Group Representations Volume 1. Part A: Background Material + Part B: Introduction to Group Representations and Characters 044488632X, 9780444886323

Table of contents :
[North-Holland Mathematics Studies 175]Group Representations, Volume 1 Part A_Background Material, Gregory Karpilovsky, 1992, lxii+620p, North-Holland, 044488632X
Copyright
Preface
Contents
Bibliography
Notation
Index
Part I. Background Material
Chapter 1. Rings and Modules
1.1. Notation and terminology
1.2. Preliminary results
1.3. Artinian and noetherian modules and rings
1.4. Semisimple modules
1.5. The radical and socle of modules and rings
A. The radical and socle of modules
B. The Jacobson radical
C. The Jacobson radical and idempotents
1.6. Idempotent lifting theory
A. General results
B. Semiregular rings
C. Algebras over complete rings
1.7. Azumaya's theorems
1.8. Local rings
1.9. Endomorphism algebras
1.10. Strongly indecomposable modules
A. Basic properties and characterizations
B. Azumaya's decomposition theorem
C. The Krull-Schmidt theorem
1.11. Direct decompositions and blocks
1.12. Matrix rings
Chapter 2. Artinian and Semilocal Rings
2.1. Semiprimitive artinian rings
2.2. Semilocal rings
2.3. Artinian rings
A. A characterization
B. Principal indecomposable modules
C. Blocks and Cartan matrices
D. The Loewy and socle series
2.4. Representations of algebras
Chapter 3. Homological Algebra
3.1. Tensor products of modules
3.2. Tensor product of algebras
3.3. Flat modules
3.4. Properties of Hom_R(V, W)
3.5. Change of coefficient rings
3.6. Projective modules
A. Basic properties
B. Generators and progenerators
C. Endomorphism rings
3.7. Pushouts and pullbacks
3.8. Injective modules
A. Basic properties
B. Injective hulls
C. Endomorphism rings
3.9. Dual modules, bilinear forms and Frobenius algebras
A. Dual modules
B. Bilinear forms
C. Frobenius algebras
Chapter 4. Restriction, Induction and Coinduction
4.1. General information
4.2. Universal characterizations
4.3. The splitting of canonical homomorphisms
4.4. Another characterization of induced modules
4.5. Induction and semisimplicity
4.6. Annihilators of induced modules
4.7. Exact sequences of induced modules
4.8. Normal subalgebras
4.9. Induction and relative projectivity
4.10. Coinduction and relative injectivity
4.11. Relative injective modules for algebras
Chapter 5. Semiperfect Rings
5.1. Projective covers
5.2. Characterizations and fundamental properties
5.3. Some classes of semiperfect rings
5.4. Block decompositions
5.5. Properties of idempotents
5.6. Semiperfect endomorphism rings
5.7. Basic rings
Morita-equivalent
5.8. Points in semiperfect rings
Chapter 6. Complexes, Homology and Resolutions
6.1. Notation and terminology
6.2. Fundamental properties of complexes
6.3. Homotopy
6.4. Resolutions
6.5. Categories and functors
6.6. Universal functors and sattelites
6.7. Derived functors
6.8. Ext and Tor
6.9. Ext^1 and extensions
Chapter 7. Heller Operators
7.1. Heller operators
7.2. Minimal resolutions
7.3. Self-injective artinian rings
7.4. Projective homomorphisms
7.5. Heller operators and dual modules
Chapter 8. Group Algebras
8.1. Definitions and elementary properties
8.2. Localization
8.3. Support of central idempotents
8.4. Some module isomorphisms
Chapter 9. Group Cohomology
9.1. Preliminaries
9.2. The standard resolution
9.3. Change of groups
9.4. Restriction, corestriction and inflation
9.5. Stable submodules
9.6. Swan's theorem
9.7. The Hochschild-Serre exact sequence
9.8. Group extensions and cohomology
Chapter 10. Graded Algebras and Crossed Products
10.1. Group-graded algebras
10.2. Crossed products
10.3. Simple crossed products
10.4. Semilinear monomial representations
10.5. Crossed products over simple rings
10.6. Equivalent crossed products
Chapter 11. Algebras over Fields
11.1. Splitting fields
11.2. Intertwining numbers and semisimplicity
11.3. Separable algebras
11.4. Blocks, characters and Cartan matrices
11.5. The Deuring-Noether theorem
Chapter 12. The Brauer Group
12.1. Central simple algebras
12.2. The Brauer group
12.3. Tensor product of division algebras
12.4. The Brauer group and crossed products
Chapter 13. Indecomposable Modules and Ground Field Extensions
13.1. Preliminary results
13.2. Homogeneous components
13.3. Idempotent liftings and group actions
13.4. Behaviour of indecomposable modules underground field extension
Chapter 14. The Schur Index
14.1. Preliminary results
14.2. Behaviour of simple modules under ground field extensions
14.3. The Witt-Fein's theorem
14.4. The Schur index
14.5. Linear independence of characters
Chapter 15. Frobenius and Symmetric Algebras
15.1. Elementary properties of Frobenius and symmetric algebras
15.2. Cogenerators
15.3. Quasi-Frobenius algebras
15.4. Frobenius algebras
15.5. Symmetric algebras
Chapter 16. Dedekind Domains and Discrete Valuation Rings
16.1. Integrally closed domains
16.2. Dedekind domains
16.3. Discrete valuation rings
16.4. Completions
16.5. Complete discrete valuation rings
[North-Holland Mathematics Studies 175]Group Representations, Volume 1 Part B_Introduction to Group Representations and Characters, Gregory Karpilovsky, 1992, xiii+654p, North-Holland, 044488632X
Copyright Page
Contents
Bibliography
Notation
Index
Part II. Introduction to Group Representations
Chapter 17. Generalities
17.1. Definitions and elementary properties
17.2. Splitting fields
17.3. Counting simple modules over splittiiig fields
17.4. Brauer’s permutation lemma
17.5. Counting simple modules over arbitrary fields
17.6. The socle and Reynolds ideal
17.7. Inner and outer tensor products
17.8. Representations of direct products
17.9. Changing the characteristic
17.10. Dimensions of absolutely simple modules
Chapter 18. Induced Modules
18.1. Restriction and induction
18.2. Induction and semisimplicity
18.3. Induction of dual and contragredient modules
18.4. Reciprocity theorems
18.5. Tensor products
18.6. Mackey’s theorems
18.7. Counting induced modules
18.8. The relative trace map
18.9. Induction and relative projectivity
18.10. An application: Knörr’s theorem
18.11. Clifford’s theorem
18.12. Monomial modules
Part III. Introduction to Characters
Chapter 19. An Invitation to Characters
19.1. Induced characters
19.2. Orthogonality relations
A. Preliminary results
B. Orthogonality relations
C. Intertwining numbers and applications
19.3. Class functions and character rings
A. Generalities
B. Splitting fields
C. C-characters
D. Prime and maximal ideals
19.4. Representations of abelian groups
19.5. Inductive sources
Chapter 20. Induction Theorems and Applications
20.1. The Witt-Berman’s induction theorem
20.2. Brauer’s theorems
20.3. Rational valued characters
Chapter 21. Central, Faithful and Permutation Characters
21.1. Central characters
21.2. Character kernels and faithful characters
A. General properties
B. Gaschütz's theorem
C. Faithful irreducible characters
D. Extra-special $p$-groups
21.3. Permutation characters
21.4. Irreducible characters of $p$’ and $p$-power degrees
Chapter 22. Character Tables
22.1. Group information
22.2. Galois actions
22.3. Character tables for A_5 and S_5
Chapter 23. Zeros of Characters
23.1. Burnside’s theorem
23.2. Characters vanishing off subgroups
23.3. Gallagher’s theorems
23.4. Applications and related results
23.5. Žmud’s theorems
Chapter 24. Characters, Conjugate Elements and Commutators
24.1. Products of conjugate elements
24.2. Characters and commutators
Chapter 25. The Frobenius-Schur Indicator
25.1. Unitary matrices
25.2. The Frobenius-Schur indicator
Chapter 26. Characters and Hall subgroups
26.1. An excursion into group theory
26.2. Characters and Hall subgroups
26.3. Degrees of faithful characters
Chapter 27. Extensions of Characters
27.1. A general criterion
27.2. Gallagher’s theorems
27.3. Two results of Thompson
27.4. Thompson’s theorems
27.5. Character restriction property
A. Introduction
B. Preliminary results
C. Two theorems of Isaacs
Chapter 28. Irreducible Constituents and Conjugacy Classes
28.1. Irreducible constituents of induced characters
28.2. Characters and conjugacy classes
Chapter 29. Fixed-Point Spaces and Powers of Characters
29.1. Characters and fixed-point spaces
29.2. Powers of characters
Chapter 30. Determinants of Characters
30.1. Determinants of characters
30.2. Character-theoretic transfer
Chapter 31. Tensor Induction of Characters
31.1. Tensor product of n ≥2 modules
31.2. Tensor induced modules
31.3. Tensor induced characters
31.4. Tensor induced class functions
31.5. An application to characters of central products
Chapter 32. Knörr’s Generalized Character
32.1. Definition
32.2. Reduction to C_G(v)
32.3. The abelian case
32.4. The main result
Chapter 33. Characters of Centralizer Rings
33.1. Characters of eFGe
A. Characters of centralizer rings
B. Bases and block idempotents
C. Applications to degrees of characters
33.2. Characters of C_{FG}(H)
Chapter 34. Characters and Relative Normal Complements
34.1. Relative normal complements
34.2. π- Sections
34.3. A lifting operator
34.4. Theorems of Dade, Brauer and Suzuki
34.5. Some generalized characters
34.6. Generalized characters and π-sections
34.7. Complements and character extensions
Chapter 35. Isometries and Generalized Characters
35.1. π-Induction
35.2. Normal π'-subgroups
35.3. Isometries and generalized characters
Chapter 36. Exceptional Characters
36.1. Trivial intersection sets
36.2. Coherent sets of characters
A. Preliminary results
B. The main theorem and its applications
36.3. Exceptional subsets
36.4. Special classes
Chapter 37. Frobenius Groups
37.1. Preliminary results
37.2. Thompson’s criterion for $p$-nilpotence
37.3. Fixed-point-free automorphisms
37.4. Structure of Frobenius groups
37.5. Characters of Frobenius groups
37.6. Coherence
Chapter 38. Applications of Characters
38.1. Burnside’s p^a q^b theorem
38.2. Wielandt’s theorems
38.3. Generalized quaternion Sylow subgroups
38.4. The Brauer-Suzuki-Wall theorem
38.5. Applications to U(\mathbb{Z} G)
A. Preliminary results
B. Torsion units
C. The isomorphism class of U(\mathbb{Z} G)
D. Effective construction of units of \mathbb{Z} G
E. Cyclic groups

Citation preview

GROUP REPRESENTATIONS Volume 1

Part A: Background Material

NORTH-HOLLAND MATHEMATICS STUDIES 175

{Continuation of the Notas de Matematica)

Editor: Leopoldo NACHBIN Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro, Brazil and University of Rochester New York, U.S.A.

NORTH-HOLLAND - AMSTERDAM • LONDON • NEW YORK• TOKYO

GROUP REPRESENTATIONS Volume 1 Part A: Background Material

Gregory KARPILOVSKY Department of Mathematics California State University Chico, CA, U.S.A.

~ ~

ffl 1992

NORTH-HOLLAND - AMSTERDAM • LONDON • NEW YORK• TOKYO

ELSEVIER SCIENCE PUBLISHERS B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands

Library of Congress Cataloging-in-Publication Data Karpilovsky, Gregory, 1940 Group representations/ Gregory Karpilovsky. p.

cm . .. (North-Holland mathematics studies;l 75)

Includes bibliographical references and index. Contents; v., pt .. A. Background material, pt. B. Introduction to group representations and characters. ISBN 0-444 -88632-X (set) 1. Representations of groups.

I. Title. II. Series.

QAl 76.K37 1992

512' .2-dc20

92-14786 CIP

ISBN: 0 444 88632 X (Set: Part A and B) © 1992 ELSEVIER SCIENCE PUBLISHERS B.V. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher, Elsevier Science Publishers B.V., Copyright & Permissions Department, P.O. Box 521, 1000 AM Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center Inc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. This book is printed on acid-free paper. Printed in The Netherlands

To my wife Helen, who typed this lengthy manuscript and remained cheerful throughout

vii

Preface

The present volume is the first of a multi-volume treatise on group representations. The principal object of these volumes is to provide, in a selfcontained manner, a comprehensive coverage of the mainstream of group representation theory. The appropriate audience for these volumes consists of aspiring graduate students and mature mathematicians working in the field of group representations. No mathematical knowledge is presupposed beyond the rudiments of abstract algebra, set theory and field theory; however, a certain maturity in mathematical reasoning is required. Apart from a few obvious exceptions ( e.g. odd order theorem, classification of finite simple groups, etc.), the volumes are entirely self-contained. The style of the presentation is informal : the author is not afraid to repeat definitions and formulas when they are needed. Many sections begin with a nontechnical description of what is about to be done. A special effort has been made to render the exposition transparent. Due to bookbinding considerations, Volume 1 is split into two parts, Part A and Part B. However, mathematically, the material is divided into three parts. The first part provides background material for all volumes. The second part is devoted to introduction to group representations, and the third part to introduction to group characters. Many of the results are presented in their greatest generality and much of the material discussed is not conveniently available in other monographs on the subject. A systematic description of the material is supplied by the introductions to individual chapters and therefore will not be repeated here. A word about notation. As is customary, Theorem 2.3.4 denotes the fourth result in Section 3 of Chapter 2; however, for simplicity, all references to this result within Chapter 2 itself, are designated as Theorem 3.4. I am indebted to my wife Helen, who not only typed the entire volume, but encouraged me (always with patience and good humour) throughout the entire project.

California State University, Chico November, 1991

G. Karpilovsky

Contents Preface

Part I

Vll

Background Material

1. Rings and Modules

1.1. 1.2. 1.3. 1.4. 1.5.

1.6.

1.7. 1.8. 1.9. 1.10.

1.11. 1.12.

Notation and terminology Preliminary results Artinian and noetherian modules and rings Semisimple modules The radical and socle of modules and rings 1.5.A. The radical and socle of modules 1.5.B. The Jacobson radical 1.5.C. The Jacobson radical and idempotents Idempotent lifting theory 1.6.A. General results 1.6 .B. Semiregular rings 1.6.C. Algebras over complete rings Azumaya's theorems Local rings Endomorphism algebras Strongly indecomposable modules 1.10.A. Basic properties and characterizations 1.10.B. Azumaya's decomposition theorem 1.10.C. The Krull-Schmidt theorem Direct decompositions and blocks Matrix rings

2. Artinian and Semilocal Rings 2.1. Semiprimitive artinian rings 2.2. Semilocal rings 2.3. Artinian rings 2.3.A. A characterization

1

3

3 8 13 22 26 26 31 39 42 42 46 52 55 60 61 65 65 70 71 73 80

89

89 92 100 100

Contents

X

2.3.B. P~incipal indecomposable modules 2.3.C. Blocks and Cartan matrices 2.3.D. The Loewy and socle series 2.4. Representations of algebras 3. Homological Algebra

Tensor products of modules Tensor product of algebras Flat modules Properties of H omR(V, W) Change of coefficient rings Projective modules 3.6.A. Basic properties 3.6.B. Generators and progenerators 3.6.C. Endomorphism rings 3.7. Pushouts and pullbacks 3.8. Injective modules 3.8.A. Basic properties 3.8.B. Injective hulls 3.8.C. Endomorphism rings 3.9. Dual modules, bilinear forms and Frobenius algebras 3.9.A. Dual modules 3.9.B. Bilinear forms 3.9.C. Frobenius algebras

3.1. 3.2. 3.3. 3.4. 3.5. 3.6.

4. Restriction, Induction and Coinduction 4.1. 4.2. 4.3. 4.4. 4.5. 4.6. 4.7. 4.8. 4.9. 4.10.

General information Universal characterizations The splitting of canonical homomorphisms Another characterization of induced modules Induction and semisimplicity Annihilators of induced modules Exact sequences. of induced modules Normal subalgebras Induction and relative projectivity Coinduction and relative injectivity 4.11. Relative injective modules for algebras

5. Semiperfect Rings 5.1. Projective covers 5.2. Characterizations and fundamental properties

101 102 105 108 113

113 121 126 132 139 141 141 148 152 155 158 158 163 169 171 171 178 184 191

191 194 196 201 202 204 206 208 212 218 220 225

225 233

Contents

5.3. 5.4. 5.5. 5.6. 5.7. 5.8.

Xl

Some classes of semiperfect rings Block decompositions Properties of idempotents Semiperfect endomorphism rings Basic rings Points in semiperfect rings

6. Complexes, Homology and Resolutions

6.1. 6.2. 6.3. 6.4. 6.5. 6.6. 6.7. 6.8. 6.9.

Notation and terminology Fundamental properties of complexes Homotopy Resolutions Categories and functors Universal functors and sattelites Derived functors Ext and Tor Ext 1 and extensions

7. Heller Operators

7.1. 7.2. 7.3. 7.4. 7.5.

Heller operators Minimal resolutions Self-injective artinian rings Projective homomorphisms Heller operators and dual modules

8. Group Algebras

8.1. 8.2. 8.3. 8.4.

Definitions and elementary properties Localization Support of central idempotents Some module isomorphisms

9. Group Cohomology

9.1. 9.2. 9.3. 9.4. 9.5. 9.6. 9.7. 9.8.

Preliminaries The standard resolution Change of groups Restriction, corestriction and inflation Stable submodules Swan's theorem The Hochschild-Serre exact sequence Group extensions and cohomology

239 242 246 249 255 257 261

261 264 270 275 285 291 302 311 321 327

327 330 336 341 348 351

351 366 369 374 379

379 392 400 403 408 411 414 420

Contents

Xll

10. Graded Algebras and Crossed Products

10.1. 10.2. 10.3. 10.4. 10.5. 10.6.

Group-graded algebras Crossed products Simple crossed products Semilinear monomial representations Crossed products over simple rings Equivalent crossed products

11. Algebras over Fields

11.1. 11.2. 11.3. 11.4. 11.5.

Splitting fields Intertwining numbers and semisimplicity Separable algebras Blocks; characters and Cartan matrices The Deuring-Noether theorem

12. The Brauer Group

12.1. 12.2. 12.3. 12.4.

Central simple algebras The Brauer group Tensor product of division algebras The Brauer group and crossed products

13. Indecomposable Modules and Ground Field Extensions

13.1. 13.2. 13.3. 13.4.

Preliminary results Homogeneous components Idempotent liftings and group actions Behaviour of indecomposable modules under ground field extension

14. The Schur Index

14.1. 14.2. 14.3. 14.4. 14.5.

Preliminary results Behaviour of simple modules under ground field extensions The Witt-Fein's theorem The Schur index Linear independence of characters

15. Frobenius and Symmetric Algebras

15.1. Elementary properties of Frobenius and symmetric algebras

427

427 437 449 451 455 461 469

469 479 483 487 492 495

495 502 508 510 519

519 523 526 529 539

539 543 546 549 553 555

555

Contents

15.2. 15.3. 15.4. 15.5.

Xlll

Cogenerators Quasi-Frobenius algebras Frobenius algebras Symmetric algebras

16. Dedekind Domains and Discrete Valuation Rings

16.1. 16.2. 16.3. 16.4. 16.5.

Integrally closed domains Dedekind domains Discrete valuation rings Completions Complete discrete valuation rings

Bibliography

Introduction to Group Representations

17. Generalities

Definitions and elementary properties Splitting fields Counting simple modules over splitting fields Brauer's permutation lemma Counting simple modules over arbitrary fields The socle and Reynolds ideal Inner and outer tensor products Representations of direct products Changing the characteristic Dimensions of absolutely simple modules

18. Induced Modules

18.1. 18.2. 18.3. 18.4.

591 599 611 613 615 XIX

Ii

Index

17.1. 17 .2. 17 .3. 17.4. 17 .5. 17 .6. 17.7. 17.8. 17 .9. 17.10.

591

xliii

Notation

Part II

559 563 571 580

Restriction and induction Induction and semisimplicity Induction of dual and contragredient modules Reciprocity theorems

621

623

623 631 634 637 639 643 647 653 662 668 671

671 681 683 691

Contents

XlV

18.5. 18.6. 18.7. 18.8. 18.9. 18.10. 18.11. 18.12. Part III

Tensor products Mackey's theorems Counting induced modules The relative trace map Induction and relative projectivity An application: Knorr's theorem Clifford's theorem Monomial modules

699 702 709 712 716 722 725 727

Introduction to Characters

731

19. An Invitation to Characters

19 .1. Induced characters 19.2. Orthogonality relations 19.2.A. Preliminary results 19.2.B. Orthogonality relations 19.2.C. Intertwining numbers and applications 19.3. Class functions and character rings 19.3.A. Generalities 19.3.B. Splitting fields 19.3.C. (i): Suppose that V has a descending chain

(ii): Applying (i), we have for all

It follows, by induction, that all positive powers of x lie in

= Li,;l Sxi

m 2 0

Li~l Sxi.

is a finitely generated S-module. = S[x]. (iii)=> (iv): Note that Bis a faithful S[x]-module. Now take V

S[x]

Hence

(ii) => (iii): Take B

= B.

7 Azumaya 's theorems

57

(iv) => (i): Let f E Ends(V) be defined by f ( v) = xv and put I = S. Then f(V) ~IV= V and hence, by Lemma 7.1, +s1r- 1 +···+Sn= 0 for some Si E S and some n ~ 1. Since V is faithful, x satisfies (1), as required. •

r

Corollary 7 .3. Let A be an S-algebra which is a finitely generated S module. Then A .is integral over S. Proof.

Apply Lemma 7.2 (iii) for B

= A.

•

Lemma 7 .4. Let A be. an S -algebra which is integral over S. (i) If x is a unit of A, then x- 1 E S[x). (ii) For any subalgebra B of A, (a) B n U(A) = U(B). (b) B n J(A) ~ J(B). Proof. We first note that, by Proposition 5.27 (i), (b) is a consequenc(' of (a). Since (a) is obviously a consequence of (i), we are left to verify (i). To prove (i), we put y = x- 1 . Using the assumption that y is integral over S, we have

Multiplying this equation by xn and taking into account that xy = yx = 1, we obtain 1 = x(s1 + S2X + ·· · + SnXn-l) Hence x- 1

= s1 + s2x + · · · + SnXn-l

E S[x), as required. •

The following result is a strengthened version of Nakayama's lemma for modules over commutative rings.

Lemma 7 .5. Let V be. a finitely generated S-module such that MV for every maximal ideal M of S. Then V = 0.

=V

Proof. Assume by way of contradiction that V -:/: 0 and let v 1 , ••• , Vn be a minimal generating set for V. Let I be the ideal of S consisting of all s E S such that sv1 E Sv2 + ··· + Svn, where Sv2 + ··· + Svn is to be interpreted as O in case n = 1. We claim that I = S which will provide a desired contradiction.

Rings and Modules

58

Assume that 1 cf. S. Then there exists a maximal ideal M of S such that M. Ifv1 E Mv1+Sv2+ .. ·+Svn,then (l-m)v1 E Sv2+ .. ·+Svn for some m E M. Hence 1 - m E J c;;, M and 1 E M, which is impossible. Thus

Jc;,

contrary to our assumption that V

= MV = Mv1 + Mv2 + ··· + Mvn,

•

Let W be a submodule of an S-module V such that V /W is a finitely generated S-module. If V = W + MV for every maximal ideal M of S, then V = W. Corollary 7.6.

1

Proof. By hypothesis, M(V/W) = V/W for every maximal ideal M of S. Hence, by Lemma 7.5, V/W = 0 as required. •

As a final preliminary result, we next record the following lemma. Lemma 7. 7. Let A be an S -algebra which is a finitely generated S module. Then an element x E A is a left {right) unit of A if and only if, for any maximal ideal M of S, x + MA is a left (right) unit of A/MA. Proof. If x is a left (right) unit of A, then clearly x +MA is a left (right) unit of A/MA, for every maximal ideal M of S. Conversely, assume that x + MA is a left unit of A/MA, for every maximal ideal M of S. Then Ax+ MA= A for each M, so by Corollary 7.6 (with V = A and W = Ax) we have Ax = A. Hence x is a left unit of A. Similarly, if x + MA is a right unit of A/MA, then xA +MA = A and, by Corollary 7.6, xA = A. Hence x is a right unit of A. •

We have now come to the demonstration for which this section has been developed. Theorem 7.8. {Azumaya {1951)). Let S be a commutative ring and let A be an S -algebra which is a finitely generated S-module. For each maximal ideal M of S, let JM be the ideal of A defined by JM/MA = J(A/MA).

Then where M ranges over all maximal ideals of S.

7 Azumaya 's theorems

59

Proof. By Corollary 5.22 (v), x E J(A) if and only if 1 - ax E U(A) for all a EA. By Lemma 7.7, the latter is equivalent to the requirement that

(1- ax)+ MAE U(A/MA) for all a E A and all maximal ideals M of S, i.e. to x E nM JM.

•

Theorem 7.9. {Azumaya {1951)). Let S be a commutative ring and let A be an S -algebra which is a finitely generated S -module. (i) Every left or right unit of A is a unit of A. {ii) If x is a unit of A, then x-t E S[x]. {iii) For any subalgebra B of A, (a) B n U(A) = U(B). (b) B n J(A) ~ J(B). Proof. Properties (ii) and (iii) follow from Corollary 7.3 and Lemma 7.4. To prove (i), we first assume that Sis a field. If x E A is a left unit of A, then the S-linear map A ---+ A, a f-+ xa is injective and hence surjective. Therefore xa = 1 for some a E A, proving that x is a unit. A similar argument shows that if x E A is a right unit of A, then x is a unit. Turning to the general case, let M be a maximal ideal of S. If x E A is a left (right) unit of A, then x + MA is a left (right) unit of A/MA. Since A/MA is a finite-dimensional algebra over the field S / M, it follows from the special case above that x + MA is a unit of A/MA. Hence, by Lemma 7. 7, x is a unit of A. • Theorem 7.10. {Azumaya {1951)). Let S be a commutative ring, let V be a finitely generated S -module and let Vt, ... , Vn be finitely many elements ofV. {i) Vt, ... , Vn generate V if and only if, for any maximal ideal M of S, Vt + MV, ... , Vn + MV generate the S / M -module V / MV. (ii) Under the assumption that V is free, Vt, ... , Vn is a basis of V if and only if, for any maximal ideal M of S, Vt + MV, ... , Vn + MV is a basis of the S/M-module V/MV. Proof. (i) If Vt, ... , Vn generate V, then clearly Vt + MV, ... , Vn + MV generate the S / M-module V / MV, for any maximal ideal M of S. Conversely, assume that for any maximal ideal M of S, Vt + MV, ... , Vn + MV generate the S/M-module V/MV. Then V = Li=t Svi + MV for any such M, so

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60

V = Li=l Svi by virtue of Corollary 7.6. (ii) Assume that V is free and let u1, ... , Um be a basis of V. Suppose further that, for any ideal M of S, v1 + MV, . .. , Vn + MV is a basis of the SfM-module VfMV. We will show that v1 , ... ,vn is a basis of V. The converse being a consequence of Lemma 2.4, the result will follow. By Lemma 2.4, u 1 + MV, . .. , Um + MV is a basis of the Sf M-module VfMV, so since Sf Mis a field, we haven= m. Now write n

Vi=

L

j=l

OijUj

(aii ES)

Then the matrix ( Oij) is invertible modulo,M, i.e. det( Oij) ¢. 0( mod M). Since this is the case for every M, it follows that det( Oij) is a unit of S, i.e. that ( Oij) is invertible. Hence v1, ... , Vn is a basis of V, as desired. •

8

Local rings

In this brief section, we introduce local rings 'and provide their characterizations. A ring R is said to be local if Rf J(R) is a division ring. In what follows, we write U(R) for the unit group of R. Proposition 8.1. For any ring R, the following conditions are equivalent: (i) R is local. {ii) R has a unique maximal left ideal. (iii) J(R) = R - U(R). (iv) The set of nonunits of R is a left ideal. Proof. (i) ::} (ii): Let J be a maximal left ideal of R. Then J f J(R) is a proper left ideal of the division ring Rf J(R) and thus I= J(R). (ii) ::} (iii): Let J be a unique maximal left ideal of R. Since J(R) is the intersection of all maximal left ideals of R, we have J(R) = J, and so J is an ideal. The inclusion J(R) ~ R - U(R) being obvious, assume that x E R - U(R). If Rx -::J R, then Rx lies in a maximal left ideal of R, whence Rx ~ J(R), so x E J(R). On the other hand, if Rx = R then yx = 1 for some y ER. Clearly y (/. J(R), otherwise 1 = yx E J(R). Hence Ry= R, so zy = 1 for some z ER and therefore z = x. Thus x E U(R), a contradiction. (iii) ::} (iv): Obvious.

9 Endomorphism algebras

61

(iv) :::} (i): Let x ~ J(R) and let I= R - U(R). If Mis a maximal left ideal of R, then M ~ If= R, whence M =I= J(R). Hence x is a unit of Rand so x + J(R) is a unit of RfJ(R). Consequently, Rf J(R) is a division ring, as required. • Corollary 8.2. Let R be a local ring. Then {i) The only idempotents of R are O and 1. (ii) For any proper ideal I of R, Rf I is a local ring. Proof. (i) Let e f= 1 be an idempotent of R. Then e is a nonunit and hence, by Proposition 8.1, e E J(R). Thus, by Proposition 5.29 (i), e = 0. (ii) Let J f I be a maximal left ideal of Rf I. Then J is a maximal left ideal of R, hence J = J(R) by Proposition 8.1. Thus Rf I has a unique maximal left ideal and, by Proposition 8.1, Rf I is local. • Corollary 8.3. Let e be an idempotent of a ring R such that eRe is a local ring. If e E I1 +···+In for some ideals Ii, ... , In of R, then e E Ii for some i E { 1, ... , n}. Proof, It clearly suffices to treat the case where n = 2. So assume that e E I1 + h. Then e E efie + ehe and efie and ehe are ideals of the local ring eRe. Hence we may assume that R is a local ring and that e = 1. If 1 = a+ b with a E Ii and b E h, then not both a and b lie in J(R), say a ~ J(R). Hence, by Proposition 8.1, a is a unit. Thus 1 = aa- 1 E I 1, as required. • Proposition 8.4. Let R be a semiregular ring. Then R is local if and only if O and 1 are the only idempotents of R. Proof. Assume that O and 1 are the only idempotents of R. By Corollary 8.2 (i), it suffices to show that R is local. If I is a maximal left ideal of R, then by Corollary 6.15 (i), I~ J(R). Hence ,R has a unique maximal left ideal and so, by Proposition 8.1, R is local. •

9

Endomorphism algebras

Throughout this section, A denotes an algebra over a commutative ring S.Given a nonzero A-module V, we write EndA(V) for the endomorphism

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62

ring of V. We remind the reader our convention that "module" means "left module" and for any f,g E EndA(V), the product Jg is defined by (Jg)(v)

= J(g(v))

for all

v EV

Note that the ring EndA(V) is an S-algebra via (sf)(v)

= (s · l)J(v)

for all

s E S, v E V

We begin by recording the following classical result. Proposition 9.1. (Schur's lemma). Then EndA(V) is a division ring.

Let V be a simple A-module.

Proof. Let f: V -+ V be a nonzero A-homomorphism. Then J(V) is a nonzero submodule of V, so f(V) = V. Because ff. 0, I( er ff. V and so I( er f = 0. Thus f is an isomorphism, as required, • Proposition 9.2. Assume that V is an A-module and let e be an idempotent of A. (i) HomA(Ae, V) ~ eV as S-modules. In particular, if f is another idempotent of A, then

HomA(Ae,Af) ~ eAJ

as S -modules

{ii) If V is a right A-module, then HomA(eA, V) ~ Ve

as S-modules

Proof. (i) If f E HomA(Ae, V), then f(e) = f(e 2 ) = ef(e) E eV. Therefore the map f 1-+ f(e) is an S-homomorphism from HomA(Ae, V) to eV. Conversely, if v E eV, then the map 9v : xe 1-+ xv is an A-homomorphism from Ae to V. Because the map v 1-+ gv is the inverse of f 1-+ J( e ), the required assertion follows. (ii) Apply the same argument as in (i). • Proposition 9.3. Let e f. 0 be an idempotent of A. Then we have the following isomorphisms of S-algebras:

EndA(Ae) ~ (eAe) 0

and

EndA(eA) ~ eAe

9 Endomorphism algebras

63

In particular,

Proof. Applying Proposition 9.2 for V = Ae, it follows that the map f i--+ f(e) is an S-isomorphism of EndA(Ae) onto eAe. Given f,g E EndA(Ae), write f ( e) = ea1 e and g( e) = ea2e for some a1, a2 E A. Then

proving that f i--+ f (e) reverses multiplication. Because e is the identity element of the algebra eAe, the above map preserves identity elements. This establishes the first isomorphism. The second isomorphism follows from Proposition 9.2 applied to V = eA. • Proposition 9.4. Let V i- 0 be an A-module, let E let e, f be two idempotents of E. {i) HomA(e(V),f(V)) ~ f Ee as S-modules. {ii) For e i- 0, End A( e(V)) ~ eEe as S-algebras. (iii} e(V) ~ f (V) if and only if Ee ~ E f.

= EndA(V)

and

Proof. (i) Any given a E E determines an A-homomorphism .\, e(V) -+ f(V) which sends e( v) to (Jae)( v ), for any v E V. Furthermore, the map fEe-+ HomA(e(V),f(V)) { f ae i--+ Aa

is obviously a well defined injective S-homomorphism. Now let us assume that µ E Hom A (e(V), f (V)) and define a E E by a( v) = µ( e( v)) for all v EV. Then a(e(v)) = µ(e(v)), v E Vand so Aa =µ,as required. (ii) By applying (i) with f = e, it suffices to verify that the map in (i) preserves multiplication, i.e. for all a,/3 E E, Aae/3 = Ao,Af3· Since the latter is a consequence of the definition of Aa, the required assertion follows. (iii) By (i ), e(V) ~ f (V) if and only if Ao, : e(V) -+ f (V) is an isomorphism for some a E E. By (i), a typical element of H omA(f(V), e(V)) is µ13,/3 EE, where µ13(f(v)) = (ef3f)(v) for all v E V. Thus e(V) ~ f(V) if and only if there exist a, f3 E E with µ13Aa = 1 and Aaµ/3 = 1. Since the latter equalities are equivalent to (ef3f)(fae) = e and (fae)(e/3!) = f, respectively, the required assertion is a consequence of Lemma 2.6. •

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64

We are now ,ready to prove the following important result which ties together the decomposition of modules and that of their endomorphism algebras.

E

Theorem 9.5. Let A be an algebra over a commutative ring S, let where V =/. 0 is an A-module and let

= EndA(V)

(1)

(½ =/. 0) be a finite direct decomposition of V. For each i E {1, ... , r }, let 1ri : V be the projection map. (i) 1r1, •.. , 1rr are pairwise orthogonal nonzero idempotents of E with

1 = 11"1

+ ··· + 1rr

½

(2)

Conversely, if 1r1, ... , 1rr are pairwise orthogonal nonzero idempotents of E satisfying (2), then V satisfies {1) for ½ = 1ri(V), 1 :S i :S r, and each 7ri : V - ½ is the projection map. {ii) ½ is indecomposable if and only if 1ri is primitive. In particular (by taking ½ = V, 1ri = 1), V is indecomposable if and only if O and 1 are the only idempotents of E. (iii) ½ ~ Vj {.} E1ri ~ E1ri {.} Eiti ~ Eitj where 1l"i is the image of 1ri in E = E/J(E). {iv) EndA(½) ~ 1riE1ri as S-algebras. (v) EndA(½)/J(EndA(½)) ~ itiEiti as S-algebras.

Proof. (i) It is clear that 1r; = 1ri, 1 :S i :S r. If i =/. j, then 7ri7rj(V) = 1ri(Vj) = 0, and hence 7ri7rj = 0. Furthermore, we obviously have 1 1r1 + ·· · + 1rr. The converse is equally clear. (ii) If 7ri = e f, where e, f are orthogonal idempotents, then ½ e(V) EB J(V). Conversely, if ½ = V/ EB V/' and e : V - V/, f : V - V/' are projection maps, then 7ri = e + f and e, f are orthogonal idempotents. Thus ½ is indecomposable if and only if 7ri is primitive. (iii) Apply Propositions 9.4 (iii) and 5.31. (iv) Since½= 1ri(V), the required assertion follows from Proposition 9.4

+

(ii). (v) We have (by (iv))

10 Strongly indecomposable modules

65

(Lemma 2.5 (iv)) as desired. •

10

Strongly indecomposable modules

This section is devoted to a study of modules with local endomorphism rings. Throughout, R denotes an arbitrary ring and V an R-module. Following Dade (1980, p. 257), we say that Vis strongly indecomposable if EndR(V) is local. Note that, by Corollary 8.2 (i) and Theorem 9.5 (ii), if V is strongly indecomposable, then V is indecomposable. The simplest example of an indecomposable but not strongly indecomposable module is V =RR where R = '7L, . For convenience, we divide this section into three subsections.

A. Basic properties and characterizations Lemma 10.1. Let V = U EB W, let 7r : V-+ U be the projection map and let V' be any submodule of V. (i) Ker(7rlV') = V' n Wand 7r(V') = (V' + W) n U. (ii} V = V' EB W if and only if 7rjV' : V' -+ U is an isomorphism. Proof. (i) We first note that I( er( 7rlV') = V' n J( er7r = V' n W. Since 7rjU = lu and 7r(W) = 0, the equality V' +W = ((V' + W) n U) + W shows that 7r(V') = (V' + W) n U. (ii) By (i), 7rjV' is injective if and only if V' n W = 0. Again, by (i) 7rjV' is surjective if and only if U ~ V' + W. Since the latter is equivalent to V' + W = V, the result follows. • Lemma 10.2. Let f: U-+ V and g: V-+ U be homomorphisms of R-modules. Then lu + gf is a unit of EndR(U) if and only if lv + Jg is a unit of EndR(V). Proof.

(lv

Assume that lv

+ fg)- 1 = lv -

+ Jg is a unit of EndR(V). Then (lv + fgt 1 fg = lv - fg(lv + fg)- 1

Rings and Modules

66

which immediately implies that (lu+g/)- 1 = lu-g(lv+ fg)- 1 f. A similar argument shows that if lu + gf is a unit, then (lv + fg)- 1 = lv - f(lu + g!)-lg . • As a preliminary to the next observation, let us recall the following terminology. An R-homomorphism a: U-+ Vis said to be a split injection if a is an injection and a(U) is a direct summand of V. It is clear that a is · a split injection if and only if there is an R-homomorphism f3 : V-+ U with {3a = lu. We say that a is a split surjection If a is a surjection such that I( era is a direct summand of U. Obviously, a is a split surjection if and only if there is an R-homomorphism f3: V-+ U with af3 = lv. Lemma 10.3. Let U be a strongly indecomposable R-module and let V be any R-module. If a : U -+ V and f3 : V -+ U are R-homomorphisms such that af3 (/. J(EndR(V)), then a is a split injection and f3 is a split surjection. Proof. Since a/3 (/. J(EndR(V)), it follows from Corollary 5.22 (v) that there exists 'Y E EndR(V) such that 1v + "f0/3 is not a unit of EndR(V). Applying Lemma 10.2 with f = "fO, g = {3, we deduce that lu + f3'Ya is not a unit of EndR(U). But EndR(U) is local, so f3'Ya is a unit of EndR(U). Hence a is a split injection and f3 is a split surjection, as desired. •

Let V be an R-module. Following Crawley and Jonsson (1964, p. 797), we say that V has the exchange property if, for any R-modules U, W and Mi( i E I), the condition

(1) implies that there exist submodules U

Mf ~

Mi such that

= V E9 (E9ierMI)

(2)

The proof of the following lemma shows that each. Mf is a direct summand of Mi and V ~ E9ier(Mi/Mf). Lemma 10.4. Assume that an R-module V has the exchange property and that V is a direct summand of U = EflieI Mi. Then {i) V ~ E9ieI Ni where each Ni is a direct summand of Mi. {ii}IfV f:. 0 and each Mi is indecomposable, then V ~ E9jeJMj for some nonempty subset J of I.

