be a minimal subgroup of (No;(U) n No;(m))/U such that ~ ~ V and the group of automorphisms of V induced by ~ contains SL(2, p). Then, if (£ = Cl)(V),

f,/<£ ~ SL(2, p). Thus ~j<£ possesses a unique subgroup ::tj<£ of order 2; we have VI = V-I for all v E V and t E::t - <£, and ::t <l ~. Let 6 be a Sylow 2-subgroup of::t. Then ::t = <£6. By the Frattini argument.

Thus the group of automorphisms of V induced by Ns(6) is SL(2, p). By minimality of ~, ~ = VNs(6). But if 91 = N[(6), 91 = <£ n Ns(6) <l Ns (6), so 91 <l VNs(6) = ~. Since VI = V-I for all v E V and some t E 6, V n Ns (6) = 1. Thus V n 91 = 1. Put V = V91j91, ~ = ~j91. Then every linear transformation of V of determinant 1 is induced by some element of ~ and hence of Ns(6)j91. Since

and ~ is the split extension of V by Ns(6)j91, ~ ~ SA(2, p). q.e.d.

7.10 Theorem. Let m be a finite group and let p be an odd prime. Then every section of m is p-stable if and only if no section of m is isomorphic to SA(2, p).

Proof If every section of m is p-stable, then no section of m is isomorphic to SA(2, p), since SA(2, p) is not p-stable (7.5).

Suppose that the converse is false and that m is a counterexample of minimal order. Thus no section of m is isomorphic to SA(2, p), but 0) has a section which is not p-stable. By minimality of 10)1,

(1) every proper section of 0) is p-stable. Thus 0) itself is not p-stable. Let & be the set of p-subgroups 9i of 0) for which there exists g E N(!j(9i) such that [9i, g, g] = 1 and gC(!j(9i) ~ Op(N(!j(9i)jC(!j(9i)). By 7.1, [l}J is non-empty. Let ~ be an element of g> of minimal order. Thus there exists a E No; (~) such that

(2) [~, a, a] = 1 and a<£ ~ 0iN(!j(~)j<£), where <£ = C(!j('P). Hence N(!j(~) is not p-stable. By (1), N(!j(~) = m. Thus

(3) 'P <l m, <£ <l m and a<£ ~ Op(mj<£). We now make a number of routine reductions. (4) Op,(m) = 1. By 6.11, <£jOp,(m) = C(!j/op((ij)('POp,(m)jOp,(m)), and by (2),

[~Op,(m)jOp,(m), aOp,(m), aOp'(O))] = 1.

Hence O)jOp'(O)) is not p-stable. Thus Op,(O)) = 1 by (1).

(5) If ~ ~ (fj and (fj = ~[, then (fj = ~~.

Since (fj = ~[, there exist h E ~ and e E [ such that a = he. If g E~, [g, h] = [g, ae-1] = [g, a] since e E C(!j(~). Since [g, a] E~, [g, h, h] = [g, a, h] = [g, a, a] = 1. Thus h commutes with every element of [~, h] and [~, h, h] = 1.

Suppose 1)/[ = 0p(f)/<£). Since (f)/[ = ~<£/<£ = m~)[/<£ ~ ~~/(~~ n [), Op(~~/~~ n [) = (1) n ~~)/(~~ n [). Hence if hC~'ll(~) E 0p(~~/C~'ll(~))' then h E 1) and a = he E 1)[ = 1), contrary to (3). Thus hC~'J1(~) ¢ 0p(~~/C~'ll(~))' Hence ~~ is not p-stable. By (1), ~~ = (f).

(6) [is a p-group. Suppose that this is false and that q is a prime divisor of 1[1 for which

q =f. p. If ,0 E Sq([), ,0 =f. 1. By the Frattini argument, (f) = [N(!j(.,Q). By (5), (f) = ~N(!j(,Q). But [~,,Q] ~ [~, [] = 1, so ~ ~ N(!j(,Q). Thus (fj = N(!j(,Q) and ,0 <J (f). Since ,0 =f. 1, it follows that Op,(f)) =f. 1, contrary to (4). Hence [ is a p-group.

(7) ~ is a minimal normal subgroup of (f), and ~ ~ <£.

Suppose that ~1 is a minimal normal subgroup of (f) and that 1 < ~1 ~ ~. Let [1 = C(!j(~I)' [2/~1 = C(!j/'ll,(W~I)' and let 1);/[i = Op(f)/[i) (i = 1,2). Thus (1)1 n 1)2)/([1 n (2) is a normal p-subgroup of (f)/([1 n (2)' Now there is an isomorphism between (fj/[ and a group of automorphisms m of ~ which leave ~1 fixed. In this isomorphism, ([1 n (2)/[ corresponds precisely to the set of those elements of m which leave fixed each element of ~1 and of W~I' By 1,4.4, ([1 n (2)/[ is a normal p-subgroup of (f)/[. Since (1)1 n 1)2)/([1 n (2) is a p-group, (1)1 n 1)2)/<£ is a normal p-subgroup of (f)/[. Thus CDI n 1)2)/<£ ~

Op(f)/[). By (3), a ¢ 1)1 n 1)2' But (f)/~1 is a proper section of (f) and is therefore p-stable. Since [~/~1' a~I' a~l] = 1, it follows that a E Il2. Thus a¢1)I' Since [~I,a,a] ~ [~,a,a] = 1, ~IEg'I. Since ~ is an element of g'I of minimal order and ~1 ~ ~, we have ~1 = ~, and ~ is a minimal normal subgroup of (f). Hence ~ ~ [.

(8) There exists b E (f) such that (f) = ([, a, ab ).

By (3), a[ ¢ Op(fj/[). Hence by 7.8, there exists b E (f) such that (a[, ab[) is not a p-group. Hence either a[ ¢ Op( (a, ab, [)/[) or ab[ ¢ 0/ (a, ab, [)/[). But [~, a, a] = [~, ab, ab] = 1; thus (a, ab, [)

is not p-stable and (a, ab, [) = (fj, by (1). (9) ~ is an elementary Abelian group by (7), so ~ may be regarded

as a vector space V over K = GF(p). (f)/[ is isomorphic to a group of automorphisms of~, so V may be regarded as a faithful K(fj/[)-module. V is irreducible by (7), and (f)/[ =f. 1 by (3). By 1.8, V(a - 1)2 corresponds to [~, a, a]; hence V(a - I? = O. By 7.7, there exists a K-subspace U of V such that dim K U = 2 and every linear transformation of U of

§ 8. Soluble Groups with One Class of Involutions 503

determinant 1 is the restriction to U of some element of G>. By 7.9, G> has a section isomorphic to SA(2, p).

This is a contradiction. Thus if no section of G> is isomorphic to SA(2, p), every section of G> is p-stable. q.e.d.