10 Strongly indecomposable modules

67

Proof. By hypothesis, there exist R-modules W and Mf ~ Mi such that (1) and (2) holds. Hence it suffices to show that each Mf is a direct summand of Mi and V ~ EBieI(Mi/Mf). The latter is obvious since

The fact that Mf is a direct summand of Mi follows by restricting to Mi the projection map U - t Mf corresponding to the decomposition (2). • Lemma 10.5. that

Assume that U, V, Wand Mi(i E J) are R-modules such U

=V

EB W

= EBieI Mi

i- 0 to U = V

and V

Let a : U - t V be the projection corresponding EB W and, for i E J, let 1ri : U - t Mi be the projection corresponding to U = ffiiEIMi, ai : Mi - t V the restriction of a to Mi and 1ri : V - t Mi the restriction of 1ri to V. Then there exists j E J such that a;1rj ¢ J(EndR(V)). Proof. Given O i- v E V, we have v E ffi;EJM; for some finite subset J of I. Then v = LjeJ 1r;( v) and, for each j E J, a( 1rj( v )) = ( Oj7rj )( v ). Hence v = a( v) = ai1rj)( v)

(L

jEJ

and so lv - LjeJ a;1rj is a nonunit of EndR(V). Therefore LjEJ Oj7rj ¢ J(EndR(V)) and the result follows. • Let V be an R-module. Following Beck (1978, p. 396), we say that V has the mutual exchange property if, for any R-modules U, W and

Mi( i E J), the condition

= Mf S Mf' for all i E J such that U = W S (EBie1Mf') = V EB (Sie1M[)

implies that there are decompositions Mi

It is clear that if V has the mutual exchange property, then V has the exchange property. Theorem 10.6. {Warfield {1969}, Beck {1978}}. For any indecomposable R-module V, the following conditions are equivalent:

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68

(i) V is strongly indecomposable. (ii) V has the mutual exchange property. (iii) V has the exchange property. Proof. (i) => (ii): Assume that U, W and Mi( i E J) are R-modules satisfying (1). Keeping the notation of Lemma 10.5, we may find j E J such that O:j'7rj is a unit of EndR(V) (since, by hypothesis, EndR(V) is local). For any i E J with i =/- j, put Mf = Mi and Mf' = 0. Also put MJ = J( ero:j and MJ' = 7r5(V). Since O:j7rj : V ---+ V is an isomorphism, Mj = MJ EB MJ'. Thus, for all i E J, Mi = Mf EB Mf'. Furthermore, 7rj : V ---+ MJ' is an isomorphism. Hence, by Lemma 10.1 (ii), V can replace MJ' in the decomposition U

= MJ EB MJ' EB ( EBi,eiMi)

Also O:j IMJ' : MJ' ---+ V is an isomorphism. Hence, by Lemma 10.1 (ii), MJ' replaces V in the decomposition U = V EB W. Thus U

=W

EB (EBiEI MI')

=V

EB ( EBiEI MI)

as required. (ii) => (iii): This is obvious. (iii) => (i): Assume that EndR(V) is not local. Then there are two endomorphisms f and g which are not automorphisms such that 1v = f - g. Let W = Vi EB Vi where Vi, Vi are copies of V and let 7l"i : W ---+ ¼ be the projection map, i = 1, 2. We imbed V into Vi EB Vi by the homomorphism (f, g) : V ---+ Vi EB Vi and we call the image V'. We can also imbed V in W by the diagonal map (1 v, 1v ), and we denote by d(V) the image of this map. Then an easy calculation shows that W = V' EB d(V). Assume by way of contradiction that V has the exchange property. Then we have either W = Vi EB V' or W = Vi EB V', since Vi and Vi are indecomposable. In the first case, this would mean that 7r 2 (f, g) was an isomorphism. This cannot happen, since it would imply that g was an automorphism. A similar argument shows that the second alternative is also impossible, thus completing the proof. • Corollary 10.7. Let V be a strongly indecomposable R-module and assume that V is a direct summand of U = EBiEI Mi. Then (i) V is isomorphic to a direct summand of Mi for some i E J. (ii) For any submodule W of U with U = V EB W, there exists j E J and

10 Strongly indecomposable modules

a decomposition Mj

= MJ EB MJ' U =WEB MJ'

69

such that

=V

EB MJ EB (EBi:;fiMi)

In particular, if each Mi is indecomposable, then U =WEB Mj

=V

EB (EBi:;fiMi)

Proof. (i) Apply Theorem 10.6 (iii) and Lemma 10.4 (i). (ii) This was established in the proof of (i) => (ii) of Theorem 10.6. • Corollary 10.8.

Let U, V, W and Mi( i E I) be R-modules such that

U = V EB W = EBiEI Mi and assume that V =/ 0 and each Mi is strongly indecomposable. Then {i) V has a direct summand isomorphic to Mi for some i EI. {ii) If V is indecomposable, then there exists j E I such that

Proof. (i) Keeping the notation of Lemma 10.5, we may find j E I such that O'.j7rj (/. J(EndR(V)). Hence, by Lemma 10.3, O'.j : Mj -+ V is a split injection, as required. (ii) If Vis indecomposable, then by (i), V ~ Mi for some i and so Vis strongly indecomposable. Now apply Corollary 10.7 (ii). • Corollary 10.9. Let Vi and Vi be strongly indecomposable direct summands of an R-module V. If Vi ~ Vz, then there exists a direct summand U of V such that V=ViEBU=ViEBU Proof. Choose submodules U1 and U2 with V = Vi EB U1 = Vi EB U2 and let f : Vi -+ Vi be an isomorphism. By Theorem 10.6, Vi has the exchange property. If V = Vi EB U2, then we may take U = Uz. In the contrary case, the exchange property for Vi implies that Vi EB Vi EB Vi = V for some submodule Vi of Uz. Setting U = {v + f(v)lv E Vi} EB U~, it follows that U satisfies the desired property. • Corollary 10.10. (Cancellation property). Let Vi, Vi be isomorphic strongly indecomposable R-modules. Then, for any R-modules X and Y,

X EB Vi

~

Y EB Vi

implies

X

~

Y

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70

Proof. We may of course assume that X EB Vi = Y EB V2. Hence, by Corollary 10.9, Vi EB U = V2 EB U = X EB V1 = Y EB V2 for some R-module U. Thus U ~ X ~ Y, as required. •

B. Azumaya's decomposition theorem The significance of strongly indecomposable modules stems from the following classical result.

Theorem 10.11.

(Azumaya (1950)). Let V be an R-module such that

where each ¼ is strongly indecomposable and each Wj is indecomposable. Then there is a bijection (j : J -+ J such that for all i E J Proof. Let M be an indecomposable direct summand of V, let a be the cardinality of the set M(I) = {i E II½~ M} and let /3 be the cardinality of the set M(J) = {j E JIWi ~ M}. By Corollary 10.8 (i), Mis isomorphic to some ½. In particular, each Wj is strongly indecomposable and, by Corollary 10.8 (i), M is isomorphic to some Wj. Thus it suffices to verify that a = (3. If a is finite, then the assertion is true by applying the cancellation property (Corollary 10.10) and induction on a. Suppose that a is infinte. Let Trj,j E J be the projections for the decomposition V = EBjeJWj. For each i E J, put

Clearly, by Lemma 10.1 (ii), j E Si if and only if rril½ : ½ -+ Wj is an isomorphism. Given i E J, there exist j1, ... ,in E J with½ n (Wj 1 + ··· + Win) =J 0. Thus Ker(rril¼) = 0 implies j E {jl,···,jn} and so each Si is finite. If i E M(I), then clearly Si ~ M(J). Conversely, if j E M(J) then, by Corollary 10.8 (ii), j E Si for some i E M(J). Thus M(J) = UieM(I)Si. Hence i i-+ Si is a mapping from M ( I) to a set of finite subsets of M ( J) that cover M(J). Therefore, since M(I) is infinite, we have

f3

= IM(J)I ~ IJN

X

M(J)I

= IM(J)I = a,

10 Strongly indecomposable modules

proving that

71

f3 ::; a. By symmetry, a ::; f3 and the result follows. •

C. The Krull-Schmidt theorem In this section we present the classical Krull-Schmidt theorem which will be derived as a consequence of Azumaya's decomposition theorem. As a point of departure, we record the following standard fact known as the Fitting's lemma.

Lemma 10.12. Assume that an R-module V has a composition series of length n 2: 1 and let f E EndR(V). Then V = r(V) EB Kerr and f restricts to an isomorphism r(v) -+ r(v) and a nilpotent map Kerr -+ Ker fn. Furthermore, if V is indecomposable, then either fn = 0 or f is an automorphism. Proof.

We first note that

V 2 f(V) 2

J2 (V) 2 · · ·

and 0 ~ Ker f ~ Ker

J2 ~ ···

Since V has length n, it follows that r(V) = r+l(V) = · · · and Kerr = Ker r+ 1 = · · ·. In particular, r(V) = f 2 n(V) and Ker f 2 n = Kerr. Then, for any x EV, there exists y EV such that u(x) = u 2 (y), where u = fn. Hence X - u(y) E Ker fn and so X = u(y) + (x - u(y)) E fn(v) +Kerr. For any x E r(v) n Kerr, we have u(x) = 0 and x = u(y) for some y E V. Then y E f 2n(V) = r(V), whence x = u(y) = 0, proving that

V

= r(V) EB Kerr. Put Vi = r(v) and Vi = Kerr.

Then r(Vi) = 0 and f(Vi) ~ Vi, so f restricts to a nilpotent map Vi -+ V2 • Since f(Vi) = r+ 1 (V) = r(V) = Vi, the restriction g of f to Vi is surjective. Since Vi is noetherian, K ergm = K ergm+l for some m 2: 1. But for any x E J( erg, we can write x = gm(y) for some y E Vi, so O = g(x) = gm+l(y) implying O = gm(y) = x. Thus Kerg = 0 and g is an automorphism of Vi. Finally, assume that V is indecomposable and r i- 0. Then V = fn(V) and Ker fn = 0. Hence fn is an automorphism and so f is also an automorphism.•

72

Rings and Modules

Corollary 1(_).13. Assume that an indecomposable R-module V is of finite length. Then V is strongly indecomposable. Proof. Given f E EndR(V), it follows from Lemma 10.12 that either f is a unit of EndR(V) or f is nilpotent. Assume that f E EndR(V) is a nonunit and let g E Endn(V). Since gf is a nonunit, it must be nilpotent, so 1 - gf is a unit. Hence, by Corollary 5.22, f E J(EndR(V)). Thus, by Proposition 8.1, Endn(V) is local, as desired. •

Let V be an R-module. We say that V has the unique decomposition property if the following two properties hold: (i) Vis a finite direct sum of indecomposable modules. (ii) If V = EBY:; 1 ½ = EBj=l Wj, where each½, Wj is indecomposable, then m = n and, after possibly reordering the Wj, ½ ~ Wi for all i E {1, ... , m }. It is now an easy matter to prove the following classical result. Theorem 10.14. {Krull-Schmidt theorem). Let R be an arbitrary ring and let V =/- 0 be an R-module of finite length. Then V has the unique decomposition property. Proof. Since V is noetherian, it follows from Proposition 3.9 that V is a finite direct sum of indecomposable submodules. Let W be any indecomposable direct summand of V. Then W is of finite length and so, by Corollary 10.13, W is strongly indecomposable. Now apply Theorem 10.11 and the result follows. •

For future use, we finally record another consequence of Azumaya's decomposition theorem. Theorem 10.15. Let R be a noetherian ring such that each finitely generated indecomposable R-module is strongly indecomposable. Then any finitely generated nonzero R-module has the unique decomposition property. Proof. Let V f=. 0 be a finitely generated R-module. Since R is noetherian, it follows from Corollary 3.7 that Vis noetherian. Hence, by Proposition 3.9, V is a finite direct sum of indecomposable submodules. The desired conclusion is now a consequence of Theorem 10.11. •

11 Direct decompositions and blocks

11

73

Direct decompositions and blocks

Let R be an arbitrary ring. Two ideals I and J of Rare said to be relatively prime if I + J = R. It is clear that I and J are relatively prime if and only if no maximal ideal of R contains both I and J. In particular, any two distinct maximal ideals of R are relatively prime. Proposition 11.1.

Let I and J be relatively prime ideals of R. Then

(i) In J = I J + JI. (ii) For all integers n, m

~

1, Im and Jn are relatively prime.

Proof. (i) It is clear that I J + JI ~ I n J. To prove the opposite containment, write 1 = i + j for some i EI, j E J. If x EI n J, then

x=xi+xjEJI+IJ, as required. (ii) If I, J and I, J' are relatively prime ideals, then I and J J' are relatively prime since the image of J J' in R/ I is the product of the images of J and J' both of which are R/ I. In particular, Im and Jn are relatively prime for all n,m ~ 1. • Proposition 11.2. Let V be an R-module and let Ii, ... , In be pairwise relatively prime ideals of R. Then the map

is a surjective R-homomorphism. Proof. It clearly suffices to show that the given map is surjective; To this end, fix i E {1, ... ,n}. For each j E {1, ... ,n}, let Xi+ Yi= 1 with xi E Ii,Yi E Ii. Then Yi l(modli), Yi O(modii)· Let ri = Yl···Yi-1Yi+1···Yn· Then Ti= l(modli), Ti= O(modii) for j-:/- i. If Vi E V, let v = Ei=l TiVi. Then v ~ vi(modli) for all i E {1, ... , n}, as required. •

=

Corollary 11.3.

=

(Chinese remainder theorem). Let I1, ... ,In ·be pair-

74

Rings and Modules

wise relatively prime ideals of R. Then the map

is a surjective ring homomorphism. In particular, n

R/(nf=i Ii) ~

IT (RI Ii)

i=i

Proof. The given map is obviously a ring homomorphism with kernel nf=ih Since, by Proposition 11.2 (with V = R), this map is surjective, the result follows. •

Let {Rili E I} be a family of rings and let R be their direct product. Then the natural projections 7ri : R - Ri are ring homomorphisms, while the natural injections Ai : Ri - R preserve addition and multiplication, but not the identity elements, and so are not ring homomorphisms. Of course, the latter implies that the images of the Ri are not subrings, although they are ideals of R. Proposition 11.4. Let Ri, R2, . .. , Rn be arbitrary rings. {i) If R = Ri X • · • X Rn, then there exists a decomposition of 1:

as a sum of pairwise orthogonal central idempotents of R such that

{ii} If ei, ... , en are pairwise orthogonal central idempotents of a ring R with 1 = Li=i ei, then R = Rei EB··· EB Ren and R ~ Rei X · · · X Ren. (iii) If Ii, ... , In are ideals in a ring R such that R = Ii EB · · · EB In and 1 = I:~i ei, then ei, ... , en are pairwise orthogonal central idempotents of R such that Ii = Rei, 1 :=:; i :=:; n. In particular, by (ii}, R ~ Ii x · · · x In. Proof. (i) Let Ai : Ri - R be the natural injection, and let ei be the image of the identity element of Ri, 1 :=:; i :=:; n. Then ei, ... , en are pairwise orthogonal central idempotents of R with 1 = I:~i ei, Hence R = Rei EB · · · EB Ren. Since I mAi = Rei and ei is the identity element of

11 Direct decompositions and blocks

75

the ring Rei, the required assertion follows. (ii) The first statement is obvious. The map R ---t Re1 X • · • X Ren, r t-t (re1, ... , ren) is obviously an injective homomorphism of rings. Since for any r1, ... , rn in R, r1 e1 + ·· · + rnen is mapped to ( r1 e1, ... , rnen), the map is an isomorphism. (iii) By Proposition 3.10 (ii) and its analogue for right modules, it follows that Ii = Rei = eiR, l ~ i ~ n and e1, ... ,en are pairwise orthogonal idempotents of R. Hence xei = eix = x for all x E h 1 ~ i ~ n. Since Iiij ~ Ii n Ij 0, we deduce that each ei is a central idempotent, as we wished to show. •

=

Corollary 11.5. Let R1, ... , Rn be arbitrary rings and let R = R1 X Rn. Then R is artinian (noetherian) if and only if each Ri is artinian (noetherian).

••• X

Proof.

Apply Proposition 11.4 and Corollary 3.6. •

Let I be an ideal of a ring R. We say that I is indecomposable if for any ideals X and Y of R, I = X E0 Y implies X = 0 or Y = 0. The case I= R leads to the notion of an indecomposable ring. Corollary 11.6. The following conditions are equivalent: (i) R is indecomposable. {ii) Z(R) is indecomposable. {iii) R is not isomorphic to a direct product of n ~ 2 rings. (iv) 0 and 1 are the only central idempotents of R. Proof.

This is a direct consequence of Proposition 11.4. •

An ideal I of a ring R is said to be a direct summand of R if R = I E0 J for some ideal J of R. By a block of R, we understand an ideal of R which is an indecomposable direct summand of R. By a block idempotent of R, we mean a primitive idempotent of Z(R). Corollary 11. 7. Let I be an ideal of a ring R. (i) I is a direct summand of R if and only if I = Re for some (necessarily unique) central idempotent e of R. {ii) I is a block of R if and only if I= Re for some (necessarily unique)

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76

block idempotent _e of R. Proof.

Apply Proposition 11.4. •

A nonzero left ideal X of R is said to be indecomposable if X is an indecomposable submodule of RR, i.e. if X = Y ffi Z, Y, Z left ideals of R, implies Y = 0 or Z = 0. Proposition 11.8. Let I1, ... Jn be ideals of R such that R = Ii ffi · · · ffi In and let 1 = e1 +···+en with ei E h 1 :S: i :S: n. Then (i) Z(R) = Z(R)e1 ffi · · · ffi Z(R)en. (ii) Z(R)ei = Ii n Z(R) and J(Z(R)ei) = J(Z(R))ei, 1 ~ i :S: n. (iii) For any R-module V, V = ffif= 1ei V is a direct decomposition of V. In particular, if V is indecomposable, then there exists i E {1, ... n} such that ei V = V and ej V = 0 for all j f= i

(iv) For each left ideal X of R, we have X

= EBf= 1(X n Ii)

In particular, each indecomposable left ideal lies in exactly one of the Ii. Proof. (i) By Proposition 11.4 (iii), e1, . .. , en are pairwise orthogonal central idempotents of R such that 1 = Li=l ei. Now apply Proposition 11.4 (ii). (ii) By Proposition 11.4 (iii), Ii = Rei and each ei E Z(R). Hence Z(R)ei ~ Ii n Z(R), 1 :S: i :S: n. Conversely, assume that z E Ii n Z(R). Since ei is the identity element of Ii, we have z = zei E Z(R)ei, proving that Ii n Z(R) ~ Z(R)ei. The second equality is a consequence of Proposition 5.29 (ii). (iii) Because ei E Z(R), eiV is a submodule of V. Since the e/s are pairwise orthogonal, we also have n

Lei V

i=l

= ffif= 1ei V

Taking into account that V

n

n

i=l

i=l

= 1 · V = (L ei)V ~Lei V

~ V,

11 Direct decompositions and blocks

the required assertion follows. (iv) By (iii), it suffices to show that eiX

eiX

~

X

77

= X n Ji, 1 ::; i::; n.

Since

n Ji = X n eiR ~ eiX

the desired conclusion is established. • Proposition 11.9. Assume that R = Ii EB··· EB In where each Ii is a block of R and let 1 = e1 +···+en with ei E Ji, 1 ::; i ::; n.

(i) If a nonzero ideal I of R is a direct summand of R, then I is a direct sum of some Ii. In particular, 11, ... , In are all blocks of R. (ii) e1, ... , en are all block idempotents of R. (iii) The central idempotents of R are precisely those of the form Lies ei, where S is a subset of {1, ... , n }. Proof. (i) By hypothesis, R = I EB J for some ideal J of R. Also, by Proposition 11.8, I= EElf= 1 (Jn Ji) and J = EEli':: 1 (J n Ji)· It follows that

and therefore

(l~i::;n) Since Ji is indecomposable, In Ji = 0 or In Ji = h Thus I is a direct sum of all Ii contained in J. (ii) By Proposition 11.4 (iii), e1, ... , en are central idempotents of R such that Ii = Rei, 1 ::; i ::; n. Hence, by Corollary 11.7 (ii), each ei is a block idempotent of R. Let e be any block idempotent of R. Then Re is a block of R (Corollary 11.7 (ii)) and, by (i) and Corollary 11.7 (ii), e = ei for some i E {1, ... , n}. (iii) It is clear that for any subset S of {1, ... , n }, L-ieS ei is a central idempotent of R. Conversely, assume that e is a central idempotent of R. Then Re is a direct summand of R (Corollary 11.7 (i)) and so, by (i), Re = R(Lies ei) for some subset S of {1, ... , n }. Thus e = Lies ei as required. • Any decomposition of R of the form R = 11 EB··· EB In where each Ji is a block of R is called a block decomposition of R. Corollary 11.10.

Then

Assume that a ring R admits a block decomposition.

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78

(i) Any two block decompositions of R are the same (up to the order of occurence). , (ii) R has only finitely many central idempotents.

Proof.

This is a direct consequence of Proposition 11.9. •

Assume that a ring R admits a block decomposition

R

= B1 EB • • • EB Bn

and write 1 = e1 +···+en with ei E Bi, 1 ::; i ::; n. Then, by Propositions 11.9 and 11.4 (iii), Bi= Rei and e1, ... ,en are all block idempotents of R. Let V be an indecomposable R-module. By Proposition 11.8 (iii), there exists i E {1, ... , n} such that

eiV=V

and

ejV=O

forall

jf:.i

In this case we say that V lies in the block Bi or V belongs to the block Bi. Of couyse, in the special case where V is an indecomposable left ideal of R, V lies in the block Bi if and only if V ~ Bi. The above gives a classification of all indecomposable (and, in particular, all simple) R-modules into blocks. Let R be an arbitrary ring. Two primitive idempotents e, f of R are said to be linked , written e"' f, if there exists a finite sequence r1

= e, r2, ... , Tm = f

of primitive idempotents of R such that for each j E {1, ... , m - 1} there exists a primitive idempotent Uj of R such that (1)

It is clear that "' is an equivalence relation on the set of all primitive idempotents of R. Note also that, by Proposition 9.2, (1) is equivalent to

Of course "' induces an equivalence relation on any given subset of the set of all primitive idempotents of R. Keeping this in mind, we now prove the following result.

11 Direct decompositions and blocks

79

Proposition 11.11. Assume that a ring R has pairwise orthogonal primitive idempotents e1, ... , en such that 1 = e1 + · ·· + en. Let

be the partition of the set { e1, ... , en} into equivalence classes of linked idempotents and, for each j E {1, ... , t}, let Bj = ffieEXjRe. Then (i) R = B1 ffi · · · ffi Bt is a block decomposition of R. (ii) Two idempotents e, f E { e1, ... , en} lie in the same block if and only if e and f are linked. (iii) Assume that every principal indecomposable R-module is isomorphic to some Rei. Then two primitive idempotents e, f of R lie in the same block if and only if they are linked. Proof. (i) If R does not admit a block decomposition, then R = I1 ffi • · ·ffilm for some nonzero ideals Ii of R with m > n. But, by Proposition 11.8 (iv), each Rei~ Ij for a unique j E {1, ... ,m}, so m ~ n, a contradiction. Thus R admits a block decomposition. It is obvious that R = ffi}= 1 Bj, By Proposition 11.9 (i), it therefore suffices to show that each B j is an ideal of R contained in a block of R. Assume that Rei ~ Bj and Rek ~ Br with j -::j:. r. If eiRek-/- O, then ei,..., ek which is impossible. Thus eiRek = 0 and therefore BjBr = 0 for j -/- r. Hence, for any j E {1, ... , t},

which shows that each Bj is an ideal of R. Assume that eRf -::j:. 0 for some primitive idempotents e, f of R. By Proposition 11.8 (iv), there exist blocks B, B' of R such that Re ~ B, Rf~ B'. It follows that O-/- (Re)(Rf) ~ BB' and hence B = B'. Therefore e, f lie in the same block which implies that if two primitive idempotents are linked, then they lie in the same block. Thus each B j is contained in a block of R, as required. (ii) This is a direct consequence of (i). (iii) By hypothesis, Re ~ Rei, Rf ~ Rej for some i,j E {1, ... ,n}. Since e and f belong to the same block (respectively, linked) if and only if the same is true for ei and ej, the required assertion follows from (ii). •

80

12

Rings and Modules

Matrix_ rings

Throughout, R denotes an arbitrary ring and, for any integer n ~ 1, Mn(R) is the ring of all n x n matrices over R. Let i,j E {1, ... , n} and let eij be the matrix with ( i,j)-th entry 1 and 0 elsewhere. The elements eij E Mn(R) are called matrix units . As is customary, we shall identify R with its image in Mn(R) consisting of all scalar matrices diag( r, r, ... , r ), r E R. As a preliminary to our first result, let us note that the matrix units satisfy the following properties: (i) eijeks = 0 if j c/- k and eijejs = eis· (ii) 1 = en + ··· + enn· (iii) The centralizer of {eij} in Mn(R) is R. (iv) R ~ e11Mn(R)e11, (v) Mn(R) is a free R-module freely generated by eij, 1 ~ i,j ~ n. Properties (i) and (ii) motivate the following definitions. A finite set {Vijli,j

= 1, ... ,n}

of elements of a ring S satisfying VijVks

=0

if

j c/- k

and

VijVjs

= Vis

for all i,j,k,s E {1, ... ,n} is called a system of matrix units in S. If, in addition, 1 = vu + ·· · + Vnn, then we refer to {Vij} as a full system of matrix units in S. Proposition 12 .1. Let S be any ring, let {Vij Ii, j = 1, ... , n} be a full system matrix units in S and let R be the centralizer of {Vij} in S. Then the map { Mn(R) ( aij)

.!+ 1---l-

S Li,j aijVij

is an isomorphism of rings and R-modules. Furthermore, Re:! v11Sv11. Proof. We know that Mn(R) is a free R-module freely generated by the matrix units eij · Hence the map eij 1---l- Vij extends to a homomorphism of R-modules, which clearly coincides with 'lj;. Applying (i), (ii) and (ii), we immediately deduce that 1/J is also a ring homomorphism. Assume that Li,jaijVij = 0 and fix k,s E {1, ... ,n}. Then, for all tE{l, ... ,n}, 0 = Vtk(L aijVij)Vst = aksVtt i,J

12 Matrix rings

81

and so 0

= L aksVtt = aks(L vu) = aks t

t

Thus ( aij) = 0 and therefore 1/; is injective. To show that 1/; is also surjective, fix s ES and, for each i,j, put

= L VkiSVjk

aij

k

Then, for all

Vrt,

we have

= L VrtVkiSVjk = VriSVjt

Vrtaij

k

and =

aijVrt

L VkiSVjkVrt = VriSVjt k

Hence

aij

commutes with all

1/;( (aij))

Vrt

and so

=

aij E

R. Moreover, by the foregoing,

L aijVij = L ViiSVjj i,j

i,j

(L Vii)s(L Vjj) = s, i

j

proving that 1/; is surjective. Finally, taking into account that 1/;(enMn(R)en)

= v11Sv11

and

e11Mn(R)e11 ~ R,

the result follows. • For any integer n 2': 1 and for any R-module V, let direct power of V. Proposition 12.2.

Proof.

vn denote the n-th

Let V be an R-module and let n 2': 1. Then

Fix i,j E {1, ... ,n} and define fij(v1,,,,,vn)

Jij

E EndR(Vn) by

= (O, .. ,,Vj,O, ... ,O)

Rings and Modules

82

where Vj is in the_ i-th place. Then clearly n

fijfks

= Djkfis

and

L fii = 1 i=l

If 'II' E EndR(vn) and 'lj,( V1, ... , Vn) = ('ll'l (V1 ), ... , 'll'n( Vn) ), then 'II' obviously centralizes all fij if and only if '11'1 = '11'2 = · · · = 'll'n· Thus the centralizer of {!ij} in EndR(Vn) is identifiable with EndR(V). The desired conclusion is now a consequence of Proposition 12.l. • Let V be an R-module and let n 2: 1. Then the R-modle vn can be viewed as an Mn(R)-module in the following natural way. We visualize the elements of vn as column vectors and, for each A E Mn(R) and X E vn, we define Ax as the matrix multiplication. We are now ready to prove the following important result. Proposition 12.3. (i) The map W r-+ wn is an isomorphism from the lattice of submodules of an R-module V onto the lattice of Mn(R)-submodules of vn. Furthermore, if V =RR, then the direct sum of n copies of vn is identifiable with the left regular Mn(R)-module. (ii) EndMn(R)(Vn) ~ EndR(V). (iii) The map V r-+ vn induces a bijective correspondence between the isomorphism classes of R and Mn(R)-modules. The inverse of this correspondence is given by W r-+ en W. Proof. (i) The map W l W" is obviously order-preserving. Thus it suffices to show that it has an inverse which is also order-preserving. Consider the projection x 1 : vn ----. V on the first factor and, for any submodule X of vn, put g(X) = x1(X). Then the correspondence X r-+ g(X) is order-preserving and clearly (gf)(W) = W. From the action of Mn(R) on vn, we see that X can be written in the form X = Vt for some submodule V1 of V. Thus (Jg)(X) = J(Vi) = Vt = X, proving the first assertion. The second assertion being obvious, (i) follows. (ii) Put S = Mn(R) and choose f E Ends(Vn). Then, by the nature of the action of S on vn, f has the same projections, say ).. f, on all factors. Conversely, any given 'lj, E EndR(V) determines an element of Ends(Vn) whose projections on all factors are equal to 'lj,. Thus the map f r-+ AJ provides the desired isomorphism. (iii) It suffices to verify that en vn ~ V and ( en Wt ~ W, where V and

12 Matrix rings

83

Ware Rand Mn(R)-modules, respectively. By the definition of en, e11 Vn

=V

X

OX

··· X

O~ V

Nate also that W

Because ek 1 w

= en WEB··· EB en1 W

as R-modules

= 0 if and only if enw = 0, 1 ~ k ~ n, w E W, the map

is well defined and is at least an R-isomorphism of W onto ( en Wt. Hence it suffices to show that

Since the latter is a consequence of the action of Mn( R) on ( e11 Wt, the result is verified. • For any ideal I of R, we put Mn(I)

= {(aij) E Mn(R)laij

EI for all

i,j E {l, ... ,n}}

Proposition 12.4. (i) The map I 1--+ Mn(I) is a bijection between the sets of ideals of R and Mn(R) (with inverse given by J 1--+ J n R). In particular, R is simple if and only if so is Mn(R). (ii) Mn(R)/Mn(I) ~ Mn(R/ I). (iii) If R = Ii EB · · · EB Is is a two-sided decomposition of R, then

is a two-sided decomposition of Mn(R). Furthermore, the ideal Ii is indecomposable if and only if so is Mn(I). (iv) R is artinian (noetherian) if and only if Mn( R) is artinian (noetherian). Proof. (i) It is clear that Mn(I) is an ideal of Mn(R) and that the given map is injective. Assume that J is an ideal of Mn(R) and put I= JnR. Then I is an ideal of R and we claim that J = Mn(I) which will prove (i).

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84

Assume that ·Li,j rijeij E J for some rij E R. Then, for all k, s, t E {1, ... ,n}, Li,j rijekseijetk E J. Hence rstekk E J and so n

rst

= L rstekk

E J

k=l

for all s, t E {1, ... , n }, as required. (ii) The natural homomorphism R -+ Rf I induces a surjective homomorphism Mn(R)-+ Mn(Rf I) whose kernel is Mn(I). (iii) This follows from the facts that Z(Mn(R)) = Z(R) and that, for any r ER, Mn(R)r = Mn(Rr). (iv) Apply Proposition 12.3 (i). • Proposition 12.5. For any ring R and any positive integer n, (i) J(Mn(R)) = Mn(J(R)). (ii) Soc(Mn(R)) = Mn(Soc(R)). Proof. (i) Let V be an R-module and let vn be the Mn(R)-module as in Proposition 12.3. Then ann(Vn) = Mn(ann(V)), by the definition of vn. Now apply Propositions 12.3 and 5.15 (i). (ii) Let V =RR. Then, by Proposition 12.3 (i), the direct sum nVn of n copies of vn is the regular Mn(R)-module. Hence Soc(Mn(R)) = Soc(nVn). On the other hand, by Proposition 5.5 (iii) and 12.3 (i),

Soc(nVn)

= nSoc(Vn) = n(Soc(Rt) = Mn(Soc(R))

as required. • Corollary 12.6. Let R be a finite direct product of full matrix rings over division rings. Then R is semiprimitive artinian. Proof. That R is semiprimitive follows from Propositions 12.5 (i) and 5.24. To prove that R is artinian, we may assume, by Corollary 11.5, that R = Mn(D) for some n ~ 1 and some division ring D. Now apply Proposition 12.4 (iv). • Corollary 12.7. Assume that R = Re 1 EB··· EB Ren for some idempotents e = e1, e2, ... , en of R such that Re~ Rei for all i E {1, ... , n }. Then (i) R ~ Mn( eRe ). (ii) fl~ Mn(efle), where e is the image of e in fl= Rf J(R).

12 Matrix rings

Proof.

85

(i) By Propositions 9.3 and 12.2, we have

as required. (ii) By (i) and Propositions 12.4 (ii) and 12.5 (i), R ~ Mn(eRe/J(eRe)). Since, by Proposition 5.29 (ii), J(eRe) = eJ(R)e, the desired conclusion follows by virtue of Lemma 2.5 (iv). • Proposition 12.8. Let {eijli,j = 1,2, ... , n} be a full system of matrix units in a ring A and let R be the centralizer of {eij} in A. {i) A = Ae11 EB · · · EB Aenn and Aeii ~ Aejj for all i, j. {ii) End A ( Aeii) ~ R 0 , 1 :S i :S n. In particular, Aeii is indecomposable (respectively, strongly indecomposable) if and only if O and 1 are the only idempotents of R (respectively, if and only if R is local). Conversely, if A = Ii EB · · · EB In is a direct decomposition of AA and the A-modules Ii and Ij are isomorphic for all i,j, then there exists a full system of matrix units { €ij li,j = 1, ... , n} such that Ii = Aeii, 1 :S i :S n. Proof. (i) This is a direct consequence of Proposition 3.10 (i) and Lemma 2.6. (ii) By (i) together with Propositions 12.1 and 9.3, we have

Conversely, assume that A = 11 EB··· EB In with Ji ~ Ij for all i,j. Then there exist orthogonal idempotents €jj, j = 1, ... , n, such that Ij = Aejj for each j. Since Ii ~ Ij, j = 1, ... , n, it follows from Lemma 2.6 that there exist e1j,ej1 such that e11e1j€jj = e1j, €jjej1e11 = ejr, e1jej1 == en and ejl e1j = e.ii. If we set eij = ei1 e1j, i, j = 1, ... , n, then a straightforward calculation shows that {eij Ii, j = 1, ... , n} is a full system of matrix units, as desired. • Proposition 12.9. Let I be an ideal of a ring R with I~ J(R) and assume that R has two systems {eijli,j = 1, ... ,n} and {fijli,j = 1, ... ,n} of matrix units such that eij = fij( mod I) for all i,j E {1, ... , n}. Then there exists x E J such that

(1-x)- 1eij(l-x)=fij

forall

i,jE{l, ... ,n}

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86

Proof. Put e::,: Ef= 1 eii, f = I:~ 1 Iii, and x = e + f - ef - Ek=l ekiflk, For each r E R, let f be the image of r in R/ I. Then €ij = Iii and e = J for all i,j E {1, ... ,n}. Hence x = 0 and so x E J. Furthermore,

which shows that to show.•

€ij(l-

x)

= (1- x )fij for all i,j E {1, ... , n}, as we wished

Proposition 12.10. Let I be an ideal of a semiregular ring R with I~ J(R), let {Eijli,j = 1, ... ,n} be a system of matrix units in R = R/I and, for each x E R, let x be the image of x in fl. Then (i) There exists a system {eiili,j = 1, ... ,n} of matrix units in R such that for all i, j E { 1, ... , n} (ii) If

I= £11 + · · · + Enn,

then 1 =en+···+ enn·

Proof. By hypothesis, R is semiregular and £11, ... , Enn are pairwise orthogonal idempotents of R. Hence, by Corollary 6.15 (iii), there exist pairwise orthogonal idempotents e1, . .. , en of R such that ei = €ii, 1 ::S: i ::S: n, and such that if I= £11 + · · · + Enn, then 1 = e1 +···+en, We next note that

and Applying Proposition 5.31 (ii) (with e = e1, f = ei, i =/. 1, a = €1i, f3 = Ei1), we may find e1i, ei1 E R with €i1 = £1i, €i1 = Ei1 and such that for all i =/. 1,

and Now put en= e1 and eij = ei1e1j for (i,j) =/. (1, 1). Then €ii= ei for every i, and {eijli,j = 1, ... ,n} is a desired system of matrix units in R, as can be readily verified. • The following result is essentially due to Azumaya (1951).

12 Matrix rings

87

Theorem 12.11. Let I be an ideal of a semiregular ring R with I ~ J(R) and assume that Rf I~ Mn(T) for some ring T and some n ~ l. Then there exists a subring S of R such that:

{i) R

~

Mn(S).

(ii} Sf(I n S) ~ T. {iii) Sf J(S) ~ T if I= J(R).

Proof. By hypothesis, there exists a full system of matrix units { cij I1 ~ i, j ~ n} in .R = Rf I such that the centralizer K of {€ij} in .R is isomorphic to T. Hence, by Proposition 12.10, there exists a full system {eijll ~ i,j ~ n} of matrix units in R with eij + I = €ij for all i, j. Setting S to be the centralizer of {eij} in R, it follows from Proposition 12.1 that R = ffii,j S eij ~

Mn(S). Now R = ffii,jSeij and, by Proposition 12.4 (i), I = ffii,j(I n S)eij, so Rf I= ffii,j[(S + I)f I]( eij + I). On the other hand, since eij +I= €ij for all i,j, we have Rf I= ffii,jK(eii + I) and (S + I)f I~ K. Thus (S + I)f I= K and therefore Sf(In S) ~ K ~ T. Finally, if I= J(R) then, by Proposition 12.5 (i), In S = J(S) as required. •

Chapter 2

Artinian and Semilocal Rings The aim of this chapter is to record a number of basic properties of artinian and semilocal rings. These properties will frequently arise in our subsequent investigations. The chapter culminates with a link between the structure of semilocal algebras and the endomorphism rings of their simple modules.

1

Semiprimitive artinian rings

In this section, we provide various characterizations of semiprimitive artinian rings. Recall that a ring R is said to be semisimple if the regular module RR is semisimple.

Theorem 1.1. For any ring R, the following conditions are equivalent: (i) R is semisimple. (ii) R is semiprimitive artinian. (iii) R is regular noetherian. (iv) Every R-module is semisimple. (v) RR is a finite direct sum of simple submodules. Proof. The equivalence of (ii) and (v) follows from Proposition 1.5.10 applied to V =RR, while the equivalence of (i) and (iv) is the content of Corollary 1.4.4. That (i) is equivalent to ( v) is a consequence of Corollary 1.4.3. Thus (i), (ii), (iv) and (v) are equivalent. If R is noetherian, then every left ideal of R is finitely generated. Hence, by Lemma 1.6.12, (i) and

89

Artinian and Semilocal Rings

90

(iii) are equivalent, as desired. • The following celebrated theorem provides the structure of rings satisfying the equivalent conditions of the preceding result. Theorem 1.2. (Wedderburn - Artin). Let R be a semiprimitive artinian ring. (i) There exist only finitely many, say Vi, ... , Vr, nonisomorphic simple R-modules and furthermore ¼ 3! Rei for some primitive idempotent ei of R, 1 ~ i ~ r. (ii} RR 3! ni Vi EB • • • EB nr Vr for some positive integers ni, 1 ~ i ~ r, where ni ¼ is a direct sum of ni copies of¼. (iii} R 3! ili=i Mn;(Di), where Di = EndR(¼) 0 3! eiRei is a division ring, 1 ~ i ~ r. (iv) If R 3! ili=i Mkj(Tj) where each Ti is a division ring, then after possible renumbering of the direct factors, r = s, ni = ki and Di ~ Ti for all iE{l, ... ,r}. Proof. (i) and (ii). By Corollary 1.3.12, there exists a finite set { e1, .. . , en} of pairwise orthogonal primitive idempotents of R such that

RR

= Rei EB • • • EB Ren

Furthermore, by Theorem 1.1, each Rei is a simple R-module. We may assume that Rei, . .. , Rer, 1 ~ r ~ n, are all nonisomorphic modules of the set {Rei, ... ,Ren}, If Vis any &imple R-module, then by the JordanHolder theorem (Corollary 1.3.15) and Lemma 1.5.13, V 3! Rei for some i E {1, ... ,r}. This establishes (i) and (ii). (iii) Put Di = EndR(¼ )0 , 1 ~ i ~ r. Then we have

R0

3!

EndR(RR)

(by Proposition 1.9.3)

r

~

IT EndR(ni¼)

(by Corollary 1.4.9)

i=i r

~

IT Mn;(D'!)

i=l and thus

r

R 3!

IT Mn;(Di)

i=i

(by Proposition 1.12.2)

1 Semiprimitive artinian rings

91

Since, by Proposition 1.9.3, EndR(¼) ~ (eiRei) 0 we have Di ~ eiRei, 1 ::S; i ::S: r. Finally, by Schur's lemma, each Di is a division ring, proving (iii). (iv) Assume that R ~ TIJ=t Mk)Tj), where each Tj is a division ring. Then, by Proposition 1.11.4 (i), there exist central idempotents u 1 , ••• , Us of R such that R = Ru1 EB · · · EB Rus and Ruj ~ Mk;(Tj), 1 ::S: j ::S: s. Hence, by Proposition 1.12.8, Ruj is a direct sum of kj copies of a simple R-module. It is clear that Rui, Ruj, i =/ j, have no common composition factors. Thus Ru 1, . .. , Ru 8 are all homogeneous components of the semisimple R-module RR. Hence r = s and, by renumbering the Uj, Rui ~ ni ¼, 1 ::S; i ::S; r. The latter implies, by the Jordan-Holder theorem, that ki = ni for all i E {1, ... ,r}. Finally, fix i E {1, ... , r }, put A = Mn/Ti) and let V be a simple Amodule. Then, by Proposition 1.12.8, EndA(V) ~ Ti° and, since Rui ~ ni¼, V ~ ¼. Consequently, Ti° ~ EndR(¼) ~ Df and therefore Ti

~

Di, as desired. •

Corollary 1.3. For any ring R, the following conditions are equivalent: (i) R is simple artinian. (ii) R is primitive artinian. {iii) R ~ Mn(D) for some n ~ 1 and some division ring D. Proof. (i) ::::} (ii): This follows from the obvious fact that any simple ring is primitive. (ii) ::::} (iii): Our hypothesis implies that R is semiprimitive artinian. Hence, by Theorem 1.2, R ~ Tii=t Mn;(Di) for some r ~ 1, some division rings Di and some ni ~ 1. It follows from Proposition 1.11.4 (i) that R = I1 EB··· EB Ir for some nonzero ideals Ii of R. Hence, by Proposition 1.11.8 (iii), r = 1 as required. (iii) ::::} (i): Apply Proposition 1.12.4 (i) and Corollary 1.12.6. • Corollary 1.4. For any ring R, the following conditions are equivalent: {i) R is semiprimitive artinian. {ii) R is a finite direct product of full matrix rings over division rings. {iii) R is. a finite direct product of simple artinian rings. (iv) R admits a block decomposition and each block of R is a simple artinian ring.

Artinian and Semilocal Rings

92

Proof. The-equivalence of (i) and (ii) is a consequence of Theorem 1.2 and Corollary 1.12.6, while the equivalence of (ii) and (iii) follows from Corollary 1.3. That (iii) is equivalent to (iv) is a consequence of the fact that each simple ring is indecomposable. •

2

Semilocal rings

A ring R is said to be semilocal if Rf J(R) is artinian. Of course, any artinian ring is semilocal. By Theorem 1.1, R is semilocal if and only if Rf J(R) is semisimple. In this section, we present a number of general properties of semilocal rings. The main results are two classical theorems due to Azumaya, namely Theorems 2.9 and 2.13. Proposition 2.1. Assume that a ring R is semilocal. Then (i) An ideal I of R is primitive if and only if I is maximal. (ii) There are only finitely many maximal ideals of R and J(R) is their intersection. (iii} If Ii, ... , In are all distinct maximal ideals of R, then each Rf Ii is a simple artinian ring and the map

{ Rf J(R)-+ Tif= 1 (Rf Ii) r + J(R) i-+ (r + Ii) is a ring isomorphism. Proof. (i) If I is maximal, then J is obviously primitive. Conversely, assume that I is primitive. Then, by Proposition 1.5.15 (i), J(R) ~ I and so Rf I is primitive artinian. Hence, by Corollary 1.3, Rf I is simple which implies that I is maximal. (ii) Since every simple R-module is annihilated by J(R), it follows from Theorem 1.2 that there exist only finitely many nonisomorphic simple Rmodules. The desired conclusion is now a consequence of (i). (iii) This is a direct consequence of (ii) and Corollary 1.11.3. • Proposition 2.2. Assume that R is a commutative ring. Then R is semilocal if and only if R has only finitely many maximal ideals. Proof. 11, ...

Jn·

Assume that R has only finitely many maximal ideals, say Then J(R) = nf= 1 Ji and, by Corollary 1.11.3, Rf J(R) ~ Tif= 1 (Rf Ii).

2 Semilocal rings

93

But any simple commutative ring is a field and so Rf J(R) is artinian. The converse being a consequence of Proposition 2.1 (ii), the result follows. Ill Proposition 2.3. Let A be an algebra over a commutative semilocal ring S such that A is a finitely generated S -module. Then (i) Af J( S)A is artinian. (ii} A is semilocal. (iii) J(Ar ~ J(S)A for some n 2'. 1. Proof. (i) Since A= Af J(S)A is a finitely generated Sf J(S)-module and Sf J(S) is artinian, it follows from Corollary 1.3.8 that A is artinian. (ii) By Proposition 1.5.26, Af J(A) ~ Af J(A) and so, by (i), A is semilocal. (iii) By Proposition 1.5.26, J(A) = J(A)f J(S)A and , by Proposition 1.5.25, J(A) is nilpotent. Thus J(Ar ~ J(S)A for some n 2'. 1, as we wished to show. II Proposition 2.4. Let V be a module over a semilocal ring R. Then (i} J(R)V = J(V). (ii} V is semisimple if and only if J(R)V = 0. (iii} J(R)V is the unique minimal submodule W of V such that VfW is semisimple. (iv) Soc(V) = {v E VJJ(R)v = O}. Proof. (i) By Proposition 1.5.18, J(R)V ~ J(V). To prove the opposite inclusion, it suffices, by Corollary 1.5.4 (ii), to prove that J(Vf J(R)V) = 0. Since R is semilocal, it follows from Theorem 1.1, that every Rf J(R)module is semisimple. In particular, the Rf J(R)-module Vf J(R)V is semisimple. Hence V f J(R)V is a semisimple R-module and so, by Proposition 1.5.1 (iv), J(VfJ(R)V) = 0. (ii) If Vis semisimple, then J(R)V = 0 by Proposition 1.5.15 (i). Conversely, assume that J(R)V = 0. Then V can be regarded as a module over the semiprimitive artinian ring Rf J ( R). Hence, by Theorem 1.1, V is a semisimple Rf J(R)-module. Thus Vis a semisimple R-module, as required. (iii) Since J(R)(Vf J(R)V) = 0, it follows from (ii) that Vf J(R)V is semisimple. Let W be any submodule of V with VfW semisimple. Then J(R)(VfW) = 0 and so J(R)V ~ W, as desired. (iv) Setting W = {v E VJJ(R)v = O}, we see that Wis a submodule of V

Artinian and Semilocal Rings

94

such that J(R)W ,= 0. Hence, by (ii), Wis semisimple and so W ~ Soc(V). On the other hand, Soc(V) is semisimple and so J(R)Soc(V) = 0. Thus S oc(V) ~ W as desired. • Corollary 2.5.

Let R be a semilocal ring. Then Soc(R)

Proof.

= {r E RIJ(R)r = O}

Apply Proposition 2.4 (iv) for V =RR. •

Lemma 2.6. Let I be an ideal of a semiprimitive artinian ring R. Then In = I for all n ~ 1 and R/ I is semiprimitive artinian. Proof. By Theorem 1.1 (iii), I= Re for some idempotent e of R and hence = I for all n ~ 1. Let J/I = J(R/I). Since R is artinian, so is R/ I and therefore J / I is nilpotent (Proposition 1.5.25). Hence Jn ~ I for some n ~ 1. But Jn = J, so J/I = 0 as required.•

r

Proposition 2. 7. Assume that R is a semilocal ring. {i) If f : R -+ S is a surjective ring homomorphism, then

J(S)

= f(J(R))

{ii) For any ideal I of R, J(R/I)

= (J(R) + I)/I

Proof. (i) By Proposition 1.5.11, f(J(R)) ~ J(S). Hence, by Corollary 1.5.12 (iv), it suffices to show that J(S/ f(J(R)) = 0. But S/ f(J(R)) is isomorphic to a factor ring of a semiprimitive artinian ring R/ J(R). Hence, by Lemma 2.6, S/ f(J(R)) is semiprimitive, as required. (ii) This is a direct consequence of (i). • Proposition 2.8. Let I be an ideal of a ring R, let n be a positive integer and let e =f. 0 be an idempotent of R. {i) If R is semilocal, then so are R/ I and eRe. (ii) R is semilocal if and only if so is Mn(R). Proof. (i) Assume that R is semilocal. Setting S = R/ I, it follows from Proposition 2. 7 (ii) that S/ J ( S) is a homomorphic image of the artinian ring

2 Semilocal rings

95

Rf J(R). Thus Rf I is semilocal. Now let e be the image of e in (ii) and Lemma 1.2.5 (iv),

fl= Rf J(R). Then, by Proposition 1.5.29

eRef J(eRe) ~

efle

Since fl is semiprimitive artinian, so is efle. Hence eRe is semilocal, as required. (ii) By Propositions 1.12.4 (ii) and 1.12.5 (i), Mn(R)fJ(Mn(R)) ~ Mn(RfJ(R))

The desired conclusion is now a consequence of Proposition 1.12.4 (iv). • Theorem 2.9. (Azumaya {1951)). Let R be a semilocal ring, let {eijli,j 1, ... ,n} and {fiili,j = 1, ... ,n} be two systems of matrix units of Rand let n

e=

I:eii i=l

n

and

f =

I:!ii i=l

If the R-modules Re and Rf are isomorphic, then there exists a unit u of R such that u- 1 eijU = fij for all i,j E {1, ... ,n} Proof. For any x E R, let x be the image of x in fl = Rf J(R). Then {eij} and {fij} are two systems of matrix units of fl and, by Proposition 1.5.31 (ii), Re ~ Rf. Assume that the result holds for fl. Then, by Corollary 1.5.23 (ii), there exists a unit v of R with v- 1 eijV = fij for all

i,j E {1, ... , n}. Hence, by Proposition 1.12.9, there exists a unit w of R such that w- 1 cv- 1 eijV)w = fij for all i,j. Setting u = vw, the required assertion follows. By the foregoing, we may assume that J(R) = O, i.e. that R is semiprimitive artinian. Now

and, by Lemma 1.2.6, Ren ~ Reii and Rfu ~ Rfii for all i E {1, ... , n }. Hence, by the Jordan-Holder theorem, Ren ~ Rfu. Furthermore, since R is semisimple, Re ~ Rf and R = Re EB R(l - e) = Rf EB R(l - J)

=

Artinian and Semilocal Rings

96

which implies R(J - e) ~ R(l - f) (again, by the Jordan-Holder theorem). Hence, by Lemma 1.2.6, there exist a1 E enRfn, b1 E f11Re11 and a2 E (1 - e)R(l - !), b2 E (1 - J)R(l - e) such that

and Now put

n

u

= L ei1aif1i + a2

n

and

v

i=l

= L fi1b1e1i + b2 i=l

Then an easy verification shows that uv = vu = 1 and eijU =

= ufij

ei1 aif1j

for all

i,j E {1, ... , n}

as required. • Corollary 2.10. Let R be a semilocal ring, let {eijli,j = 1, ... ,n} and {fijli,j = 1, ... , m} be two full systems of matrix units of R and let C and D be the centralizers of {eij} and {fij}, respectively. Then m = n and there exists a unit u of R such that U

-l

eijU

= f ij

for all

i, j E { 1, ... , n}

and Proof. The last assertion is a consequence of the preceding assertions. By Theorem 2.9 (with e = f = 1), it suffices to show that m = n. We may clearly assume that J(R) = 0, i.e. that R is semiprimitive artinian. Then

and, for all i E {1, ... , n }, j E {1, ... , m },

whence m

= n, as required.

•

Corollary 2.11. Let R and S be semilocal rings and let n, k be positive integers such that Mn(R) ~ Mk(S). Then n = k and R ~ S.

2 Semilocal rings

Proof.

97

Apply Corollary 2.10 and Proposition 2.8 (ii). •

Corollary 2.12. Let R = Mn(S) where S is a commutative semilocal ring. Then every S-endomorphism of the S-algebra R is an inner automorphism. Proof. Let f be an S-endomorphism of the S-algebra R, let {eijli,j = 1, ... ,n} be the matrix units of Rand let Jij = f(eij), 1::; i,j::; n. By Corollary 2.10, there exists a unit u of R such that

u- 1 eijU Hence, for x

= fij

= E Sijeij, Sij

for all

i,j E {1, ... ,n}

E S, we have

as desired. •

In what follows,

vn denotes the n-th direct power of V.

Theorem 2.13. (Azumaya {1951)). Let V be a module over an arbitrary ring R such that EndR(V) is semilocal, let W be any R-module and let n 2: 1. If Vn ~ wn, then V ~ W. Proof. Our assumption implies the existence of an R-module U with U = V1 EB · · · EB Vn = W1 EB · · · EB Wn where ½ ~ V, Wi ~ W for all i E {1, ... , n}. By the proof of Proposition 1.12.2, we may find two full systems of matrix units { eij} and {Jij} in EndR( U) such that eii(U) = ½, fii(U) = Wi

(1 ::; i ::; n)

Since EndR(U) ~ Mn(EndR(V)) is semilocal (Proposition 2.8 (ii)), it follows from Corollary 2.10 that there exists a unit u of EndR(U) such that u- 1 eii~

= !ii

for all

i E {1, ... , n}

But then, by Corollary 1.2.8, EndR(U)eii ~ EndR(U)fii and so, by Proposition 1.9.4, ½ ~ Wi for each i E {1, ... , n }. • Proposition 2.14. Let R be a semilocal ring. Then (i) Every nonzero finitely generated R-module is a finite direct sum of

Artinian and Semilocal Rings

98

indecomposable R~modules. (ii) Every nonzero idempotent of R is a sum of pairwise orthogonal primitive idempotents. In particular, by Proposition 1.11.11, R admits a block decomposition. {iii) If I is a left or right ideal of R and e E I is a nonzero idempotent, then I contains a primitive idempotent.

Proof. (i) Let Vi Obe a finitely generated R-module. By Nakayama's lemma, J(R)V i V. Hence, by Theorem 1.1, VJ J(R)V i O is a finitely generated semisimple Rf J(R)-module. It follows, from Corollary 1.4.3, that V/J(R)V is a finite direct sum of, say n, simple R/J(R)-modules. Now assume that

V

= Ef:li=1¼

where each¼ is a nonzero submodule of V. By the foregoing,

VI J(R)V ~ Ef:)i=l ¼/ J(R)¼ where each ¼/ J(R)¼ i O and is a finite direct sum of simple Rf J(R)modules. Since the composition length of VJ J(R)V is equal ton, we deduce that r :'.S: n. Thus Vis a finite direct sum of indecomposable R-modules. (ii) Let e i O be an idempotent of R. Then, by (i), Re= Vi Ef:) · · ·Ef:l Vn for some finitely many indecomposable submodules Vi, ... , Vn of Re. Writing e = e1 + ··· + en with ei E ¼, it follows from Proposition 1.3.10 (ii), (iii) that e1, ... , en are pairwise orthogonal primitive idempotents of R. (iii) By (ii), we may write e = e1 +···+en where ei, ... , en are pairwise orthogonal primitive idempotents of R. Since e E I and ee 1 = e1e = e1, it follows that e1 E I as required. • Our final aim is to show that for any ideal I of a semilocal ring R, the units of Rf I can be lifted to R. This will be achieved with the aid of the following two observations.

Let V be a finite-dimensional right vector space over a division ring D and let R = Endn(V). Given a right ideal I of R, put Lemma 2.15.

IV= {. E R satisfies I+ >.R = R, then I + >. contains a unit of R. Proof. By Corollary 1.5.23 (ii), an element of Risa unit if and only if it is a unit modulo J(R). Hence, after passing to R/ J(R), we may harmlessly assume that J(R) = 0. Then Risa finite direct product of full matrix rings over division rings. Therefore, we can further assume that R = Endv(V), where V is a finite-dimensional right vector space over a division ring D. In this case, it follows from Lemma 2.15 that

I

= {

.(V) = V. Choose subspaces Vo~ IV and U ~ V such that

V = Vo EB >.(V) = Ker>. EB U Then >. induces an isomorphism from U to >.(V), so Ker>. ~ Vo. Choose r E R so that r( U) = 0 and r induces an isomorphism from Ker>. to Vo. Then r(V) = V0 ~ IV, so r E I. Moreover, r + >. E I+ >. is clearly an automorphism of V. • Corollary 2.17. If I is an ideal of a semilocal ring R, then the natural homomorphism U(R)---+ U(R/ I) is surjective. Proof. Assume that >. E R is invertible modulo R. Then I+ >.R = R and so, by Lemma 2.16, I+ >. contains a unit of R, as required. •

100

3

Artinian and Semilocal Rings

Artinian rings

This section provides some preliminary information on the special class of semilocal rings, namely artinian rings. The results presented here are of an elementary nature. Some deeper properties of artinian rings will be exhibited in our future treatment of a more general class of rings, namely semiperfect rings.

A. A characterization The following result provides an important characterization of artinian rings.

Theorem 3.1. For any ring R, the following conditions are equivalent: {i) R is artinian. (ii) R is semilocal noetherian and J( R) is nilpotent. Proof. (i) => (ii): By Proposition 1.5.25, J(R) is nilpotent. Since R is artinian, it is also semilocal. To prove that R is noetherian, we argue by induction on the nilpotency index n of J(R). If n = 1, then J(R) = 0 and so, by Theorem 1.1, R is noetherian. Consider the exact sequence 0-+ J(Rr- 1 -+ R-+ Rf J(Rr- 1 -+ 0

Since, by Corollary 1.5.12 (ii), J(Rf J(Rt- 1 ) = J(R)f J(Rt-1, it follows from the induction hypothesis that Rf J(Rt- 1 is noetherian. Hence, by Proposition 1.3.5, it suffices to verify that J(Rt- 1 is noetherian. Since J(Rt- 1 is a submodule of the artinian module RR, J(Rt- 1 is artinian. Furthermore, since J(Rt = 0, it follows from Proposition 2.4 (ii) that J(Rt- 1 is semisimple. Thus, by Corollary 1.4.3, J(Rt- 1 is noetherian, as required. (ii) => (i): We argue by induction on the nilpotency index n of J(R). If n = 1, then there is nothing to prove. By induction hypothesis, Rf J(Rt- 1 is artinian. On the other hand, J(R)n-l is noetherian since it is a submodule of the noetherian module RR, Because J(Rt = 0, and R is semilocal, it follows from Proposition 2.4 (ii) that J(Rt- 1 is semisimple noetherian. Thus, by Corollary 1.4.3, J(Rt- 1 is artinian. Hence, by Proposition 1.3.5, R is artinian. •

3 Artinian rings

101

Corollary 3.2. Let V -:/ 0 be a finitely generated module over an artinian ring R. Then (i) V is artinian and noetherian (hence, by Proposition 1.3.16, V has a composition series). (ii} V has the unique decomposition property. Proof. (i) Apply Theorem 3.1 and Corollary 1.3.7. (ii) Apply (i) and Theorem 1.10.14. a Corollary 3.3. Let V be a finitely genemted indecomposable module over an artinian ring R. Then V is strongly indecomposable.

Proof.

Apply Corollaries 3.2 (i) and 1.10.13. •

B. Principal indecomposable modules Our main purpose is to tie together principal indecomposable and simple modules of artinian rings. Proposition 3.4.

Let R be an artinian ring and, for each x E R, let

x be the image of x in fl= Rf J(R). (i) An idempotent e of R is primitive if and only if e is a primitive idempotent of fl. (ii} For any primitive idempotent e of R, Ref J( R)e is a simple R-module isomorphic to Re. In particular, J(R)e is a unique maximal submodule of Re. (iii} For any two idempotents e and f of R, Re~ Rf{:=> RefJ(R)e

~

Rf fJ(R)f

(iv) There exist pairwise orthogonal idempotents e1, ... , en of R with 1 = e1 + ··· + en. Furthermore R = EElf= 1 Rei and R = EElf= 1 Rei where each Rei is an indecomposable R-module and each Rei is a simple fl-module. ( v) If Re 1, ... , Rer, r :s; n are all non isomorphic modules of the set { Re 1, ... , Ren}, then Re 1, ... , Rer are all nonisomorphic principal indecomposable R-modules and ReifJ(R)e1, ... ,RerfJ(R)er are all nonisomorphic simple R-modules.

Artinian and Semilocal Rings

102

Proof. (i) ~ince J(R) is nilpotent, it follows from Lemma 1.6.1 that every idempotent of R/ J(R) can be lifted to R. Now apply Theorem 1.6.3

(i). (ii) By Lemma 1.2.5 (iii), Re/ J(R)e ~ Re and, by (i), Re is a simple R-module. Hence J(R)e is a maximal submodule of Re. But, by Proposition 1.5.30, J(Re) = J(R)e and so J(R)e is a unique maximal submodule of Re. (iii) By Lemma 1.2.5 (iii), Re/ J(R)e ~ Re and Rf/ J(R)f ~ Rf. Now apply Proposition 1.5.31 (ii). (iv) Apply Corollary 3.2 (ii) (with V =RR); Proposition 1.3.10 and (ii). (v) This is a consequence of (ii), (iii) and Corollary 3.2 (ii). • Proposition 3.5. Let V be a finitely generated module over an artinian ring R and let e be a primitive idempotent of R. Then the following conditions are equivalent: (i) Re/J(R)e is a composition factor of V. (ii) H omR(Re, V) =J 0. (iii) eV =J 0. Proof. The equivalence of (ii) and (iii) is a consequence of Proposition 1.9.2 (i). Assume that (ii) holds and let / : Re -+ V be a nonzero R-homomorphism. If I= Ker/ and W = Imf, then Re/I~ Wand, by Proposition 3.4 (ii), I ~ J(R)e. Hence Re/ J(R)e is isomorphic to a factor module of W, proving (i). Finally, if eV = 0 then eW = 0 for any composition factor W of V. Hence Re/ J(R)e cannot be a composition factor of V, as desired. • Corollary 3.6. Let e be a primitive idempotent of an artinian ring R and let f be an arbitrary idempotent of R. Then eRf -/- 0 if and only if Re/ J(R)e is a composition factor of Rf. Proof.

Apply Proposition 3.5 for V

= Rf. •

C. Blocks and Cartan matrices Proposition 3.7. Let R be an artinian ring. Then R admits a block decomposition and, for any two primitive idempotents e, f of R, the following conditions are equivalent: (i) e and f are linked.

3 Artinian rings

103

{ii) e and f lie in the same block. (iii) There exists a finite sequence e

= fi,f2, ... ,Jm = f

of primitive idempotents of R such that for each j E { 1, ... , m - 1}, RJj and Rfj+i have a common composition factor. Proof. By Propositions 3.4 (iv) and 1.11.11 (i), R admits a block decomposition. Furthermore, by Proposition 1.11.11 (iii) and Corollary 3.2 (ii), (i) and (ii) are equivalent. The equivalence of (ii) and (iii) being a consequence of Corollary 3.6, the result follows. •

The following notation will be fixed for the remainder of this section. Let R be an artinian ring and let e1, ... , er be primitive idempotents of R such that Re1, ... , Rer are all nonisomorphic principal indecomposable Rmodules. Then, by Proposition 3.4 (v), Rei/J(R)e1, ... ,Rer/J(R)er are all nonisomorphic simple R-modules. Given i,j E {1, ... , r }, let Cij be the multiplicity of Re1 /J(R)e 1 as a composition factor of Rei. The nonnegative integers Cij are called the Cartan invariants of R, and the r x r matrix 1:'.Si,j:Sr is called the Cartan matrix of R. The Cartan matrix C is determined only up to a permutation of its rows and columns ( depending on the numbering of e1, ... , er). Note that, by defnition, Cii ~ 1 for all i E {1, ... , r }. Observe also that if R is semiprimitive artinian, then Cii = 1 for each i and Cij = 0 for i f: j, i.e. C is the identity matrix. The next simplest case is recorded in the following observation. Proposition 3.8. Let C be the Cartan matrix of an artinian ring R. Then the following conditions are equivalent: {i) C is a diagonal matrix. {ii} Each block of R contains a unique isomorphism class of principal indecomposable R-modules. (iii} Each block of R contains a unique isomorphism class of simple Rmodules. (iv) For each block B of R, B/J(B) is a simple ring. Proof. (i) ::} (ii): Since C is a diagonal matrix, Rei/ J(R)ei is unique, up to isomorphism, composition factor of Rei, 1::; i::; r. Thus, for all if: j,

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in {1, ... , r }, Rei and Rej have no composition factor in common. It follows that ei, ej are linked if and only if Rei~ Rej. The desired conclusion is now a consequence of Proposition 3.7. (ii) => (iii): Let B be a block of R and let Red J(R)ei, Rej/ J(R)ej belong to B for some i, j E { 1, ... , r}. Then Rei and Re j belong to B, hence i = j as asserted. (iii) => (iv): Let Vi and Vi be simple B-modules. Then Vi and Vi are simple R-modules belonging to B. Hence Vi ~ Vz as R-modules and therefore Vi~ Vz as B-modules. Thus B/J(B) is a semiprimitive artinian ring whose all simple modules are isomorphic. Hence, by Theorem 1.2, B / J(B) is simple. (iv)=> (i): Suppose that Rej/J(R)ej is a composition factor of Rei for some i,j E {1, ... , r }. Then Rei and Rej have a common composition factor Rej/J(R)ej, so Rei and Rej are linked and therefore belong to the same block, say B. Hence Reif J(R)ei and Rej/J(R)ej are simple B-modules and so they are isomorphic as B-modules. It follows that Red J(R)ei ~ Rej/ J(R)ej as R-modules, proving that i = j. Thus C is a diagonal matrix, as desired. • Proposition 3.9. Let B1, ... , Bm be all blocks of an artinian ring R and let C = (Cij) be the Cartan matrix of R. {i) Cij -=/- 0 if and only if ejRei -=/- 0. (ii} Upon renumbering the e1, ... , er, if necessary, we have

C

= diag(C1,C2, ... ,Cm)

where Ci is the Cartan matrix of Bi, 1 ::; i ::; m. {iii} Each Ci cannot be written in the form diag( Ci1, Ci2) with square matrices Ci1 and Ci2. Proof. (i) This is a direct consequence of Corollary 3.6. (ii) It suffices to show that if Rei and Rej, 1::; i,j::; r, do not lie in the same block, then Cij = 0. Suppose that Rei ~ B and Rej ~ B' for some distinct blocks B and B' of R. Then

eJ·Re·i _ C (Re J·)(Re·) CB'· B i -

=0

and so, by (i), Cij = 0. (iii) Fix i E {1, ... ,m} and assume by way of contradiction that an ordering of the ej in Bi is possible for which C = diag( Ci1, Ci2) with square

3 Artinian rings

105

matrices Ci1 and Ci2· Then we have an arrangement e1, ... ,ek, ek+1, ... ,e8 of primitive idempotents in Bi such that if i ::; k < j ::S s, then Cij = Cji = 0. By Proposition 3.7, e1 and e2 are linked; hence there are integers n,t such that n ::S k < t ::S s and Ren, Ret have a common composition factor, say Re>..f J(R)e;..; then .X :S s. If .X::; k, then Ct>. i- 0, a contradiciton. If .X > k, then Cn>. i- 0, a contradiction. • Proposition 3.10. Let C be the Cartan matrix of an artinian ring R. If detC i- 0, then any two finitely generated projective R-modules with the same composition factors (counting multiplicities) are isomorphic. Proof. Let e 1, ... , er be primitive idempotents of R such that Re 1, ... , Rer are all nonisomorphic principal indecomposable R-modules. Put Pi = Rei and ½ = Rei/ J(R)ei, 1 :S i :S r. Then, by Proposition 3.4 (iv), V1, ... , Vr are all nonisomorphic simple R-modules. If Vis a finitely generated projective R-module then by Corollary 3.2 (ii), V can be written uniquely as

where niPi is a direct sum of ni ~ 0 copies of Pi. To each R-module V, we now associate f (V) = (n1, ... , nr) where EBi=l ni ½ is the direct sum of the composition factors of V. Then C is defined by the property f( EBi=l niPi) = (n1, ... , nr )C Since, by hypothesis, C is invertible (over CQ) we deduce that f is one-to-one on finitely generated projective R-modules, as desired. • D. The Loewy and socle series

Throughout, R denotes an artinian ring and all R-modules are assumed to be finitely generated. For any ideal I of R, we put I 0 = R. Let V i- 0 be an R-module. Then the descending chain V 2 J(R)V 2 J(R)2V 2 · · ·

of submodules of V is called the Loewy series of V. Because J(R) is nilpotent, there is an integer k ~ 1, called the Loewy length of V such that J(R/- 1v i- O but J(RlV = 0

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Note that if J(R)iV = J(R)i+l V for O ~ i < k, then by Nakayama's lemma J(R)iV = 0, implying J(Rl- 1V = 0, a contradiction. The conclusion is that, if k is the Loewy length of V, then

V :J J(R)V :J J(R)2V :J · · · :J J(Rt- 1v

::::> 0

The semisimple R-modules

V/ J(R)V, J(R)V/ J(R) 2V, • • •, J(Rl- 1V/ J(Rtv

= J(R)k-l V

are called the Loewy factors of V. Let us write

where ¼1 , ... , Vir; are simple R-modules. Then the Loewy factors of V are described by the two following array of simple R-modules:

Vir1 ½r2

V11, V21,

Again, let V

f-

0 be an R-module. Put

So(V) and, for each i

~

= 0, S1 (V) = S oc(V)

2, define the i-th socle Si(V) of V by

Si(V)/ Si-1(V)

If V =RR, then we put Si( R) of R. Lemma 3.11.

Let V

f-

= Soc(V/ Si-1(V))

= Si(V)

and refer to Si( R) as the i-th socle

0 be an R-module. Then for all i

~

0

Proof. The case i = 0 being trivial, we argue by induction on i. By Proposition 2.4 (iv) and the induction hypothesis, we have

Si(V)

{v E VIJ(R)v ~ Si-1(V)} {v E VIJ(Ri- 1(J(R)v) = O} {v E VIJ(R)iv = O}

3 Artinian rings

107

as required. • Let V -::J O be an R-module. Then, by Lemma 3.11, there exists an integer n 2: 1 such that

Sn-1 (V) -::/ V

but

Sn(V)

=V

in which case we have an ascending chain: 0 C S1(V) C · · · C Sn-1(V) CV

We refer to n as the socle length of V and to the chain above as the socle series of V. Proposition 3.12. Let V -::JO be an R-module and let k and n be the Loewy and socle lengths of V, respectively. Then k = n and

J(R)iV ~ Sk-i(V)

for all i E {O, 1, ... , k}

Proof. Since J(RlV = 0, it follows from Lemma 3.11 that Sk(V) = V and son :S k. We claim that J(RiV ~ Sn-i(V) for all i E {0,1, ... ,n}; if sustained, it will follow that J(RtV = So(V) = 0 and k :S n, which will complete the proof. To prove J(R)iV ~ Sn-i(V), 0 :S i :S n, we argue by induction on i. The case i = 0 is a consequence of the fact that Sn(V) = V. If J(Ri- 1V ~ Sn-i+i(V), then by Lemma 3.11

as desired. • Proposition 3.13. Let V -::J O be an R-module, let i 2: 1 and let e be a primitive idempotent of R. Then the following conditions are equivalent: (i) Re/J(R)e is a composition factor of Si(V)/Si-1(V). (ii) There exists a submodule W of V such that W has Loewy length i and W/ J(R)W ~ Re/ J(R)e Proof. (i) => (ii): By Proposition 3.5, there exists a nonzero ~Rhomomorphism f : Re ---+ Si(V)/ Si-1 (V). Let v E Si(V) - Si-1 (V) be such that f( e) = V + si-1 (V). Then V + Si-1 (V) = ev + si-1 (V) and we put

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108

Since, J(R)iW = 0 and J(R)i- 1 ev =/- 0, W has the Loewy length i. Furthermore, the map Re-+ W/ J(R)W, re t--t rev+ J(R)W induces the desired isomorphism Re/J(R)e ~ W/J(R)W. (ii) => (i): Since W has Loewy length i, we have J(R)iW = 0 and so by Lemma 3.11, W ~ Si(V), Si-1(W) = W n Si-1(V) and J(R)W ~ Si-1(W). Furthermore, Si-i(W) =/- W by Proposition 3.12. But Re/ J(R)e is simple, so J(R)W is a maximal submodule of Wand hence J(R)W = Si-1(W). It follows that

W

= Rev.

(W

+ Si-1(V))/ Si-1(V)

~

W/(W

n Si-1(V))

W/J(R)W ~

Re/J(R)e

as required. •

4

Representations of algebras

Throughout, A denotes an algebra over a commutative ring R. Any given A-module V will also be regarded as an R-module via rv

= (r · l)v

for all

r E R, v E V

and, for any a E A, la will denote the element of EndR(V) defined by la(v)=av

for all v EV

By a representation of A, we understand a pair (V, p) where V is an Rmodule and p : A -+ EndR(V) is a homomorphism of R-algebras; by abuse of language, we shall often refer top itself as a representation of A. Two representations (Vi,p1) and (Vi,P2) of A are said to be equivalent if there exists an R-isomorphism f: Vi -+ V2 such that forall

aEA

Let V be an A-module and let pv : A-+ Endn(V) be defined by pv(a)

= la

for all

a EA

Then (V, pv) is a representation of A which will be simply denoted by pv and to which we refer as the representation of A afforded by V. The

4 Representations of algebras

109

representation of A afforded by the regular module V =A A is called the regular representation of A. In what follows, we write pv "' pw to indicate that pv is equivalent to PW. Proposition 4.1. The map V ~ pv is a bijective correspondence between A-modules and representations of A such that V ~W

if and only if Pv "' PW

Proof. It is clear that if V, Ware A-modules such that pv = pw, then V and Ware identical A-modules. Conversely, any representation (V,p) of A determines Vas an A-module via av= p(a)v for all a E A, v E V which provides the inverse of the given map. Assume that f : V ---. W is an isomorphism of A-modules. Then f is also an R-isomorphism such that for all a EA, x EV, (pw(a)J)(x)

= af(x) = J(ax) = (Jpv(a))(x)

which shows that Pv "' PW. Conversely, assume that pv "' PW and let isomorphism such that Pw(a)f

= f Pv(a)

for all

J

V ---. W be an R-

a EA

Then, for all a EA, x EV, f(ax)

= fpv(a)(x) = pw(a)f(x) = af(x)

as required. • Let pv be a representation of A afforded by an A-module V. We say that pv is irreducible (indecomposable, completely reducible) if V is simple (indecomposable, semisimple). By Proposition 4.1, if pv is irreducible (indecomposable, completely reducible) then any representation equivalent to pv is also irreducible (indecomposable, completely reducible). If p : A ---. EndR(V) is a representation of A and V is free of rank n, then n is called the degree of p and is denoted by deg(p ). By a matrix representation of A, we understand any R-algebra homomorphism .X : A---. Mn(R) for some n 2': 1

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110

The integer n ;:::: J is called the degree of the representation .X and is denoted by deg(.X). Assume that V is an R-module which is free of finite rank n. Then any choice of an R-basis of V leads to the identification of EndR(V) with the R-algebra Mn(R) of all n X n matrices with entries in R. Hence, if p : A - Endn(V) is a representation of A, then by using one of such identifications we obtain a matrix representation p*: A- Mn(R)

of A to which we refer as a matrix representation of A corresponding to p. Conversely, any such matrix representation p* determines a representation p : A - Endn(V) where V is the R-module of all n x 1 matrices over R and p(a)v = p*(a)v for all a EA, v EV. Two matrix representations P1 : A - Mn(R) and p2 : A - Mt(R) are called equivalent, written Pl "' P2, if n = t and there exists an invertible matrix X in Mn(R) such that for all

a EA

Let V be an R-module which is free of finite rank. For any representation p : A - Endn(V), denote by {p} the equivalence class of p and by {p*} the equivalence class of a matrix representation p* of A corresponding to p. Proposition 4.2. The map {p} t-t {p*} is a bijective correspondence between the equivalence classes of representations of the form p : A Endn(V), where V is a free R-module of finite rank, and the equivalence classes of matrix representations of A. Proof. It clearly suffices to verify that P1 "' P2 if and only if pi "' p 2. Since the latter is an obvious consequence of the definitions of equivalences, the result follows. •

Let V be an R-module which is free of finite rank and let p : A - Endn(V)

be a representation of A. For each f E Endn(V), let tr(!) denote the trace of the linear transformation f. Then the R-linear map

Xp: A- R

4 Representations of algebras

111

given by

Xp(a)

= tr(p(a))

a EA

for all

is called the character of p. The degree deg(xp) of XP is defined by deg(xp) = deg(p). The character X.\ of a matrix representation .X of A is defined by X.\ (a) = tr( .X( a)) where tr( .X( a)) is the trace of .X( a). The degree deg(x.\) of X.\ is defined by deg(x.\) = deg(>.). It is an immediate consequence of the definitions that: (i) Equivalent representations of the form p: A--+ EndR(V), where Vis free of finite rank, have the same characters. (ii) Xp = Xp• and deg(xp) = deg(xp• ), where p* is any matrix representation of A corresponding to p. (iii) deg(p) = xp(l) if charR = 0. Let Pi : A --+ EndR(Vi) be a representation of A, where ¼ is a free Rmodule of finite rank, 1 ~ i ~ n, and let Xi be the character of Pi· We say that X1,X2,,,,,Xn are R-linear independent if, for any given r1, ... , rn in R,

r1x1(a)

+ ... + TnXn(a) = 0

for all

a EA

implies

r1

= r2 = · · · = rn = 0

Proposition 4.3. Let A be a semilocal algebra over a commutative ring R. Then (i) A has only finitely many, say Vi, ... , v;., nonisomorphic simple modules. {ii) If Pi : A --+ EndR(Vi) is the representation of A afforded by ¼ and Ei = EndA(Vi), then¼ is a free Ei-module of finite rank ni and

Pi(A)

= EndE;(¼) ~ Mn;(Ef)

as R-algebras

{iii) The map { A/J(A) a+ J(A) is an isomorphism of R-algebras.

--+ r-+

I1i= 1 EndE;(Vi) (p1(a), ... ,Pr(a))

Artinian and Semilocal Rings

112

Proof. (i) Since the isomorphism classes of simple A-modules are in bijective correspondence with the isomorphism classes of simple A/ J(A)modules, the desired conclusion is a consequence of Theorem 1.2 (i). (ii) Since J( erpi is a primitive ideal of A, it follows from Proposition 2.1 that A/ I( erpi is a simple artinian ring. Hence we may assume that A is a simple artinian ring, in which case (by Theorem 1.2) the A-module A is a direct sum of, say ni copies of¼. Accordingly, by Proposition 1.2.2, Pi(A) = EndE;(¼) and ¼ is a free Ei-module of rank ni. Finally, EndE;(¼) ~ Mn;(E'!) by Propositions 1.12.2 and 1.9.3. (iii) Apply (ii) and Proposition 2.1 (iii). • Corollary 4.4. Let R be a commutative semilocal ring and let A be an R-algebra which is a finitely generated R-module, say A = Li=l Rai. Let p : A --+ EndR(V) be the representation of A afforded by a simple Amodule V and let D = EndA(V) (so that, by Propositions 2.3 and 4.3, p(A) = Endv(V)). Identify Endv(V) with Mk(D) for some k ~ 1 and let di ED be the trace of p(ai) E Mk(D). Then

Proof. Given d ED, choose any matrix CE Mk(D) with traced. We know that C = p(a) for some a EA. Writing a= Li=l riai with Ti ER, we have n

C = p(a) = I>ip(ai) i=l

Taking traces of both sides gives d

= Li=l ridi, as required.

•

Chapter 3

Homological Algebra This chapter records some essential information on topics in homological algebra which will be required for future use. Special attention is drawn to a detailed discussion of projective and injective modules. For example, we give a precise description of the endomorphism rings of such modules, prove that every injective indecomposable module is strongly indecomposable, provide necessary and sufficient conditions for an injective module to be indecomposable and demonstrate that the endomorphism rings of injective modules are semiregular. The final section contains a number of applications of previous results. These include a detailed discussion of dual modules and bilinear forms together with the introduction of Frobenius algebras (which are defined over arbitrary commutative rings following the work of Eilenberg and Nakayama (1955)). A thorough treatment of Frobenius algebras over fields will be presented as part of the general theory of algebras over fields.

1

Tensor products of modules

Throughout this section, R denotes an arbitrary ring. Given a right Rmodule V and a left R-module W, the 'D, -module

called the tensor product of V and W, is defined as follows. Let F be a free 'D, -module with V X W as a basis; then each element of 113

Homological Algebra

114

F can be uniquely written in the form: (Zij

E

z

,Vi

E V,Wj E W)

with finitely many Zij distinct from 0. Let T be the submodule of F generated by all elements of the form:

+ v2,w) (v,Wt + w2) (vt

(v,rw)

(vt,w)- (v2,w) (V, Wt) - ( V, W2) (vr,w)

where v,vt,V2 EV, w,w1,w2 E W, and r ER. Then V ®n Wis defined to be the factor module F /T. The image of (v, w) under the natural homomorphism F - F /T is denoted by v ® w. With this notation, the 7'L -module V ®n W consists of all finite sums (vi E V,wi E W) and the elements v ® w satisfy the relations:

+ v2) ® w v ® (w1 + w2) (v1

v®rw

Vt®w+v2®w V@ Wt+ V@ W2 vr®w

(1)

(2) (3)

for all v,v1,v2 EV, w,wt,W2 E W, r ER. Given a right R-module V, a left R-module Wand a Z -module U, a map f : V X W - U is said to be R-balanced if it satisfies the following three properties: (i) f(v1 + v2,w)= f(vi,w) + f(v2,w) (ii) f( v, Wt + w2) = f( v, w1) + f( v, w2) (iii) f(vr,w) = f(v,rw) for all v,v1,v2 EV, w,wt,W2 E W, r ER. Proposition 1.1. Let V be a right R-module and let W be a left Rmodule. {i) The map f : V x W - V ®n W, ( v, w) 1-+ v ® w is R-balanced. (ii) If U is any 7'L -module and the map g : V X W - U is R-balanced, then there exists a unique 7'L -homomorphism 'lj;: V ®R W - U such that

'lj;( V @ W)

= g( V, W)

for all

v E V, w E W

1 Tensor products of modules

115

Proof. (i) This is a direct consequence of properties (1), (2) and (3). · (ii) Let F be a free 7l: -module freely generated by V x W. Then the map (v,w) 1-+ g(v,w) extends to a Z-homomorphism F--+ U. Since g is R-balanced, the submodule Tin the definition of V ®R Wis mapped to 0. This shows that the map 'lj; : V ®R W --+ U defined by '1/J( v ® w) = g( v, w) is a 7l: -homomorphism. The uniqueness of such 'lj; is a consequence of the fact that the elements v ® w, v EV, w E W generate the 7l: -module V ®R W. • Assume that Vis an (S, R)-bimodule. Then, for each s E S, the map

'1/Js : V

X

W

--+

V ®R W

defined by

'1/Js( v, w)

= sv ® w

is R-balanced. It follows from Proposition 1.1 (ii) that V ®R W is a left S-module with

s( V

@ W)

= SV @ W

(sES,vEV,wEW)

In particular, if V and W are modules over a commutative ring R, then V ®R Wis an R-module via r( v ® w)

= rv ® w

(r E R,v E V,w E W)

Similarly, if Wis an (R, S)-bimodule, then V ®R Wis a right S-module via (V

Q9

W

)s

= V Q9 WS

( v E V, w E W, s E S)

In what follows, we regard Ras an (R, R)-bimodule. Proposition L2. (i) For any right R-module V, the map V--+ V®RR, v 1-+ v@ 1 is an isomorphism of right R-modules whose inverse is v ® r 1-+ vr, v E V,r ER, (ii) For any left R-module V, the map V --+ R ®RV, v 1-+ 1 ® v is an isomorphism of left R-modules whose inverse is r ® v 1-+ rv, r E R, v E V. Proof. (i) It is clear that the given map is a homomorphism of right R-modules. Since the map V x R--+ V, (v,r) 1-+ vr is obviously R-balanced, it follows from Proposition 1.1 (ii) that the map V ®RR--+ V, v ® r 1-+ vr is a 7l: -homomorphism which is the inverse of v 1-+ v ® 1, as required. (ii) Apply the same argument as in (i). •

Homological Algebra

116

Proposition 1.3. Let V be a right R-module, W an (R, S)-bimodule, and U a left S-m·odule where R and S are arbitrary rings. Then V ®R W is a right S-module, W ®s U is a left R-module, and we have

(V ®R W) ®s U ~ V ®R (W ®s U) Proof.

We leave to the reader to verify that the mappings

A

(v®w)@uf--+v@(w®u)

µ

v®(w®u)f--+(v®w)®u

define hom.om.orphism.s for which Aµ = 1 and µA = 1. • Let R be an arbitrary ring, let V, V' be two right R-m.odules and let W, W' be two left R-modules. If f : V --+ V' and g : W --+ W' are Rhom.omorphisms, then the map

{

VxW

--+

(v,w)

f--+

V' ®R W' f(v)®g(w)

is R-balanced. Hence, by Proposition 1.1 (ii), there exists a unique 7.l hom.omorphism h: V ®R w--+ V' ®R W' such that h(v ® w)

= f(v) ® g(w)

(v E V,w E W)

The map h is denoted by f ® g and is called the tensor product g. Proposition 1.4.

of f and

Assume that

V' ~ V

!