7.11 Remarks. a) GLAUBERMAN [6, Theorem 4.1] has proved the following. Suppose that m is an elementary Abelian p-group, where p is odd,

and that G> is a subgroup of Aut m for which Cm(G» = 1 and em, G>] = m. Suppose that Op(G» = 1. Suppose that Wl is a maximal subgroup of G>, 6 E Sp(Wl), m is a non-identity Abelian subgroup of 6 and em, m, m] = 1. Suppose also that <m, g) = G> for all g E G> - Wl. Then G> ~ SL(2, pn) for some n, and m is the direct product of G>-invariant GF(pn)-spaces on each of which G> acts as SL(2, pn). Further, n = 1 if

Iml = p. b) Theorem 7.4 is a consequence of Theorem 7.10. For if p > 3,

SA(2, p) is not p-soluble, and the Sylow 2-subgroup of SA(2, 3), that is, the quaternion group of order 8, is non-Abelian.

Exercises

4) Let n = {2, 3}. Show that if a is a transposition in the symmetric group 6n' then <a, ab ) is a n-group for every b E 6n' but that for n ~ 5, a ¢ O,,(G».

§ 8. Soluble Groups with One Class of Involutions

In this section, we shall prove that the 2-length of a soluble group G> is 1 if all involutions in G> are conjugate and the Sylow 2-subgroup contains more than one involution. To do this we need two preliminary results. The first one is number-theoretical; it arises also elsewhere in group-theory (e.g. Chapter XII, § 7).

8.1 Lemma. Let ¢n be the n-th cyclotomic polynomial over the field of rational numbers. Let q be a prime, let a be an integer prime to q and let f be the order of a modulo q. For each non-zero integer x, let wq(x) = max {llql divides x}.

(1) For q > 2, the following hold. a) wq(¢Aa)) > O. b) Wi¢jq,(a)) = 1 for all i ~ 1.

c) wi<Pm(a)) = 0 for all other m ~ 1. d) Wq(an - 1) = 0 iff does not divide n. e) wian - 1) = wq(a f - 1) + win) iff divides n.

(2) Also

W2(<P2,(a)) = 1 for i ~ 2.

w2(<Pn(a)) = 0 for n =j:. 2i (i = 0,1, ... ).

Proof (ARTIN [1,2]). (1) a) As f is the order of a modulo q, wiai - 1) = 0 iff F Thus wi<pi(a)) = 0 iff +i. From

<Pf(a) TI <Pi (a) = af - 1, ilf.i#!

it follows that wi<pf(a)) = wia f - 1) > O. b) Put n = fqi, r = fqi-l. Then

an - 1 ((a r - 1) + l)q - 1

ar - 1 ar - 1

= (a r _ l)q-l + q(a r - 1)q-2 + ... + (1)(a r - 1) + q.

Since ar - 1 == 0 (q) and m == 0 (q),

thus

As is well-known,

Thus

( an - 1) Wq ar _ 1 = 1.

an - 1 ar - 1

TI <pia). din. drr

L Wi¢d(a)) = 1. din. dr'

Hence there exists a divisor d of n such that dtr and wi<pia)) = 1. Thus wiad - 1) > 0 and f I d. The only integer d satisfying these conditions is d = n; thus wq(<Pn(a)) = 1.

c) If wi<Pm(a)) =#: 0, wiam - 1) =#: 0 and f divides m. Thus it is to be shown that wq(<Pm(a)) = 0 if m = fqi[ with i ~ 0, qt1 and [ > 1. If r = fqi, <Pm(a) is a divisor of (am - l)j(a' - 1). Since a' == 1 (q),

am - 1 «a' - 1) + 1)1 - 1 a' - 1 a' - 1

= (a' - 1)1-1 + [(a' - 1)1-2 + ... + [ == [ =1= 0 (q).

Thus wq(<Pm(a)) = O. d) is obvious. e) We have

an - 1 -f-1 = n <pia). a - dln,aU

It follows from this and c) that

( an - 1) Wq -f--1 = L wi<pJq'(a)). a - fqiln,i>O

Since f divides q - 1, q does not divide f. Hence the range of values of i in this sum is {ill ~ i ~ win)}. Thus by b),

wian - 1) - wq(a f - 1) = win).

(2) Since q = 2, f = 1. Write n = 2i l with i ~ 0, [ odd and I ~ 1. If 1 > 1, w2(<pn(a)) = 0 exactly as in c): w2(<Pn(a)) ~ w2«an - l)j(a2i - 1)) and

Also

a:i - 1 = (a2i _ 1)1-1 + [(a 2i - W- 2 + ... + I =1= 0 (2). a - 1

so for i ~ 2,


4>2i(a) == 2 (4)

and W2(4)2i(a)) = 1. q.e.d.

8.2 Lemma. Let IPn(x, y) = y4>(n) 4>n(x/y), where 4> is the Euler function and 4>n is the n-th cyclotomic polynomial. Let

where a, b run through all complex numbers/or which lal ~ Ibl + 1 and Ibl ~ 1. Then L(n) > nplnP, where p runs through all prime divisors ofn, except when n = 1,2,3 or 6.

Proof (ARTIN [1, 2]). The assertion will be proved by induction on n. a) If p divides n, L(np) ~ L(n) and L(np) ~ (1 + p)4>(n). Since p divides n, IPnp(a, b) = IPn(aP, bP). Since lal ;::: Ibl + 1 and

Ibl ;::: 1, lalp ;::: IW + 1 and IW ;::: 1, so L(np) ;::: L(n). Also

IPn(a, b) = n (a - eb), t

where e runs through the 4>(n) primitive n-th roots of unity. This gives

IIPn(a, b)1 ;::: (Ial - Ibl)4>(n).

But for lal ;::: Ibl + 1 and Ibl ;::: 1,

lal P - Ibl P = (lal - Ibl + Ibl)P - Ibl P

;::: (Ial - Ibl)P + p(lal - Ibl)p-llbl ;::: 1 + p.

Hence IIPn(aP, bP)1 ~ (1 + p)4>(n) and L(np) ;::: (1 + p)4>(n). b) For p ;::: 5, L(p) > 2p. We have

where x = lal, y = Ibl. Thus x ;::: y + 1 and

(xP - yP)(1 + 2y) = x(1 + Y)(XP-l - (1 + y)P-l) + y(xP - (1 + y)P)

+ yP(x - (1 + y)) + (x + y)((l + y)P - yP)

;::: (x + y)((1 + y)P - yP).

Hence

1<1> (a, b)1 ~ xP - yP ~ (1 + y)P - yp. P x + y 1 + 2y

But Y = 1 b 1 ~ 1, so

P

(1 + y)P - 2P = L (~)(yi - 1) ~ yP - 1 i=O

and

P

(1 + y)P - 2Py = L (~)(yi - y) ~ yP - y. i=O

Therefore

3(1 + y)P - 2P(1 + 2y) = ((1 + y)P - 2P) + 2((1 + y)P - 2Py)

~ yP - 1 + 2(yP - y) = 3yP - (1 + 2y).