V" --+ 0

W' 2. W .!.,. W" --+ 0 are exact sequences of right and left R-modules, respectively. Then 0--+ Im(a ® 1) + Jm(l ® 1 )--+ V ®R W /3~} V" ®R W"--+ 0

is an exact sequence of 7.l -modules

1 Tensor products of modules

117

Proof. Put I( Im(a 0 1) + Im(l 0 ,). It is clear that /3 0 8 is surjective and that I( ~ I( er(/3 ® 6). Consider the map f : V" x W" (V ®R W)/ I( given by f(f3(v), 6( w)) = v ® w +K. Then f is obviously well defined and R-balanced. Hence the map f* : V" ®R W" - (V 0R W)/ K, f3(v) 0 6(w) i---+ v 0 w + I( is a 7Z-homomorphisms which is inverse to the 7Z -homomorphism (V 0R W)/ I( - V" ®R W" induced by /3 ® 6. •

=

Corollary 1.5. Let W' 2+ W !+ W" - 0 be an exact sequence of left R-modules and let V be a right R-module. Then

V ®R W'

S.,, V

1

®R W

1~ 8

V ®R W" - 0

is an exact sequence of 7Z -modules. Proof. This is a special case of Proposition 1.4 in which V' V" and /3 is the identity map. •

= 0, V =

Corollary 1.6. Let V' ~ V ! V" - 0 be an exact sequence of right R-modules and let W be a left R-module. Then

V' ®R W

a,~l

V ®R W

,B~I

V" ®R W - 0

is an exact sequence of 7Z -modules. Proof. map.•

Apply Proposition 1.4 for W'

= 0, W = W" and 6 the identity

Let V' be a submodule of a right R-module V, let W be a left R-module and let a : V' - V be the inclusion map. Then the homomorphism

a 0 1 : V' ®R W - V ®R W is not necessarily injective. Hence we must distinguish between the element v ® w, v E V', w E W calculated in V' ®R W and the element v ®w calculated in V ®R W. For this reason, we write I m(V' ® R W) for the image of V' ®R W under a® 1. Similarly, if W' is a submodule of a left R-module Wand 1': W' - Wis the inclusion map, we write Im(V ®R W') to denote the image of V ®R W' under 1 ®, : V ®R W' - V ®R W. Finally, we refer to a® 1, 10, and a® 1 as the natural homomorphisms.

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Corollary 1. 7. Let V' be a submodule of a right R-module V and let W' be a submodule of a left R-module W. Then the map

{

(V ®R W)/(Im(V' fi>R W) + Im(V fi>R W')) (v ® w) + Im(V' ®R W) + Im(V ®n W')

-+ i---+

(V /V') ® R (W /W') ( V + V') ® ( w + W')

is a 'D, -isomorphism. Proof.

Applying Proposition 1.4 for the natural exact sequences O-+ V' ~ V _£ V/V'-+ O

and 0 -+ W'

:!.+ W !.+ W /W' -+ 0

the result follows. • Let V' be a submodule of a right R-module V and let W' be a submodule of a left R-module W. In case the natural homomorphism V' fi>R W' - V fi>R

w

is injective, we shall identify V' fi>R W' with its image in V fi>R W. Proposition 1.8. Let V = EBiEI ¼ be a right R-module and let W = be a left R-module. Then (i) For all i E J, j E J, the natural homomorphisms¼ fi>R Wj -+ V fi>R W are injective. (ii) V fi>R W = EB(i,j)ElxJ(¼ fi>R Wj).

EEljEJ Wj

Proof. For each i E I, j E J, let Ei : ¼ -+ V, Dj : Wj -+ W be the inclusion maps and let 'lri : V -+ ¼, Aj : W-+ Wj be the projection maps. Then

as required. (ii) It is clear that V®nW are such that

= I:i,j(¼®nWj)-

Assume that Xij E ¼®RWj

1 Tensor products of modules

119

with finitely many Xij -::/: 0. Then, applying that each Xij is zero, as required. •

7ri

® Aj to both sides, we deduce

Proposition 1.9. Let V be a right R-module and let W be a left Rmodule. (i) If W is free with basis {Wj Jj E J}, then each element of V ®R W can be written uniquely in the form EjEJ Vj ® Wj with Vj E V and with finitely many Vj-::/: 0. (ii} If V is free with basis { viii E /}, then each element of V ®R W can be written uniquely in the form EiEI Vi ® Wi with Wi E W and with finitely many Wi-::/: 0. (iii) If V is free with basis {viii EI} and Wis free with basis {wjlj E J}, then each element of V ®R W can be written uniquely in the form

L

(virij) ® Wj

(i,j)ElxJ

with rij E R and with finitely many rij -::/: 0. Proof. (i) For each j E J, put Wj = Rwj. Then W = ffijEJWj and, by Proposition 1.2 (i), each element of V ®R Wj can be uniquely written in the form Vj ® Wj with Vj EV. Now apply Proposition 1.8 with½ = V. (ii) Apply the same argument as in (i). (iii) Direct consequence of (i) and (ii). • Corollary 1.10. Let R be a subring of a ring S. (i) IfW is a free left R-module with basis {wjlj E J}, then S ®R Wis a free left S -module with basis {l ® Wj Jj E J}. (ii) If V is a free right R-module with basis { Vi Ii E /}, then V ® R S is a free right S -module with basis {Vi ® 1Ii E /}. Proof.

Apply Proposition 1.9 (i), (ii). •

Corollary 1.11. Let V and W be free modules over a commutative ring R with bases {Vi Ii E I} and {Wj Ii E J}, respectively. Then V ® R W is a free R-module with basis {vi® Wjli E I,j E J}. Proof.

Apply Proposition 1.9 (iii). •

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Proposition _1.12. Let V and W be modules over a commutative ring R. Then the map V ® R W --+ W ®RV, v ® w 1-+ w ® v is an R-isomorphism. Proof. The map V x W--+ W ®RV, (v,w) 1-+ w ®vis R-balanced. Hence, by Proposition 1.1 (ii), the map v®w 1-+ w®v is a 7Z -homomorphism. Since this map obviously preserves the action of R, it is in fact an Rhomomorphism. A similar argument shows that the map W ®RV ---+ V ® RW, w ® v 1-+ v ® w is also an R-homomorphism, as required. • Proposition 1.13. Let R be a ring with no zero divisors, let V be a free right R-module and let W be a free left R-module. If v E V and w E W are such that v ® w is the zero element of V ®R W, then v = 0 or w = 0. Proof. Let {Vi Ii E I} and {Wj Ii E J} be bases of V and W, respectively. Then

v

= L Viri, w = LµjWj

(ri,µj ER)

j

and so v ®w

= 1:(viriµj) ® Wj i,j

Since v ® w = 0, it follows from Proposition 1.9 (iii) that riµj = 0 for all i E J,j E J. Hence, if v =J. 0, that is Ti =J. 0 for some i EI, then µj = 0 for all j E J, whence w = 0. •

In what follows, given an ideal I of R and a right R-module V, we write VI for the submodule of V consisting of all finite sums I: Vi Xi with Vi E V,xi E J. Since the ring R/I can be regarded as an (R,R)-bimodule in a natural way, the tensor product V ®R (R/I)

is a right R-module via ( v ® (r

+ I) )r1· = v ® (rr1 + I)

for all

v E V, r, r1 E R

With this information at our disposal, we are now ready to prove the following result.

2 Tensor product of algebras

121

Proposition 1.14. Let I be an ideal of a ring R and let V be a right R-module. Then the map

.!+

{ V (i!)R (R/I) v ® ( r + I)

1-+

V/VI vr + VI

is an R-isomorphism. Proof. The map V x (R/I) ---+ V/VI, (v,r + I) 1-+ vr + VI is well defined and R-balanced. Hence, by Proposition 1.1 (ii), the map 1/; is a 7Z homomorphism. It is clear that 1/; preserves the action of R, so 1/; is in fact an R-homomorphism. Let f : V/V I---+ V ®R (R/ I) be defined by

f( v +VI)

If v'

= v + I: Vi Xi( Vi

= v ® (1 + I)

for all

v EV

E V, Xi E I), then

v' ® (1 +I)= v ® (1 +I)+

L Vi® (xi+ I)= v ® (1 + I)

which shows that f is well defined. Since f is a homomorphism which is the inverse of 1/;, the result follows. • Proposition 1.15. Let I be an ideal of a ring R and let V and W be right and left R-modules, respectively, such that VI = IW = 0. Then V and W can be regarded as a right and left R / I-modules, respectively, and the map

{ V ®R w v®w

.t 1-+

V ® R/ I w v®w

is a 7Z -isomorphism. Proof. The map V x W---+ V ®R/I W, (v,w) 1-+ v ® w is obviously Rbalanced. Hence, by Proposition 1.1 (ii), the map 1/; is a 7Z -homomorphism. By considering the R/I-balanced map V x W---+ V ®R W, (v,w) 1-+ v ® w, we obtain the inverse of 1/;, as required. •

2

Tensor product of algebras

Throughout this section, R denotes a commutative ring. Let A be an R-algebra and let B, C be commuting subalgebras of A,

Homological Algebra

122

i.e. B, C are subalgebras such that

be= cb

for all

b E B, c E C

Then the set BC consisting of all finite sums I: bici with bi E B, Ci E C is the smallest subalgebra of A containing Band C. If A= BC, then we say that A is generated by B and C. Assume that A and B are R-algebras. Because they are R-modules, A 0R B is an R-module, and we assert that this R-module is also an Ralgebra. Proposition 2.1. Let A and B be R-algebras. (i) A 0R B is an R-algebra with multiplication given by

(a1 0 b1)(a2 0 b2)

= (a1a2) 0

(b1b2)

(a1,a2 E A,b1,b2 EB)

(ii) The maps A--+ A 0R B, a 1-l- a 01 and B--+ A 0R B, b 1-l- 10 bare R-algebra homomorphisms which are injective if A and B are R-free and R is an integral domain. (iii) A0l and l0B are commuting subalgebras of A0RB which generate A0RB. Proof. (i) Let F be a free R-module freely generated by AX B. Define multiplication in F distributively using

Then Fis obviously an R-algebra. Moreover, the submodule T of Fin the definition of A0RB forms an ideal of F. Hence A0RB = F/T inherits the R-algebra structure. In particular, since (a,b) maps to a 0 b, we have

for all a1, a2 EA, b1, b2 E B, as required. (ii) It is clear that the given maps are R-algebra homomorphisms. Furthermore, by Proposition 1.13, they are injective provided A and B are R-free and R is an integral domain. (iii) This follows from the definition of multiplication in A 0 R B. • We shall refer to the homomorphisms A --+ A 0 R B, a 1-l- a 0 1 and B --+ A 0R B, b 1-l- 10 bas canonical. The next result provides a universal characterization of the tensor product of algebras.

2 Tensor product of algebras

123

Proposition 2.2. Let A1, A 2 and A be R-algebras, let 'lj;i : Ai -+ A be an R-algebra homomorphism, i = 1,2 such that 'lf;1(A1) and 'lf;2(A2) are commuting subalgebras of A, and let Ji : Ai -+ A1 ®R A2 be the canonical homomorphism, i = 1, 2. Then the map

is a homomorphism of R-algebras whose image is 'lj;1(A1)'lf; 2(A2). Furthermore, h is a unique R-algebra homomorphism with 'lj;i = ho Ji, i = 1, 2.

Proof. The map A1 x A2-+ A, (a1,a2) i---+ 'lf;1(a1)'lf;2(a2) is R-balanced and so his a homomorphism of R-modules. Moreover, if x = a 1 ® a2 and y = ai ® a~, then

h(xy)

h(a1ai ® a2a;)

= 'lf;1(a1)'lf;1(ai)'lf;2(a2)'lf;2(a;)

['lf;1 (a1 )'lf;2( a2 )]['lf;1 ( ai )'lf;2( a~)] h( X )h(y ), proving that his a homomorphism of R-algebras whose image is 'lf;1 (A1)'lf;2(A2). Since h(fi(ai)) = 'lj;i(ai), ai E Ai, we have 'lj;i = ho Ji, i = 1, 2. Finally, the uniqueness of h is a consequence of the fact that any R-algebra homomorphism A1 ®R A 2 -+ A is uniquely determined by its restriction to {a1 ® a2Ja1 E A,a2 E A2}. • Proposition 2.3.

Let A and B be R-algebras. Then the map

A ®RB-+ B ®RA, a® bi---+ b ® a is an isomorphism of R-algebras.

Proof. The given map is an R-isomorphism, by virtue of Proposition 1.12. Since this map obviously preserves multiplication, the result is established. • Proposition 2.4. Let A and B be R-algebras. Then {i) A ® R B is an ( A, B )-bimodule via

Homological Algebra

124

{ii) If B is R-free with 'basis {bili E J}, then A ®R B is a free left Amodule with basis {1 ® bili E J}. (iii) If A is R-free with basis {ai Ii E J}, then A ®R B is a free right B-module with basis {ai ® 1lj E J}.

Proof. (i) Since A is an (A, R)-bimodule, the formula a1(a®b) = a1a®b defines a left A-module structure on A®RB. Similarly, since Bis an (R, B)bimodule, the formula (a® b)b1 = a® bb1 defines a right B-module structure on A®RB, Since the given actions of A and Bon A®RB obviously commute, the required assertion follows. (ii) Apply (i) and Proposition 1.9 (i). (iii) Apply (i) and Proposition 1.9 (ii). • Proposition 2.5. Let A and B be R-algebras and let I and J be ideals of A and B, respectively. Then the map {

(A ®R B)/(Im(I ®RB)+ lm(A ®R J)) a® b + Im(! ®RB)+ lm(A ®R J)

--+

-

(A/I)® (B/J) (a+ I)® (b + J)

is an isomorphism of R-algebras.

Proof. By Corollary 1.7, the given map is a Z -isomorphism. Since it obviously preserves the action of R and multiplication, the result follows. • Proposition 2.6. Let {Aili E J} and {Bili E J} be two finite families of R-algebras. Then the map { flie1Ai®RCTjeJBj (ai)®(bj)

--+

-

n(i,j)ElxAAi®RBj) (ai®bj)

is an isomorphism of R-algebras.

Proof. Since the sets l,J are finite, we have CTieiAi = EBie1Ai and CTjeJ Bi = EBieJBj, Hence, by the external form of Proposition 1.8 (ii), the given map is a Z -isomorphism. Since it obviously preserves the action of R and multiplication, the result follows. • Proposition 2. 7. integers. Then

Let A and B be R-algebras and let m, n be positive

2 Tensor product of algebras

125

(i) Mn(A) 0R B ~ Mn(A 0R B). (ii} Mn(A) 0R Mm(B) ~ Mnm(A 0R B). Proof.

(i) Let eij,l :S i,j :Sn, be the matrix units of Mn(A). Then

and the right-hand side can be viewed as the full matrix ring over A 0R B with eij 0 1, 1 :S i,j :Sn as the matrix units. (ii) Applying (i) and Proposition 2.3, we have Mn(A) 0R Mm(B)

C:,!

Mn(A 0R Mm(B))

C:,!

Mn(Mm(A 0R B))

C:,!

Mnm(A0R B)

as required. • Proposition 2.8. Then the map {

Let A be an R-algebm and let I be an ideal of R. A0R(R/I) a 0 ( r + I)

--+

A/A·I

1--+

ar

+ A ·I

is an isomorphism of R-algebras. Proof. By Proposition 1.14, the given map is an R-isomorphism. Since it obviously preserves multiplication, the result follows. • Proposition 2.9. Then the map

Let I and J be two ideals of a commutative ring R.

--+

R/(I + J) r1r2 +(I+ J)

is an isomorphism of R-algebras. Proof. Setting A Proposition 2.8. •

=

R/J, we have A· I

=

(I+ J)/J. Now apply

Proposition 2.10. Let A and B be two R-algebras and let I be an ideal of R such that I· B = 0. Then A 0R B is an R/I-algebm which is isomorphic to (A/IA) 0R/I B.

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126

Proof. It is .clear that A ®RB is an R/J-algebra via (r + I)(a ® b) = ra®b. Furthermore the map A®RB-+ (A(IA)®R/IB, a®b r-t (a+IA)®b is an isomorphism of R/ I-algebras. •

Let A be an R-algebra and let U be an A-module. Then U can be regarded as an R-module via rv = (r · l)v. With this view in mind, we finally record Proposition 2.11. Let A and B be two R-algebras, let U be an Amodule and let V be a B-module. Then U ®RV is an A ®RB module via

(a® b)( u ® v)

= au® bv

( a E A, b E B, u E U, v E V)

Proof. Given a E A,b EB, the map Ux V-+ U®R V, (u,v) r-t au®bv is R-balanced, so u ® v r-t au ® bv is an endomorphism of U ®R V. We may therefore define the map

f: Ax B-+ End'D, (U ®RV) by J(a,b)(u®v) and so the map

= au®bv, a E A,b EB, u E U,v EV. { A ®RB a® b

.!+ r-t

Then J is R-balanced

End'D, (U ®RV) J(a,b)

is a 'D, -homomorphism. Since 'ljJ obviously preserves multiplication, it follows that 'ljJ is a ring homomorphism, as desired. •

3

Flat modules

Throughout, R denotes an arbitrary ring. Let V be a right R-module and let W be a left R-module. We say that Vis W-flat if for every left R-module W' and for every injective homomorphism f : W' -+ W, the homomorphism 1 ®f

: V ®R W'

-+

V ®R W

is injective. The module Vis called flat if Vis W-flat for any left R-module W. A flat left R-module is defined in a similar fashion. In order that the r'ight R-module V to be flat, it is necessary and sufficient for V to be flat as a left R 0 -module.

3 Flat modules

127

Proposition 3.1. Let V be a right R-module and let W be a left Rmodule. Then the following conditions are equivalent: (i) V is W -fiat.

(ii) If O --t W' then 0

LW

--t

!!..,, W"

V ® R W'

1~

--t

O is an exact sequence of left R-modules,

V ®R W 1!Ef V ®R W"

-+

0

is an exact sequence of '7L, -modules. (iii} For every submodule W' of W, the natural homomorphism V ®R W' --t V ®R W is injective. (iv) For every finitely generated submodule W' of W, the natural homomorphism V ®R W' --t V ®R W is injective. Proof. (i) {:} (ii): Apply Corollary 1.5. (i) => (iii): This is obvious. (iii)=> (i): Let f : W' -+ W be an injective homomorphism. If x = LVi ® Wi E Ker(l ® f), then LVi ® f(wi) = 0 in V ®R Wand hence, by hypothesis, I: Vi® f (wi) = 0 in V ®R f(W'). But then x = 0, since the map 1 ® f: V ®R W' --t V ®R f(W') is obviously an isomorphism. (iii) => (iv): This is obvious. (iv) => (iii): Let W' be an arbitrary submodule of W. Assume that the image of x = I: vi® Wi E V ®R W' in V ®R W is equal to zero and let W" be the submodule of W' generated by all Wi. Since, by hypothesis, the composite map V ®R W" --t V ®R W'-+ V W

®;_

is injective, the element x' = L Vi® Wi of V ®R W" is equal to zero. But, since x is the image of x', we have x = 0, as required. • Corollary 3.2. Let V be a right R-module. Then the following conditions are equivalent: (i) V is fiat.

(ii) If O --t W' then 0-

LW

!!..,, W"

V ® R W'

1~

--t

O is an exact sequence of left R-modules,

V ®R w

!E? V

1

® R W"

-+

0

is an exact sequence of '7L, -modules. (iii} For any left R-module W and for every submodule W' of W, the natural homomorphism V ®R W'-+ V ®R W is injective. (iv) For every left R-module W and for every finitely generated submodule

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128

W' of W, the natural homomorphism V ®R W' __,. V ®R W is injective. (v) If W' L W !L. W" is an exact sequence of left R-modules, then V ®R W'

!.E! V ®R W

1

1~

V ®R W"

is an exact sequence of ~ -modules.

Proof.

Properties (i) - (iv) are equivalent, by virtue of Proposition 3.1.

It is clear that (v) implies (ii). Assume that (ii) holds and let W' L W !L. W" be an exact sequence ofleft R-modules. Consider the natural exact sequence 0 __,. Imf ..!... W

f.+ !mg__,. 0

where i is the inclusion map and g' is induced by g. Then, by (ii),

Ker(l ® g) = Ker(l ® g') = Jm(l ® i) = Jm(l ® f), as required. • Lemma 3.3. Let f : W' __,. W be an injective homomorphism of left R-modules such that f (W') is a direct summand of W. Then, for any right R-module V, 1 ® f : V ®R W' __,. V ®R W

is an injective homomorphism and its image is a direct summand of V ® R W.

Proof. Let µ : V ®R f(W') __,. V ®R W be the natural homomorphism. By Proposition 1.8, µ is injective and Imµ is a direct summand of V ®R W. But Imµ= Im(l ® f) and, sinceµ is injective, so is 1 ® f. • Let V be a right R-module and let W be a left R-module such that V is W-flat. If W' is a submodule or a factor module of W, then V is W' -flat. Lemma 3.4.

Proof. Assume that W' is a submodule of W. If W" is a submodule of W', then the composite map V ®R W" - V ®R W' - V ®R w is injective and so V®RW" __,. V®RW' is also injective. Thus, by Proposition 3.1, V is W'-flat.

3 Flat modules

129

Let W' = W/U' for some submodule U' of Wand let W" = U/U' be a submodule of W'. Since Vis U-flat and W-flat, we may identify V rzm U' with its image in V 0R U and V 0R U with its image in V 0R W. With these identifications, we have

Now assume that x = I: Vi 0 (Ui + U') E V 0 R W", Vi E V, Ui E U, is such that its image x' = I: Vi 0 ( Ui + U') E V 0R W' is zero. To prove that x = 0, it suffices to show that y = I:vi®Ui E V®RU', for then x = (lv®f)(y) = 0 where f: U--+ U/U' is the natural homomorphism. Let g: W--+ W/U' be the natural homomorphism. Then (lv0g)(y) = x' = 0 and so, by Corollary 1.5, y E V 0R U' as required. • Lemma 3.5. Let W = El\oWi be a left R-module and let V be a right R-module. Then V is W -flat if and only if V is Wi-flat for all i E J.

Proof. If V is W-flat then, by Lemma 3.4, V is Wi-flat for all i E J. Conversely, assume that V is Wi-flat for all i E J. We first consider the case where I is a finite set of n elements. By induction on n, we may then take I = {1, 2}. Let 7ri : W --+ Wi be the projection map, i = 1, 2, and let 1r~ = 1r2IW'. By Corollary 1.5, we have a commutative diagram V®(W'nW1)

10i

j

107r~

j 10 j

V@W

V 0 1r2(W')

j1 ®

1®J

1 ®g

V@W1

V®W'

10 7r2

h

V®W2

(i,j,g,f,h are inclusion maps) with exact rows. Since Vis W1 and Wr flat, both 10 g and 10 h are injective. Moreover, by Lemma 3.3, 10 j is also injective. Hence, if x E J( er(l 0 f), then x E J m(l 0 i) by the commutativity of the right square. But then x = 0, by the commutativity of the left square. Turning to the general case, let W' be a finitely generated submodule of W. Then there exists a finite subset J of I such that W' ~ M = EBjEJWj. Hence, by the finite case proved above, V is M-flat and so the natural homomorphism V®RW'-V@RM

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130

is injective. On .the other hand, since M is a direct summand of W, the natural homomorphism V ®n M-+ V ®R Wis also injective (Lemma 3.3). Hence the natural homomorphism V ®R W'-+ V ®R Wis injective and, by Proposition 3.1, V is W-flat. • Let V be a right R-module and let I be a left ideal of R. Then the 1Z homomorphism V ®nl-+ V which sends v® x to vx, v EV, x EI, is called canonical; its image is the subgroup VI consisting of all finite sums E Vi Xi with Vi EV, Xi EI. The induced homomorphism V ®RI-+ VI is called canonical. By Proposition 1.2 (i), the canonical map V ®RR-+ Vis an isomorphism ofright R-modules. Note that if Vis an (S, R)-bimodule, then all canonical maps are homomorphisms of S-modules. We are now-ready to provide the following characterization of flat modules. Theorem 3.6. Let V be a right R-module. Then the following conditions are equivalent: {i) V is flat. {ii) V is RR-flat. {iii) For any left ideal I of R, the canonical map V ® R I -+ VI is an isomorphism. {iv) For any finitely generated left ideal I of R, the canonical map V ®R I -+ VI is an isomorphism. Proof. Let I be a left ideal of R. Then the canonical map V ®nl--+ VI is an isomorphism if and only if the natural homomorphism V ®nl -+ V ® RR is injective. Hence, by Corollary 3.2, (ii), (iii) and (iv) are equivalent. Since (i) obviously implies (ii), we are left to verify that (ii) implies (i). Assume that Vis RR-flat and let W be any left R-module. Then W ~ Ff K for some free left R-module F and some submodule K of F. By Lemma 3.5, Vis F-flat and hence, by Lemma 3.4, Vis W-flat as required. •

Another useful property of flat modules is contained in the following result. Proposition 3.7. Let V if and only if each Vi is flat. Proof.

= EBieiVi

be a right R-module. Then V is flat

Let W' be a submodule of a left R-module W. By Proposition

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131

1.8, each ¼ ®R W (respectively, ¼ ®R W') is identifiable with its image in V ®R W (respectively, V ®R W') and with these identifications,

It follows that the natural homomorphism V ®R W' - V ®R Wis injective if and only if, for all i EI, the natural homomorphism¼ ®R W' - ¼ ®R W is injective. Thus Vis W-flat if and only if each ¼ is W-flat, as required. • Let V be -a flat right R-module and let W' be a submodule of a left R-module W. Then the natural homomorphism V ®R W' - V ®R Wis injective and, in what follows, we shall identify V ®R W' with its image in V®RW.

Let V be a flat right R-module. (i) If f : X - Y is a homomorphism of left R-modules, then

Proposition 3.8.

Ker(lv ® f)

=V

®R (Ker!)

and Im(lv ® f)

= V ®R (Im!)

(ii} If W', W" are two submodules of a left R-module W, then

V ®R (W' n W")

= (V ®R W') n (V ®R W")

Proof. (i) Consider the exact sequence O - J( er f - X 1+ Imf - 0. Since Vis flat, it follows from Corollary 3.2 that the sequence

is also exact, proving (i). (ii) Consider the diagram 0----+ W'

n W"----+

j 0----+ W"----+ W'

W'----+ W' /(W'

n W")----+

0

j + W"----+ (W' + W")/W"---+ 0

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in which j is the canonical isomorphism and all other arrows represent canonical injective and surjective homomorphisms. This diagram is commutative with exact rows and yields the following commutative diagram: V ® (W'n W")

V®W'

j V®W"

f

V ® (W' /(W' n W"))

j

19 V ® (W'

1 ®j

+ W")

h

V ® ((W'

+ W")/W")

in which the rows are exact (Corollary 1.5) and 1 ® j is an isomorphism. Then

V ® (W' n W")

Ker f

= K er(l ® j) o f)

Ker(h o g)

= (V ® W') n (V ® W")

as required. •

4

Properties of HomR(V, W)

Let R be an arbitrary ring and let V, W be R-modules. Then the set H omR(V, W) of all R-homomorphisms of V into W has the structure of a 7Z -module with(!+ g)(v) = f(v) +g(v) for all f,g E HomR(V, W), v EV. In what follows, unless explicitly stated otherwise, a "module" means a "left module". Lemma 4.1. (i) IfV is an (R,S)-bimodule and Wan R-module, then HomR(V, W) is a left S-module with (sf)(v) = f(vs) for alls ES, v EV, f E HomR(V, W). (ii) IfV is a right R-module and W is an (S, R)-bimodule, then H omR(V, W) is a left S-module with (sf)(v) = s(f(v)) for all s E S, v E V, f E HomR(V,W). (iii) If V is an ( S, R)-bimodule and W a right R-module, then H omR(V, W) is a right S-module with (f s)(v) = f(sv) for all s E S, v E V, f E HomR(V, W). (iv) IfV is an R-module and Wis an (R, S)-bimodule, then H omR(V, W) is a right S-module with (fs)(v) = f(v)s for alls E S, v E V, f E

4 Properties of H omR(V, W)

133

HomR(V,W). (v) If V is an R-module, then H omR(R, V)

~

V.

Proof. The verification of (i) - (iv) is straightforward. To prove (v), note that, by Proposition 1.9.2 (i), H omR(R, V) - V, f f--+ /(1) is a 7Z isomorphism. Since this map obviously preserves the action of R, the result follows. • It must be noted that, in case R is a commutative ring, for any Rmodules V and W, H omR(V, W) is also an R-module via ( r !)( v) = r f( v) for all r E R, v E V, f E H omR(V, W). This assertion is a consequence of Lemma 4.1 (i). In what follows, the S-structure of H omR(V, W) is defined with the aid of Lemma 4.1.

Proposition 4.2. Let V = EBiEIVi and W be R-modules and, for any f E H omR(V, W), let Ji : ¼ - W be the restriction off to¼, i E J. Then the map

is a 7Z -isomorphism. Moreover, if each ¼ is an ( R, S)-bimodule, then this isomorphism is actually an S -isomorphism. Proof. The given map, say o:, is obviously a 7Z -homomorphism, which is also an S-homomorphism provided each ¼ is an (R, S)-bimodule. Since any set {Ji Ii E I} with Ji E H omR(¼, W) defines an f E H omR(V, W) such that Ji = fl¼, the map o: is surjective. Clearly, Ji = 0 for all i E I implies that f = 0, thus a is also injective. • Proposition 4.3. Let V and W = TiiEI Wi be R-modules and, for each i EI, let 7ri : W - Wi be the projection map. Then the map { H omR(V, W) -

f

f--+

TiiEI H omR(V, Wi) (7ri

Of)

is a 7Z -isomorphism. Moreover, if each Wi is an (R, S)-bimodule, then this isomorphism is actually an isomorphism of right S-modules. Proof. The given map, say {3, is obviously a 7Z -homomorphism, which is also a homomorphism of right S-modules provided each Wi is an (R, S)-

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bimodule. If 7ri o f = 0 for all i E I, then f = 0 and so /3 is injective. Suppose that, for any i E I, we are given Ji E H omR(V, Wi)· Then the map f : V --+ W defined by f (v) = (Ji( v)) is a homomorphism such that /3(!) = (Ji), as required. •

Corollary 4.4. Let V = EBiEIVi and W that both I and J are finite. Then

= EBjEJWj

be R-modules such

Moreover, if R is commutative, then the above isomorphism is actually an R-isomorphism. Proof.

This is a direct consequence of Propositions 4.2 and 4.3. •

Assume that U, V and Ware R-modules and let f E H omR(U, V). Then, for any g EH omR(V, W), we have go f E H omR(U, W). The map

f*:

HomR(V, W)--+ HomR(U, W)

given by

= g Of

f*(g)

is obviously a homomorphism. The reader may easily verify that for any Rmodules V', V, V", Wand for any a E HomR(V', V) and /3 E HomR(V, V"), the homomorphisms a* : H omR(V, W)--+ H omR(V', W), satisfy a*

o

/3*

/3* : H omR(V", W)--+ H omR(V, W)

= (/3 o a)*.

Proposition 4.5. Let V', V, V" be three R-modules and let a: V'--+ V, --+ V" be two R-homomorphisms. Then the sequence

/3 : V

V' ~ V

£

V" --+ 0

(1)

is exact if and only if for any R-module W, the sequence 0--+ HomR(V", W) ~ HomR(V, W) ~ HomR(V', W)

is exact.

(2)

4 Properties of H omR(V, W)

135

Proof. Assume that the sequence ( 1) is exact. If f E J( er /3*, then Jo /3 = 0. Since /3 is surjective, f = 0 and so /3* is injective. Next we note that a* o/3* = (/3oa)* = O* = 0

and therefote I m/3* ~ I( era*. To prove J( era* ~ I m/3*, assume that / E Kera*. Then O =a*(!)= f o a and Ker/3 =Ima~ Ker/. Hence /3(x) = /3(y) implies f(x) = J(y) for all x,y EV. Since /3 is surjective, we may write a typical element of V" in the form ()(x), x E V. It follows readily that the map 1 : V"-+ W, ()(x) f--+ f(x), x EV, is a well defined homomorphism. Since f = 1 0/3, we deduce that f = /3*( 1 ) ~ Im/3*, proving that (2) is exact. Conversely, assume that (2) is exact for any R-module W. Since

a* o /3* = (/3 o a)*

=0

it follows that I o /3 o a = 0 for every R-homomorphism 1 : V"-+ W. Taking W = V" and,= lw, we have /3oa = 0 and so Ima~ Ker(). Now take W = C okera and let f : V -+ W = V /Ima be the natural homomorphism. Then a*(!)= Jo a= 0 by definition and hence there exists a 1j; EH omR(V", W) with f = /3*(1/;) = 1j; o (3. This obviously implies Ima= Kerf ~ Ker/3, which proves that (1) is exact at V. Finally, let 1r: V"-+ W = V"/f3(V) be the natural homomorphism. Then /3* ( 1r) = 1r o /3 = 0, so 1r = 0 and hence V" = f3(V), as required. • Corollary 4.6. Let a : V' -+ V be a homomorphism of R-modules. Then a is surjective (bijective, zero) if and only if for any R-module W, the map a* : H omR(V, W) -+ H omR(V', W) is injective {bijective, zero).

Apply Proposition 4.5 to the case V" and V" = V, /3 = lv). • Proof.

Corollary 4. 7.

= 0 (respectively, V' = 0

Suppose we are given an exact sequence 0

-+

V' ~ V

I!.., V"

-+

0

(3)

of R-modules. If (3) splits, then for any R-module W the sequence

0-+ HomR(V", W) ~ HomR(V, W) ~ HomR(V', W)-+ 0

(4)

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136

is exact and split_s. Conversely, if for any R-module W sequence (4) is exact, then sequence {3) splits. Proof.

Assume that (3) splits. Then there exists a homomorphism o o: = 1. Hence ('Y o o: )* = a* o 7* = 1 and so, by Proposition 4.5, (4) is exact and splits. Conversely, suppose (4) is exact for W = V'. Then there exists f E H omR(V, V') such that f o a = 1, as required. • 'Y : V -+ V' such that 'Y

Assume that U, V and Ware R-modules and let f E HomR(U, V). Then, for any g E H omR(W, U), we have fog E H omR(W, V). The map

given by is obviously a homomorphism. It is an immediate consequence of the definition that for any R-modules V', V, V", Wand for any a E HomR(V', V), /3 E H omR(V, V"), the homomorphisms

a* : H omR(W, V'l-+komR(W, V), /3* : H omR(W, V) -+ H omR(W, V")

Proposition 4.8. Let V', V, V" be three R-modules and let a: V'-+ V, /3 : V -+ V" be two R-homomorphisms. Then the sequence O -+ V'

~ V !!.+ V"

(5)

is exact if and only if for any R-module W, the sequence 0-+ HomR(W, V') ~ HomR(W, V) ~ HomR(W, V")

(6)

is exact. Proof. Suppose that the sequence (5) is exact. If f E Kera*, then a of = 0. Since o: is injective, it follows that f = 0 and so a* is injective. Since

4 Properties of H omR(V, W)

137

we have Ima* ~ Ker/3*. To prove the converse, assume that I E Ker/3~. Then /3 o 1 = 0 and so J m 1 ~ 1( er/3 = I ma. But a is injective, hence the map 1 1 = a- 1 o I of W into V' is well defined. Since 1 = ai,1 ), it follows that I E Ima* and so (6) is exact. Conversely, suppose that the sequence (6) is exact for every R-module W. Since

it follows that f3 o a o 1 = 0 for every R-homomorphism 1 : W - V'. Taking W = V' and 1 = lw, we have /3 o a = 0 and so Ima ~ Ker/3. Now take W = 1( er/3 and let 'I/; : W - V be the inclusion map. Then /3*( 'I/;) = /3 o 'I/; = 0 by definition and hence there exists f E H omR(W, V') such that 'I/; = a*(J) = a of. This obviously implies that 1( er/3 ~ I ma and so Ker/3 = Ima. Finally, if 0 is the identity map of Kera, then a*(0) = 0 and so 0 = 0. Thus 1( era = 0 and the result follows. • Corollary 4.9. Let f : V - U be a homomorphism of R-modules. Then f is injective if and only if for any R-module W, the map

is injective. Proof.

V"

= U.

Apply Proposition 4.8 to the case where V'

= 0,

/3

=

f and

•

Corollary 4.10.

Suppose we are given an exact sequence 0 - V'

~ V f!+ V"

- O

(7)

of R-modules. If (7) splits, then for any R-module W, the sequence

0 - H omR(W, V') ~ H omR(W, V) ~ H omR(W, V") - 0

(8)

is exact and splits. Conversely, if for any R-module W the sequence (8) is exact, then sequence (7) splits. Proof. Assume that (7) splits. Then there exists a homomorphism 1 : V" - V such that /3 o 1 = l. Hence (/3 o 1 )* = /3* o '* = 1 and so, by Proposition 4.8, sequence (8) is exact and splits. Conversely, suppose

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138

(8) is exact for W = V". Then there exists f E HomR(V", V) such that /3*(!) = /3 of = i, as required. • For the rest of this section, A and B denote algebras over a commutative ring R. We remind the reader our convention that, unless explicitly stated otherwise, a "module" means a "left module". As a preliminary to our final result, let us recall that if V is a ( B, A)bimodule and U is a B-module, then

H omB(V, U) is an A-module via

(af)(v)

= f(va)

(a E A,v E V,f E HomB(V, U))

Let V be a (B,A)-bimodule, W an A-module and U a B-module. For each f E H omB(V ®AW, U), let f' : W--+ H omB(V, U) be defined by Proposition 4.11.

f'(w)(v)

= f(v ® w)

(v E V,w E W)

Then the map HomA(W,HomB(V, U))

!' is an R-isomorphism. Proof. It is clear that the given map is an R-homomorphism. Let

g :W

--+

H omB(V, U)

be an A-homomorphism. Then the map V x W--+ U, (v,w) i--+ g(w)(v) is A-balanced. Hence there exists a homomorphism 'l/J(g) : V ® A W --+ U such that

'l/J(g)(v ® w)

= g(w)(v)

for all

v E V, w E W

Since g( w) is a B-homomorphism, so is 'l/J(g). It follows that g homomorphism which is easily seen to be the inverse of ..(b,f): B 0R V---+ B 0R W,

c@v

1-+

cb@f(v)

is at least 'D, -homomorphism. Furthermore, since for x, y E B, a E A and VE V, >..(b, f)[(x 0 a)(y 0 v)]

>.(b, f)(xy 0 av)= xyb 0 f(av) xyb® af(v)

= (x 0

a)>..(b,f)(y 0 v)

the map >..(b,f) is a (B 0R A)-homomorphism. This provides us with the map

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141

which is easily seen to be R-balanced. It therefore follows that 'lj; is an Rhomomorphism. The last assertion being a consequence of the definition of 'lj;, (i) follows. (ii) If V = A, then both sides are isomorphic to B ©R W and so 'I/; is an R-isomorphism. Consequently, 'lj; is an R-isomorphism if Vis a free Amodule on a finite basis. Since V is a finitely presented A-module, there is an exact sequence F1-+ Fo-+ V-+ O

where Fo, F 1 are free A-modules with finite bases. From this, by Proposition 4.5, we obtain the following exact sequence 0-+ HomA(V, W)-+ HomA(Fo, W)-+ HomA(F1, W)

Put C = B ©RA, Ff= B ©R Fi, i = 0, 1, and W' = B ©R W, V' By the above, we have the following isomorphisms

'1/;i : B ©RH omA(Fi, W)-+ H omc(Ff, W')

= B ©RV.

(i=0,1)

Since Bis a fl.at R-module, we obtain from Corollary 3.2(v) and Proposition 4.5 the following commutative diagram with exact rows 0---+ B © HomA(V, W)---+ B © HomA(Fo, W)---+ B © HomA(F1, W)

0---+ Homc(V', W')---+ Homc(F~, W')---+ Homc(F{, W')

Since '1/;o and 'I/J1 are isomorphisms, we deduce at once that 'I/; is an isomorphism.•

6

Projective modules

A. Basic properties

Throughout, R denotes an arbitrary ring. Let V be an R-module and let

(1)

142

Homological Algebra

be an exact sequence of R-modules. We know, from Proposition 4.8, that the corresponding sequence

0 -+ H omR(V, X) ~ H omR(V, Y) ~ H omR(V, Z)

-+

0

(2)

of 1L -modules is always exact except possibly at H omR(V, Z) (i.e. the map g* need not be surjective ). We say that V is projective if for any exact sequence (1), the corresponding sequence (.2) is also exact. A similar definition applies to the case where Vis a right R-module. Proposition 6.1. The following conditions are equivalent: (i) V is projective. (ii) For every surjective homomorphism g : Y -+ Z of R-modules, the induced homomorphism g* : H omR(V, Y) -+ H omR(V, Z) is surjective. (iii) For every surjective homomorphism g : Y -+ Z of R-modules and every homomorphism h : V -+ Z there exists a homomorphism t : V -+ Y such that h = g o t.

(iv) Every exact sequence O -+ X .!.,, Y ~ V -+ 0 of R-modules splits. (v) V is a direct summand of a free R-module. Proof. The equivalence of (ii) and (iii) is a consequence of the definition of g*, while the equivalence of (i) and (ii) follows from the definition of projectivity and Proposition 4.8. Assume that (iii) holds. Then, applying (iii) for Z = V and h = lv, we see that (iv) holds. Assume that (iv) holds and choose Y to be a free R-module. Then V is isomorphic to a direct summand of Y, proving (v). Finally, assume that (v) holds. Then V is a· direct summand of a free R-module F with basis {xi}, say. Let 7r: F-+ V be the natural projection, £ : V -+ F the inclusion map and let g : Y -+ Z and h : V -+ Z be respectively a surjective homomorphism and a homomorphism of R-modules. Then, for each Xi, h1r(xi) E Zand because g is surjective, h1r(xi) = g(yi) for some Yi E Y. Hence the homomorphism a : F -+ Y defined by a(xi) = Yi satisfies goa = ho1r. Setting t: V-+ Y to bet= a£, it follows that got= h. This proves (iii) and hence the result. • Corollary 6.2. Let V be a finitely generated R-module. Then V is projective if and only if V is a direct summand of a finitely generated free R-module.