Hence

3(1 + y)P - 3yP ~ (2P - 1)(1 + 2y)

and

1<1> (a b)1 > (1 + y)P - yP > 2P - 1. P' - 1 + 2y - 3

2P - 1 Hence L(p) ~ -3- and for p ~ 5, L(p) > 2p.

c) For p ~ 5, L(2p) > 2p. This follows from <1>2p(a, b) = <1>p(a, -b) and b). d) If p does not divide n, L(np) ~ L(n)p-1. Since p does not divide n,

<1>np(a, b) = Il <1>n(a, be). ,p=l",

e) If x ~ 3 and y ~ 3, then Xy-1 ~ xy. By elementary calculus, x y - 1 - xy is an increasing function of y for

y ~ 3. Thus

for x ~ 3.

f) If n is divisible by the square of a prime p, then L(n) ~ L (~) by a).

The assertion then follows at once from the inductive hypothesis unless

'!. is 1,2,3 or 6. Since p divides '!., the only possibilities are n = 4,9, 12 p p or 18. But by the second assertion of a), L(4) ~ 3, L(9) ~ 16, L(12) ~ 9 and L(18) ~ 16.

Suppose then that n is square-free. Lei p be the greatest prime divisor of n and write n = pm. By hypothesis, p ~ 5. If m = 1, the assertion follows from b); if m = 2, it follows from c). If m = 3, n = 3p and L(n) ~ L(pf > 4p2 > 3p by d) and b). Similarly if m = 6, L(n) ~ L(2p)2 > 4p2 > 6p by d) and c). For other values of m, L(m) > TIqlm q by the inductive hypothesis. Thus L(m) ~ 3, for m cannot be a power of 2 since m is square-free. Hence bye), L(m)P-l ~ pL(m). By d), it follows that L(n) ~ L(m)P-l ~ pL(m) > TIq In q. q.e.d.

8.3 Theorem (ZSIGMONDY [1]). Let a, n be integers greater than 1. Then except in the cases n = 2, a = 2b - 1 and n = 6, a = 2, there is a prime q with the following properties.

(1) q divides an - 1. (2) q does not divide ai - 1 whenever ° < i < n. (3) q does not divide n.

In particular, n is the order of a modulo q.

Proof (ARTIN). Suppose that for each prime divisor q of an - 1, there exists i such that ° < i < nand q divides a i - 1. Since ai - 1 = TIdli ¢ia), it follows in particular that for each prime divisor q of ¢n(a), there exists d < n such that q divides ¢ia). Let f be the order of a modulo q. Since wq(¢n(a» > ° and wq(¢ia» > 0, it follows from 8.1 that if q > 2, n = fqk and d = fqi, where ° ~ j < k. Thus q divides n. Also wq(¢n(a» = 1 by 8.1. Ifw2(¢n(a» > 0, we see from 8.1(2) that n = 2i and, if n > 2, w2(¢n(a» = 1. It follows that if n > 2,

l¢n(a)1 = TI qwq(",,(a)) ~ TI q.

q qln

By 8.2, however,

1 ¢n(a) 1 = l<Pn(a, 1)1 > TI q, qln

except when n is 1,2, 3 or 6. Further,

for all a > 1, and

for a ~ 3. Thus only the cases n = 2 and a = 2, n = 6 remain. If the assertion is false for n = 2, each prime divisor of a2 - 1 divides a - 1. It follows at once that 2 is the only prime divisor of a + 1, since (a + 1, a-I) ::;; 2; thus a = 2b - 1.

Apart from the stated exceptions, then, there is always a prime q such that q divides an - 1 and q does not divide ai - 1 for 0 1. There exists a prime q > n such that q divides pn - 1, except in the case p = 3, n = 2.

Proof First suppose that there exists a prime q such that ql(pn - 1) but qf(pi - 1) for 0 n.

Secondly, suppose that every prime divisor of pn - 1 divides pi - 1 for some i with 0 < i < n. By 8.3, either p = 2 and n = 6, or n = 2. In the case p = 2, n = 6, the required prime is 7. If n = 2, suppose that p2 _ 1 is not divisible by any prime greater than 2. Then p2 - 1 is a power of 2. Hence so are p - 1 and p + 1. Thus p = 3. q.e.d.

Our second preliminary result is the following.

8.5 Theorem. (ITO [1]). Suppose that <!> is a p-soluble group and that <!> has a faithful representation of degree n in a field K, where char K does not divide I<!>I and n < p. Then the Sylow p-subgroups of <!> are Abelian. Further, either the Sylow p-subgroup of <!> is normal, or I<!>I is even and n = p - 1 = 2m for some positive integer m.

Proof It is clear that we may suppose that K is algebraically closed. Let V be a faithful K<!>-module of dimension n over K.

a) The Sylow p-subgroups of <!> are Abelian. Suppose IS E Sp(<!». By V, 12. llb), the K-dimension of every irre

ducible KG-submodule of V is a power of p. Since n < p, it follows that the K-dimension of every irreducible KIS-submodule of V is 1. Since

char K does not divide I (fj I, V is the direct sum of irreducible K 5-submodules, so 5 is represented on V by diagonal linear transformations. Since V is faithful, 5 is Abelian.

Suppose that the Sylow p-subgroup of (fj is not normal, and let (fjo be a subgroup of (fj of the smallest order for which the Sylow p-subgroup of (fjo is not normal.

b) For some prime q, (fjo has a normal q-subgroup 9l such that (fjo/9l is a p-group.

Since (fjo is p-soluble and the Sylow p-subgroups of (fjo are Abelian, (fjo is of p-Iength 1, by VI, 6.6a). Also, by minimality of I (fjo I, (fjo has no proper normal subgroup of index prime to p. Thus (fjo = 0p',p(fjo)' Let 50 be a Sylow p-subgroup of (fjo. Since 50 is not normal in (fjo, there exists a prime q such that N(!j.(50) contains no Sylow q-subgroup of (fjo. If 911 E Sq(Op,(fjo)), (fjo = Op,(fjo)N(!j.(9ld, by the Frattini argument. Hence I (fjo : N(!j.(9l1)1 is prime to p and N(!jo(9l1) contains a Sylow p-subgroup of (fjo. Hence N(!jo (9l1) ~ 5~ for some x and N(!j.(9l) ~ 50' where 9l = 9lr l

• Thus 5 09l is a subgroup of (fjo, but 50 is not normal in 5 09l, since 9l E Sq(fjo). By minimality of I (fjol, (fjo = 5 09l·

c) There exists ~ <J (fjo such that if ~ = (fjo/~, ~ has a faithful irreducible representation on a subspace U of V, and the Sylow p-subgroup of ~ is not normal.

Since char K does not divide I (fjo I, V is the direct sum of irreducible K(fjo-submodules U1 , U2 , •••• Let ~i be the kernel ofthe representation of (fjo on Ui . Since (fjo is represented faithfully on V, ni ~i = 1. Thus there is a monomorphism of (fjo into the direct product of the (fjO/~i' Hence the Sylow p-subgroup of this direct product is not normal, and it follows that the Sylow p-subgroup of some (fjO/~i is not normal. Thus ~ = ~i and U = Ui have the stated properties.

By b), ~ has a unique Sylow q-subgroup .0. Suppose ~ E Sp(~).

d) U is the direct sum of isomorphic irreducible K.Q-submodules. By V, 17.3, U is the direct sum of K.Q-submodules WI' W2 , ••• , Wk ,

such that each Wi is the direct sum of isomorphic irreducible K.Q-modules and there is a transitive permutation representation of ~ of degree k. Since k :::;;; n < p, it follows that k = 1.

e) I (fjl is even and n = p - 1 = 2m for some m. Since ~ ;tl ~, ~ 1, Cf,(.Q). But if .01 < .0 and ~ :::;;; Nf,(.Qt), then

~.Ql = (fjt!~ for some proper subgroup (fjl of (fjo. By minimality of I (fjo I, the Sylow p-subgroup of (fjl is normal, so ~ :::;;; Cf,(.Ql)' It follows from III, 13.5 that .0 is a special q-group, 4>(.0) is centralized by ~ and there is an irreducible representation of ~ on .0/4>(.0).

By d), U is the direct sum of isomorphic irreducible K.Q-modules, so Z(.Q) is represented on U by scalar multiples of the identity mapping.

Hence Z(.Q) ~ Z(~) and Z(.Q) is cyclic. Since ~ ~ ~,.Q i Z(~); hence .0 is non-Abelian. Thus .0 is extraspecial and 1.01 = q2m+1 for some m.

Since ~ is represented fixed point freely on .QjcJi(.Q) and l.QjcJi(.Q)1 = q2m, q2m _ 1 is divisible by p. By V, 16.14, the degree of any faithful irreducible representation of.Q in K is qm. Hence by d), dimKU = jqm for some integer j. Thenjqm ~ n < p, so (qm - 1, p) = 1. Thus p divides qm + 1; indeed p = qm + 1 since qm < p. Thus q = 2 and I (f; I is even. Fromjqm ~ n < p = qm + 1, it follows thatj = 1 and n = qm = p - 1.

q.e.d.

8.6 Theorem (THOMPSON). Suppose that (f; is a soluble group of even order and that the Sylow 2-subgroup of (f; contains more than one involution. Suppose that all the involutions in (f; are conjugate. Then the 2-1ength of (f; is 1, and the Sylow 2-subgroups of (f; are either homocyclic or Suzuki 2-groups.

Proof Since (f;j02,(f;) satisfies the hypotheses, we may suppose that 02,(f;) = 1. Let:! = 02(f;); thus:! :f. 1. Let 3 = Q 1(Z(:!)). Since all involutions in (f; are conjugate and 3 :f. 1, 3 contains all involutions in (f;. Let 131 = q = 2n. Thus n > 1.

a) There exists a prime divisor p of 2n - 1 such that p > nand 6 i Cm(3), where 6 E Sp(f;).

(f; has 131 - 1 = 2n - 1 involutions and they are all conjugate; hence I (fj : Cm(t)1 = 2n - 1 for any t E 3 - {1}. By 8.4, there exists a prime divisor p of 2n - 1 such that p > n. Suppose 6 E Sp(fj). Since 161 does not divide ICm(t)I, 6 -t Cm(t)· Thus 6 -t Cm(3).

Let .0 be a Hall 2' -subgroup of (fj such that .0 ~ 6. b) If t1, t2 are involutions, there exists x E.Q such that tf = t2 . .0 and Cm(t 1) are of coprime indices in (fj, so by I, 2.13, (f; = Cm(t 1).0.

But t2 = t~ for some g E (fj, and if g = yx with y E Cm(t1)' x E.Q, then

t2 = tf· c) Any subgroup of even order normalised by .0 contains 3. If 91 is such a subgroup, 91 contains an involution t 1. If t2 E 3 - {1},

then by b), t2 = tf for some x E .0, so t2 E 91x = 91. d) Any non-identity 2-subgroup 91 normalised by .0 is either homo

cyclic or a Suzuki 2-group. By b), the set of involutions in 91 is permuted transitively by the

group of automorphisms x ~ x Y (x E 91, y E .0) of 91 induced by .0. If 91 is non-Abelian, 91 is a Suzuki 2-group by VIII, 7.1. If 91 is Abelian, 91 is homocyclic, by VIII, 5.8.

e) 6 is a normal Abelian subgroup of.Q. By d), :! = 02(f;) is either homo cyclic or a Suzuki 2-group. If:! is

homo cyclic, I:!: cJi(:!) I = q. If:! is a Suzuki 2-group, then by VIII, 7.9,

I'!: cl>('!) I is q or q2. By 1.4 and 1.6, C(jj('!/cl>(,!)) = '!. Thus .0 has a faithful representation on U = ,!/cl>(,!); the degree of this is n or 2n. Since n < p, e) follows at once from 8.5 when the degree is n. Suppose then that it is 2n.

Let F = GF(2). If U is a reducible F,Q-module, U = U1 EB Uz for F,Q-submodules Ui (i = 1, 2). If Ui = ,!;/cl>(,!), each '!i is either homocyclic or a Suzuki 2-group, by d). Thus I'!il is a power of q and indeed I'!il = qZ. Thus if Sli is the kernel of the representation of,Q on Ui, the Sylow p-subgroup of ,Q/Sli is normal and Abelian, by 8.5. Since Sll n Slz = 1,.0 is isomorphic to a subgroup of (,Q/Sll) x (,Q/Slz), so 6 is a normal Abelian subgroup of .0.

If U is an irreducible F,Q-module, let K be a finite extension of F which is a splitting field for .0, and let V = U· ®F K. Thus V is a K.,Qmodule. By V, 13.3, the irreducible components of V are all of the same degree m, and by V, 12.11 b), m is odd. Since m divides 2n, we have m ::;; n. Hence 8.5 may be applied to each component, and in the same way we find that 6 is a normal Abelian subgroup of ,0.

f) '! E Sz(m). Let 9{ = .0 n 02,2,(m), Thus 9{ is a Hall 2'-subgroup of 02,2,(m)

and 0z,z,(m) = '!9{. By 1.3, 0z,z,(m) ~ C(jj(Oz,z,(m)!'!); thus CQ (9{) ::;; .0 n 0z z,(m) = 9{. Bye), 6 is a normal Abelian subgroup of .0. Hence by III, 13.4,

6 = (6 n Z(69{)) x [6,9{J.