6 Projective modules

143

Proof. Let v1 , ••. , Vn be a finite generating set for V and let F be a free R-module freely generated by x 1 , ... , Xn. Then the map Xi I-* Vi determines a surjective homomorphism F -+ V. Thus if V is projective, then by Proposition 6.1, Vis a direct summand of F. The converse being true by virtue of Proposition 6.1, the result follows. • Corollary 6.3. Let V = EBiei¼ be an R-module. Then V is projective and only if each ½ is projective. Proof. The module Vis a direct summand of a free module if and only if each ½ is a direct summand of a free module. Now apply Proposition 6.1 and the result follows. • Proposition 6.4. exact sequence

An R-module V is projective if and only if for any

of R-modules, the corresponding sequence H omR(V, X) ~ H omR(V, Y) ~ H omR(V, Z) of :lZ -modules is also exact.

It is clear that the given condition implies that V is projective. Conversely, assume that V is projective and let X 1+ Y .!.+ Z be an exact sequence of R-modules. Consider the exact sequence Proof.

.

h

0-+ Imf ~ Y-+ Img-+ O where i is the inclusion map and h is induced by g. Since V is projective, we have as required. • Proposition 6.5. An R-module V is projective if and only if there exists a family { viii E I} of elements in Vanda family {!iii E I} of elements in H omR(V, R) such that every v E V can be written in the form

v=I:fi(v)vi iEI

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144

with finitely many Ji( v) distinct from 0. Moreover, if V is projective, then { viii E I}can be, any generating set. Proof. Let {Vi Ii E I} be any generating set for V, let F be a free Rmodule freely generated by {xiii E J}, and let f : F-+ V be the surjective homomorphism defined by f(xi) = Vi for all i E J. Owing to Proposition 6.1, Vis projective if and only if there exists a homomorphism g : V -+ F with fog= 1. Assume that V is projective and choose g E H omR(V, F) with fog = 1. If v E V, then write

g( V)

= :~:::>iXi iEI

with Ti ER depending on v and with finitely many Ti f 0. Then the family {!iii E J} with Ji : V -+ R, v 1-+ Ti satisfies

v

= fg(v) = Lfi(v)vi iEI

Conversely, let {viii E J} and {!iii E J} be as described in the theorem. Then {viii E J} generates V and f : F -+ V, Xi 1-+ Vi is a surjective homomorphism. Now define g: V-+ F by g(v) = Lie! fi(v)xi. Then g is a homomorphism with f o g = 1, as required. • Propos_ition 6.6. Let V be, a finitely generated projective module over a local ring R. Then V is free (in fact, if v1 + J(R)V, ... , Vn + J(R)V is a basis of the Rf J(R)-module Vf J(R)V, then v1, ... , Vn is a basis of V ).

' Proof. Since Vf J(R)V is a finitely generated module over the division ririg Rf J(R), there exists a finite basis, say v1 + J(R)V, ... , Vn + J(R)V, of Vf J(R)V. If Wis a submodule of V generated by v1, ... ,vn, then V = W + J(R)V. Hence, by Nakayama's lemma, V =Wand so v1 , ... ,vn is a generating set for V. Furthermore, it is clear that no proper subset of { v1, ... , vn} generates V. Suppose that we are given a relation

If Ti r/. J(R) for some i E {1, ... , n }, then Ti E U(R) since R is local. But then we can write Vi is an R-linear combination of the vjs with j f i, which is impossible. Thus all ri E J(R). Let W be a free R-module freely generated

6 Projective modules

145

by w1, ... , Wn and map W onto V by sending Wi to Vi. It suffices to show that the kernel, say U, of this map is zero. By the foregoing, U ~ J(R)W. By hypothesis, V is projective and so, by Proposition 6.1, W = U EB V' with V' ~ V. Since U ~ W/V', it must be finitely generated. Moreover, J(R)W = J(R)U EB J(R)V' and hence the inclusion U ~ J(R)W forces U = J(R)U. Thus, by Nakayama's lemma, U = 0 and the result follows. • Proposition 6. 7.

Let V be a projective right R-module. Then V is

flat. Proof. By the analogue of Proposition 6.1 for right modules, V is a direct summand of a free R-module. Hence, by Proposition 3. 7, we may assume that V = RR. Let I be any left ideal of R and let f : V ®RI ---+VI= I be the canonical homomorphism. Then, by Proposition 1.2 (ii), f is an isomorphism. Thus, by Theorem 3.6, Vis flat. • Proposition 6.8. Let R be a commutative ring and let V, W be finitely generated projective R-modules. Then H omR(V, W) is a finitely generated projective R-module. Proof. By Corollary 6.2, we may assume that there are two finitely generated free R-modules F and Fi such that F = V EB V' and Fi = WEB W'. Since HomR(R,R) ~ R, it follows from Corollary 4.4 that HomR(F,F1 ) is a finitely generated free R-module. On the other hand, by Corollary 4.4, H omR(V, W) is a direct summand of H omR( F, F1 ). Hence HomR(V, W) is finitely generated and, by Proposition 6.1, it is also projective. • Proposition 6.9. Let V be an (S, R)-bimodule and let W be an Rmodule. If V is a projective S -module and W is a projective R-module, then

V ®R W is a projective S-module. Proof. If W is free then, by Proposition 1.8 and 1.2 (i), V ®R Wis a direct sum of copies of V. Hence, by Corollary 6.3, V ®R Wis a projective S-module. In the general case, we may assume that F =WEB W' for some free R-module F. Then, by Proposition 1.8, V ®RF= V ®R WEB V ®R W'. Hence, by Corollary 6.3, V ®R Wis a projective S-module. • Corollary 6.10.

Let R be a commutative ring. If V and W are two

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146

projective R-modules, then V ®R W is a projective R-module. Proof. This is a special case of Proposition 6.9 in which S is commutative. •

=R

and R

Let R be an arbitrary ring and let V be an R-module. Then, by Lemma 4.1 (iv), V* = H omR(V, R) is a right R-module via

(fr)(v) Lemma 6.11.

= J(v)r

for all

r E R, v E V, f E V*

Let V and W be R-modules. Then the map a : V* ®R W

--+

H omR(V, W)

given by

a(J ® w)(v)

= J(v)w

(J E V*, w E W, v E V)

is a 'IL, -homomorphism. Furthermore, if R is commutative, then a zs an R-homomorphism. Proof. The map 0 : V* x W --+ H omR(V, W), 0(!, w )( v) = J( v )w is R-balanced and so a is a 'IL, -homomorphism. If R is commutative, then a obviously preserves the action of R, as required. •

We shall refer to the homomorphism & of Lemma 6.11 as the canonical homomorphism. Proposition 6.12. Let V be a finitely generated projective R-module. Then for any R-module W, the canonical homomorphism a: V* ®R W--+ HomR(V, W)

is a 'IL, -isomorphism. Furthermore, if R is commutative, then a is an Risomorphism. Proof. We first assume that V is free, freely generated by v 1 , ... , Vn. Then V* is a free R-module with basis Ji, ... , f n, where Ji( Vj) = 0 if i -/- j and Ji( vi) = 1. Hence each element of V* ®R W can be uniquely written

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147

in the form Ef= 1 fi 0 Wi with Wi E W. However, a(E~ 1 Ji 0 wi) = f is a homomorphism V --+ W with f( vi) = Wi, 1 ~ i ~ n. Since V is free, any homomorphism! : V --+ W is uniquely determined by its values f( vi)Hence a is an isomorphism. Turning to the general case, we may write F = V EB U where F is a finitely generated free R-module. Then F* = V* EB U* and F* 0R W = V* 0R WEB U* 0R W, where we identify V* (respectively, U*) with the subgroup of F* consisting of all f E F* with f(U) = 0 (respectively, with J(V) = 0). With a similar identification, we also have

H omR( F, W) Let /3 : F* 0R W

--+

= H omR(V, W) EB H omR( U, W)

H omR(F, W) be the canonical isomorphism. Then

/J(V* 0R W) ~ HomR(V, W)

and /3(U* 0R W) ~ HomR(U, W)

and hence f3(V* 0R W) = H omR(V, W). Since the restriction of /3 to V* 0R W coincides with a, it follows that a is a 7Z -isomorphism. The last assertion being a consequence of Lemma 6.11, the result follows. B Proposition 6.13. Let V and W be modules over a commutative ring R. Then the map f: V 0R W--+ HomR(V*, W) given by

[J(v 0 w)](a)

= a(v)w

( v E V, w E W, a E V*)

is an R-homomorphism. Furthermore, if V is finitely generated and projective, then f is an R-isomorphism. Proof. Apply a similar argument to that used in the proof of Proposition 6.12. B Proposition 6.14. Let V and W be modules over a commutative ring R. If V is finitely generated and projective, then the map

'I/;: V* 0R W*--+ (V 0R W)* given by

'1/;(f 0 g)(v 0 w) is an R-isomorphism.

= f(v)g(w)

(f E V*,g E W*,v E V,w E W)

148

Homological Algebra

Proof.

Let_,: W* ®RV*-+ (V ®R W)* be the composite map W* ®RV*-:\ HomR(W, V*)

.! (V ®R W)*

where a is the R-isomorphism given by Proposition 6.12 and /3 is the inverse of the R-isomorphism given by Proposition 4.11 (with A = B = U = R). Then

1 (g ® f)(v ® w)

= g(w)f(v)

Since 'lj; = 1 o A, where A : V* ®R W* Proposition 1.12, the result follows. •

(v E V,w E W,g E W*,J EV*) -+

W* ®RV* is the isomorphism of

B. Generators and progenerators

Let R be an arbitrary ring. A left (respectively, right) R-module Vis said to be a generator if every left (respectively, right) R-module is a quotient of a direct sum of copies of V. A (left or right) R-module V is called a progenerator if V is a finitely generated projective generator. Lemma 6.15. For any (left or right) R-module V, the following conditions are equivalent: (i) V is a generator. (ii} R is a direct summand of a finite direct sum of copies of V. (iii} There exist finitely many R-homomorphisms Ji : V -+ R, 1 ~ i ~ n, such that 1 = Li=l fi( vi) for some V1, •.• , Vn in V. Proof.

(i) {:} (ii) : Assume that V is a generator. Then there is a

surjective R-homomorphism W .LR where W = EBiEIWi and each Wi ~ V. Now let w E W be such that f(w) = 1. Then w E U = EBf= 1 ¼ for some n ~ 1 where each "Vj E {Wi}. The restriction off to U gives a surjective homomorphism U -+ R. Since R is projective, R is a direct summand of U, proving (ii). Conversely, assume that W = REBX where Wis a finite direct sum of copies of V. Let Y be an arbitrary R-module. Then there is a free R-module F and a surjective homomorphism F -+ Y. Hence there exists a direct sum U of copies of W and a surjective homomorphism U -+ Y, proving (i ). (ii){:} (iii): Assume that f1~ 1 ¼ = REBX where each¼= V. Put U = ITi=l ¼ and let 7r : U -+ R be the projection map. Choose v1, ... , Vn E V

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149

such that 7r( v1, ... , vn) = 1 and let Ji E H omR(V, R) be given by Ji = 1l" o Si, where Si : V-+ U is the natural injection, 1 ::; i ::; n. Then I:i=l Ji( vi) = 1, proving (iii). Finally assume that (iii) holds. Put U = Tii=l ½ where each v; = V. Then the map f : U -+ R, J( v1, ... , vn) = I:i=l Ji( Vi) is a surjective Rhomomorphism. Since R is projective, it follows that R is a direct summand of U, proving (ii). • Assume that P is a right R-module and let S = EndR(P). Then P is an (S, R)-bimodule, where the left S-module structure on P is given by sx = s(x) for all s E S, x E P. Similarly, P* = H omR(P, R) is an (R, S)bimodule with the actions of Rand S given by ( r f)(x) = r f(x) and Js = f os for all r ER, x E P, f E P*, s ES. Thus, for any left R-module V, P ®RV is a left S-module with the action of S given by

s( X

Q9

V)

= SX Q9 V

(s E S, x E P, v EV)

Similarly, for any left S-module W, P* ®s Wis a left R-module via

r(f ® w)

= rf

( r E R, f E P*, w E W)

®w

With this information at our disposal, we now define the following two important maps:

{

P®RP* O(x ® f)(y)

{

P*®sP f®x

0 -+

s xf(y)

(1)

(x,y E P,f E P*)

(f E P*, x E

(iii) : Obvious. (iii) => (iv) : Let g : Y - Z, h : V - Z and t' : VB - YB be as in (iv) and let h =got'. By hypothesis, WA= V EB V' for some A-module V'. Consider the maps o: : WA - Z and f3 : WA - Y defined by

Define

f:

= h(v),/3( v + v') = t'( v) Y by f(a ® w) = af3(1 ® w),

o:( v + v')

( v E V, v' E V')

wA -

a EA, w E W. The map

f is obviously a well defined A-homomorphism. Furthermore, for all a EA, wEW,

= agf3(1@ w) = ao:(1@ w) = o:(a@ w) Hence, setting t = f IV, we have h = g o t. (gf)(a@ w)

(iv) => (i) : Assume that O - X ~ Y ..!.+ V - 0 is an exact sequence of A-modules such that

is a split exact sequence of B-modules. Setting Z = V and h = lv, it follows that h =got' for some B-homomorphism t' : VB - YB. Hence, by hypothesis, g is a split surjection. • Corollary 9.3. Let V = EBiEIVi be an A-module. projective if and only if each ½ is B-projective.

Then V is B-

Proof. Assume that Vis B-projective. Then, by Proposition 9.2 (iii), Vis a direct summand of WA for some B-module W. Hence each½ is a direct summand of WA and so, by Proposition 9.2 (iii), each½ is B-projective.

9 Induction and relative projectivity

215

Conversely, assume that each ¼ is B-projective. Then, by Proposition 9.2 (iii), for each i E J, there exists a B-module Wi such that ¼ is a direct summand of W/. Hence, by Proposition 1.1 (iii), V is isomorphic to a direct summand of EBiEJW/ ~ ( EBiEI~)A. Thus, by Proposition 9.2 (iii), V is B-projective. • Corollary 9.4. Assume that A is a projective (left) B-module and let V be an A-module such that VB is projective. Then the following conditions

are equivalent : (i) V is B-projective. (ii} For any pair of A-modules U, W such that UB, WB are projective B-modules the exact sequence

o-w..£.uLv-o splits provided

is a split exact sequence of B-modules.

*

Proof. (i) (ii) : Obvious. (ii) (i) : Since VB is projective, so is U = (VB)A by Proposition 1.1 (v). Hence, by Proposition 1.1 (vi), UB is projective. Let f : U -+ V be the canonical homomorphism and let W = I( er f. Then we have an exact sequence

*

where g is the inclusion map. By Corollary 3.3, WB is projective. Since, by Proposition 3.2, the corresponding sequence

splits, it follows from hypothesis that f : U -+ Vis a split surjection. Hence, by Proposition 9.2 (ii), V is B-projective. • Lemma 9.5. Assume that we are given the following commutative diagram of A-modules with exact rows:

Restriction, Induction and Coinduction

216

v ____ w ()

u __a_ _ µ

>.

x __, __ y

. o

. and Q" = K erh, we see that Q' ~ P and, since g = f oh, Q" ~ K erg. Since g( Q') = g( Q) = V, g restricts to a surjective homomorphism g' : Q' -+ V. Furthermore, Q' being a direct summand of Q, must also be projective. Finally, since g>. = fh>. = f, it follows that Kerg' = >.(Ker!). But, by Lemma 1.1 (i), Ker f is a superfluous submodule of P, hence K erg' is a superfluous submodule of >.(P) = Q'. Thus, by Lemma 1.1 (i), g': Q'-+ V is a projective cover of V. •

g =

The following result shows that if a module does have a projective cover, then it has (essentially) only one. More precisely, the following important property holds. Corollary 1.4.

{Uniqueness of projective cover). A°ssume that

Ji : Pi

-+

Vi and h : P2

-+

½

are projective covers of R-modules Vi and ½. If o: : Vi -+ V2 is an Risomorphism, then there is an R-homomorphism () : P1 -+ P2 such that h o () = o: o Ji and each such homomorphism is an isomorphism. Proof. Keeping the notation of the proof of Proposition 1.3, put P = P2, Q = Pi, V = Vi, f = h, g = o:ofi, and()= h. Then(): Pi-+ P2 satisfies

h

o ()

= o:

o

Ji.

Conversely, any such () is surjective, since

h

is essential.

Semiperfect Rings

228

Also Ker/3 ~ Kerfi, whence Ker/3 is a superfluous direct summand of Pi which forces K er/3 = 0. Thus any such /3 is an isomorphism, as desired. • A link between a projective cover of a module and its homomorphic image is given by the following easy consequence of Proposition 1.3. Corollary 1.5. Assume that an R-module V has a projective cover P and let M be an R-module such that V is a homomorphic image of M. If M has a projective cover Q, then P is isomorphic to a direct summand of

Q. Proof. By hypothesis, there are surjective homomorphisms a : Q -+ M and /3: M-+ V. Setting g = /3 o a, it follows that g: Q -+Vis a surjective homomorphism. Now apply Proposition 1.3. •

Turning to projective covers of direct sums, we next record the following property. Proposition 1.6. Assume that V is an R-module such that V = Vi E0 • • • E0 Vn and each ¼ has a projective cover. Then an R-homomorphism

J : P -+ V is a projective cover of V if and only if P has a decomposition P = Pi E0 · · · E0 Pn such that JIPi : Pi -+ ¼ is a projective cover of½, 1 ~ i ~ n. In particular, if Pi is a projective cover of¼, then E0f= 1 Pi is a projective cover of V. Proof.

To prove sufficiency, note that Ker f

J is obviously surjective. Hence, by Lemma 1.1,

= E0f=1 K er(f lPi) and that

f : P -+ V is a projective cover of V. Conversely, assume that f : P -+ V is a projective cover of V. Let Yi : Qi -+ ¼ be a projective cover of¼. Setting Q = EBf= 1 Qi, it follows from the above that the induced homomorphism g : Q -+ V is a projective cover of V. Hence, by Corollary 1.4, there is an R-isomorphism h : Q -+ P such that g = Joh. Setting Pi= h(Qi), we theri. have P = E0f= 1 Pi and J(Pi) = ¼, 1 ~ i ~ n. Since K er(JIPi) = h(K ergi) is a superfluous submodule of Pi, the result follows. •

Let

j : U -+ V be a homomorphism of R-modules. We say that a

1 Projective covers

229

decomposition V = EBf= 1 ¼ can be lifted to U with respect to f, if there exists a decomposition

such that f(Ui) = ¼ for all i E {1, ... , n }. In case V = U/W for some submodule W of U and f : U -+ V is the natural homomorphism, the qualifier "with respect to f" will be omitted. Corollary 1. 7. Let V = EBf=l ¼ and assume that each ¼ has a projective cover. If f : P -+ V is a projective cover of V, then the decomposition V = EBf= 1 ¼ can be lifted to P with respect to f. Proof. By Proposition 1.6., there exists a decomposition P = EBf= 1 Pi such that f(Pi) =¼for each i E {l, ... ,n}. • Lemma 1.8. Let I be an ideal of R and let P be a projective R-module. Then P /IP is a projective Rf I-module. Proof. Hence

Since Pis projective, F = P EB P' for some free R-module P.

F/IF ~ (P/IP) EB (P'/IP') and so, by Lemma 1.2.4, Pf IP is a projective R/I-module. • Proposition 1.9. Let I be an ideal of R and let V be an Rf I-module. If f : P -+ V is a projective cover of the R-module V, then the induced homomorphism f* : P /IP -+ V is a projective cover of the Rf I-module V. Proof. Since IV :a: 0, we have Ker f ] IP. Thus f induces a surjective homomorphism f*: Pf IP-+ V given by f*(x+IP) = f(x) for all x E P. By Lemmas 1.1 (i) and 1.8, we are left to verify that Ker f* is superfluous. To this end, put K = Kerf and let P/IP = (S/IP)+ (K/IP) for a submodule S of P containing IP. Then P = S +Kand Kerf* = K/IP. Since K is superfluous, P = S and so Ker f* is superfluous. • Lemma 1.10. Let e be an idempotent of R and let I be a left ideal of R with I ~ J(R). Then the natural homomorphism Re-+ Re/le zs a projective cover of Re/ I e.

Semiperfect Rings

230

Proof. The tiatural homomorphism Re -, Re/ I e has kernel I e and, by Proposition 1.5.30, le~ J(R)e = J(Re). Since, by Proposition 1.5.7, J(Re) is superfluous, the desired conclusion follows. • Proposition 1.11. A cyclic R-module V has a projective cover if and only if V ~ Re/ le for some idempotent e of R and some left ideal I of R with I~ J(R). Proof. Assume that f : P -, V is a projective cover of V. Since V is cyclic, there is an epimorphism g : R -, V. Hence, by Proposition 1.3, we may assume that R = PffiP' with f = gjP. Thus there exists an idempotent e of R with P = Re and I = I( er f is a superfluous submodule of Re. It follows from Porpositions 1.5.7 (i) and 1.5.30 that J ~ J(Re) = J(R)e ~ J(R) and V ~ Re/le. The converse being a consequence of Lemma 1.10, the result is established. • Proposition 1.12. Let I be an ideal of R with I ~ J(R). Then the following conditions are equivalent : (i) Each idempotent of Rf I can be lifted to R. (ii} Each direct summand of the left R-module Rf I has a projective cover. Proof. (i) ::::} (ii) : Every direct summand V of Rf I is of the form V = (Rf I)£ where £ is an idempotent of Rf I. By hypothesis, we may therefore write V = (R/I)(e +I)= (Re+ I)/I for some idempotent e of R. Since ( Re + I)/ I ~ Re/ ( I n Re) = Re/ I e

the desired conclusion follows from Lemma 1.10. (ii) ::::} (i) : Let E1 be an idempotent of Rf I and let

£2

=1-

£1.

Then

Rf I= (Rf !)£1 ffi (Rf I)£2

and, by Lemma 1.10, the natural map 1r : R -, Rf I is a projective cover of Rf I. By hypothesis, both (Rf I)E1 and (R/ I)£2 have a projective cover. Hence, by Corollary 1. 7, there exists idempotents e 1, e 2 of R with 1 = e 1 + e 2 and such that (R/I)Ei = 1r(Rei) = (R/I)(ei

Since the latter obviously implies Ei

+ I)

(i=l,2)

= ei + I, the result follows.

•

1 Projective covers

231

Our next aim is to provide some information on projective covers of simple modules. Proposition 1.13. For any given projective R-module P, the following conditions are equivalent : {i) P is the projective cover of a simple R-module. {ii} J(P) is a superfluous maximal submodule of P. {iii} There exists a primitive idempotent e of R such that P ~ Re, Re/J(R)e is a simple R-module and Pis the projective cover of Re/J(R)e. (iv) P is strongly indecomposable. Proof. (i) =:> (ii) : By hypothesis, P has a superfluous maximal submodule. But J(P) is contained in every maximal submodule of P and, by Proposition 1.5.7 (i), J(P) contains every superfluous submodule of P. Thus J(P) is a superfluous maximal submodule of P. (ii) =:> (iii) : By hypothesis, P is the projective cover of the simple Rmodule Pf J(P). Since every simple R-module is a homomorphic image of R, it follows from Proposition 1.3 (with Q = R) that P ~ Re for some idempotent e of R. Since J(P) is a unique maximal submodule of P and P is finitely generated, P is indecomposable (Proposition 1.3.4). Hence e is a primitive idempotent. Since P/J(P) ~ Re/J(Re) and, by Proposition 1.5.30, J(Re) = J(R)e, the required assertion follows. (iii) =:> (iv) : By (i) =:, (ii), J(P) is a superfluous maximal submodule of P. Hence, by Schur's lemma, EndR(P/J(P)) is a division ring and, by Proposition 3.6.19 and Lemma 3.6.18,

EndR(P)/ J(EndR(P)) ~ EndR(P/ J(P)), as required. (iv) =:> (i) : By hypothesis, EndR(P) is a local ring (in particular, P =j; 0) and, by Proposition 3.6.20, there is a maximal submodule M of P. It therefore suffices to verify that M is a superfluous submodule of P (for then Pis the projective cover of the simple R-module P/M). Assume that P = K + M for some submodule K of P. Then

P/M

= (K + M)/M

~

K/K n M)

and therefore there is a nonzero homomorphism f : P -+ K/(K n M). Since P is projective, there is an endomorphism >. : P -+ K ~ P such that f = 1r o >., where 1r : K -+ K/(K n M) is the natural homomorphism.

Semiperfect Rings

232

Since O i- f = 1r o A, ImA Cf: M, whence ImA is not superfluous in P. Thus, by Proposition 3.6.19, A r/. J(EndR(P)) which implies that A is an automorphism of P. Hence I( = P and so M is a superfluous submodule of P, as desired. • Corollary 1.14. For any idempotent e of a ring R, the following conditions are equivalent : (i) Re/J(R)e is simple. (ii) J(R)e is the unique maximal submodule of Re. (iii) Re is strongly indecomposable. (iv) eJ(R) is the unique maximal submodule of eR. (v) eR/eJ(R) is simple. Proof. (i) :::} (ii) : It is clear that J(R)e is a maximal submodule of Re. Since, by Proposition 1.5.30, J(Re) = J(R)e it must be unique. (ii) :::} (iii) : Setting P = Re, we have J(P) = J(R)e is a maximal submodule of P. On the other hand, by Proposition 1.5.7 (ii), J(P) is superfluous. Hence, by Proposiiton 1.13, P is strongly indecomposable. (iii) :::} (i) : Applying Proposition 1.13 for P = Re, we see that J(P) = J(R)e is a maximal submodule of P. Hence Re/ J(R)e is simple, as required. By the arguments above, we see that (iv) and (v) are equivalent to eR being strongly indecomposable. Since, by Proposition 1.9.3, the latter is equivalent to (iii), the result follows. •

We close by proving three further results which will be needed later. Proposition 1.15. Suppose that P is a projective R-module such that H omR( P, V) i- 0 for any simple R-module V. Then P is a generator. Proof. Put E = H omR(P, R) and let I = LJEE f(P). Then I is a left ideal of R and, by Lemma 3.6.14 (iii), it suffices to show that I = R. Assume by way of contradiction that I i- R. Then there is a maximal left ideal M of R with I ~ M. Then R/M is simple, so there is a nonzero R-homomorphism f : P ---+ R/M. Since P is projective, there is an Rhomomorphism g: P---+ R with f = hog where h: R---+ R/M is the natural homomorphism. This provides a contradiction, since I mg ~ I ~ M. • Proposition 1.16.

Let P and Q be finitely generated projective R-

2 Characterizations and fundamental properties

233

modules and let I be a left ideal of R with I~ J(R). Then (i) P is a projective cover of P /IP. (ii) P ~ Q if and only if P/IP ~ Q/IQ. Proof. (i) Since IP ~ J(R)P and, by Nakayama's lemma, J(R)P is superfluous, it follows that IP is superfluous. Thus the natural homomorphism P - P /IP is essential, as required. (ii) It is clear that P ~ Q implies P /IP ~ Q /IQ. Conversely, assume that Pf IP~ Q/IQ. By (i), P and Q are projective covers of Pf IP and Q /IQ, respectively. Hence, by Corollary 1.4, P ~ Q. • Corollary 1.17. Let P and Q be finitely generated projective R-modules _ and let I be an ideal of R with I~ J(R). Then the following conditions are equivalent : (i)P~Q. (ii) P /IP ~ Q / IQ as R-modules. (iii) P /IP ~ Q / IQ as Rf I-modules. Proof. 1.16. •

2

It is clear that (ii) is equivalent to (iii). Now apply Proposition

Characterizations and fundamental properties

We continue to use our convention that all modules are assumed to be left modules. Our concern so far has been almost exclusively with the properties of projective covers under the assumption that they exist. Since in future applications, finitely generated modules will play a major role, we are confronted with the task of characterizing those rings over which all finitely generated modules have a projective cover. We refer to such rings as semiperfect rings. With the properties of projective covers exhibited in the previous section, we are now in a position to undertake the proof of the following major result. Theorem 2.1. For any given ring R, the following conditions are equivalent : (i) Every simple R-module has a projective cover. (ii) R is semiperfect. (iii} R is semilocal and every idempotent of R/J(R) can be lifted to R.

Semiperfect Rings

234

(iv) There exist pairwise orthogonal primitive idempotents e1, ... , en of R with 1 = Li=l ei and with each Rei -.strongly indecomposable. (v) R is idempotent-lifting and semilocal. (vi) R is semiregular and semilocal.

Proof. (i) => (ii) : Let {¼Ii E l} be a full set of representatives of the isomorphism classes of simple R-modules and, for each i E l, let Pi be the projective cover of¼. By Corollary 3.6.3, Q = EBiEIPi is projective and, by Proposition 1.15, Q is a generator. Given a finitely generated R-module V, we may therefore choose finitely many, say P1, ... , Pn, projective covers of simple R-modules and a surjective homomorphism P =

L V.

Since f(J(R)P) = J(R)V, there is a surjective homomorphism Et)i= 1 (Pd J(R)Pi) - V/ J(R)V. · But each Pd J(R)Pi is simple by Proposition 1.13 (ii) and Lemma 3.6.18. Hence, by Propositions 1.5.10 and 1.3.5, V / J (R) V is a finite direct sum of simple R-modules. Thus, by Proposition 1.6, V/J(R)V has a projective cover. But J(R)V is superfluous by Nakayama's lemma, so V- V/ J(R)V is an essential epimorphism. Hence, by Propositions 1.2 and 3:6.1 (iii), V has a projective cover. (ii) => (iii) : By hypothesis, every direct summand of Rf J(R) has a projective cover. Hence, by Proposition 1.12, every idempotent of Rf J(R) can be lifted to R. To see that R is semilocal, let L be a left ideal of R with J(R) ~ L. Then; since the cyclic R-module R/L has a projective cover, it follows from Proposition 1.11 that

Et)i=lpi

R/L ~ Ref le for some idempotent e of Rand some left ideal l of R with le~ J(R)e. But then J(R)(Re/le) ~ J(R)(R/L) = 0 and so J(R)e ~ le. Thus le= J(R)e and Rf L ~ Ref J(R)e ~ (Rf J(R))( e + J(R)) is projective over Rf J(R). Therefore L/ J(R) is a direct summand of Rf J(R), proving that R/J(R) is semisimple and hence that R is semilocal. (iii) => (iv) : Choose orthogonal primitive idempotents .s 1 , ... , En of Rf J(R) whose sum is 1. Then, by Theorem 1.6.3, there exist pairwise orthogonal primitive idempotents e1, ... , en of R with 1 = e 1 + ··· + en and Ei = ei + J(R), 1 ~ i ~ n. Since, by Lemma 1.2.5 (iii), each Reif J(R)ei is simple, the desired 'assertion follows from Corollary 1.14. (iv) => (i) : By Corollary 1.14, each Rei/ J(R)ei is simple, while by Proposition 1.11, it has a projective cover. But each simple R-module is

2 Characterizations and fundamental properties

235

isomorphic to a submodule of Rf J(R) ~ Reif J(R)e1 E9 · · · E9 Ren/ J(R)en, and so is isomorphic to one of the Reif J(R)e;. (iii) ~ (v) ~ (vi) : Apply Proposition 1.6.14 and Theorem 2.1.1. • Note that Theorem 2.1 (iii) shows that R is semiperfect if and only if so is R 0 • Thus conditions (i) and (ii) of Theorem 2.1 for right R-modules also serve to characterize semiperfect rings. Corollary 2.2. Let R be a semiperfect ring. Then (i) For any ideal I of R, Rf I is semiperfect. (ii) RR has the unique decomposition property. (iii) R is local if and only if O and 1 are the only idempotents of R. Proof. (i) Apply Proposition 1.9. (ii) Apply Theorem 2.1 (iv) and Azumaya's decomposition theorem (Theorem 1.10.11). (iii) Apply Theorem 2.1 (iv). •

The following theorem provides some of the fundamental properties of semiperfect rings. Theorem 2.3. Let R be a semiperfect ring. Then there exist pairwise orthogonal primitive idempotents e1, . .. , es of R with each Re; strongly indecomposable and such that the following properties hold : (i) Reif J(R)e1, ... , Res/ J(R)es are all nonisomorphic simple R-modules and Re; is the projective cover of Re;/ J(R)e;, 1 ~ i ~ s. (ii} Let V be a finitely generated R-module and write

for some integers n; 2: 0. Then an R-module P is a projective cover of V if and only if P ~ E&f= 1 n;(Re;) (iii) For any given finitely generated projective R-module P, P ~ E9f= 1 n;( Re;) for some uniquely determined integers n; 2: 0. (iv) Re 1, . .. , Res are all nonisomorphic principal indecomposable R-modules. (v) Re 1 , .. . , Res are all nonisomorphic projective indecomposable R-modules.

236

Semiperfect Rings

Proof. (i) We may order the idempotents e1, . .. , en in Theorem 2.1 (iv) in such a way that each Rei, 1 :s; i :s; n, is isomorphic to precisely one of the Re 1 , ••• , Res, s ~ n. Since each Rei is strongly indecomposable, it follows from Corollary 1.14 and Lemma 1.10 that each Rei/ J(R)ei is simple and Rei is the projective cover of Rei/ J(R)ei, By Corollary 1.4, the latter implies that each Red J(R)ei, 1 ~ i ~ n, is isomorphic to precisely one of the Rei/ J(R)e1, ... , Res/ J(R)es, Since Rf J(R) ~ ffif=l Rei/ J(R)ei, (i) is established. (ii) Put Q = ffif= 1ni(Rei). By Corollary 1.4, It suffices to show that Q is a projective cover of V. To this end, note that Q is finitely generated projective and Q/J(R)Q ~ V/J(R)V. Since, by Propositions 1.5.7 and 1.5.8, J(R)Q is a superfluous submodule of Q, the natural homomorphism Q -+ Q / J(R)Q is a projective cover of Q / J(R)Q. Also, since J(R)V is a superfluous submodule of V, the natural homomorphism V-+ V/J(R)V is essential. Hence, by Propositions 1.2 and 3.6.1 (iii), Q is a projective cover of V. (iii) Since Pf J(R)P is a finitely generated module over Rf J(R), it follows from (i) that P/J(R)P ~ ffif= 1 ni(Rei/J(R)ei) for some uniquely determined integers ni ~ 0. Setting Q = ffif= 1ni(Rei), it follows that Q and Pare projective covers of P/J(R)P. Hence, by Corollary 1.4, P ~ Q as required. (iv) This is a direct consequence of (iii). (v) By (iv), it suffices to show that every projective indecomposable Rmodule V is principal indecomposable. To this end, write F = V EB V' for some free R-module F. By Theorem 2.1 (iv), Fis a direct sum of strongly indecomposable R-modules of the form Rei for some i E {1, ... , s }. Hence, by Corollary 1.10.8 (i), V ~ Rei for some i E {1, ... , s }, as desired. •

Recall that an R-module V is said to be projective-free if V has no nonzero projective direct summands. The importance of this notion for modules over semiperfect rings resides in the following result. Theorem 2.4. Let V f:. 0 be a finitely generated module over a semiperfect ring R. Then there is a decomposition V = P EB W where P is projective and W is projective-free. Furthermore, if V = P' ffi W' is another such decomposition, then P ~ P' and W ~ W'. Proof. By Proposition 2.2.14 (i), V is a finite direct sum of indecomposable submodules. By separating off projective direct summands, we

2 Characterizations and fundamental properties

237

therefore obtain a decomposition V = P E0 ( E0i=I ½) where each ½ is indecomposable nonprojective and P is projective. Put W = E0f= 1 ½ and note that if W has a nonzero projective direct summand, then W has an indecomposable projective direct summand, say Q. By Theorem 2.3 (v), Q is strongly indecomposable. But then, by Lemma 1.10.4 and Theorem 1.10.6, Q is a direct sum of certain ½, which is impossible. Thus W is projectivefree. Assume that V = P' E0 W' is another decomposition where P' is projective and W' is projective-free. We argue by induction on the number n of indecomposable direct summands of P. If n = O, then P = P' = 0 and W = W'. Let Q be an indecomposable direct summand of P. Then Q is strongly indecomposable (Theorem 2.3 (v)) and Q is isomorphic to a direct summand of P' (Lemma 1.10.4 and Theorem 1.10.6). The desired conclusion now follows by induction together with the cancellation property ( Corollary 1.10.10). • The following two theorems provide an important information· on projective modules over semiperfect rings. Theorem 2.5. Let P be a finitely generated projective module over a semiperfect ring R and let I be a left ideal of R with I~ J(R). Then (i) P is indecomposable if and only if P /IP is indecomposable. (ii) Every decomposition P /IP = E0f= 1 ½ can be lifted to P, i.e. there is a decomposition P = E0f= 1 Pi with (Pi+ IP)/IP = ½, 1 ~ i ~ n.Furthermore, (a) Pi/IPi ~ ½, 1 ~ i ~ n. (b) ½ is indecomposable if and only if Pi is indecomposable, 1 ~ i ~ n. (c) Pi~ Pj if and only if½~ Vj, i,j E {1, ... ,n}. Proof.

(i) If P

= P1 E0 P2 with P1 =/ 0, P2 =/ 0, then:

where both direct summands are nonzero by Nakayama's lemma. Conversely, assume that P /IP = V1 E0 ½ with V1 =/ O, ½ =/ 0. By Proposition 1.16 (i ), the natural homomorphism f: P-+ P/IP is the projective cover of P/IP. Thus, by Proposition 1.6, there exists a decomposition P = P 1 E0 P2 with !(Pi)= V1 and J(P2) = ½ and hence with P1 =/ 0, P2 =/ 0. (ii) Let f: P-+ P/ IP be the natural homomorphism. Then, by Proposiiton 1.16 (i) and 1.6, there exists a decomposition P = P1 E0 · • · EB Pn with

Semiperfect Rings

238

J(Pi) = ¼, 1 :S (:Sn. Since ¼=(Pi+ IP)/IP ~ Pi/(Pi n IP)= Pi/IPi

(1 :Si :Sn)

property (a) holds. Property (b) follows from (a) and (i), while (c) is a consequence of (a) and Proposition 1.16 (ii). • Theorem 2.6. Let I be an ideal of a semiperfect ring R with I~ J(R). For each R-module V, let the R/ I-module V be defined by V = V /IV. Then the map P 1-+ P induces a bijection between the isomorphism classes of finitely generated projective R-modules and finitely generated projective Rf I-modules. Proof. By Lemma 1.8, for every projective R-module P, Pis a projective R/ I-module. If P and Q are finitely generated projective R-modules, then by Corollary 1.17, P ~ Q if and only if P ~ Q. Thus it suffices to show that every finitely generated projective R/ I-module V is isomorphic to P for some finitely generated projective R-module P. By Propositions 2.2.8 and 2.2.14 (i), Vis a finite direct sum of indecomposable modules, say V = EB~ 1 Qi· If the required assertion is true for each Qi, then it is certainly true for V. Thus we may assume that Vis indecomposable. Hence, by Corollary 2.2 (i) and Theorem 2.3, we may also assume that V = (R/ I)£ for some primitive idempotent £ of R/ I. By Theorem 2.1 (v), we may write£= e +I for some idempotent e of R. Setting P = Re, it follows from Lemma 1.2.5 (iii) that

P = Re/Ie

~ (R/I)c

=V

as required. • A ring R is said to be primary if R/ J( R) is a simple artinian ring. An R-module V is called homogeneous if V is a finite direct sum of isomorphic indecomposable modules. Theorem 2.7. Let R be a semiperfect ring. Then the following conditions are equivalent : (i) R is primary. (ii) All principal indecomposable R-modules are isomorphic. (iii) All simple R-modules are isomorphic. (iv) All projective indecomposable R-modules are isomorphic.

3 Some classes of semiperfect rings

239

(v) The regular module RR is homogeneous. (vi) R ~ Mn(S) for some local ring S and some n ~ 1. Furthermore, n is uniquely determined by R, while S is determined up to isomorphism. Proof. The implications (i) :::} (ii), (ii) :::} (iii), (iii) :::} (iv) and (iv) :::} (v) follow from Theorems 2.1 and 2.3. The implication (vi) :::} (i) being obvious, we are left to verify (by Corollary 2.2.11) that (v) implies that R ~ Mn(S) for some local ring S and some n ~ 1. Assume that R = Rei EB · · · EB Ren for some primitive idempotents ei, ... , en of R such that Rei ~ Rei for all i E {1, ... , n }. Setting S = eiRei, it follows from Corollary 1.12.7 (i) that R ~ Mn(S). Since, by Theorem 2.3 (iv), Rei is strongly indecomposable, the rings S must be local, as we wished to show. •

3

Some classes of semiperfect rings

Our aim here is to provide some important classes of semiperfect rings. All conventions and notation adopted in previous sections remain in force.

Let R be a semilocal ring satisfying at least one of the following two conditions : {i) J(R) is nil (e.g. R is artinian). {ii} R is complete at J(R). Then R is a semiperfect ring. Proposition 3.1.

Proof. If R satisfies (i) or (ii), then by Lemmas 1.6.1 and 1.6.2 (ii), every idempotent of R/J(R) can be lifted to R. Now apply Theorem 2.1

(iii) . • A ring R is said to be self-injective if the left regular module RR is injective.

Assume that R is a self-injective ring. Then R is semiperfect if and only if R is semilocal. Proposition 3.2.