But 6 n Z(69i) ::;; CQ (9i) ::;; 9i and [6, 9i] ::;; 9i since 9{ <l .0. Hence 6 ::;; 9{. But by a), 6 $, C(jj(3). Thus [9{,3] "# 1. Since [9i,3] ::;; 3 and 9{ n 3 = 1, it follows that [9{, 3] $, 9{, or 3 $, N(jj(9{). But .0 certainly normalises N(jj(9{), since .0 ::;; N(jj(9{). It follows from c) that I N(jj(9i) I is odd. But by the Frattini argument applied to the Hall 2'-subgroup 9i of 0z,z,(m),

Hence'! E Sz(m). It follows at once that the 2-length of m is 1, and by d), '! is either

homocyclic or a Suzuki 2-group. q.e.d.

The following is a similar theorem for p odd.

8.7 Theorem (GASCHUTZ and YEN [1]). Suppose that m is a p-soluble group, where p is an odd prime divisor of Iml. If the subgroups of m of order p are permuted transitively by Aut m, the p-length of m is 1.

Proof Let <» be a counterexample of minimal order. Then Op'(<») = 1, for <»/Op'(<») satisfies the hypotheses ofthe theorem. Let Wl be a maximal characteristic subgroup of <». Then <»/Wl is either a pi_group or an elementary Abelian p-group. Since <» is not of p-Iength 1, Wl is not a pi -group. Hence the subgroups of Wl of order p are permuted transitively by Aut Wl. By minimality of 1<»1, the p-Iength of Wl is 1. Hence <»/Wl is a p-group. Also, since Op'(<») = 1, Op'(Wl) = 1. The Sylow p-subgroup 'l3 of Wl is therefore normal. Let 3 = Ql(Z('l3)). Then 3 #- 1, so it follows from the hypothesis that 3 is a minimal characteristic subgroup of <» and that 3 is the set of all elements of <» of order a divisor of p.

By the Schur-Zassenhaus theorem, 'l3 has a complement .0 in Wl and all such complements are conjugate in Wl. Thus by the Frattini argument, <» = No;(.Q)Wl = No;(.Q)'l3. Hence No;(.Q) #- .0 and No;(.Q) contains an element x of order p. Let 6 be a Sylow p-subgroup of <» and let y be an element of Z(6) of order p. Then <x) = <y)~ for some automorphism ~ of <», so x E Z(6~). Also x E 3 n No;(.Q) = N3(.Q) and

Thus Co;(x) ~ <6~,.Q) = <», so x E Z(<»). Hence Z(<») n 3 #- 1, and since 3 is a minimal characteristic subgroup of <», 3 ::;; Z(<»).

Of course, 1<»/31 is divisible by p. We show that the subgroups of <»/3 of order p are permuted transitively by Aut (<»/3). To do this, let u3 be an element of order p in Z(6/3); thus

[<u), 6] ::;; 3.

Now let v3 be any element of <»/3 of order p. Then uP, vP are elements of <» of order p and uP = vPi( = (viOP for some automorphism ( of <» and some integer i with (i, p) = 1. Now w = (viOa E 6 for some a E <» and wP = (viOP = uP since 3 ::;; Z(<»). Then [u, w] E 3 and

since p is odd. Then uw-1 E 3 and

Hence (v i3)(' = u3 for some automorphism (' of <»/3. It follows from the minimality of 1<»1 that <»/3 is of p-Iength 1. If 0p·(<»/3) = '!/3, '! = 3 x U for some U, since 3 ::;; Z(<»). Thus U <J <» and U ::;; 0p'(<») = 1. Hence 0p.(<»/3) = 1 and 6/3 <J <»/3. Thus 6 <J <». q.e.d.


It follows easily from 8.7 that the Sylow p-subgroup e of (fj has the property that its subgroups of order p are permuted transitively by Aut e. As was remarked in VIII, 7.11, this implies that e is Abelian.

Notes on Chapter IX

§ 1: This formulation of 1.10 was given in a private communication from G. Higman. § 2: We thank K. Johnsen for the proof of 2.3. § 3: The proof of Lemma 3.11 follows a suggestion of K. Johnsen. § 5: Theorem 5.2 for p odd was proved by J. G. THOMPSON [6, Lemma 5.22]. For the version given here we are indebted to K.-U. Schaller. Sections 5.10-5.14 are due to M. B. Powell, whom we thank for a written communication of these results. § 6: We are grateful to H. Kurzweil for suggesting several improvements to this section. § 8: The proof of 8.2 follows Artin, but includes an improvement due to R. M. Bryant.

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Index of Names

Adyan, S. I. 385 Alperin, J. L. 299, 364, 365 Amayo, R. K. 239, 349 Artin, E. 504,506,508,514

Baer, R. 169,500 Baker, H. F. 239, 348 Basev, V. A. 71 Bauman, S. F. 491 Bender, H. 462 Berger, T. R. 406,429,467, 492 Berman, S. D. 41 Birkhoff, G. 366, 367 Blackburn, N. 275,404 Bourbaki, N. 15,20,96,404 Brauer, R. 2,3,21,31,37,38,69, 180,

196, 198,237,315 Brenner, S. 71 Bryant, R. M. 396,397,400,403,404,

514 Burkhardt, R. 193 Burnside, W. 362, 385, 405, 449, 460,

463

Cartan, E. 2, 162, 164, 166 Cartier, P. 370 Clifford, A. H. 2, 118, 123 Conlon, S. 137 Cossey, J. 185, 186, 188, 237 Curtis, C. 13, 63, 168

Dade, E. C. 69, 144,237,406,428,429, 458,492

Deuring, M. 13, 26, 237 Dickson, L. E. 91,406 Dress, A. 24,237

Eckmann, B. 161 Engel, F. 388 Evens, L. 404

Feit, W. 138 Felsch, W. 345

Fermat, P. 424 Fischer, I. 404 Fitting, H. 72 Fong, P. 108, 186, 187, 188, 230, 237 Fox, R. H. 379, 384 Frobenius, G. 165, 166, 169, 172, 173,

237,266

Gaschiitz, W. 82, 86, 185, 186, 187, 188, 204,205,207,237,512

Gauss, C. F. 384 Glauberman, G. 503 Glover, D. J. 237 Goldschmidt, D. M. 299 Golod, E. S. 392, 404 Gorenstein, D. 299, 363, 484, 487, 491 Gow, R. 106, 128,237 Green, J. A. 74, 93, 142, 209, 223 Gross, F. 279,299,315,443,452,456,

467,482,483,485

Hall, M. 419 Hall, P. 363,405,407,417,422,426,428,

439,440,453,458,460,470 Hartley, B. 404, 492 Hausdorff, F. 239,348 Heineken, H. 404 Heller, A. 71 Herstein, I. N. 363,484,487 Higman, D. G. 46, 63, 68, 84, 85, 86, 237 Higman, G. 299,313,361,362,366,404,

405,407,411,417,422,426,428,439, 440,453,458,460,470,514

Hilbert, D. 27 Hill, E. T. 261 Hill, R. 209 Hoare, A. H. M. 448, 452 Hopkins, C. 73, 349 Huppert, B. 121, 133,237 Hurley, T. C. 384

Isaacs, I. M. 128 Ito, N. 509

526

Jacobi, C. G. 1. 238,316 Jacobson, N. 1, 348 James, G. D. 178 Janko, Z. 145,463 Jennings, S. 252, 254, 258, 259, 404 Johnsen, K. 514

Kaplansky, I. 73, 90 Kasch, F. 68, 237 Killing, W. 166 Kneser, M. 68, 237 Kostrikin, A. I. 361,366,388,404 Kovacs, L. G. 396, 397,400,403 Kreknin, V. A. 356, 361, 366,404 Kupisch, H. 68,237 Kurzweil, H. 406,492,514

Landrock, P. 145,315 Lazard, M. 404 Leedham-Green, C. R. 346 Lie, S. 316,326,370 Liebeck, H. 404 Loewy, W. 157, 261

Macdonald, I. D. 345 MacLane, S. 3, 57, 161 McLaughlin, 1. F. 32 Magnus, W. 239,366,380,381,385,404 Mathieu, E. 178 Meixner, T. 361 Mersenne, M. 424 Michler, G. 145, 196, 233

Nakayama, T. 44, 50, 56, 124, 152, 237 Nesbitt, C. 38, 237 Neubiiser, J. 345 Neumann, P. M. 345, 346 Noether, E. 26,237 Norton, S. 32 Novikov, P. S. 385

Osima, M. 182

Passman, D. S. 182,404 Plesken, W. 345 Pliicker,1. 118 Pollatsek, H. 116 Powell, M. B. 476,514

Rae, A. 492 Reiner, I. 13, 63, 71, 168

Ringel, C. M. 71 Rips, I. A. 384 Robinson, D. J. S. 404 Roiter, A. V. 71 Roth, R. L. 140,237

Safarevic, I. R. 392, 404 Sanov, I. N. 459 Schaller, K.-u. 514 Schopf, A. 161 Schreier, O. 414 Schumann, H. G. 384 Schur, I. 21

Index of Names

Schwarz, W. 140,231,236,237 Scimemi, B. 361 Serre,1.-P. 27 Shaw, D. 299, 313, 404 Shult, E. E. 299,315,482,483,491 Sjogren, J. A. 384 Srinivasan, B. 69, 128, 233 Stewart, I. 239, 349 Stonehewer, S. 93 Struik, R. R. 404 Suzuki, M. 116,238, 299 Swan, R. G. 32, 143, 144

Thompson,1. G. 406,429,472,492,511, 514

Tsushima, Y. 173

Vaughan-Lee, M. R. 332,341,345,404 Villamayor, O. E. 93 Voigt, D. 237

Waerden, B. L. van der 76,77,116 Wall, G. E. 389, 396 Walter,1. H. 462 Ward, H. N. 126, 133 Ward,1. N. 487,491 Wedderburn, 1. H. M. 10, 13, 18,28,443 Wiegold, 1. 346, 404 Wielandt, H. 266 Willems, W. 62,93, 107, 114, 120, 126,

128, 133, 146, 189, 192, 193, 195, 196, 203, 229, 236, 237

Witt, E. 238, 239, 366, 367, 381,404

Yen, T. 512

Zassenhaus, H. 1. 264, 404 Zsigmondy, K. 508

Index

Abelian normal subgroup of a Sylow or Hall subgroup 464, 465, 472

Abelian Lie algebra 323 Abelian Sylow subgroup 462 absolutely indecomposable module 73,

223,229 - and extension of ground-field 74ff. -, decomposition as direct sum of 78 absolutely irreducible module 27 adjoint mapping 338 algebra, enveloping 372 -, Frobenius 165ff. -, Lie 316ff. -, Magnus 385 -, quasi-Frobenius 86, 169 -, symmetric 166 algebraic conjugate of a module 15 analogue of Theorem B 477 annihilator 155 associative basis 370 augmentation ideal 65, 252 -, basis of factors of powers of 254 - of group-ring of a free group 378ff. automorphism - of order 2 483 - of order 3 350 -, fixed point free of order 4 484, 487 automorphism group induced on the

Frattini factor group of a p-group 396, 403

automorphism of Lie algebra 324 - and derived series 355-357 - and lower central series 360 -, ideal generated by fixed points 349ff.

Baer's lemma on injective modules 169 Baer's theorem on Op(fj) 500 Baker-Hausdorff formula 239, 348 basis of factors of powers of

augmentation ideal 254 bilinear form, (fj-invariant 106 bilinear mappings on fields 288, 289 Birkhoff-Witt theorem 366,367

bisecting automorphisms 429, 434 block 1, 180 -, kernels of 194-198 -, number of 178 -, principal 180, 184-189, 198 block character 180 block ideal 180 block idempotent 180 Burnside problem 385, 388, 394, 405,

458ff. - for exponent 4 459 - for exponent 6 417,418 - for linear groups 463

Cartan matrix 2,162,179 -, symmetry of 164,170 - of 6 4 214, 218 - of 215 221, 222, 223 central idempotent 174, 182 central product 276 central series, factors of 244-247, 363 - with elementary Abelian factors 242,

248 centralizer of factors of upper n-series

408,409 characters in prime characteristic 14 chief factors, as (fj-modules 209 chieffactors and blocks 188 class of nilpotent Lie algebra 323 Clifford type theorem 2, 118, 123ff. cohomology group 61 coinduced module 51ff. -, universal property of 55 commutation as bilinear mapping 238,

286 commuting idempotents 283 completely reducible factor module 12 composition factors, multiplicity of 162 conjugacy classes, maximal size in

p-groups 341, 345, 346 conjugate module of a normal subgroup

124 contragredient representation 98

528

counting argument 275 critical group 419 -, representation of 422 cyclic normal subgroup of a Sylow

subgroup 465 cyclic Sylow subgroup 2, 68-71 cyclotomic polynomial 503, 506

decomposition of algebra, as right module 151

- into two-sided ideals 174 degree less than p 509 degree of irreducible module 144 derived length, and fixed points of

automorphisms 364-365 - and p-length 467 - and that of Sylow subgroups 470 - of Lie algebra 323 derived series, and fixed points of

automorphisms 355-357 - of Lie algebras 323 dimension subgroup problem 239, 384 direct product, representation of 136ff. direct summand, independent of ground-

field 25 dual module 97ff. - and tensor products 100 - of projective modules 116

eigen function 285 Engel commutator 264 Engel condition 388 enveloping algebra 372 exceptional linear transformation 437 exponent and p-length 451,452 extending a K9l-module to a K<V-module