Proof. If R is semiperfect, then by Theorem 2.1 (iii), R is semilocal. Conversely, assume that R is self-injective and semilocal. By Proposition 2.2.14, R is a finite direct sum of indecomposable injective modules. Since

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240

each indecomposable injective R-module is strongly indecomposable (Corollary 3.8.22), R is semiperfect by virtue of Theorem 2.1 (iv). • Lemma 3.3. Let R be a commutative noetherian ring and let V be a finitely generated R-module. Then the R-algebra EndR(V) is a finitely generated R-module. Proof. Write V = Rv1 + · ·· + Rvn for some Vi E V and let W be a free R-module freely generated by wi, ... , Wn. Let i.p : W --+ V be the R-homomorphism such that i.p( wi) = Vi and set I( = I( eri.p. Since EndR(W) ~ Mn(R), EndR(W) is a finitely generated R-module. Put E = {f E E"!'._dR(W)lf(K) ~ K}. If f ~ Ewith J(wi) = EJ= 1 aijWj, then f induces an f E EndR(V) given by f(vi) = EJ= 1 aijVj. For any g E EndR(V), write n

g( vi)

= L aijVj

(ai; ER)

j=l

and define g E EndR(W) by n

g( Wi)

= L aijWj j=l

0 = g(L b;v;) j whence g(°I:'bjWj) j

= L bjUjkVk j,k

= L bjajkWk EK, j,k

proving that g E E. It follows that g f-+ g is a surjective R-homomorphism E --+ EndR(V). Now EndR(W) is a finitely generated R-module and R is noetherian. Hence E and EndR(V) are finitely generated R-modules, as required. • We are now ready to provide another important class of semiperfect rings. Theorem 3.4. Let I be an ideal of a commutative ring R such that R is complete at I and R/ I is artinian, and let A be an R-algebra which

3 Some classes of semiperfect rings

241

is a finitely genemted R-module. Then A is semiperfect under either of the following hypotheses : {i) A is R-free. {ii) R is noetherian. Proof. By Corollary 1.6.19, A is complete at I A. Hence, by Lemma 1.6.2 (i), IA ~ J(A). Now A/IA is a finitely generated module over an artinian ring Rf I and so A/ I A is artinian. Hence A/ J(A) is artinian, i.e. A is semilocal. Since J(A)/IA = J(A/IA), we have J(Al ~ IA for some k 2: 1. Hence, by Lemma 1.6.2 (iii), A is complete at J(A). Now apply Proposition 3.1 (ii). • Corollary 3.5. Let R be a complete commutative semilocal ring and let A be an R-algebra which is a finitely generated R-module. Then A is semiperfect under either of the following hypotheses : (i) A is R-free. {ii) R is noetherian. Proof.

Apply Theorem 3.4 for I= J(R). •

Theorem 3.6. Let I be an ideal of a noetherian commutative ring R such that R is complete at I and Rf I is artinian, and let A be an R-algebra which is a finitely generated R-module (e.g. let A be as in Corollary 3.5 satisfying (ii)). Let V i- 0 be a finitely generated A-module. Then {i) EndA(V) is semiperfect. (ii) If V is indecomposable, then V is strongly indecomposable. {iii) V has the unique decomposition property. Proof. (i) Note that each EndA(V) is an R-submodule of EndR(V). Hence, by Lemma 3.3 and Corollary 1.3.7, the R-algebra EndA(V) is a finitely generated R-module. Thus, by Theorem 3.4, EndA(V) is semiperfect. (ii) Assume that V is indecomposable. Then O and 1 are only idempotents of the semiperfect ring EndA(V). Hence, by Corollary 2.2 (iii), End A (V) is local. (iii) Since A is a finitely generated module over the noetherian ring R, A is noetherian (Corollary 1.3.8). Now apply (ii) and Theorem 1.10.15. •

242

4

Semiperfect Rings

Block decompositions

Let R be an arbitrary ring and let V be an R-module. By a subquotient of V, we understand any module of the form U /W where W ~ U are submodules of V. Note that if V has a composition series, then the simple sub quotients of V are precisely the composition factors of V. In general, however, a module can have a simple subquotient without having a composition series. Let us now observe that, by Theorem 2.3 (iv) and Corollary 1.14, the hypothesis of the lemma below is satisfied by any primitive idempotent in a semiperfect ring R. Lemma 4.1. Let e be an idempotent of a ring R such that Re/J(R)e is simple. Then, for any given R-module V, the following conditions are equivalent : {i) eV-:/ 0. {ii} HomR(Re, V)-:/ 0. (iii} V has a subquotient isomorphic to Re/J(R)e.

Proof. (i) {:} (ii) : Apply Proposition 1.9.2 (i). (i) {:} (iii) : If eV = 0, then eQ = 0 for any subquotient Q of V. Hence Re/ J( R)e cannot be a sub quotient of V. Conversely, assume that e V -:/ 0. Then there is a nonzero R-homomorphism f : Re ---+ V. If I = I( er f and W = Imf, then Re/I~ W. Furthermore, by Corollary 1.14 (ii), I~ J(R)e. Hence Re/J(R)e is isomorphic to a factor module of W, as required.• Proposition 4.2. Let R be a semiperfect ring. Then R admits a block decomposition and, for any two primitive idempotents e, f of R, the following conditions are equivalent : (i) e and f are linked. {ii) e and f lie in the same block. {iii} There exist primitive idempotents e1 = e, e2, ... , en = f such that the members of each consecutive pair Rei, Rei+l, i E {1, ... , n - l}, have a common simple subquotient. Proof. By Theorems 2.1 (iv) and 2.3(iv) and by Proposition 1.11.11, it suffices to verify that (i) and (iii) are equivalent. The latter will follow provided we show that Re, Rf have a common simple subquotient if and only if there exists a primitive idempotent u of R with uRe-:/ 0 and uRf-:/ 0. The

4 Block decompositions

243

desired conclusion is therefore a consequence of Lemma 4.1 and Theorem 2.3

(i) . • We next exhibit circumstances under which there is a bijection between the blocks of R and its factor ring. The following preliminary result will clear our path. Lemma 4.3. Let I be an ideal of a ring R such that n~=l In = 0 and let J be an ideal of R with J ~ I 2 • (i) If e is an idempotent of R, then e E Z(R) if and only if e + J E Z(R/J). (ii} The natural map R --+ R/ J induces an injection of the set of central idempotents of R into those of R/ J. Proof. (i) If e E Z(R), then obviously e + J E Z(R/ J). Conversely, assume that e+J E Z(R/J). Then e+I 2 E Z(R/I 2 ), whence eR(l-e) ~ I 2 and el~ le+ I 2 • We claim that eR(l - e) ~ rn and (1- e)Re ~ rn for all n 2". 1; if sustained, it will follow that eR(l - e) = (1 - e)Re = 0, whence e E Z(R) by Lemma 1.2.9. Assume, by induction on n, that eR(l - e) ~ In. Then

eR(l - e)

C

er(l - e) ~(le+ I 2 )r- 1 (1 - e)

C IeR(l - e) + r+i(l - e) ~ r+i and, by a similar argument, (1 - e )Re~ rn. (ii) Let e, f be central idempotents of R with e + J (e - ef) 2

=e -

= f + J.

Then

ef E J ~ I

Hence e - ef E n~= 1In = 0, proving that e = ef. By the same argument, f = ef and so e = f, as desired. • Proposition 4.4. Let I be an ideal of a semiperfect ring R such that n~=l rn = 0 and let J be an ideal of R with J ~ I 2 • Then the natural map R --+ R/ J induces a bijection of the set of central idempotents of R onto those of R/ J and the block decomposition R = EBi=l Bi of R induces the block decomposition

of R/J.

244

Semiperfect Rings

Proof. The first assertion follows from Lemma 4.3 by applying the fact that R is idempotent-lifting (Theorem 2.1 (v)). Note also that I contains no nonzero idempotent and, by Theorem 2.1 (vi), R is semiregular. Hence, by Corollary 1.6.15, I~ J(R) and so J ~ J(R). Because R is semiperfect, so is R/ J by virtue of Corollary 2.2 (i). Hence, by Proposition 4.2, both R and R/ J admit a block decomposition. Let E1, ... , En be all block idempotents of R/ J (hence 1 = E1 + ·· · + En). Since J ~ J(R), it follows from Theorem 1.6.3 that there exist pairwise orthogonal idempotents ei, ... , en of R with 1 = e1 + · · ·+en and C:i = ei + J, 1 ~ i ~ n. By Lemma 4.3 (i), each ei E Z(R). If ei = ei + ei' for some orthogonal idempotents ei,ef, then Ei = (ei + J) + (ei' + J) where ei + J, ef + J are orthogonal central idempotents of R/ J. Hence ei + J = 0 or ef + J = 0 which, by Lemma 4.3 (ii), implies that ei = 0 or ei' = 0. Thus each ei is a block idempotent of R and the result follows. •

By Proposition 1.5.26, the assumption (a) below holds whenever R is a finitely generated Z(R)-module. Note also that (b) holds under the stronger assumption that R is a noetherian Z(R)-module. Proposition 4.5. Let a semiperfect ring R satisfy the following two conditions : (a) J(Z(R)) ~ J(R). (b) For any idempotent e of R, the Z(R)-module eR(l - e) is finitely generated. Then, for any ideal I of R with I ~ R · J(Z(R)), the natural map R ......+ R/ I induces a bijection between the central idempotents of R and those of R/ I and the block decomposition R = EBf= 1 Bi induces the block decomposition

of R/I. Proof. The assumptions (a) and I~ R·J(Z(R)) ensure that I~ J(R). If e, f are central idempotents of R with e + I = f + I, then e = f by Proposition 1.5.3 (i). Conversely, let E be an arbitrary central idempotent of R/ I. Since R is idempotent-lifting, E = e + I for some idempotent e of R. It will next be shown that e E Z(R). Since e + I is a central idempotent of R/ I, we have eR(l - e) ~ I ~

4 Block decompositions

245

R · J(Z(R)). Hence

eR(l - e)

eI(l - e)

~

eJ(Z(R)) · R(l - e)

J(Z(R))eR(l - e)

Invoking (b) and Nakayama's lemma, we deduce that eR( 1 - e) = 0 and, replacing e by 1 - e in the above argument, (1 - e )Re = 0. Thus, by Lemma 1.2.9, e E Z(R). This shows that the natural map R ---+ Rf I induces a bijection between the central idempotents of R and those of Rf I. The second assertion regarding the induced block decomposition follows by the same argument as in the proof of Proposition 4.4. • Corollary 4.6. Let R be a semiperfect ring such that J(R) = R · J(Z(R)) and such that for any idempotent e of R, the Z(R)-module eR(l-e) is finitely generated. Then each block of R is primary. Proof. tion 4.5,

Let R

= EBf= 1 Bi be the block decomposition of R. R/J(R)

By Proposi-

= EBf= 1 (Bi + J(R))/J(R)

is the block decomposition of the semiprimitive artinian ring Rf J(R). Since each block of Rf J(R) is simple artinian and (Bi+ J(R))/J(R) ~ Bi/(Bi n J(R))

= Bif J(Bi),

(1 :Si :Sn)

the result follows. • Corollary 4. 7. Let R be a complete commutative noetherian semilocal ring and let A be an R-algebm which is a finitely generated R-module. Then (i) For any ideal I of A with I~ A·J(Z(A)), the natural map A---+ A/I induces a bijection between the central idempotents of A and those of A/ I and the block decomposition A = EBf= 1 Bi of A induces the block decomposition A/I= EBf= 1 (Bi + !)JI of A/I. {ii} If J(A) =A· J(Z(A)), then each block of A is primary. Proof. (i) By Corollary 1.3.7, A is a noetherian R-module. Hence A is a noetherian Z(A)-module and so both assumptions ( a) and (b) in Proposition 4.5 hold (see Propositions 1.5.26 and 1.3.3). Since, by Theorem 3.4 (ii), A is semiperfect, (i) follows from Proposition 4.5. (ii) Apply Corollary 4.6. •

246

Semiperfect Rings

5

Propert!es of idempotents

In this section, we collect together a number of important properties of idempotents of semiperfect rings, Proposition 5.1.

Let R be a semiperfect ring and let {e1, ... , en} and

{Ji, ... , fm} be two sets of pairwise orthogonal primitive idempotents of R with 1=

n

m

i=l

j=l

I:ei = I:!i

Then n = m and, upon renumbering the fj if necessary, there exists a unit u of R such that fi

Ji

= U -1 €iU

for all i E {1, ... , n}

Proof. Owing to Corollary 2.2 (ii), n = m and, upon renumbering the if necessary, Rei ~ Rfi for all i E {1, ... , n }. Now apply Lemma 1.2.7. •

Proposition 5 .2. Let e, f be two primitive idempotents of a semiperfect ring R. Then Re~ Rf if and only if f = u- 1eu for some u E U(R). Proof.

R(l- e)

~

By Corollary 1.2.8, it suffices to verify that Re

~

Rf implies

R(l - f). Since R = Re EB R(l - e) = Rf EB R(l - f), the required

assertion follows from Corollary 2.2 (ii). • Proposition 5.3. Let e be a primitive idempotent of a semiperfect ring R. If e E Ii+···+ In for some ideals Ii, ... ,In of R, then e E Ji for some iE{l, ... ,n}. Proof.

Apply Corollary 1.8.3 and Theorem 2.3 (iv). •

Proposition 5.4. Let I be a left or right ideal of a semiperfect ring R. Then the following conditions are equivalent :

(i) I~ J(R). (ii) I contains no nonzero idempotents. (iii) I contains no primitive idempotents. Furthermore, if I is an ideal of R, then each of the above conditions is equivalent to :

5 Properties of idempotents

247

{iv) There exist pairwise orthogonal primitive idempotents e1 , ... , en of R with 1 = Li=l ei and with ei rJ. I for all i E {1, ... , n }. Proof. (i) ¢> (ii) : Apply Corollary 1.6.15 (i) and Theorem 2.1 (vi). (ii) ¢> (iii) : Apply Proposition 2.2.14 (iii) and Theorem 2.1 (iii). (iii) => (iv) : Obvious. (iv) => (iii) : Let f be a primitive idempotent of R. By Proposition 2.2.14 (i) and Theorem 2.1 (iii), there exist pairwise orthogonal primitive idempotents Ji = f, h, ... , fm with 1 = Lj=l fj. Hence, by Proposition 5.1, f = u- 1 eiu for some i E {1, ... , n }. But, by hypothesis, I is an ideal of R and ei rJ. I. Hence f rJ. I and the result follows. • Proposition 5.5. Let R be a semiperfect ring and let I be a left ideal of R. Assume that x1, ... , Xn E R are such that

Xi 1 ~ i,j ~ n, such that

i-/-

= x;

( mod!)

and

Xix j

=0

( mod!)

j. Then there exist orthogonal idempotents e1, ... ,en of R

(1:::; i:::; n) Moreover, if I~ J(R) and X1 +· ··+ Xn

=1

( mod!), then e1 +···+en

= 1.

Proof. Owing to Theorem 2.1 (v), R is idempotent-lifting. The desired conclusion is therefore a consequence of Theorem 1.6.11. •

Let us note that, since a semi perfect ring is idempotent-lifting, one of the major results of idempotent-lifting theory, namely Theorem 1.6.3, is valid for any semiperfect ring. Proposition 5.6. Let I be an ideal of a semiperfect ring R. {i) If e is a primitive idempotent of R with e rJ. I, then e +I is a primitive idempotent of R/I. (ii) If E is a primitive idempotent of R/ I, then E can be lifted to a primitive idempotent of R. Proof. (i) Let J ;2 I be the ideal of R with J/I = J(R/J). By Corollary 1.5.12 (i), J ;2 J(R) and, by Theorem 1.6.3, it suffices to show that e+J

Semiperfect Rings

248

is a primitive idempotent of Rf J. Now Rf J(R) is semiprimitive artinian, so there is an ideal X of R with X 2 J(R) such that R/J(R)

= J/J(R) EB X/J(R)

Since e (/. I and J / I contains no nonzero idempotents, we have e + I and the first equality follows. The second equality is established by the same argument.•

Let N be a normal subgroup of G. Then R(G/N) is an RG-module via for all

g,g1 E G

The following result provides circumstances under which R( G / N) is a direct summand of RG. Proposition 1.15. Let N be a finite normal subgroup of G such that INI is a unit of R. Then e = INl- 1 ExeN x is a central idempotent of RG and the map R(G/N) { RGe ge 1-t gN (g E G)

is an isomorphism of R-algebras and RG-modules. In particular, the RGmodule R( G / N) is a direct summand of RG. Proof. Since N is a union of some finite conjugacy classes of G, e is central. It is immediate to verify that e is an idempotent. Let f : RG R( G / N) be the natural homomorphism. Since RG = RGe EB RG(l - e) and, by Proposiiton 1.7, I( er f = RG · I(N), it suffices to show that RG · I(N) = RG(l - e). Since f( e) = 1, we have 1 - e E J( er f and therefore RG(l - e) ~ RG · I(N). Given n E N, we also have ne = e so that (n - l)e = 0 and RG · I(N)e = 0. Thus RG · I(N) ~ RG(l - e) and the result follows. •

Our next aim is to examine direct decompositions of group algebras. Proposition 1.16. Let {Rili E J} be a family of commutative rings. (i) If R is a subdirect product of Ri, i EI, then RG is a subdirect product

Group Algebras

360

of RiG, i E J. (ii) If R ~

ILe 1 Ri,

then RG ~

ILeJCRiG).

Proof. (i) By hypothesis, for any i E J, there is a surjective homomorphism /i : R -+ Ri such that nK er Ii = 0. Let 'lri : RG -+ RiG be the homomorphisI? induced by Ji. Then Kerrri = (Kerfi)G and hence nie1Ker1ri

= nie1(Kerfi)G = (nioKerfi)G = O

as required. (ii) It suffices to show that for any (Yi) in ILeJCRiG), there exists x E RG such that rrµ(x) = Yµ for anyµ E J. Write

The hypothesis on R ensures that for any g E G, there exists r9 E R such that for anyµ E J, fµ(r 9 ) = r 9 µ, Therefore, for anyµ E J,

as asserted. • Corollary 1.1 7. Let R be a co-mmutative noetherian ring. Then there exist finitely many indecomposable rings R1, ... , Rn such that

Proof. Since R is noetherian, R is a finite direct product of indecomposable rings. Now apply Proposition 1.16 (ii). •

Let us now examine criteria for RG to be artinian, noetherian and semilocal. Proposition 1.18. For any finite group G, the following properties hold: (i) RG is artinian if and only if R is artinian. (ii) RG is noetherian if and only if R is noetherian. (iii) RG is semilocal if and only if R is semilocal. Proof. (i) If RG is artinian, then so is R since R ~ RG / I( G). The converse is a consequence of Corollary 1.3.8.

1 Definitions and elementary properties

361

(ii) If RG is noetherian, then so is R since R 3:! RG / I( G). The converse follows from Corollary 1.3.8. (iii) If RG is semilocal, then (by Proposition 2.2.8 (i)) so is any homomorphic image of RG. Hence R 3:! RG / I( G) is semilocal. The converse is a consequence of Proposition 2.2.3 (ii). • Corollary 1.19. Let G be a finite group, let R be a commutative semilocal ring and let I be an ideal of RG. Then the natural homomorphism U(RG)-+ U(RG/I) is surjective. Proof. 2.2.17. •

By Proposition 1.18, RG is semilocal. Now apply Corollary

We next provide necessary and sufficient conditions for RG to be semisimple. If Risa field, then the following result reduces to the classic.µ Maschke's theorem. Proposition 1.20. Let G be a finite group and let R be a commutative ring. Then RG is semisimple if and only if R is semisimple and JGJ zs a unit of R. Proof. Assume that RG is semisimple. Since R 3:! RG / I( G), R must be semisimple. Let n be the order of an element g E G. By Theorem 2.1.1, RG is regular. Hence there exists x ERG such that (1- g)x(l - g) = 1- g. Therefore (1 - g)(l - x(l - g)) = 0. The latter, by Corollary 1.14, implies that 1- x(l - g) = (1 + g + · · · + gn-l )y

for some y E RG. Hence, by taking the augmentation of both sides, we obtain 1 = n aug(y), as required. Conversely, assume that JGJ is a unit of R and that R is semisimple. Let W be a submodule of an RG-module V. Since R is semisimple, we have V = WEB W' for some R-submodule W' of V. Let 0 : V -+ W be the projection map and let 'I/> : V -+ W be defined by 1/>(v)

= 1a1-1

L x0x- v 1

(v EV)

xEG

Since for all v EV and y E G, 1/>(yv)

= 1a1- 1 I: x0x- 1 yv = 1a1-1 I,:(yz)0(yz)- 1 yv xEG

zEG

-

Group Algebras

362

= 1a1-1 L

yzoz- 1 v

= y'lj;(v)

zEG

and since 'ljJ is R-linear, 'ljJ is an RG-homomorphism. Suppose now that v E W. Then, for any x E G, x- 1 v E W, so O(x- 1 v) = x- 1 v. Accordingly, xOx- 1 v = v and 'lj;( v) = v. Thus V =WEB J( er'lj; and the result follows. • As a preliminary to the next result, we record the following useful observation.

Let R be a commutative local ring such that R/J(R) is of prime characteristic p and let G be a finite p-group. Then (i) J(RG) = {x E RGjaug(x) E J(R)}. (ii) RG/J(RG) ~ R/J(R). (iii) U(RG) = {x E RGjaug(x) E U(R)}. Lemma 1.21.

Proof.

(i) Put F

= R/J(R).

We first show, by induction on IGI, that

J(FG)

= I(G)

The case IGI = 1 being obvious, assume that IGI > 1 and choose a central element z in G of order p. Setting N =< z >, it follows from Proposition 1.12 that FG ·I(N) = FG(z-1). Since (z- l)P = zP -1 = 0, FG ·I(N) is a nilpotent ideal of FG. Now the image of J(G) in F(G/N) under the natural map FG --t F( G / N) is J( G / N) and the latter is nilpotent, by the induction hypothesis. This means that J(G) is nilpotent modulo FG ·I(N) and hence is nilpotent. Since FG/I(G) ~ F, we deduce that J(FG) = I(G). By Proposition 1.5.25, J(R)G ~ J(RG) and hence by Corollary 1.5.12

(ii),

= J(RG/J(R)G) RG/J(R)G --t FG, "f:,x 9 g + J(R)G t--t "f:,(x 9 +J(R))g is an J(RG)/J(R)G

Since the map isomorphism, it follows from the previous paragraph that "f:, x 9 g E J( RG) if and only if "f:, x 9 E J(R), as required. (ii) By (i), J(RG) is the kernel of the surjective homomorphism RG --t

aug(x) + J(R). (iii) By (ii), RG is local and so x E U(RG) if and only if x . This is impossible since g - 1 E I( G) = J(FG) and J(FG) is nilpotent. Thus G is a p-group and the result follows. • Corollary 1.23. Let R be a commutative ring such that Rf J(R) is a field of prime characteristic p, and let P be a Sylow p-subgroup of a finite group G. If V is a finitely generated projective RC-module, then V is R-free of finite rank n such that n is divisible by IPl. Proof. Since RP is R-free of rank IPJ, it suffices to show that V is RP-free of finite rank. It clear that V is a finitely generated RP-module. Furthermore, by Propositions 4.1.1 (vi) and 1.10, V is a projective RPmodule. Since, by Proposition 1.22, RP is local it follows from Proposition 3.6.6 that V is RP-free of finite rank, as desired. •

Let A be an R-algebra and let EndR(A) be the algebra of all R-endomorphisms of A. For any given a E A, let la denote the map "left multiplication by a ". The map r given by {

A a

-+ f--+

la

is an injective homomorphism of R-algebras. We shall refer to r as the regular representation of A. Suppose now that A is R-free on a finite basis. Then we can define the character x of r by putting x( a) = tracer( a) for all a E A. The following lemma illustrates how the character of the regular representation of RG can be brought into argument.

Group Algebras

364

Lemma 1.24. Let G be a finite group and let x be the character of the regular representation of_RG. Then

x(L x 9 g) = IGlx1 gEG

Proof. By taking G as an R-basis for RG, we see that x(g), g E G, is just the number of elements in G fixed under multiplication by g. Hence x(g) = 0 for g =fi 1 and x(l) = IGI. Because xis an R-linear map, the result follows. • 0

The above result implies that in case G is finite x(x), x ERG, is a fixed scalar multiple of x1, the identity coefficient of x. Following Passman (1977), we define, for an arbitrary group G, a map tr: RG- R

called the trace map by

-

It is clear that the trace map is R-linear. Before illustrating the significance of the trace map, let us recall the following piece of information. Let A be an R-algebra which is a finitely generated projective R-module. Owing to Proposition 3.9.25, A is a symmetric R-algebra if and only if there exists 'lj; E A* = H omR( A, R) such that the following three properties hold : (i) 'lj;(xy) = 'lf;(yx) for all x, y E A. (ii) K er'lj; contains no nonzero left ideal. (iii) Every element of A* is of the form a'lj; for some a EA, where a'lj; EA* is defined by (a'lf;)(x) = 'lj;(xa), x EA. For the sake of brevity, the above situation will be described by saying that (A, 'lj;) is a symmetric R-algebra (a similar terminology will in future be applied to Frobenius algebras). Proposition 1.25. (Eilenberg and Nakayama {1955)). For any commutative ring Rand anyfinite group G, (RG,tr) is a symmetric R-algebra. Proof.

If x =

L x 9 g and

tr(xy)

y=

I: y9 g are in RG, then

= L x 9 yg-1 = L gEG

gEG

y9 xg-1

= tr(yx ),

1 Definitions and elementary properties

365

proving (i). Assume that I is a lfet ideal of RG with tr(I) = 0. If x = is in I, then for any h E G,

and thus I= 0, proving (ii). Now let f : A -+ R be an R-homomorphism and let a Then, for any h E G, tr(ha)

L x9g

= LgEG f(g)g- 1 •

= tr(L f(g)hg- 1 ) = f(h), gEG

proving (iii) and hence the result. • Corollary 1.26. given by

If G is a finite group, then the map

a(a)(x)

= tr(xa)

a : RG-+ (RG)*

a,x ERG

is an isomorphism of (RG, RG)-bimodules. Proof. Since, by Proposition 1.25_, the trace map satisfies condition (ii) in Proposition -3.9.25, the result follo~s by virtue of _the last paragraph in the proof of Proposition 3.9.25. •

In what follows, id(V) denotes the injective dimension of V. Corollary 1.27.

Let G be a finite group. Then for any R-module V, id(RG ®RV)= id(V)

where RG ®RV is regarded -as an RG-module. Pr·oof. The inequality id(RG ®RV) ~ id(V) follows from Corollary 6.8.17 and Proposition 1.25. The opposite inequality is a consequence of Proposition 6.8.16 and the fact that each R-module can be regarded as an ll.G-module -via the augmentation map RG-+ R. • Corollary 1.28. For any given commutative ring R and any finite group G, the group algebra RG is self-injective if and only if R is selfinjective. Proof.

Apply Corollary 1.27 for V == R. •-

366

2

Group Algebras

Localization

This section pertains purely to commutative ring theory and will be used to investigate support of central idempotents in group algebras. Throughout, R denotes a commutative ring. A subset S of R is said to be multiplicative if O n we have from Lemma 3.2

)Pk-n+l

x

Pk-n+l

xEX

where a E [RG,RG] and each rx ER. By Lemma 3.3, tg(a) = O. Choose k sufficiently large so that pk-n+I is greater than the p-part of the order of k any element of X. We now see that tg( e) = tg( eP ) = 0. • We have now come to the demonstration of the first major result in this section. The following theorem was furnished by Cliff in a private communication to the author. Theorem 3.5. (Cliff {1990)). Let R be an integral domain of characteristic O in which the prime p is a nonunit , let G be a finite group and let e be a central idempotent of RG. Then each element in Suppe is p-regular.

3 Support of central idempotents

373

Proof. Replacing R by the supporting subring of e, if necessary, we may assume that R is noetherian. Suppose that g E Suppe and write e = LteG ett with et E R. Since e is central, then if h E G is conjugate to g, we have eh = e9 by virtue of Proposition 1.9. Hence, because R has characteristic 0, we deduce that tg(e) 'I 0. Note also that, by Corollary 2.6, n~= 1 pnR = 0. Thus there exists n ~ 1 with tg(e) (/. pnR. Now let 1r: RG--+ (R/pnR)G be the natural map. Then 1r(e) is a central idempotent of (R/pnR)G, and it is clear that tg(1r(e)) = 1r(t9 (e)) 'I 0. It therefore follows from Lemma 3.4 that g is p-regular, as desired. • In.what follows, Ca(P) and Na(P) denote the centralizer and normalizer of Pin G, respectively, while Cl(G) denotes the set of all conjugacy classes of G. For any given subset X of G, we write x+ for the sum of all elements in X. Let RG be the group algebra of a finite group G over a commutative ring R of prime characteristic p. If P is a p-subgroup of G, S = Ca(P) and H a subgroup of G with S ~ H ~ Na(P), then the projection map 1r: RG--+ RS induces a ring homomorphism of Z(RG) into Z(RH) whose kernel is the R-linear span of all c+ with C E Cl(G) and Lemma 3.6.

C

n S = 0.

Proof. Let C1 , C 2 , ••• , Cr be all conjugacy classes of G, let Xi = Ci n S and assume that Xk 'I 0 fork= 1, ... , t and Xk = 0 fork= t +1, ... , r. We know, from Proposition 1.9, that the elements ct, C;j", .. . , C;t constitute an R-basis for Z(RG). Observe that r

1r(E >..ict) i=l

t

= I: >..ixt i=l

and that X 1 , ••• , Xt are mutually disjoint. Furthermore, each Xi, 1 ~ i ~ t, is a union of conjugacy classes of H. Thus 1r induces an R-homomorphism from Z(RG) to Z(RH) whose kernel is of the required form. We are therefore left to verify that 1r(CtCf) = 1r(Ct)1r(Cf) for all i,j E {1, ... , r }. To this end, fix i,j E {1, ... , r} and, for any given s E S, put T(s) = {(x,y) E Ci X CjJxy = s}, T = UsesT(s)

If T = 0, then either Xi = 0 or Xj = 0, in which case 1r(CtCf) = 0

=

Group Algebras

374

rr( Ct)rr( Cf). Assume that T =/= 0. Then rr(Ctct)

=

L

xy

(x,y)ET

=L

L

xy

sES (x,y)ET(s)

Since for all h E P ands ES, h- 1 sh = s, Pacts as a permutation group on T( s) via ( x, y)h = (h- 1 xh, h- 1 yh). Since char R = p, the sum of all elements xy where (x, y) ranges all orbits of T(s) of size =f: 1, is equal to zero. On the other hand, an orbit of T(s) has size 1 if and only if it is of the form {(x, y)} with x,y ES and hence with x E Xi, y E Xj. It follows that

rr(C;Cf)

=

L

xy

= rr(C;)rr(Cf)

(x,y)EX;XXj

as required. • We shall refer to the homomorphism Z(RG) -+ Z(RH) constructed in Lemma 3.6 as the Brauer homomorphism. The following result for the case where Risa field of characteristic p > 0 is due to Osima (1955) (see also Passman (1969)). Theorem 3.7. Let R be a commutative ring of prime characteristic p, let G be a finite group and let e be a central idempotent of RG. Then each element in Suppe is p-regular. Proof. Assume by way of contradiction that g E Suppe, g = xy = yx, where x =/= 1 has order of a power of p and where the order of y is prime top. Let P be the subgroup of G generated by x, let S = Cc(P) and let rr: Z(RG)-+ Z(RS) be the Brauer homomorphism. Then rr(e) is a central idempotent of RS. Furthermore, since g E S, we also have g E Supprr(e). Consequently, we may assume that g E Z( G), in which case tg( e) =f: 0. Since the latter contradicts Lemma 3.4, the result follows. •

4

Some module isomorphisms

In this section, we exhibit a number of isomorphisms of modules over group algebras. Throughout, RG denotes the group algebra of an arbitrary group Gover a commutative ring R. The following proposition can also be derived as a consequence of Theorem 4.2. However, we feel it deserves a separate proof since the latter is very instructive.

4 Some module isomorphisms

Proposition 4.1.

375

Let G be a group and let I( G)

{i) G/G' e:! {ii) "f:,xg(g-1) = Tigx 9 - l(modJ(G) 2 ) (iii) G n (1 + J(G) 2 ) = G'.

= I(7l G).

Then

I(G)/I(G) 2 •

Proof.

(x 9 E 7l ).

(i) Consider the map 'ljJ: G-+ I( G)/ I( G) 2 defined by

'l/J(g)=g-l+I(G) 2 Then the identity

ab - 1 = (a - 1) ( b - 1) + (a - 1) + (b - 1)

(a,bEG)

ensures that 'ljJ is a homomorphism. Hence 'ljJ induces a homomorphism

{ G/G' gG'

L f-*

I(G)/I(G) 2 (g - 1) + J(G) 2

Since J(G) is a free abelian group with basis {g - 111 :p g E G}, the map g - 1 f-t gG' extends to a homomorphism ..v) = g(>..v) = 9 >..gv = 9 >..p(g)v for all r

>.. E F, g E G, v E V. This shows that each p(g) E GS(V), the general semilinear group of V. Here the automorphism of E determined by p(g) coincides with that determined by g. Moreover, by the definition of p, we have p(x )p(y) - a(x, y)p(xy) for all x, y E G. For each g E G, let r p(g) be the matrix of p(g). Then one easily verifies that

where xr p(y) is obtained from r p(y) by applying x to each entry. Taking the determinants of both sides and setting µ(g) = detr p(g), we obtain for all

x, y E G

Thus ar is a coboundary. Hence, by Theorem 4.7, [AY

= 1.

•

The order of [A] E Br(F) is called the exponent of [A]. For any central simple F-algebra A, its exponent is defined to be the exponent of its Brauer class [A]. Thus Lemma 4.9 may be expressed by saying that for any Brauer class, the exponent divides the index. Although it is not true in general that the exponent is equal to the index, we do have the following general result.

For any [A] E Br(F), the index and exponent of [A] have the same prime factors. Theorem 4.10.

Proof. Let [A] have index r and exponent n, so that nlr by Lemma 4.9. It therefore suffices to verify that, for any prime p, plr implies pin. In

4 The Brauer group and crossed products

517

the notation of Corollary 4.8, we may assume that A = E0tG. Let S be a Sylow p-subgroup of G and let E 0 be the fixed field of S. Then (Eo : F) = ( G : S) = m is prime to p, while ISi = pk for some k ~ 1. Let µ denote the index of A ®F Eo, Then, by Proposition 2.4, µlrlµm. Moreover, since A is split by E, we have

Bearing in mind that ( E : E 0 ) = pk, we deduce that µlpk. Hence µ is a power of p. Ifµ = 1, then rim, but plr and pf m, a contradiction. Thus µ is a positive power of p.Let AJL be the tensor product over F ofµ copies of A and let (A ®F E 0 )µ be the tensor product over Eo ofµ copies of A@p E 0 . Then Aµ ®F E 0 ~ (A ®F E 0 )µ and, by Lemma 4.9. [A ®F E 0 ]µ = 1. Hence the exponent of A is divisible by p, as desired. • We now derive two consequences of interest. Corollary 4.11. Let D be a central simple division F-algebra of index n = p~ 1p; 2 ... p:• where the Pi are distinct primes. Then

where Di is a central simple division F-algebra of index pf;, 1 Proof. exponent

~

i ~ s.

By Theorem 4.10 and Lemma 4.9, the Brauer class [D] has

m -_ pm1pm2 1 2

pm•

"· s

where O < mi ~ ni, 1 ~ i ~ s. By the structure theorem for abelian groups, [D] can be written as a product of Brauer classes which are powers of [D] with prime power exponent. Let Di be a division algebra similar to a power of [D] with exponent Pi;, 1 ~ i ~ s. Then

Owing to Theorem 4.10, the Di have coprime indices. Hence, by Theorem 3.2, we have a division algebra on the right, and therefore the two sides are isomorphic. • A central simple F-algebra is called primary if it is -:/ F and contains no proper central simple F-subalgebra.

The Brauer Group

518

Corollary 4.12. (i) Every central simple F -algebra is a tensor product of primary ones. (ii) A primary F-algebra is either a division algebra of prime power index or of the form Mp(F), where p is a prime. Proof. (i) Direct consequence of Theorem 1.9. (ii) Apply Corollary 4.11. •

We close by remarking that the converse of Corollary 4.12 (ii) does not hold in general : a division algebra of prime power index need not be primary.

Chapter 13

Indecomposable Modules and Ground Field Extensions In this chapter we examine in detail the behaviour of indecomposable modules under ground field extensions. In future we shall exhibit a striking similarity between seemingly unrelated processes, namely between restriction of modules to normal subgroups and decomposition of modules under ground field extensions. The role of a conjugate of a module will be played by a Galois conjugate, while the Schur index will play the role of the ramification index. The main results are based on an important work of Thevenaz ( 1983 a) and some of the theorems recorded will be used in the next chapter which deals with the Schur index.

1

Preliminary results

All algebras below are assumed to be finite-dimensional over their ground fields and all modules over algebras are assumed to be finitely generated. Some of the results recorded below will be used in the next chapter. Let F be any field and let G be a group of automorphisms of F. Recall that the set of all elements in F fixed under all automorphisms in G is a subfield of F called the fixed field of G. In what follows, we use some standard field-theoretic facts which can be found in Karpilovsky (1988). Lemma 1.1.

Let E / F be a normal extension of fields of characteristic 519

Indecomposable Modules and Ground Field Extensions

520

p > 0 and let G = Gal(E/ F). If K is the fixed field of G, then E/ K is Galois and K / F is purely inseparable.

Proof. Since Gal(E/F) = Gal(E/K), it follows that if). EE is such that o-(A) =). for all o- E Gal(E/K), then). EK. Thus E/K is Galois. Let E be an algebraic closure of E and let o- : K -+ E be an F-homomorphism. Then o- extends to an automorphism of E. Its restriction to E is an Fautomorphism of E, since E / F is normal. Thus o-( x) = x for all x E J(, as required. • Lemma 1.2. Let Z be the centre of a division algebra D over a field F and let A be an F-algebra. Then the map

is a bijection between ideals of D ®FA and those of Z ®FA.

Proof. Let {ai} be a basis for A over F. Then, upon usual identifications, { ai} is a basis for Z ®FA as vector space over Zand D ®FA as vector space over D. Hence we may identify D ®FA with D ®z (Z ®FA). Let G be the group of all inner automorphisms of D. Then any o- E G determines an automorphism i7 of D ®FA given by

=

{ulo- E G} and show that the result will follow provided we We put G prove that the map V f-+ Vn(Z®FA) is a bijection between the collection of subspaces of D ®FA over D invariant under G and all subspaces of Z ®FA over Z. Indeed, assume that the above map is a required bijection. If I is an ideal of D ®FA, evidently I is a D-subspace mapped into itself by every element of G. Hence If-+ In (Z ®FA) is an injection of the set of ideals of D ®FA into the set of ideals of Z ®FA. On the other hand, if J is an ideal of Z ®FA, then I= D ®z J is an ideal of D ®FA, since the elements of D ® 1 and those of 1 ® J commute. Hence I is invariant under left or right multiplication by the elements of D as well as those of I. Therefore I f-+ In (Z ®FA) is a bijection between the set of ideals of D ®FA and Z®FA. By the foregoing, we are left to verify that V f-+ V n (Z ®FA) is a bijection between the G-invariant D-subspaces of D ®F A and all Z-subspaces of Z ®FA. To this end, note that Z ®FA consists precisely of those elements

1 Preliminary results

521

of D @FA which are fixed by all elements of G. Put M = Z @FA and denote by V any subspace of D@z M = D @FA which is mapped into itself by every ii E G. Then V n Mis a Z-subspace of Z @FA and we are left to show that V=D@z(VnM) Let {ekik E K} be a Z-basis of M. Then we can choose a subset L of I( such that the subspace W = LkEL Dek is a complement of Vin D @z M, that is D@zM = VEBW Let C be the complement of Lin J( and write for c EC, ec =le+ Uc where le EV, Uc E W. Then one easily verifies that {lclc EC} is a Z-basis for V. If

.(x1) + >.(x2) and obviously >.(µ( a, cp)) = µ>.( a, cp) for all µ E F. Thus >. E B*. Now assume that b = (a, cp) is such that >.(bB) = 0. Then, for all x E A and '1/J EA*, we have (a'I/J + cpx)(l) = 0. Hence for all x E A, '1/J E A* '1/J( a) + cp( X) = 0 In particular, taking x = O, we obtain '1/J(a) = 0 for all '1/J E A* and hence a = 0. Therefore cp = 0 also and the result follows. • If Xi

2 Cogenerators

2

559

Cogenerators

In this section, we fix a finite-dimensional algebra A over a field F. As usual, all A-modules below are assumed to be finitely generated. We also assume (unless explicitly stated otherwise) that all A-modules are left A-modules. Let V be a (left or right) A-module. Then V is called a cogenerator if for any A-module U, nJEHomA(U,V)J( er f = 0 Lemma 2.1. Let V be a (left or right) A-module. Then V is a cogenerator if and only if for any simple A-module W, the injective hull of W is isomorphic to a submodule of V. Proof. We shall treat the case where Vis a left A-module. A similar argument will establish the case where Vis a right A-module. Assume that Vis a cogenerator and let W be a simple A-module with injective hull I(W). Since V is a cogenerator, nJEHomA(I(W),V)J( er f

=0

and hence there exists f E HomA(I(W), V) with W i Kerf. Since Wis simple, we must have W n J( er f = 0. But Wis an essential submodule of I(W), hence J( er f = 0 and so I(W) ~ f(I(W)). Conversely, assume that for any simple A-module W, the injective hull of Wis isomorphic to a submodule of V. Let U f:. 0 be an arbitrary A-module and choose Of:. u E U. Denote by M a maximal submodule of Au and let W = Au/M. If I(W) is an injective hull of W, then by hypothesis there exists an injective A-homomorphism 'ljJ: I(W)---+ V. Let

1r: Au---+ Au/M be the natural homomorphism. Then 1r( u) = u + M f:. 0. Since I(W) is an injective module, the homomorphism 7r: Au---+ I(W) can be extended to a homomorphism .X : U---+ I(W), (see Proposition 3.8.1 (iv)). In particular, .X( u) = 1r( u) f:. 0. Setting f = 'ljJ o .X, it follows that f E Hom A ( U, V) and u ¢ Kerf. Thus nJEHomA(U,V)J( er f = 0, proving_ that Vis a cogenerator. • Lemma 2.2. Let Vi, ... , v;. be all nonisomorphic simple A-modules and let I(¼) be the injective hull of½,, 1 ~ i ~ r. Then

Frobenius and Symmetric Algebras

560

(i) I(Vi), ... , I(Vr) are all nonisomorphic injective indecomposable Amodules. (ii) If an A-module V is a cogenemtor, then each I(¼) is isomorphic to an indecomposable direct summand of V. {iii) If an A-module V is an injective cogenerator, then V ~ EBi=t nil(¼)

for some r 2': 1 and some ni 2". 1, 1 ::;

i::; r.