128 extension of ground-field 4 - and absolute indecomposability 74ff. -, behaviour of complete reducibility

12 -, behaviour ofradical 9 - for Lie algebras 324 -, Galois 15ff., 287, 289 extraspecial group 419, 421, 422, 425

factor module, completely reducible 12 factors of central series 244-247, 363 faithful irreducible representations 267,

268 Fermat prime 424,476 -, exceptional role in Hall-Higman

theory 453 Fitting height 482 -, bounds on 492 Fitting subgroup of odd order 443

Index

fixed point free automorphism 349, 356, 361,362,482,485

- of order 4 484 fixed point free group of automorphisms

of type (p, p) 491 fixed point subgroup and derived length

364-365 form, invariant 109 Frattini factor group 280-282 - and automorphisms of p-groups 396,

403 free associative algebra 370 free group, and regular representation of

a group 400 -, group-ring of 378, 380 -, lower central series of 366, 380ff. -, subgroups of 414 free Lie algebra 371, 372 - and regular representation of a group

397 Frobenius algebra 165ff. - as injective module 169 Frobenius group 266 Frobenius reciprocity 50ff.

generators, of minimal simple group 476

- in Lie algebras 317-319,330 group-basis of a free group 366 group, in which every automorphism is

inner 404 - of class 2 regarded as Lie ring 347 - of degree less than p 509 - of exponent 6 417,418 - of prime exponent 386-389 group-ring, ideals of 166, 167 - is symmetric 166 - of p-group 65, 252ff. group, with cyclic Sylow subgroups 2,

66-71 - with one class of involutions 511

half-transitivity 266 Hall and Higman's Theorem B 405,

419ff. -, elementary form of 411 Hall-Higman methods 405 Hall subgroup 412 head of module 12 - is isomorphic to soc1e for projective

modules 170 Hom, as K<V-module 104, 119 homocyclic p-groups 279 -, p'-automorphisms of 280 homogeneous component of a free

associative algebra 372

Index

homogeneous component of a module 18

homomorphism of Lie algebra 317

3-adic ring 148 3-adic topology 148 ideal of Lie algebra 317 idempotents, central 182, 187 -, commuting 283 - in local rings 72-73 indecomposability of tensor products

133 indecomposable Abelian operator group

280-282 indecomposable group-ring 186 indecomposable module, induced module

from 126 -, number of 63ff. - of groups with cyclic Sylow subgroups

66-71 - of large dimension 63, 68 -, principal projective 203, 208 indecomposable two-sided ideal 174ff.,

234 induced module 2,44ff. - and projective module 91 - from indecomposable module 126 -, universal property of 49 inertia subgroup 124 inflation 91, 208 injective envelope 161 injective module 1, 4 -, same as projective for group-rings

86 invariant form 106 involution, group with unique class of

511 -, number in a 2-group 275-277 -, permuted transitively by auto-

morphisms 266, 269, 279, 296, 299ff.

irreducibility of tensor products 133 irreducible Frattini factor group of

Abelian p-group 280-282 irreducible module, degree of 144 - for GL(2, p) 43 - for SL(2, p) 38 -, number of 32,37,41-43

Jacobi identity 238,316,326 Jacobson radical 1, 71 - of K9l for 9l ~ (f) 93-95 Janko's simple group 31 145,463 Jennings' formula 259

kernels of blocks 194-198

529

Lie algebra 316ff. - associated with associative algebra

318 -, automorphism of 324 -, power of 320-323, 330, 360, 373 Lie basis 370 Lie homomorphism 317 Lie ring 316 - associated with strongly central

series 327 Lie ring method, for nilpotent groups

331 lifting idempotents 1, 148, 149 linear group, of finite exponent 463 - of small degree 509 local ring 71 -, idempotents in 72-73 Loewy length 157 Loewy series 157,172,207,261,262 - of 6 4 214 lower central series, of free groups 380-

384 - of free Lie algebras 373,375,377 - of groups 239 - of Lie algebras 321

McLaughlin's simple group 32 Magnus algebra 385 Mathieu group 9Jl22 178 maximal class, p-group of 251,258 maximal normaln-subgroup 407 maximaln-factor group 407 maximal size of conjugacy class 341,

345,346 Mersenne prime 424 minimal polynomial of p-element 411,

422,426,429 minimal simple group 476 minimal subring ~ with ~2 = 92 338 modular representation 1 - of SL(2, 3) 210££. - of 6 4 214ff. module, absolutely indecomposable 73,

223,229 -, coinduced 51 -, dual 97ff. - for direct products 136ff. -, (f)-conjugate 124 -, induced 2, 44ff., 126 -, relatively injective 82 -, relatively projective 81 -, self-dual 106ff. multiplicity of composition factors 162

Nakayama's lemma 152 net 253

530

nilpotent groups and Lie ring method 331

nilpotent Lie algebra 323 normal Abelian subgroup of Sylow or

Hall subgroup 464,465, 472 normal subgroup, representation of

123ff. number of blocks 178 number of involutions in a 2-group

275-277 number of irreducible modules 32,37,

41-43

orbits of equal length 266 order of an integer modulo a prime

508 orthogonal group 116

p-chief factors as modules 2 p-constrained group 186,408 p-group, group-ring of 65, 252ff. - of maximal class 251, 258 - with all normal subgroups

characteristic 404 p-length and derived length 467 p-length and regularity of Sylow

p-subgroups 456 p-length bounded by p-exponent 413,

451-452 p-nilpotent group, modular representa

tion of 233 power of a product 239, 264 prime characteristic, projective modules

in 90 -, characters in 14 -, splitting field in 31 prime exponent 386-396 primitive permutation group 266 principal block 180,184-189,198 principal indecomposable projective

module 203-208 projective envelope 230 -, dimension of 231 projective module 1, 3, 4 - and induced modules 91 - and tensor products 92 -, dual of 117 -, general properties 153ff. -, head of 157 - in prime characteristic 90 - over an algebra 156 -, principal indecomposable 203-208 -, same as injective for group-rings 86 -, socle of 159 p-stable group 492

quadratic form, invariant 109 quasi-Frobenius algebra 86, 169

radical, Jacobson 71

Index

reciprocity theorems 2, 50, 56, 59 regular automorphism 349,361,362 regularity and p-length 456 relatively injective module 82 relatively projective module 81 representation, absolutely indecom-

posable 73 -, absolutely irreducible 27 -, faithful irreducible 267, 268 -, modular 1 - of normal subgroup 123ff. - of small degree 509 restricted Burnside problem 389, 405,