Proof. (i) This is a special case of Lemma 7.3.1. (ii) By Lemma 2.1, V contains a copy of each I(¼) and any such copy being an injective submodule of Vis a direct summand of V. Since, by (i), each I(¼) is indecomposable, the required assertion follows. (iii) Apply (i) and (ii). • In what follows, in accordance with our previous notation, we put A* H omp(A, F). We remind the reader that A* is an (A, A)-bimodule via

(af)(x) (fa)(x)

=

=

=

f(xa) f(ax)

for all x, a E A and f E A*. Unless explicitly stated otherwise, we regard A* as a left A-module. Lemma 2.3. {i) Soc(A*) = {1P E A*IJ(A) ~ Ker7P}. (ii) A* is an injective hull of Soc(A*) and A/ J(A) ~ Soc(A*). {iii) A* is an injective cogenemtor.

Proof. (i) Note that Soc(A*) = {1P E A*IJ(A)1P = O}. Hence it suffices to show that J(A)1P = 0 if and only if J(A) ~ 1( er 'Ip. If J(A)1P = 0, then for all a E J(A), (a1P)(l) = 1P(a) = 0 and so J(A) ~ Ker7P. Conversely, assume that J(A) ~ Ker7P. Then, for all a E J(A), x EA, 0 = 1P(xa) = (a1P)(x) and therefore J(A)1P = 0, as required. (ii) We know, from Proposition 3.9.11 (iii), that A* is injective. Since Soc(A*) is an essential submodule of A*, it follows from Proposition 3.8.12 that A* is an injective hull of Soc(A*). By (i), the map

L

(A/ J(A))*

I-+

if;

2 Cogenerators

561

where '¢;(a+ J(A)) = 7/J(a), is well defined. It is clear that f is an injective A-homomorphism. Furthermore, for any µ E (A/J(A))* and the natural map 1r: A-+ A/ J(A), we haveµ o 1r E Soc(A*) andµ o 7r = µ. Thus f is an A-isomorphism. Now A/J(A) is symmetric, by virtue of Proposition 1.4. Hence there is an A/ J(A)-isomorphism g: (A/ J(A))*-+ A/ J(A)

We may clearly view gas an A-isomorphism, in which case go f: Soc(A*)-+ A/ J(A)

is a required A-isomorphism. (iii) By (ii), Soc(A*) ~ A/ J(A) and hence A* contains a copy of any simple A-module. Since A* is injective, it must also contain a copy of an injective hull of any simple A-module. The desired conclusion is therefore a consequence of Lemma 2.1. • We close by proving two auxiliary results, which will be used in the next section. In what follows, l(X) and r(X) denote the left and right annihilators of X, respectively. Lemma 2.4.

If A is an injective left A-module, then for any left ideals

X,Y of A, r(X n Y) Proof. It is clear that r(X) that a E r(X n Y). Then the map

= r(X) + r(Y)

+ r(Y)

{ X+Y X +y

~

r(X n Y). Conversely, assume

L

A

1-->

ya

is well defined, since for x, x1 E X, y, Y1 E Y, x + y = x1 + Y1 implies x - x1 = Y1 - y E X n Y and so Y1 a = (Y1 - y )a + ya = ya. It is clear that f is an A-homomorphism. Since A is injective, f extends to an Ahomomorphism A -+ A (see Proposition 3.8.1). Hence there exists z E A such that

f(x+y)=(x+y)z=ya

forall

xEX,yEY

Frobenius and Symmetric Algebras

562

In particular, taking y = 0, we obtain z E r(X), and taking x = 0 we obtain f(y) = yz = ya for all y E Y. The latter implies a - z E r(Y) and hence a= z + (a - z) E r(X) + r(Y), as required.•

Assume that A satisfies the following two conditions : (a) For any left ideals X, Y of A, r(X n Y) = r(X) + r(Y). (b) For any right ideal X of A, r(l(X)) = X. Then AA is injective. Lemma 2.5.

Proof. Let X be a left ideal of A and let f : X ---+ A be an Ahomomorphism. By Proposition 3.8.1, it suffices to verify that f can be extended to an A-homomorphism A ---+ A, or equivalently that there exists a EA such that f(x) = xa for all x EX. We argue by induction on the number n of generators of X. First assume that n = 1, say X = Axo for some xo EX. Since yxo = 0 for y EA implies f(yxo) = yf(xo) = 0, we have l(xo) ~ l(f(xo)). Thus l(xoA) ~ l(f(xo)A) and, by (b), xoA = r(l(xoA)) ;;;2 r(l(f(xo)A)) = f(xo)A

Hence there exists a E A with f(xo) = xoa and so f(yxo) = (yxo)a for all y E A, proving the case n = 1. Let X = I:7~/ Axi for some Xi E X. Then, by induction hypothesis, there exist z1,z2 E A such that /(I:'f= 1 YiXi) = (Li=l Yixi)z1 and f(Yn+1Xn+1) = Yn+1Xn+1z2 for all Y1, ... , Yn+I in A. Applying (a), we have n z1 -

E

z2

n

r(L Axi n Axn+I) = r(L Axi) + r(Axn+1) i=l

and hence z1 a= z1 - c = z2

i=l

= c - d with c E d, we obtain

z2 -

n+l

f(L YiXi) i=l

r(I:~ 1 Axi),

d E r(Axn+1). Setting

n

f(L YiXi) + f(Yn+1Xn+1) i=l n

(L YiXi)(z1 - c) + Yn+1Xn+1(z2 i=l

n+l

(L Yi Xi )a, i=l

as required. •

d)

3 Quasi-Frobenius algebras

3

563

Quasi-Frobenius algebras

Throughout this section, A denotes a finite-dimensional algebra over a field F. All left and right A-modules are assumed to be finitely generated. Recall that, by definition, A is a Frobenius algebra if A ~ A* as left A-modules. A weaker requirement is that if A1, ... , An and B1, . .. , Bk are all nonisomorphic indecomposable direct summands of A and A*, respectively, then n = k and, after possibly reordering the Bi, Ai ~ Bi for all i E { 1, ... , n}. If this requirement holds, then we say that A is a quasi-Frobenius algebra . In this section, we provide a number of characterizations of quasi-Frobenius algebras. Theorem 3.1. The following conditions are equivalent : (i) A is a quasi-Frobenius algebra. (ii) AA is a cogenerator. {iii) A is self-injective (i.e. AA is injective). (iv) AA is a cogenerator. (v) AA is injective. (vi) For any left ideal X of A and any right ideal Y of A,

l(r(X))

=X

and

r(l(Y))

=Y

*

Proof. (i) (ii) : By Lemma 2.3.(iii), A* is an injective cogenerator. Hence in the notation of Lemma 2.2 (iii),

(1) for some positive integers ni, 1 ~ i ~ r. By hypothesis, A is a quasiFrobenius algebra, hence each I(½) is isomorphic to a submodule of AA. Hence, by Lemma 2.1, AA is a cogenerator. (ii) (iii) : By Lemma 2.2 (ii), each I(½) is isomorphic to an indecomposable direct summand of AA, Since the number of nonisomorphic indecomposable direct summands of AA is precisely r (Proposition 2.3.4), we conclude that all indecomposable direct summands of AA are injective. Hence AA is also injective. (iii) =} (i) : Since AA is injective, so is any indecomposable direct summand of AA. But the number of such nonisomorphic summands is precisely r, hence by Lemma 2.2 (i), I(½), ... ,I(Vr) are all nonisomorphic indecomposable direct summands of A. Now apply (1). (ii) (vi) : Let X be a left ideal of A. Then X is a submodule of AA

*

*

Frobenius and Symmetric Algebras

564

and we obviously have X ~ l(r(X)). In particular, we may assume that A-:/- X, in which case we can choose a EA, a(/ X. Since AA is a cogenerator, there exists a homomorphism f : A/ X ---+ A for which J( a+ X) -:/- 0. If 7r: A---+ A/Xis the natural homomorphism, then O = J01r(X) = X(f 01r(l)). Hence (Jo 1r)(l) E r(X). Moreover,

a(! o 1r)(l) =(Jo 1r)(a)

= J(a + X)-:/- 0

and so a(/ l(r(X)). This proves that l(r(X)) = X. Let Y be any right ideal of A and write Y = Li=I YiA for some Yi E Y. Then l(Y) = nf= 1l(yiA) and, applying Lemma 2.4, we have n

n

r(l(Y)) = r(l(LYiA)) =

r(nf= 1 l(yiA)) = L r(l(yiA)) i=l

i=l

It therefore suffices to show that

r(l(yA))

= yA

for all

y EA

It is clear that yA ~ r(l(yA)). Assume that x E r(l(yA)). Then l(y) ~ l(x) and so the map Ay ---+ A, ay 1-t ax, a E A, is an A-homomorphism. Since A is injective, there exists z E A such that yz = x, i.e. x E yA. Thus r(l(yA)) ~ yA, as required. (vi) => (iii) : By Lemma 2.5, it suffices to show that for any left ideals X, Y of A, r(X n Y) = r(X) r(Y). By hypothesis,

+

l(r(X n Y))

X nY l(r(X)

= l(r(X)) n l(r(Y))

+ r(Y))

Applying right annihilators to both sides of the above equality, the required assertion follows. (iii) {:} ( v) : Let e1, ... , en be primitive idempotents of A such that Ae 1 , ••. , Aen are all nonisomorphic principal indecomposable A-modules. Then, by Lemma 1.2.6, e1 A, ... , enA are all nonisomorphic principal indecomposable right A-modules. By Proposition 3.9.11, (Ae1)*, ... , (Aen)* are nonisomorphic indecomposable right A-modules and any given Aei is injective if and only if (Aei)* is projective. This shows that (iii) implies (v) and the reverse implication follows by a similar argument. (iv) {:} (v) : Apply the right analogue of arguments employed in the proof of (ii) {:} (iii). •

3 Quasi-Frobenius algebras

565

Corollary 3.2. The following conditions are equivalent : (i) A is a quasi-Frobenius algebra. (ii) Every projective left A-module is injective. (iii) Every injective left A-module is projective. (iv) Every projective right A-module is injective. (v) Every injective right A-module is projective. (vi) For any left ideals X, Y of A and for any right ideal Z of A,

r(X

n Y) = r(X) + r(Y) and r(l(Z)) = Z

Proof. Note that AA (respectively, AA) is injective if and only if every projective left (respectively, right) A-module is injective. Hence, by Theorem 3.1, (i) and (ii) a.re equivalent. By Lemma. 2.2 (i), the number of nonisomorphic (left or right) injective indecomposable A-modules is equal to the number of nonisomorphic projective indecomposable A-modules. This shows that (ii) is equivalent to (iii) and (iv) is equivalent to (v). The implication (ii) :::} (vi) is a. consequence of Lemma. 2.4 and Theorem 3.1, while the implication (vi) :::} (ii) follows from Lemma. 2.5. •

Let V be a. (left) A-module. Then H omA(V, A) is a. right A-module via. (fa)(v)

= f(v)a

(f E HomA(V,A), v EV, a EA)

Similarly, if V is a. right A-module, then H omA(V, A) is a. left A-module via. (af)(v)

= af(v)

(f E HomA(V,A), v EV, a EA)

Lemma 3.3. Let A be a quasi-Frobenius algebra, let V be a (left or right) A-module and let U be a submodule of V. If

U1-

= {f E H omA(V, A)lf(U) = O}

then the map { HomA(V,A)/U1f+UJ_

~ HomA(U,A) f-+

JIU

is an A-isomorphism. Proof. Assume for definiteness that V is a. left A-module (the same argument will establish the case where V is a. right A-module). Then, by

Frobenius and Symmetric Algebras

566

definition, ,\ is an injective A-homomorphism. By Theorem 3.1, AA is injective and so, by Proposition 3.8.1, any A-homomorphism 1jJ: U - A can be extended to an A-homomorphism f: V - A. Since .X(f + u.1.) =JIU= 1/J, the result follows. • Theorem 3.4. The following conditions are equivalent : (i) A is a quasi-Frobenius algebra. (ii) For any simple left and right A-modules Vi and Vi, respectively, HomA(½,A) is simple, i = 1,2. (iii) For any left and right A-modules Vi and Vi, respectively, the composition length of½ is equal to that of HomA(½,A), i = 1,2. Proof. (i) => (ii) : Let V be a simple left A-module. By Theorem 3.1, AA is a cogenerator. Hence there exists an injective A-homomorphism f : V - A and so H omA(V, A) f= 0. Fix O f= o: E H omA(V, A). We will show that the right A-module H omA(V, A) is generated by o: which will imply that HomA(V,A) is simple. Because AA is injective (Theorem 3.1), for any f3 E H omA(V, A) there exists an A-homomorphism (3' : A - A such that f3 = (3' o o:. Since (3' is a right multiplication by some element a E A, we have f3 = o:a. A similar argument establishes the case where Vis a simple right A-module. (ii) => (i) : Let X and Y be left and right ideals of A, respectively. By '.Theorem 3.1, it suffices to show that l(r(X)) = X and r(l(Y)) = Y. Let X1 ~ X2 be left ideals of A such that X2/ X 1 is a simple A-module. We first claim that either r(X1)/r(X2) is simple or equal to zero. The same argument will prove an analogous assertion for right ideals of A. Consider the map f: r(X1)/r(X2) - HomA(X2/X1,A) given by

This map is clearly well defined and is an injective homomorphism of right A-modules. Since, by hypothesis, H omA(X2/ X 1, A) is simple, the claim is verified. There exists a composition series of AA containing X, say 0

= Xo C X1

C · · · C Xt

=A

(2)

Now consider the chain A= r(Xo) ;;2 r(X1) ;;2 · · · ;;2 r(Xt)

=0

(3)

567

3 Quasi-Frobenius algebras

By the claim proved above, c(AA) :::; c(AA), where c(V) is the composition length of an A-module V. By symmetry, c(AA) :::; c(AA) and so c(AA) = c(AA)· Hence (3) is a composition series of AA. But then 0 = l(r(Xo)) C · · · C l(r(Xt)) = A

is also a composition series of AA. Since (2) is a composition series for AA and Xi ~ l(r(Xi)), we have Xi = l(r(Xi)), 1 :::; i :::; t. In particular, l(r(X)) = X. A similar argument proves that r(l(Y)) = Y for any right ideal Y of A. (iii) => (ii) : Obvious. (ii)=> (iii): Let V be a left A-module. To prove that V and H omA(V, A) have the same composition length, we argue by induction on n = c(V). By hypothesis, the required assertion is true for n = 1. Assume that the result is true for all V with c(V) :::; n and let W be a left A-module with c(W) = n+ 1. Denote by U a simple submodule of W. Then, by induction hypothesis, c(HomA(W/U,A)) = n. Put u1- = {f E HomA(W,A)lf(U) = O}. Then obviously HomA(W/U,A) ~ u1- and hence c(U1-) = n. Now A is a quasi-Frobenius algebra (by (ii) => (i)) and hence, by Lemma 3.3, the map HomA(W,A)/U1- HomA(U,A) { f + UJ_ ~ JIU is an isomorphism. Since, by hypothesis, H omA(U, A) is simple, we deduce that c(HomA(W,A)) = n+l. A similar argument establishes the case where V is a right A-module. • Theorem 3.5. The following conditions are, equivalent : (i) A is a quasi-Frobenius algebra. {ii) For any primitive idempotent e of A, Soc(Ae) and Soc(eA) are simple and Soc(AA) (respectively, Soc(AA)) contains an isomorphic copy of all simple left (respectively, right} A-modules. (iii} For any primitive idempotent e of A, Soc(Ae) and Soc(eA) are simple and Soc(AA) = Soc(AA). (iv) If e 1, . .. , ek are primitive idempotents of A such that Ae1, ... , Aek are all nonisomorphic principal indecomposable A-modules, then there exists a permutation 1r of {1, ... , k} such that

Soc(Aei) ~ Ae1r(i) where ei is the image of ei in

and S0c(e1r(i)A) ~ ei.A

A = A/ J(A).

(1:::; i:::; k)

Frobenius and Symmetric Algebras

568

*

Proof. (i) (ii) : Let e be a primitive idempotent of A and let V be a simple submodule of Ae. Since Ae is indecomposable and, by Corollary 3.2, it is also injective, it follows from Proposition 3.8.17 that Ae is an injective hull of V. Hence V is an essential submodule of Ae, which implies that V = Soc( Ae ). A similar argument shows that Soc( eA) is simple. Since, by Theorem 3.1, both AA and AA are cogenerators, the remaining assertion follows by virtue of Lemma 2.1. (iii) : Let e be a primitive idempotent of A. We will show that (ii) Soc(AA) contains Soc(eA), which will imply that Soc(AA) ~ Soc(AA). A similar argument will show the opposite inclusion. By hypothesis, Soc(AA) contains an isomorphic copy of Ae and hence eSoc(AA) -::/ 0. Since Soc(AA) is a two-sided ideal, it follows that Soc(AA) 2 eSoc(AA)-::/ 0. But Soc(eA) is simple and is an essential submodule of eA. Hence Soc(AA) 2 eSoc(AA) 2 Soc(eA), as required. (iii)* (ii): Let e be a primitive idempotent of A. Since O-/:- Soc(eA) = eSoc(eA), it follows that O -:/ eSoc(AA) = eSoc(AA). Hence there exists x E Soc(AA) such that Aex is simple. Therefore A e .~ Aex which shows that Soc(AA) contains an isomorphic copy of all simple left A-modules. A similar argument shows that Soc(AA) contains an isomorphic copy of all simple right A-modules. (iii) * (iv) : Since Soc(Aei) is simple for each i E {1, ... , k }, there exists 11"( i) E {1, ... , k} such that

*

(1::;

i::; k)

(4)

By (ii), Soc(AA) contains an isomorphic copy of all simple left A-modules which readily implies that 'II" is a permutation of {1, ... , k }. Choose ai E A such that Soc( Aei) = Aaiei, 1 ::; i ::; k. It follows from ( 4) that e1r(i)aiei -::/ 0 and hence

Soc(Aei)

= Ae1r(i)aiei

(5)

= Soc(AA), we have e1r(i)aieiA ~ Soc(AA) n e1r(i)A = S0c(e1r(i)A)

Because O-:/ e1r(i)aiei E Soc(AA)

and, since S0c(e1r(i)A) is simple, we infer that

S0c(e1r(i)A)

= e1r(i)aieiA

But then the surjective homomorphism

S0c(e1r(i)A) { eiA eia 1-+ e1r(i)aieia

(6)

3 Quasi-Frobenius algebras

569

induces an isomorphism ei A~ S0c(e1r(i)A), as required. (iv) =} (ii) : By the Krull-Schmidt theorem, we may assume that the idempotent e in (ii) coincides with ei in (iv) for some i E { 1, ... , k}. Thus (ii) is a consequence of (iv). (ii) =} (i) : Owing to Theorem 3.4, it suffices to show that for any simple left and right A-modules V and W, respectively, HomA(V,A) and H omA(W, A) are simple. By hypothesis, the latter will follow provided (in the notation of (iv)) we show that the modules H omA(Soc(Aei), A) and HomA(Soc(eiA),A) are simple, 1 ~ i ~ k. By symmetry, it suffices to show that H omA(Soc(Aei), A) is simple. Consider a nonzero A-homomorphism f : Soc(Aei) -+ A. Since (ii) implies (iii), it follows from (5) that for any a E A,

and that for x

= f(e1r(i)aiei),

Ae1r(i)x

0 cf e1r(i)xA ~ Soc(AA)

= f(Soc(Aei)) is simple. n e1r(i)A

Now

= S0c(e1r(i)A)

and, by hypothesis, Soc( e1r(i)A) is simple. Hence, by (6),

= S0c(e1r(i)A) = e1r(i)aieiA and therefore there exists e1ao E eiA such that e1r(i)X = e1r(i)aieiao. e1r(i)xA

It follows

that for all a E A,

which means that f(y) = yeiao for ally E Soc(Aei) = Ae1r(i)aiei. Hence the map { eiA :I:,, HomA(Soc(Aei),A) eia 1--+ r a where ra(Y)

= yeia, y E Soc(Aei) is a surjective A-homomorphism. 0

Since

= Soc(AA)J(A) = Soc(AA)J(A)

we have eiJ(A) ~ Ker'lj;. But eJ(A) is a maximal submodule of eiA, hence H omA(Soc(Aei), A) is simple, as required. • We next provide the following application of Theorem 3.1.

Frobenius and Symmetric Algebras

570

Theorem 3.6. Let A be a quasi-Frobenius algebra. Then the map X f-r r(X) is a duality of the lattice of left ideals of A onto the lattice of right ideals of A, that is X f-r r(X) is a bijection with

r(X1 + X2)

= r(X1) n r(X2)

and

r(X1

n X2) = r(X1) + r(X2)

for all left ideals X 1 , X 2 of A. Similarly, X f-r l( X) is a duality of the lattice of right ideals of A onto the lattice of left ideals of A. Proof. By Theorem 3.1 (vi), the maps X f-r r(X), X f-r l(X) are bijections. We obviously have r(X1 + X2) = r(X1) n r(X2). Furthermore, by Lemma 2.4, r(X 1 n X 2) = r(X1) + r(X 2), proving the first assertion. To prove the second assertion, we first note that l(X1 + X2) = l(X1) n l(X2) for any right ideals X 1 , X 2 of A. Furthermore, by Theorem 3.1 (vi),

l(r(l(X1)) n r(l(X2))) l(r(l(X1) l(Xi)

+ l(X2))

+ l(X2),

as required. • As an easy application of Corollary 3.2, we next record the following useful result. Theorem 3. 7. Let A be a quasi-Frobenius algebra and let n be the nilpotency index of J(A) (i.e. J(At = O and J(At- 1 =/= 0). Then each indecomposable A-module V such that J(At- 1V =/= 0 is projective. Proof. Choose a simple submodule W of J(At- 1V and let I(W) be the injective hull of W. Note that, by Theorem 7.5.1 (iii), I(W) is a finitely generated A-module. Hence, by Corollary 3.2, I(W) is a finitely generated projective A-module. Since I(W) is injective, there is an A-homomorphism f : V --+ I(W) with f( w) = w for all w E W (see Proposition 3.8.1 (iii)). Now I(W) is projective and, by Corollary 3.8.20, it is also indecomposable. Hence J(A)I(W) is a unique maximal submodule of I(W) (Proposition 2.3.4 (ii)). If f(V) ~ J(A)I(W), then

f(J(Ar- 1 V)

= J(Ar- 1f(V)

~ J(At I(W)

=0

which is impossible, since W ~ J(At- 1V and f(w) = w for all w E W. Thus f is surjective and so I(W) is a direct summand of V. Since V is indecomposable, we conclude that V = I(W) is projective. •

4 Frobenius algebras

571

Corollary 3.8. Let A be a quasi-Frobenius algebra and let n be the nilpotency index of J(A). Then A is of finite representation type if and only if A/ J(At- 1 is of finite representation type. Proof. Assume that A/J(At- 1 is of finite representation type. If V is an indecomposable A-module which is not an A/J(At- 1 -module (i.e. if J(At- 1 V f:. 0), then V is projective, by Theorem 3.7. Since there are only finitely many such nonisomorphic V, we deduce that A is of finite representation type. The converse being obvious, the result follows. •

4

Frobenius algebras

In what follows, A denotes a finite-dimensional algebra over a field F and all A-modules are assumed to be finitely generated. For each subset X of A, l( X) and r( X) denote the left and right annihilators of X, i.e.

= O} {a E AIXa = O}

l(X)

{a E AlaX

r(X)

Suppose that (A,'lj,) is a Frobenius algebra and that Xis a subset of A. Define the subsets .L X and X .L of A by .LX

{aEAl'lj,(aX)=O}

XJ.

{a E Al'lj,(Xa) = O}

It is plain that if X is a subspace of A, then so are J. X and X .L. Lemma 4.1. Let ( A, 'lj,) be a Frobenius algebra and let X be an Fsubspace of A. (i) (.LXl = J.(XJ.) = X anddimFXJ. = dimFJ.X = dimFA-dimFX, (ii} If X is a left (respectively, right} ideal of A, then r(X) = X J. (respectively, l(X) = J. X). In particular, for every idempotent e of A

(Ael A/(Ael dimFeA

(1- e)A C>1

eA

as right A-modules

dimFAe

(iii} If X is a left ideal of A and Y is a right ideal of A, then dimpA

= dimpX + dimFr(X) = dimF(Y) + dimpl(Y)

Frobenius and Symmetric Algebras

572

m

~

Proof. (i) Let {a 1 , ... , an} be an F-basis of A such that {a1, ... , am}, n, is an F-basis of X. The map

{ A a

F x F x · · · x F ( n times) ('1j;(a1a), 'lj;(a2a), ... ,'lj;(ana))

--+ 1---+

is an isomorphism of F-spaces such that the image of X .L consists of aJl (A1, ... , An) with A1 = ···=Am = 0. Thus

and, by a similar argument,

It follows that dimF(.LX)-1 = dimFX. Since (.lX)-1 ;;;2 X, we deduce that (.LX)-1 = X. A similar argument shows that .1(x.1) = X, as required. (ii) Suppose that X is a right ideal of A. Then a E .l X if and only if 'lj;( aX) = 0. Because the latter equality is equivalent to aX = 0, we have l(X) = .L X. The equality r(X) = X.L for a left ideal X of A follows by the same argument. Now assume that e is an idempotent of A. Then, for all x EA,

x E ( Ae

l {::} Aex = 0 {::} ex = 0 {::} x E (1 -

e )A

which implies (Ae)-1 = (1 - e)A and A/(Ae)-1 ~ eA. Finally, dimFAe dimFeA, by applying (i) and the fact that A/(Ae).L ~ eA. (iii) This is a direct consequence of (i) and (ii). • We next provide the following characterizations of Frobenius algebras. Theorem 4.2. The following conditions are equivalent : (i) A is a Frobenius algebra. (ii) A is a quasi-Frobenius algebra such that

Soc(AA)

~

A/ J(A)

and Soc(AA)

~

A/ J(A)

(iii) A is a quasi-Frobenius algebra such that Soc(AA)

~

A/J(A)

or Soc(AA)

~

A/J(A)

=

4 Frobenius algebras

573

(iv} If e1, ... , ek are primitive idempotents of A such that Ae 1 , ••• , Aek are all nonisomorphic principal indecomposable A-modules and A~ EBf= 1 niAei, then there exists a permutation 1r of {1, ... , k} such that Soc(Aei) ~ Ai\·(i), S0c(e1r(i)A) ~ ei.A

and

ni

= n:ir(i)

(1

~

i ~ k)

where ei is the image of ei in A= A/ J(A). (v) Soc(AA) ~ A/J(A) and Soc(AA) ~ A/J(A). (vi) For all left ideals X and right ideals Y in A, l(r(X))

= X,

r(l(Y))

=Y

and dimFA

= dimFr(X) + dimFX = dimFl(Y) + dimFY

Proof. (ii) :::} (iii) : 0 bvious. (iii) :::} (iv) : By Theorem 3.5, there exists a permutation such that

(1

~

1r

of {1, ... , k}

i ~ k)

(1) and A~ EBf= 1 niei A

and

Soc(AA) ~ EBf= 1 nie1r-1(i).A

(2)

Hence, if Soc(AA) ~ A, then by (1), ni = n:ir(i), while if Soc(AA) ~ A, then by (2) ni = n:ir-l(i) for each i E {1, ... , k }. Thus in both cases ni = n:ir(i) for each i, as required. (iv) :::} (ii) : By Theorem 3.5, A is a quasi-Frobenius algebra. Furthermore, since ni = n:ir(i) for each i, it follows from (1) and (2) that Soc(AA) ~ A/ J(A) and Soc(AA) ~ A/ J(A). (ii) ~ (v) : It is clear that (ii) implies (v). Conversely, assume that (v) holds. Then Soc(AA) contains a copy of any simple left A-module and Soc(AA) contains a copy of any simple right A-module. By Theorem 3.5 (ii), we are therefore left to verify that Soc( Aei) and Soc( eiA) are simple for all i E {1, ... ,k}, where e 1 , ••. ,ek are as in (iv). We have

Frobenius and Symmetric Algebras

574

for some positive integers ni. Since A~ Soc(AA), we then have

Soc(AA) ~

EB7= 1 niSoc(Aei) ~ EB7= 1 niAei

and so each Soc( Aei) is simple. A similar argument shows that each Soc( eiA) is also simple, as required. (i) :::} (iii) : Since A is a Froben:i@ algebra, it is also a quasi-Frobenius algebra. Furthermore, since the left A-modules A and A* are isomorphic, it follows from Lemma 2.4 (ii) that A/J(A) ~ Soc(AA), as required. (ii):::} (i): By Theorem 3.1, AA is injective. Since Soc(AA) is an essential submodule of A, it follows from Proposition 3.8.12 that A is an injective hull of Soc(AA). On the other hand, by Lemma 2.4 (ii), A* is an injective hull of a copy of A/ J(A). Since, by hypothesis, Soc(AA) ~ A/ J(A), we deduce that the left A-modules A and A* are isomorphic. Thus A is a Frobenius algebra. (i) ::::,. (vi) : Apply Lemma 4.1 (iii) and Theorem 3.1 (vi). (vi):::} (i): By Theorem 3.1 (vi), A is a quasi-Frobenius algebra. Since (i) is equivalent to (iv), it suffices, by Theorem 3.5 (iv), to show that ni = n1r(i), 1 :S i :S k, in the notation of (iv). To this end, we fix i E { 1, ... , k} and put

S = Soc(A), D = ei Aei

and

D' = e1r(i)Ae1r(i)

Then D and D' are division algebras such that ni is equal to the right dimension of A ei over D and also to the left dimension of ei A over D. We know, by Theorem 3.5 (iv), that

and therefore as F-spaces. Thus

(3) Using the hypothesis concerning the dimensions of the annihilators, we also have

dimFe1r(i)S dim FA - dimF(l( e1r(i)S)) dimFA - dimF(J(A)e1r(i) dimF(A e1r(i))

+ A(l -

e1r(i)))

(4)

4 Frobenius algebras

575

On the other hand,

and dimp(ei A)

Thus, by (3) and (4), ni

= dimD(ei A)dimpD = nidimpD

= n1r(i)

and the result follows. •

Corollary 4.3. Let A be a Frobenius algebra, let e1 , ... , ek be primitive idempotents of A such that Ae1, ... , Aek are all nonisomorphic principal indecomposable A-modules and let A ~ EBf= 1 niAei for some positive integer ni, 1 ~ i ~ k. Then there exists a permutation 1r of { 1, ... , k} such that (i) Soc(Aei) ~ Ae1r(i), S0c(e1r(i)A) ~ ei A and ni = n1r(i) (1 ~ i ~ k). (ii) Aei ~ ( e1r(i)A)*, 1 ~ i ~ k where ei is the image of ei in A= A/ J(A). Proof. By Theorem 4.2 (iv), we may choose such 1r which satisfies (i). To prove that (ii) also holds, we first note that A ~ EBf= 1 ni( eiA) as right A-modules. Hence

( as left A- modules) where (e1A)*, ... ,(ekA)* are pairwise nonisomorphic. Since A~ A*, it follows from the Krull-Schmidt theorem that there exists a permutation a of {1, ... , k} su_:h that Aei ~ ( ea(i)A)*, 1 ~ i ~ k. It will next be shown that Soc( Aei) ~ A ea(i) for all i E {1, ... , k }, which by (i) will imply that a = 1r, as desired. Note that ea(i)A has a unique maximal A-submodule ea(i)J(A). Hence, by Proposition 3.9.11, (ea(i)A)* has a unique minimal A-submodule ( ea(i)J(A))J_ which is isomorphic to (ea(i)A)*. On the other hand,A ea(i) and (ea(i)A)* belong to the same simple component of A and hence are isomorphic. Thus Soc(Aei) ~ A ea(i) and the result follows. • Theorem 4.4. Let ( A, 'ljJ) be a Frobenius algebra. Then there exists an F-algebra automorphism er of A such that : (i) 'I/J( xa) = 'I/J( er( a )x) for all a, x E A. (ii} If I is a left ideal of A, then there are A-isomorphisms I~ [A/er(r(I))]*, A/I~ [er(r(I))]*

Frobenius and Symmetric Algebras

576

(iii) If 1r is a permutation given by Corollary 4.3, then

(1

~

i ~ k)

(iv) If A is a symmetric algebra, then a is the identity automorphism (in particular, 1r is the identity permutation). Proof. (i) As we have seen in the proof of Proposition 3.9.24, the map ---+ A* given by f (a)( x) = '¢( xa) for all x, a E A, is an isomorphism

f :A

of left A-modules. Similarly, we have an isomorphism g : A ---+ A* of right A-modules given by g(a)(x) = '¢(ax) for all x,a E A.Thus, for any given a E A, there exists a unique a(a) E A such that f(a) = g(a(a)). Hence 'l/J(xa) = 'l/J(a(a)x) for all a,x EA. By the foregoing, we are left to verify that a is an F-algebra automorphism of A. It is clear that a is an F-automorphism of the vector space A. Since

'l/J(a(ab)x)

= 'l/J((xa)b) = 'l/J(a(b)xa) = 'l/J(a(a)a(b)x),

(a,b,xEA)

we must also have a(ab) = a(a)a(b) for all a,b EA, as required. (ii) Consider the A-isomorphism f : A ---+ A* given in (i) and keep the notation of Proposition 3.9.11. We identify A** with A by means of

a(x)

= x(a)

forall

aEA,xEA*

Then we have

f(Il

{a E AJa(f(J)) = O} {a E AIJ(I)(a) = O} {a E AJ'l/J(aJ) = O} J_I

We now claim that J_ J = a(r(I)); if sustained, the required isomorphisms will follow by virtue of Proposition 3.9.11 (iv). To substantiate our claim, we first note that a(r(J)) = r(a(I)). Hence, by (i) and Lemma 4.1 (ii),

a E a(r(J))

'l/J(a(I)a)

= 0 '¢(al)= 0 a E

as required. (iii) Consider the left ideal I= Aei. Then r(J)

J_

I,

= (1-ei)A and a(r(I)) =

4 Frobenius algebras

577

( 1 - o-( ei ))A. Hence A/ o-( r(I)) ~ o-( ei )A and therefore, by (ii), Aei ~ (o-(ei)A)*. On the other hand, by Corollary 4.3, Aei ~ (e1r(i)A)*, which proves that o-( ei)A ~ e1r(i)A by Proposition 3.9.11 (ii). Thus, by Lemma 1.2.6, Ao-( ei) ~ Ae1r(i) as asserted. (iv) Assume that (A,'¢,) is symmetric. Then, in the notation of (i), f = g and so o- is the identity automorphism. • The automorphism o- in Theorem 4.4 is called the Nakayama automorphism . Turning to factor algebras, we now prove the following useful result due to Nakayama (1939).

Let X be an ideal of the F-algebra A. (i) Assume that (A,'¢,) is a Frobenius algebra. Then A/ X is a Frobenius algebra if and only if there exists a E A such that r(X) = aA = Aa. (ii) Assume that ( A, '¢,) is a symmetric algebra. Then A/ X is a symmetric algebra if and only if there exists a E Z(A) such that r(X) = aA. Theorem 4.5.

(i) Suppose that (A/X,µ) is a Frobenius algebra, and let 1r : A ---+ A/ X be the natural map. Since (A,'¢,) is a Frobenius algebra, the map f : A ---+ A* given by f (a)( x) = '¢,( xa) for all x, a E A, is an isomorphism of left A-modules. Hence there exists a E A such that µ1r = 1Pa, where 1Pa(b) = '¢,(ba) for all b E A. It will be shown that r(X) = aA = Aa. Since 1Pa(X) = '¢,(Xa) = 0, we have a E X.L. Hence aA ~ X.L and, by Lemma 4.1 Proof.

(ii), aA

~

r(X).

= 0 and thus = 1Pa(Ab) = '¢,(Aba) = 0

Assume that b E l(aA) orb E l(Aa). Then ba

µ((A/ X)(b

+ X)) = (µ1r)(Ab)

Thus b E X and therefore l(aA) 4.2 (vi), we deduce that

aA

~

X and l(Aa)

~

X. Applying Theorem

= r(l(aA)) 2 r(X),

=

proving that r(X) aA. It follows that Aa ~ r(X) and hence l(Aa) 2 X, again by Theorem 4.2 (vi). Thus X = l(Aa) and therefore r(X) = rl(Aa) =

Aa. Conversely, suppose that r(X) = aA = Aa for some a EA. Consider the map

{

A/X ~ F b + X 1-+ '¢,(ba)

(b EA)

Frobenius and Symmetric Algebras

578

Since a E r(X), µ is well defined and obviously an F-linear map. Assume that b EA is such that µ((A/X)(b + X)) = 0. Then 'l/J(Aba) = 0, so ba = 0 and hence b E l(aA). However, r(X) = aA implies that X = l(r(X)) = l( aA); hence b E X. Thus A/ X is a Frobenius algebra, as required. (ii) Suppose that A/Xis symmetric. By (i), r(X) = aA for some a E A. On the other hand,

= 'l/Ja(xy) = 'l/Ja(yx) = 'l/J(yxa) = 'l/J(xay) Thus ay = ya for all y E A and a E Z(A).

'l/J(xya)

for all x, y E A. Conversely, assume that r(X) = aA with a E Z(A). Then, by (i), A/ X is obviously symmetric, thus completing the proof. • Let A be a Frobenius algebra and let V be an A-module. We know, from Corollary 3.2, that V is injective if and only if V is projective. We close by providing another characterization of injective A-modules, due to M. Ikeda. Again, assume that ( A, 'l/J) is a Frobenius algebra. Then the map f : A ---,. A* given by for all x, a E A f(a)(x) = 'l/J(xa) is an isomorphism of left A-modules. Let a1, ... , an be an F-basis of A and let ai, ... , a~ be the dual basis of A*, i.e. forall Setting bj that

= J- 1(a;), it then follows

i,jE{l, ... ,n}

that b1, ... ,bn is an F-basis of A such

for all

i,j E {1, ... ,n}

(5)

We shall refer to the basis b1, ... , bn of A as the 'l/J-d ual of a1, . .. , an. Lemma 4.6. Let ( A, 'l/J) be a Frobenius algebra, let a1, . .. , an be an F basis of A and let the basis b1, . .. , bn of A be the 'l/J-dual of a1, . .. , an. Then, for any A-module V and any A E EndF(V), n

L biAai E EndA(V) i=l

Proof.

For any given a E A, write n

a·a -- """'µ··(a)a· i ~ iJ J j=l

µij (a) E F, 1 s; i

s; n

(6)

4 Frobenius algebras

579

We claim that

n

abi

= Lµji(a)bj

1 ::s; i ::s; n

(7)

j=l

Indeed, write abi

Lj=l Oji(a)bj with

{1, ... ,n}, aiabk =

Oji(a) E F.

n

n

j=l

j=l

Then for any k E

L µij( a )ajbk = L Ojkaibj

Hence, applying (5), we obtain 1/J(aiabk) = µik(a) = Oik(a),

proving (7). Now fix an A-module V and A E EndF(V). Then, applying (6) and (7), we obtain for all a E A, n n n a(LbiAai) = LLµji(a)bjAai i=l

and

n

i=lj=l

n

n

(L biAai)a =LL µij(a)biAaj i=l

i=lj=l

Because the two double summands are the same, the result is established. • Theorem 4. 7. Let ( A, 1/J) be a Frobenius algebra, let a1, ... , an be an F-basis of A and let the basis b1, ... , bn of A be the 1/J-dual of a1, ... , an. Then, for any given A-module V, the following conditions are equivalent : (i) V is projective. (ii) V is injective. (iii) There exists A E EndF(V) such that Li=l biAai = lv, where lv is the identity map on V ._ Proof. By Corollary 3.2, (i) and (ii) are equivalent. In what follows we prove that (ii) is equivalent to (iii). (ii) => (iii) : Assume that Vis injective and put W = A ®F V. Then the F-space W may be vfowed as a left A-module in a natural way. Note that each element of W can be uniquely written in the form E bi ® Vi with Vi E V. Applying (6) and (7), it follows that the map

Frobenius and Symmetric Algebras

580

is an injective A-homomorphism. Because V is injective, we see that a(V) is a direct summand of W and there exists /3 E EndA(W) which projects W onto a(V). Now write 1 = E 1'illi with 1'i E F. Define 6 E EndF(W) by setting

We claim that

(8) If sustained, then setting >. = a- 1 f36(3a one immediately verifies that >. satisfies (iii). To prove (8), fix v E V and note that, by (6) and (7), Lbi6(Lbkµkj(ai) ® v) k

Lbi(L 1 ® 1'kµkj(ai)v) Lbi ® (L1'kµkj(ai)v) k

i

k

L 1'k'I/J( akaibj) k

'1/J(E 1kakaibj) k

'lj;( aibj)

= 6ij

Thus we must have (Ebi6ai)(bj ® v) = bj ® v, as desired. (iii) => (ii) : Suppose that ). E EndF(V) satisfies (iii). To prove that V is injective, assume that V is a submodule of W and let a E EndF(W) be a projection of W upon V. Define /3 E EndF(W) by /3 = E bi>.aai, By Lemma 4.6, /3 E EndA(V) and one easily verifies that /3 projects W upon V. It follows that V is a direct summand of W and thus V is injective. •

5

Symmetric algebras

As before, A denotes a finite-dimensional algebra over a field F and all Amodules are assumed to be finitely generated. Since any symmetric algebra

5 Symmetric algebras

581

is Frobenius and quasi-Frobenius, the results of the preceding two sections are valid for symmetric algebras. Our aim here is to present a number of additional properties arising from the stronger assumption that A is a symmetric algebra. Theorem 5.1. Assume that (A, 'lj,) is a symmetric algebra. Then (i) r(X) = l(X) for any two-sided ideal X of A. (ii} For any primitive idempotent e of A,

Soc(Ae) ~ Ae/J(A)e, Soc(eA) ~ eA/eJ(A) (iii} For any idempotent e of A, Ae

~

(eA)*

and eA

~

(Ae)*

(iv) There exists z E Z(A) such that Soc(A) = Az and J(A) = l(z). (v) If P is a projective A-module, then P is a projective cover and an injective hull of Soc(P). (vi) i(P, V) = i(V, P) for any projective A-module P and any A-module V. (vii) The Cartan matrix of A is symmetric, provided that F is a splitting field for A. Proof. (i) Since A is symmetric, .L X = X .L for any subset X of A. Hence, by Lemma 4.1 (ii), r(X) = X.L = .L X = l(X) for any ideal X of A. (ii) By Theorem 4.4, the permutation 11" of Corollary 4.3 is the identity permutation. Since Ae/J(A)e ~ Ae and eA/eJ(A) ~ e.A, where e is the image of e in A= A/J(A), (ii) follows by virtue of Corollary 4.3 (i). (iii) We may clearly assume that e is primitive, in which case Ae ~ ( eA)* by Corollary 4.3 (ii). Taking the duals of both sides yields eA ~ (Ae)*. (iv) By hypothesis, A is symmetric and, by Proposition 1.4, A/ J(A) is also symmetric. Hence, by Theorem 4.5 (ii), r(J(A)) = Az for some z E Z(A). Since Soc(A) = r(J(A)), it follows that Soc(A) = Az. By Theorem 4.2 (vi), we also have

J(A)

= l(r(J(A))) = l(Az) = l(z),

as required. (v) We may assume, by Propositions 3.8.15 and 5.1.6, that P = Ae for some primitive idempotent e of A. Hence, by (ii), P is a projective cover

Frobenius and Symmetric Algebras

582

of Soc(P). The fact that P is an injective hull of Soc(P) holds under a weaker assumption that A is quasi-Frobenius. Indeed, by Corollary 3.2, P is injective. Hence, by Proposition 3.8.17, Pis an injective hull of Soc(P). (vi) We may assume that P = Ae for some primitive idempotent e of A. If V is simple, then i(P, V)

= i(Ae/J(A)e, V) = i(V,Ae/J(A)e) = i(V,P)

where the last equality follows by (ii). We now argue by induction on dimp V. Assume that W f:. 0 is a proper submodule of V and consider the exact sequence 0--+ W--+ V--+ V/W--+ 0 Since Pis projective, we have the exact sequence 0--+ HomA(P, W)--+ HomA(P, V)--+ HomA(P, V/W)--+ 0

Because P must also be injective, we have another exact sequence 0--+ HomA(V/W,P)--+ HomA(V.P)--+ HomA(W,P)--+ 0

It follows that i(P, V)

= i(P, W) + i(P, V/W)

i(V,P)

= i(W,P) + i(V/W,P)

and This proves (vi), by applying the induction hypothesis to Wand V/W. ( vii) Let e1, . .. , em be primitive idempotents of A such that Ae1, ... , Aem are all nonisomorphic principal indecomposable A-modules. Let C = (Ckt), 1 ::; k, t ::; m, be the Cartan matrix of A. Since F is a splitting field for A, it follows from Lemma 11.4.7 that Ckt = i(Aet,Aek). Hence, by (vi),

as desired. • Corollary 5.2. Let A be a symmetric algebra and let e1 , ... , em be primitive idempotents of A such that Ae 1, ... , Aem are all nonisomorphic principal indecomposable A-modules. If C = (cij), 1 ::; i,j ::; m, is the Cartan matrix of A, then for any given i E {1, ... , m}, either Cii 2". 2 or Aei is simple, Cii = 1 and Cij = 0 for all j f:. i.