406, 458ff. ring, Lie 316 -, local 71

Schur index 2,21 second Loewy factor 207 self-dual module 106ff. semisimplicity 1 separable algebra 13 socle of module 12, 155 -, isomorphic to head for projective

modules 170 soluble Lie algebra 323 special affine group 494, 500 splitting field 27 - in prime characteristic 31 square mapping, in 2-groups 294,297,

313-315 strongly central series 326 -, associated Lie ring 327 subgroup of free group 414ff. subgroups of order p permuted transi-

tively by automorphisms 266, 269, 279, 280, 296, 299ff.

subnormal subgroup 223 sums of powers of 2 271-274 Suzuki's simple groups 116 Suzuki 2-group 238, 299 Sylow subgroup of symplectic group

432-433 symmetric algebra 1, 166 symmetric form, ffi-invariant 106ff. symplectic group 116

tensor product, and blocks 188 -, indecomposability 133 -, irreducibility 133 - of dual modules 100

Index

- of extensions of a field 7 - of projective modules 92 Theorem B of Hall and Higman 405,

426 transitive linear groups 266ff. transitively permuted involutions 266,

269, 279, 296, 299ff.

universal enveloping algebra of a Lie algebra 372

upper Loewy series 172 upper p-series 407

Witt identity 238 Witt's formula 375, 384

531

Grundlehren der mathematischen Wissenschaften A Series ojComprehensive Studies in Mathematics

A Selection

162. Nevanlinna: Analytic Functions 163. Stoer/Witzgall: Convexity and Optimization in Finite Dimensions I 164. Sario/Nakai: Classification Theory of Riemann Surfaces 165. MitrinoviclVasic: Analytic Inequalities 166. GrothendieckiDieudonne: Elements de Geometrie Algebrique I 167. Chandrasekharan: Arithmetical Functions 168. Palamodov: Linear Differential Operators with Constant Coefficients 169. Rademacher: Topics in Analytic Number Theory 170. Lions: Optimal Control of Systems Governed by Partial Differential Equations 171. Singer: Best Approximation in Normed Linear Spaces by Elements of Linear

Subspaces 172. Buhlmann: Mathematical Methods in Risk Theory 173. Maeda/Maeda: Theory of Symmetric Lattices 174. Stiefel/Scheifele: Linear and Regular Celestial Mechanic. Perturbed Two-body

Motion - Numerical Methods - Canonical Theory 175. Larsen: An Introduction to the Theory of Multipliers 176. GrauertiRemmert: Analytische Stellenalgebren 177. Augge: Practical Quantum Mechanics I 178. Augge: Practical Quantum Mechanics II 179. Giraud: Cohomologie non abeIienne 180. Landkof: Foundations of Modem Potential Theory 181. Lions/Magenes: Non-Homogeneous Boundary Value Problems and

Applications I 182. Lions/Magenes: Non-Homogeneous Boundary Value Problems and

Applications II 183. Lions/Magenes: Non-Homogeneous Boundary Value Problems and

Applications III 184. Rosenblatt: Markov Processes. Structure and Asymptotic Behavior 185. Rubinowicz: Sommerfeldsche Polynommethode 186. Handbook for Automatic Computation. Vol. 2. Wilkinson/Reinsch: Linear Algebra 187. Siegel/Moser: Lectures on Celestial Mechanics 188. Warner: Harmonic Analysis on Semi-Simple Lie Groups I 189. Warner: Harmonic Analysis on Semi-Simple Lie Groups II 190. Faith: Algebra: Rings, Modules, and Categories I 191. Faith: Algebra II, Ring Theory 192. Mallcev: Algebraic Systems 193. P6lya/Szeg6: Problems and Theorems in Analysis I 194. Igusa: Theta Functions 195. Berberian: Baer*-Rings 196. Athreya/Ney: Branching Processes 197. Benz: Vorlesungen uber Geometric der Algebren 198. Gaal: Linear Analysis and Representation Theory 199. Nitsche: Vorlesungen tiber Minimalfliichen 200. Dold: Lectures on Algebraic Topology

201. Beck: Continuous Flows in the Plane 202. Schmetterer: Introduction to Mathematical Statistics 203. Schoeneberg: Elliptic Modular Functions 204. Popov: Hyperstability of Control Systems 205. Nikollskii: Approximation of Functions of Several Variables and Imbedding

Theorems 206. Andre: Homologie des Algebres Commutatives 207. Donoghue: Monotone Matrix Functions and Analytic Continuation 208. Lacey: The Isometric Theory ofOassical Banach Spaces 209. Ringel: Map Color Theorem 210. GihmanlSkorohod: The Theory of Stochastic Processes I 211. Comfort/Negrepontis: The Theory ofUltrafilters 212. Switzer: Algebraic Topology - Homotopy and Homology 214. van der Waerden: Group Theory and Quantum Mechanics 215. Schaefer: Banach Lattices and Positive Operators 216. P6Iya/Szegt): Problems and Theorems in Analysis II 217. Stenstrom: Rings of Quotients 218. GihmanlSkorohod: The Theory of Stochastic Processes II 219. DuvantlLions: Inequalities in Mechanics and Physics 220. Kirillov: Elements of the Theory of Representations 221. Mumford: Algebraic Geometry I: Complex Projective Varieties 222. Lang: Introduction to Modular Forms 223. Bergh/LOfstrom: Interpolation Spaces. An Introduction 224. Gilbarg/Trudinger: Elliptic Partial Differential Equations of Second Order 225. SchUtte: Proof Theory 226. Karoubi: K-Theory. An Introduction 227. Grauert/Remmert: Theorie der Steinschen Raume 228. Segal/Kunze: Integrals and Operators 229. Hasse: Number Theory 230. Klingenberg: Lectures on Oosed Geodesics 231. Lang: Elliptic Curves: Diophantine Analysis 232. Gihman/Skorohod: The Theory of Stochastic Processes III 233. StroocklVaradhan: Multi-dimensional Diffusion Processes 234. Aigner: Combinatorial Theory 235. DynkinlYushkevich: Markov Control Processes and Their Applications 236. GrauertlRemmert: Theory of Stein Spaces 237. KOthe: Topological Vector-Spaces II 238. Graham/McGehee: Essays in Commutative Harmonic Analysis 239. Elliott: Probabilistic Number Theory I 240. Elliott: Probabilistic Number Theory II 241. Rudin: Function Theory in the Unit Ball of[" 242. HuppertlBlackburn: Finite Groups II 243. HuppertlBlackburn: Finite Groups III 244. KubertlLang: Modular Units 245. Comfeld/Fomin/Sinai: Ergodic Theory 246. Naimark: Theory of Group Representations 247. Suzuki: Group Theory I 248. Suzuki: Group Theory II 249. Chung: Lectures from Markov Processes to Brownian Motion 250. Arnold: Additional Chapters from the Theory of Ordinary Differential Equations 251. Chow/Hale: Methods of Bifurcation Theory

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