5 Symmetric algebras

Proof.

583

Fix i E {1, ... , n} and assume that Cii

< 2. By Theorem 5.1

(ii), Soc(Aei) ~ Aeif J(A)ei ~

A ei

where ei is the image of ei in A= A/ J(A). Hence Cii = 1 and Aei is simple. Since Aei is simple, we obviously have Cij = 0 for j =J i, as asserted. • Proposition 5.3. Let V be a projective-free A-module and let !l(V) be the Heller module of V. (i) If A is a quasi-Frobenius algebra, then Soc(!l(V)) = Soc(P(V)). (ii) If A is a symmetric algebra, then Soc(!l(V)) ~ V/J(A)V. Proof. (i) Apply Theorem 3.1 (iii) and Proposition 7.3.2 (iii). (ii) It is a consequence of Theorem 5.1 (v) that for P = P(V), Soc(P) ~ P/ J(A)P. Since !l(V) ~ J(A)P, the isomorphism V ~ P/!l(V) induces an isomorphism V/J(A)V~P/J(A)P Hence, by (i), Soc(!l(V))

= Soc(P) ~ V/ J(A)V

as asserted. • Let [A, A] be the commutator subspace of A, i.e. the F-linear span of all Lie products [x,y] = xy- yx with x, y E A. As before, if ( A, 'Ip) is a Frobenius algebra and X is a subset of A, we put .1X = {a E Al'lf'(aX) = O} and

X.1 = {a E Al'lf'(Xa) = O} Lemma 5 .4. (i) If ( A, 'Ip) is a Frobenius algebra, then for any subset X of A, l(X) is the largest left ideal of A contained in .l X. (ii) If ( A, 'Ip) is a symmetric algebra, then

Z(A)

= [A, A].1

Proof. (i) It is a consequence of the definition of .l X that l(X) ~ Denote by L any left ideal of A contained in .l X. Then 'lf'(ALX)

= '!f'(LX) = 0

.l X.

Frobenius and Symmetric Algebras

584

= 0 and thus L ~ l(X), as asserted. (ii) If z EA, then z E [A, A]l. if and only if for all x, y E A

so LX

'lj;((xy - yx)z)

= 'lj;(xyz)- 'lj;(yxz) = 'lj;(xyz) -

'lj;(xzy)

=0

or, equivalently, if and only if

'lj;(A(yz - zy))

=0

for all

Since the latter is equivalent to yz - zy established. •

yEA

= 0 for

Let A be an F-algebra and assume that charF

T(A)

= {a E AlaPn E [A,A]

all y E A, the result is

= p > 0.

for some

Then we put

n ~ 1}

Note that, by Proposition 8.3.1, T(A) is an F-subspace of A such that

T(A) 2 [A, A]. Theorem 5.5. (Kulshammer {1981)). Let A be an algebra over a field F of prime characteristic p. {i) T(A) = [A, A]+ J(A) and J(A) is the largest left ideal of A contained in T(A). (ii) If A is a Frobenius algebra, then Soc(A) = T(A)l. A. (iii) If A is a symmetric algebra, then (a) T(Al = Z(A) n Soc(A) and, in particular, T(Al is an ideal of Z(A). {b) Soc(A) = (Z(A) n Soc(A))A. (c) (T(A)L)2 = Z(B1)EB···EBZ(Br), where B1, ... ,Br are all blocks of A which are simple algebras. Proof.

(i) We first treat the case where A is simple. Because for any

z E Z(A) and x, y EA, we have (xy- yx)z

= x(yz) -

(yz)x E [A,A]

both [A, A] and T(A) may be viewed as algebras over the field Z(A). Thus we may assume that Z(A) = F, i.e. that A is a central simple F-algebra. Denote by Ethe algebraic closure of F. Then

5 Symmetric algebras

for some n

~

585

l. It is clear that

and

(1) Bearing in mind that

and dimpA

= dimEAE,

we infer that dimp[A, A] = dimpA - l. If it were true that A = T(A), then since T(AE) is an E-subspace of AE, we would have AE = T(AE), contrary to ( 1). Thus A -:/ T( A) and, since T( A) 2 [A, A], we must have T(A) = [A, A]. Since A is simple, (A, 7/i) is a symmetric algebra for some 7/i E A* (Proposition 1.4). But then [A,A] ~ Ker'lj,

and therefore T(A) contains no nonzero left ideals of A. This proves the case where A is a simple algebra. Turning to the general case, put A= A/J(A) and write

A = A1 EB .. · EB Am where the Ai are pairwise orthogonal simple F-algebras. By the special case proved above, we have T(Ai) = [Ai, Ai], 1 :Si :Sm. This implies that

([A, A]+ J(A))/ J(A)

[A,A] [A1, A1] EB•·· EB [Am, Am] T(A1) EB··· EB T(Am) T(A)

Since [A,A] + J(A) ~ T(A) and T(A)/J(A) c T(A), we have T(A) T(A)/ J(A) and T(A) = [A, A]+ J(A). Finally, let L be a left ideal of A contained in T(A). Then (L+J(A))/ J(A) is a left ideal of A contained in T(.A) and hence in [A, A]. But A is semisimple, so A is symmetric (Proposition 1.4) and thus, by the previous argument, [A, A] contains no nonzero left ideals of A. Consequently, L ~ J(A) and the

Frobenius and Symmetric Algebras

586

required assertion follows. (ii) Assume that A is a Frobenius algebra. Then, by Lemma 4.1 (i),

J_(T(A)J_)

= T(A)

(2)

while by Theorem 4.2 (vi),

r(l(T(Al A))= T(A)J_ A

(3)

Owing to (i) and (2), J(A) is the largest left ideal of A contained in J_(T(A)J_). Applying Lemma 5.4 (i), we conclude that

(4) Hence, by (3) and ( 4), we must have

Soc(A)

= r(J(A)) = T(A)J_ A

as desired. (iii) Assume that A is a symmetric algebra. Owing to (ii) and Lemma 5.4 (ii), to prove (a), we need only show that

T(Al

= [A, A]J_ n T(Al A

Now l(T(A)J_ A) is a left ideal of A and so

(l(T(Al A))J_

= r(l(T(Al A))

(5)

by Lemma 4.1 (ii). Thus (by (i))

=

[A, A]J_ n [l(T(Al A)JJ_

(by ( 4))

=

[A, A]J_ n r(l(T(Al A))

(by (5))

=

[A, A]J_ n T(A)J_ A

(by (3))

proving (a). Property (b) being a consequence of (a) and (ii), we are left to verify ( c).

5 Symmetric algebras

587

To prove ( c), we may harmlessly assume that A is indecomposable, in which case it suffices to show that (T(Al )2

={

if if

Z(A~

J(A) =/- O J(A) = O

Suppose that J(A) = 0. Then 1 E Soc(A) n Z(A) = T(A).1. Because T(A).l is an ideal of Z(A), we have T(Af = Z(A) and hence (T(Al)2

= Z(A)

Now assume that J(A) =/- 0. Then 1 1. A simple inductive argument shows that if a product of more then two factors is divisible by a prime p, then so is one of the factors. Thus a1lbk for some k, say k = 1 (by renumbering the bi's) and so a1 = b1u for some unit u of R. Dividing a1 ... ar = b1 ... bs by b1, we obtain ua2 ... ar = b2 ... bs. It therefore follows, by induction, that r - 1 = s - 1 and that we can renumber the bi's so that ai is associated to bi, 2 ::; i ::; r. Because this also holds for i = 1, the uniqueness of factorization is proved. (ii) :::} (iii) : As has been observed earlier, R is a GC D-domain. Taking into account that there are only finitely many of pairwise nonassociated divisors of x E R, we infer that ( x) is contained in only finitely many principal ideals of R. Thus ACCP is satisfied in R. (iii) :::} (i) : By Theorem 1.7 (i), every irreducible element of R is a prime. Thus, given a nonunit x i- 0, it suffices to show that x is a product of irreducible elements. We first show that x has an irreducible divisor. If x is irreducible, there is nothing to prove. Otherwise, let x = X1Y1 where X1, Y1 are non units. Then either x1 is irreducible or x1 = X2Y2 where X2Y2 are nonunits. Continuing this process, we obtain the ascending chain ( x) ~ (x1) ~ (x2) ~ · · ·. By hypothesis, this chain must terminate; if Xn is the last term, Xn is irreducible, and Xnlx,

1 Integrally closed domains

597

We now let Xn = P1 and write x = Pl c1. If c1 is irreducible, then the proof is complete. Otherwise, we have c1 = p2c2 where p 2 is irreducible. Continuing in this way, we obtain the ascending chain ( x) ~ (c1 ) ~ ( c2 ) ~ · · ·. This terminates with an irreducible element en = Pn+I· Thus there exist irreducible elements p1,P2, ... ,Pn+l such that X = P1C1

= P1P2C2 = · · · = PIP2 · · •Pn+l

and the result follows. • Corollary 1.9. Proof.

Every PI D is a U FD.

Apply Theorems 1.7 (ii), (iii) and 1.8 (iii). •

For future use, we next record the following elementary properties. Proposition 1.10. {i) Let R ~ S be integral domains with S integral over R. Then S is a field if and only if R is a field. Furthermore, if I is a nonzero ideal of S, then In R is a nonzero ideal of R. (ii) Let R ~ S be commutative rings with S integral over R, and let Q be a prime ideal of S. Then Q is maximal if and only if Q n R is maximal. Proof. (i) Assume that R is a field and let O i= s E S. If sn + r1s-n-l + · · · + rn = 0, Ti E R, is an equation of integral dependence for s of smallest degree, then rn i= 0 (since S is an integral domain). Hence s- 1 = -r; 1(sn-l + r 1sn- 2 + ·· · + rn-I) ES and therefore Sis a field. Conversely, assume that Sis a field and let O i= r E R. Then r- 1 E Sis integral over R and so we have an equation r-m + r~ r-m+l + ··· + r:n = 0, ri ER. It follows that ~-I= -(r~ + r;r + · · · + r:nrm-l) is in R, proving the first assertion. Let I i= 0 be an ideal of S and let O i= s E I. Then, in the notation of the first paragraph, Tn

= -(sn + r1sn-l + ··· + Tn-1s) EJ n R,

proving (i). (ii) By Lemma 1.5 (i), S/Q is integral over R/(R n Q) and both these rings are integral domains. Now apply (i). • Corollary 1.11. Let R be an integral domain with quotient field F, let E / F be a finite field extension and let S be the integral closure of R in E.

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Dedekind Domains and Discrete Valuation Rings

Then (i) E is the quotient field of S and S is integrally closed. (ii) If every nonzero prime ideal of R is maximal, then the same is true for S. Proof. (i) Given ).. E E, we may write ,xn + a1.xn-l + · · · + an = 0 for some ai E F, n 2: 1. Writing ai = rif ri with Ti, ri E R, ri -/:- 0 and multiplying the above equation by rn where r = r{ · · · r~ E R, we see that >.r is integral over R. Hence Ar E S and E is the quotient field of S. Since, by Corollary 1.4, Sis integrally closed in E, the required assertion follows. (ii) Let Q -/:- 0 be a prime ideal of S. Then Q n R is a prime ideal of R which is nonzero by Proposition 1.10 (i). Hence, by assumption Q n R is a maximal ideal of R. Therefore, by Proposition 1.10 (ii), Q is a maximal ideal of S. •

As a preliminary to the proof of our next result, we recall the following elementary field-theoretic fact. Let E / F be a finite separable field extension and let a1, ... , O"n be all distinct F-homomorphisms of E into its algebraic closure. Then the trace map Tr

= TrE/F : E

-+

F

is given by n

Tr()..)= Eai(A) i=l

A major significance of this map stems from the fact that if u 1 , ••• F-basis of E, then there exists a second basis v 1 , ••• ,vn such that Tr(u·v·) -- b·· i J tJ

(1

~ i,j ~

n)

, Un

is an

(4)

Theorem 1.12. Let R be an integrally closed integral domain with quotient field F, let E / F be a finite separable field extension of degree n and let S be the integral closure of R in E. Then S is an R-submodule of a free R-module of rank n. In particular, if R is noetherian then S is a finitely generated R-module (hence S is noetherian). Proof. Let u1, ... , Un and v1, ... , Vn be two F-bases of E for which ( 4) holds. By Corollary 1.11 (i), we may assume that each Ui E S. Note also that ifs E S, then each ai( s) is integral over R, hence Tr( s) E F n S = R.

2 Dedekind domains

599

We now claim that S ~ L-j=l Rvj, which will complete the proof. Fix s ES and writes= L-j=I >..jVj with >..j E F. For any i E {1, ... ,n}, we have UiS E S and therefore Tr( UiS) E R. It follows that for each i E

{1, ... ,n}, n

Tr( UiS)

= Tr(L >..jUiVj) = >..i E R j=l

as required. • Corollary 1. 13. Let R be a PI D with quotient field F, let E / F be a finite separable field extension of degree n and let S be the integral closure of R in E. Then S is a free R-module of rank n. Proof. By Theorems 1.7 (iii) and 1.12, Sis an R-submodule of a free Rmodule ofrank n. Furthermore, by Proposition 6.8.14 (i), any R-submodule of a free R-module of rank n is free of rank ::; n. Since, by the proof of Theorem 1.12, S contains an F-basis of E, the result follows. •

2

Dedekind domains

An integral domain R is called a Dedekind domain if it is noetherian, integrally closed and every nonzero prime ideal of R is maximal. All the discussion below reduces to triviality if R is a field. For this reason, in many of the proofs below we tacitly assume that R is not a field. We begin by providing some important classes of Dedekind domains. Proposition 2.1.

Any principal ideal domain is a Dedekind domain.

Proof. Let R be a PID. By Theorem 1.7 (iii), it suffices to show that any given nonzero prime ideal (p) of R is maximal. Let (q) be a prime ideal of R with (p) ~ (q). Then p,q are primes and p = qt for some t ER. But a prime is always irreducible, hence t E U(R) and (p) = (q), as required. •

The main supply of Dedekind domains is provided by the following result. Theorem 2.2. Let R be a Dedekind domain with quotient field F, let E / F be a finite separable field extension and let S be the integral closure of R in E. Then

600

Dedekind Domains and Discrete Valuation Rings

(i) S is a Dedekind domain and a finitely genemted R-module. (ii) For any nonzero prime ideal Q of S, the field S / Q is a finite extension of its subfield R/(Q n R). Proof. (i) Apply Corollary 1.11 and Theorem 1.12. (ii) Since S is a Dedekind domain, Q is maximal and so S / Q is a field. Furthermore, by Proposition 1.10 (ii), R(Q n R) is a subfield of S/Q. Since Sis a finitely generated R-module, S/Q is a finite extension of R/(Q n R), as required. •

It can be shown that Sis a Dedekind domain without the assumtion that E /Fis separable. This can be achieved by considering the purely inseparable case and subsequently applying Theorem 2.2. However, one can present a much easier proof by assuming the following standard fact of commutative ring theory : Let R be a noetherian integral domain with quotient field F and let S be any ring between R and F. If every nonzero prime ideal of R is maximal, then S is noetherian (see Kaplansky (1974, Theorem 9.3)). Although the following result will not be needed, we shall present a proof for expository reasons. Theorem 2.3. Let R be a Dedekind domain with quotient field F, let E / F be a finite field extension and let S be the integml closure of R in E. Then S is a Dedekind domain. Proof. By Corollary 1.11, it suffices to show that Sis noetherian. By Corollary 1.11 (i ), E is the quotient field of S, so we may choose a basis s1, ... ,sn of E over F with each Si ES. Setting So= R[s1, ... ,sn], it follows from Lemma 1.1 that S 0 is a finitely generated R-module. Hence So is noetherian. Furthermore, since S0 is integral over R, any nonzero prime ideal of So is maximal (Proposition 1.10). Since S lies between So and its quotient field E, Sis noetherian. •

Let R be an integral domain and let F be the quotient field of R. By a fractional ideal of R, we understand any nonzero R-submodule I of F for which there exists a nonzero d in R such that d · I ~ R. Expressed otherwise, this means that the elements of I have a "common denominator" d E R. The ordinary nonzero ideals of R are fractional ideals ( with d = 1).

2 Dedekind domains

601

However, F itself is not a fractional ideal (unless R = F), since otherwise there exists O id ER such that F = R[d- 1]. Then d- 2 = rd- 1 with r ER, and so d- 1 = r ER, showing that F = R[d- 1] = R. Any nonzero finitely generated R-submodule I of Fis a fractional ideal. This follows from the fact that, if I = Rx1 + · · · + Rxn for some Xi = ri/di(ri, di E R), then d = d1d2 ... dn is a common denominator for the elements of I. Conversely, if R is noetherian, then every fractional ideal J of R is a finitely generated R-submodule of F. This is so since I~ d- 1 R for some O id ER and since the R-modules d- 1Rand Rare isomorphic. We define the product I· J of two fractional ideals I and J as the set of all finite sums 'E, XiYi where Xi E J and Yi E J. If d1 and d 2 are the common denominators for I and J, then d1 d2 is a common denominator for the R-submodule I· J of F. Hence I· J is a fractional ideal and therefore the fractional ideals of R constitute a multiplicative commutative monoid whose identity element is R. This monoid will be denoted by I( R). The invertible elements of I(R), also called invertible fractional ideals of R, constitute a group. Every principal fractional ideal of R, i.e. an ideal of the form Rx with O i x E F is invertible. Moreover, the principal fractional ideals of R form a group which we denote by P(R). It will be shown below that Dedekind domains are precisely those integral domains R for which I(R) is a group. To achieve this, we need some preliminary results. Lemma 2 .4. Let R be a noetherian ring and let I i R be an ideal of R. Then there exist prime ideals P1, ... , Pr of R such that Pi 2 I, 1 :'.S i :'.S r, and I 2 PiP2 ···Pr. Proof. The assertion holds when I is a prime ideal. If I is not prime, we may find a,a' (/. I such that aa' E J. Put J =I+ Ra and J' =I+ Ra', so J J' ~ I, but IC J, IC J'; hence J i Rand J' i R. If the statement is already true for the ideals J, J' then it is true for I; namely, there exist prime ideals A, ... , Ps containing J ::) I and Q1, . .. , Qt containing J' ::) I such that A·· ·Ps ~ J, Q1 ···Qt~ J', so

If however the statement fails for I, for example, then I is not a prime ideal and we may repeat the argument. This process must terminate, otherwise it would give rise to a strictly ascending infinite chain of ideals of R, which is impossible since R is noetherian. •

602

Dedekind Domains and Discrete Valuation Rings

Lemma 2.5. Let R be a Dedekind domain with quotient field F, let I be a nonzero ideal of Rand let 1-1 = {a E Fled~ R}. (i) 1- 1 is a fractional ideal of R containing R and 1- 1 I= R. Moreover, if I -=I- R then 1-1 :::) R. {ii) If Pi··· Pn ~Qi· .. Qm, where the {Pi} and {Qj} are nonzero prime ideals of R, then m ~ n and, upon renumbering the Pi if necessary, Q j = Pj for all j E {1, ... ,m}.

Proof. (i) If O -=I- x E I, then ax E R implies a E Rx- 1 • Thus J- 1 is an R-submodule of the finitely generated R-module Rx- 1 • Since R is noetherian, 1- 1 is a finitely generated R-module. This shows that 1- 1 is a fractional ideal of R such that 1- 1 2 R. Moreover, R- 1 = R, since if a E F is such that aR ~ R, then a E R. We now show that if I -=I- R, then 1- 1 :::) R. Choose O -=I- a E I. By Lemma 2.4, we can find nonzero prime ideals P1, ... , Pr such that PiP2 ···Pr ~ aR ~ I. Choose a set of nonzero prime ideals with minimal possible r. Because I -=I- R, we know that I ~ P for some maximal ideal P of R. Thus P1 · · · Pr ~ P and we claim that P = Pi for some i, say P = P 1. Indeed, if no Pi = P, then each Pi contains an element Oi ¢ P, and a1 · · · ar E P, which is impossible. We now have P P2 ···Pr ~ I ~ P. By the minimality of r, we know that P2 · · · Pr !l aR. Choose (3 E P2 ···Pr with (3 ¢ aR and put A= a- 1 (3. Then A¢ R, but A·l= a- 1 (3[~ a- 1(3P ~ a- 1 PP2···Pr ~ R and so.XE 1- 1 . It is now an easy matter to prove that 1-1 I = R. Put S = 1-1 I and observe that S is an ideal of R. Therefore JJ- 1 s- 1 = ss- 1 ~ R, so 1-1s- 1 ~ 1- 1. Hence, for any (3 E s-1, we have 1- 1 (3 ~ 1-1, and so 1- 1 (3m ~ 1- 1 for all m ~ 1. But then J- 1[(3] is an R-submodule of 1- 1 and therefore is finitely generated. Note further that R[.8], being a submodule of 1-1 [(3] is also finitely generated. Since R is integrally closed, it follows that (3 E R. Hence s- 1 ~ R and so s- 1 = R. Applying the preceding paragraph, we deduce that S = R, proving (i). (ii) We have P1 · · · Pn ~ Qi··· Qm ~ Q1, hence, as in (i), Q1 must be equal to some Pi, say Q1 = Pi. Then P 1- 1P1 · · · Pn Q 11Q1 · · · Qm, whence P2 · · ·Pn ~ Q2 ···Qm, and so on.•

t

Lemma 2.6. Let R be an integral domain with quotient field F and let I be a nonzero ideal of R. Then the following conditions are equivalent :

2 Dedekind domains

603

{i) I is a projective R-module. {ii) I is an invertible ideal. (iii) I is a finitely generated projective R-module.

Assume that J is invertible. Then R = 11- 1 and so 1 = E J, bi E 1- 1. Let the R-homomorphisms f: J-+ Rn and g: Rn-+ I be given by f(x) = (xb1, ... , xbn) and g(x1, ... , Xn) = Li=l Xiai, Then gf is the identity map and so I is a finitely generated projective Rmodule. Now assume that I is a projective R-module. It suffices to verify that J is invertible. By Proposition 3.6.5, there exists a family {atlt E T} of elements in J and a family {!tit E T} of elements in H omR(I, R) such that every element of I can be written in the form Proof.

Li=l aibi with ai

with finitely many ft(x) distinct from zero. Observe that for x,y E J, we have Yft(x) = ft(yx) = ft(xy) = xft(Y) and so the ratio Xt = ft(x)x- 1 E F is independent of the choice of nonzero x in J. Moreover, Xtf ~ R and so Xt E J' = {.,\ E Fl.XI ~ R}. It is clear that J' is a fractional ideal of R. Note also that for any fixed nonzero x E J, only finitely many elements ft( x) = XXt are nonzero. Thus only finitely many of xt's are nonzero, say x1, . .. , Xn- This implies that n

x

=L t=l

n

ft(x)at

=L t=l

n

XXtat

= x L Xtat t=l

and therefore 1 = I:~ 1 Xtat, Thus R ~ I'· I, whence R to show. •

= I'· I

as we wished

We are now ready to prove the following basic theorem, which includes classical results of Dedekind, Noether, Krull and Matusita. Theorem 2.7. For any integral domain R, the following conditions are equivalent : (i) R is a Dedekind domain. (ii) Every nonzero proper ideal of R is expressible in a unique way as a product of prime ideals. {iii) Every nonzero proper ideal of R is a product of prime ideals.

604

Dedekind Domains and Discrete Valuation Rings

(iv) The monoid I( R) of fractional ideals of R is a group. (v) All ideals of R are projective R-modules. Proof. (i) =} (ii) : Let I -/- 0 be a proper ideal of R. By Lemma 2.4, A··· Pr ~ I for some nonzero prime ideals A, ... , Pr of R. Choose a maximal ideal P of R with I~ P. Then, as in the proof of Lemma 2.5 (i), P coincides with some Pi, say P = P 1 . Hence, by Lemma 2.5 (i),

p-l p P2 · .. Pr

= P2 ···Pr

~ p-l J ~ R

Suppose we have shown that a nontrivial ideal of R which contains a product of fewer than r prime ideals is expressible as a product of prime ideals ( this is trivially true for r = 2 since prime ideals of R are maximal). We then deduce that p-l I= Qi··· Qn, where each Qj is prime, and so I= P p-l I= PQ 1 • • • Qn. Applying induction on r, we deduce that I is a product of prime ideals. The uniqueness of such decomposition follows from Lemma 2.5 (ii). (ii) =} (iii) : Obvious. (iii) =} (iv) : It suffices to show that every nonzero prime ideal of R is invertible. Indeed, let I be a fractional ideal of R and let O -/- d E R be such that dl ~ R. Then J = (Rd)- 1 (Rd · I) and, by hypothesis, Rd· I and Rd are products of nonzero prime ideals. Thus if each nonzero prime ideal of R is invertible, the same holds for I. Now let P-:f. 0 be a prime ideal of Rand let O-/- a E P. By assumption, Ra is a product of prime ideals, so P 2 Ra = Pi · · · Pr for some prime ideals Pi, 1 ~ i ~ r. Because Ra is invertible, then each Pi is invertible. Moreover, since P is prime, we have P 2 Pi for some i E { 1, 2, ... , r}. Thus we are left to verify that if P is an invertible prime ideal of R, then P is maximal. For this purpose, we first prove the following assertion. Let S be an integral domain, let A, ... , Pr be invertible prime ideals of S and let J = P 1 · · ·Pr. We claim that if J = Q1 · · · Q s with Q j prime ideals, then s = r and after some permutation of indices, Pi = Qi for all i E {1, 2, ... , r }. Indeed, let A be a minimal ideal among Pl,··· ,Pr. Since A 2 J = Q1 ···Qs, it follows that there exists j E {1, ... , s }, say j = 1, such that A 2 Q 1 . Similarly, Qi 2 J = Pi··· Pr, hence Qi 2 Pi for some i E {1, ... , r }. Thus A 2 Pi and, by the minimality, i = 1 and P 1 = Q 1 . Now, since A is invertible, multiplying the equality A··· Pr = J = Qi··· Qs by P 1- 1 yields P2···Pr

= pl-lJ = Q2···Qs

Hence, proceeding inductively, the claim follows. Let P be an invertible prime ideal of R. By the foregoing, we are left

2 Dedekind domains

605

to verify that P is maximal. To this end, choose a E R - P and consider the ideals J = P + Ra and J' = P + Ra 2. By hypothesis, J = Pi · · ·Pm, J' = Q1 · · · Qn, where Pi, Qj are prime ideals which must contain J, J', respectively, and hence contain P. Put S = Rf P and denote by a the image of a in S. Then Sa=

-

-

Pi ···Pm, Sa

2

-2 -2 = Qi ···Qn = P1 ···Pm

where A = Pd P, Qj = Qi/ P are invertible prime ideals of S. Hence, by the preceding- paragraph, 2m = n and, after renumbering we have

Q2i-1=Q2i=A forall iE{l, ... ,m} It follows that Q 2i-1 = Q2i = Pi, so (P P ~ P

+ Ra) 2 =

P

+ Ra 2 and thus

+ Ra 2 = (P + Ra)(P +Ra)~ P 2 + Ra

This implies that every element x E P may be written in the form x = y+za with y E P 2,-z ER. Hence za = x -y E P and, since a (iv), (iv) is equivalent to the requirement that every nonzero ideal I of R is invertible. Now apply Lemma 2.6. •

*

Corollary 2.8. Let R be a Dedekind domain. Then l(R) is a free abelian group with the collection of nonzero prime ideals of R as free generators.

Dedekind Domains and Discrete Valuation Rings

606

Proof. As we have seen in the proof of Theorem 2.7, every fractional ideal of R can be expressed as a product of prime ideals of R and their inverses. Hence the nonzero prime ideals of R generate I(R), and Theorem 2.7 (ii) shows that they generate it freely. •

Let R be a Dedekind domain. The factor group C(R)

= I(R)/ P(R)

is called the class group of R. The following result ties together Dedekind domains, U F D's and PI D's. Theorem 2.9. The following conditions are equivalent : {i) R is a Dedekind domain and C(R) = 1. {ii) R is a Dedekind domain and a U FD. {iii) R is a P ID. Proof. (i) :::} (ii) : Let r E R be a nonzero nonunit. Then (r) is a nontrivial ideal of Rand so, by Theorem 2.7, we may write (r) = (P1) ... (Pn) for some prime elements P1, ... ,Pn of R. Hence r = P1 .. ·PnU for some unit u of R, proving that r is a finite product of primes. Thus, by Theorem 1.8, Risa UFD. (ii) :::} (iii) : By Corollary 2.8, it suffices to show that every nonzero prime ideal P of R is principal. To this end, let O =/- x E P. Since x is a nonzero nonunit, it is a product of prime elements, say x = P1P2 ... Pr· Because x E Pit follows that Pi E P for some i, hence (Pi)~ P. But (Pi) is a prime ideal of R, hence it must be maximal, since R is a Dedekind domain. Thus P = (Pi) is a principal ideal. (iii) :::} (i) : Let R be a PI D. Then, by Proposition 2.1, R is a Dedekind domain. Since any nonzero prime ideal of R is principal, it follows from Corollary 2.8 that C(R) = 1, as required. •

Let I and J be ideals of a commutative ring R. Recall that, by definition, I and J are relatively prime if I+ J = R. Lemma 2.10. R. Then TI~ 1 Ii= Proof.

Let Ii,I2, ... ,In be pairwise relatively prime ideals of

n~ 1 h

The case n

= 2 is a consequence of Proposition 1.11.1

(i). Sup-

2 Dedekind domains

607

pose that n > 2 and the result is true for 11, ... , ln-1, and let J = Tii,.,;;-/ Ii = ni,;l h Because Ii + In = R, 1 :'.:S: i :'.:S: n - 1, we have Xi + Yi = 1 for some Xi E Ji, Yi E In and therefore n-1

n-1

i=l

i=l

II Xi= II (1- Yi)= l(modln)

Thus In

+I =R

and so n

II Ii= Iln =In In= ni=lli, i=l

as desired. • Let R be a Dedekind domain and let A, B be nonzero ideals of R. We say that A is divisible by B or that B divides A if A = BC for some ideal C of R. The greatest common divisor of A and B is defined to be an ideal of R which divides both A and B and is divisible by any other ideal with this property. Similarly, the least common multiple of A, B is defined as an ideal of R divisible by A and B and dividing any other such ideal. Lemma 2.11. Let A, B be nonzero ideals of a Dedekind domain R. Then B divides A if and only if B contains A. Proof. If B divides A, then A = BC for some ideal C of R and thus B 2 A. Conversely, assume that B 2 A. If B = R, then A= BA and B divides A. If Bi- R, then by Theorem 2.7, B = Q1 · · · Qm and A= Pi··· Pn, where the {Pi} and {Qj} are nonzero prime ideals. Hence, by Lemma 2.5 (ii), B divides A. •

It is now an easy matter to prove the following result. Theorem 2.12. Let A, B be nonzero ideals of a Dedekind domain R and let A = Pf! ... B = Pf 1 ••• P!n (ai' bi ~ 0)

p;n '

where the {Pi} are distinct prime ideals, and

PP is taken to be R.

(i) g.c.d.{A, B} =A+ B = Tii=l Ptin(a;,b;). (ii) l.c.m.{A, B} =An B = Tii=l Ptax(a;,b;).

Then

Dedekind Domains and Discrete Valuation Rings

608

(iii) A, B are relatively prime if and only if A, B have no common prime factor. (iv) The nonzero ideals A 1, .. . , An of R are pairwise relatively prime if and only if (1 ::; i::; n) Ai+ Aj = R

II

#i

Proof. We first observe that (iii) follows from (i), and (iv) from (iii). That g.c.d.{A,B} = A+ Bis a consequence of Lemma 2.11. This proves (i) upon noting that A divides B if and only if ai ::; bi for all i. A similar argument proves (ii). • Corollary 2.13. Let I be a nontrivial ideal of a Dedekind domain R, say I = P;_" 1 • • • Pfk, ni > 0, where P1, ... , Pk are distinct prime ideals of R.

Then the map R/I r+I

{

--+

TI7=1(R/Pt)

1--j,

(r

+ Pt;)

is a ring isomorphism. Proof. By Theorem 2.12 (iii), the ideals Pt, 1 ::; i ::; k, are relatively prime. -Hence, by Lemma 2.10, I= nf= 1 Pti. Now apply Corollary 1.11.3 and the result follows. •

Let R be a Dedekind domain, let Pi, ... ,Pn be distinct nonzero prime ideals of R and let x1, ... , Xn E R. Then, for any given nonnegative integers e1, ... , en, there exists x E R such that Corollary 2.14.

X -

X· i

E p:i - p,e;+l '

(1::;

i::; n)

'

Proof. Given i E {1, ... ,n}, we have Pt;:) Pt;+l, so there exists ai E Pt - Pt+l. By Corollary 2.13, there exists x ER such that x - (xi+ ai) E e-+1 1or .c all i· E { 1, ... , n }. T h ere1ore .c e· -Pi' e·+l , Pi• x-xi = [ x- ( Xi+ai )] +ai E Pi• as required. • Theorem 2.15.

Any semilocal Dedekind domain is a principal ideal

domain. Proof. Let R be a semilocal Dedekind domain. Then R has :finitely many, say Pi, ... , Pn distinct nonzero prime ideals. It suffices to show that

2 Dedekind domains

609

each Pi is a principal ideal. By Corollary 2.14 (with x 1 = · · · = Xn = 0, ei = 1, ej = 0 for j -=/= i), for each i E {1, ... , n }, there exists an element Yi E R such that Yi E Pi - P/ and Yi (/. Pj for j -=/= i. Thus, by Lemma 2.11, PilRYi, Pl f Ryi and, for j -=/= i, Pj Y RYi· Hence the decomposition of Ryi into prime ideals is Ryi = Pi, as required. • Since a Dedekind domain is noetherian, all its ideals are finitely generated. This can be improved by using the following result. Proposition 2 .16. Let R be, a Dedekind domain and let A, B be nonzero ideals of R. Then there exists a nonzero ideal C of R such that C is relatively prime to B and AC is a principal ideal. Proof. Write A = Pt1 · · · P~n, B = Pf 1 • • • p~n with ai 2: 0, bi 2: 0, where P1, ... , Pn are distinct prime ideals. Choose ai E Pt -Pt+l, 1 :S: i :S: n. Then, by Corollary 2.13, there exists a E R such that a= ai ( mod Pt+ 1 ), 1 :S: i :S: n. Hence a E Pt - Pt+l, so a E nf= 1 Pt = A (Lemma 2.10 and Theorem 2.12 (iii)) and aR+ AB~ A. Therefore, by Lemma 2.11, we have

aR + AB

= P{1 ... p~n

with

Ci 2: ai

But a d\Z= Pia·+l • , so Ci = ai and thus aR + AB = A. Since aR ~ A, we have aR = AC for some ideal C of R (Lemma 2.11). Therefore AC+ AB = A and multiplying by A- 1 yields C + B = R, as desired. • Corollary 2.17. Let A-=/= 0 be, an ideal of a Dedekind domain R. Then, for any nonzero r1 E A there exists r2 E A such that A= r1R + r2R. Proof. By Proposition 2.16, we may choose a nonzero ideal B of R relatively prime to r 1R and such that AB= r 2R for some r 2 ER. Then we have r1R + r2R = r 1R + AB ~ A. On the other hand, since B + r 1R = R, we have 1 = /3 + r1 a for some /3 E B, a E R. Hence, for each a E A, we have

We next record another consequence of Proposition 2.16.

Dedekind Domains and Discrete Valuation Rings

610

Corollary 2.18. Let A, B be nonzero ideals of a Dedekind domain R. Then the additive groups of R/ A and B / AB are isomorphic. In particular, for any n ~ 1, R/ A and An/ A n+l have isomorphic additive groups. Proof. Owing to Proposition 2.16, we may find a nonzero ideal C of R relatively prime to A and such that BC = r R for some nonzero r E R. Then the map 0. Assume that g1, h1 E R[X] are such that : ( a (/_ P, 0 < r < n). (a) 91 = axr +a' xr-l +... and deg( h1) ::; n - r {b) f = g1h1 and g1,h1 are relatively prime. Then there exist polynomials g, h E R[X] of degree r and n - r, respectively, with g = aXr + •·· and such that f = gh, g = 91, h = h1.

Proof. Our plan is to construct two sequences {gk} and {hk} of polynomials in R[X] such that

= axr + ···, deg(hk) :s; n- r 9k+l = 9k(modPk[X]), hk+I = hk(modPk[X]) f = gkhk( mod pk[X])

(1)

9k

(2) (3)

We claim that this will complete the proof. Indeed, write r

9k

=E

s

akmxm, hk

m=O

= E bktxt

( akm, bkt E R, s

=n -

r)

t=O

where, by (1), akr = a for each k. By (2), for any fixed m, t the sequences { akm}, {bkt} are coherent and we put am = lim llkm, bt = lim bkt, 0 ::; m ::; r', 0 :s; t :s; s. Then g = I:~=o amXm E R[X], h = I::=o btXt E R[X] and ar = lim llkr = a. Furthermore, for each k 2:: 1, g

=9k(modPk[X]), h =hk(modPk[X])

and, in particular, g = 91, h = h1. Hence, by (3), f = gh(modPk[X]) for each k 2:: 1 and so f = gh. Since deg f = deg(g) + deg( h) (by the fact that a E R - P = U(R)) and deg(g) = r, we also have deg(h) = n - r. This substantiates our claim. We proceed by induction, assuming that 91, ... , 9k and h1, ... , hk have been constructed so that they satisfy (1), (2), and (3). Clearly, for k = l, the given 91 and h1 will suffice. Let { v1, ... , vn} be a generating set for the R-module pk_ We define 9k+I

n

n

i=l

i=l

= 9k + L V1Ai, hk+i = hk + LViµi

(4)

Dedekind Domains and Discrete Valuation Rings

618

where the ,\ and µi are as yet unspecified polynomials whose degrees are less that r and not greater than n - r, respectively. It is obvious that (1) and (2) are thereby satisfied. We now determine Ai and µi so that (3) is satisfied for 9k+1 and hk+l. Applying (4), we have

=f

f - 9k+Ihk+1

- 9khk -

n

n

i=l

i,j=l

I: Vi(µi9k + Aihk) - I: ViVjAiµj

On the other hand, by (3), f - gkhk = Ef:1 ViO:i, where the O:i are polynomials whose degrees may be assumed to be at most n, since the same is true of f - gkhk, It follows that n

f - 9k+thk+1

= :L:Vi(ai - µi9k -

Aihk)(modPk+l[X])

i=l

In order to obtain (3) for 9k+I and hk+I, we therefore need only show that Ai and µi can be determined so that O:i = µi9k + Aihk( mod P[X]), or equivalently (since 9k g1(modP[X]) and hk h1(modP[X]) so that

=

=

(1 ::; i ::; n) But the latter is possible, since 91 and h1 are relatively prime mod P[X]. Furthermore, since deg(g 1 ) = r, deg(h 1)::; n-r, we may choose Ai of degree less than rand µi of degree not greater than n- r. This completes the proof of the lemma. • We have now come to the demonstration of the main result of this section. Theorem 5.3. (Cohen {1946)). Let R be a complete noetherian local ring which is contained in an integral domain S such that S is a finitely generated R-module. Then S is a complete noetherian local ring and J(S)

= {x

E Slxk E J(R)S

for some

k ~ 1}

Proof. By Lemma 1.1, S is integral over R. Hence, by Lemma 5.1 (ii), PS -/:- S where P = J(R). Setting J = {x E Slxk E PS for some k ~ 1}, we have J -/:- S. We now show that J contains all nonunits of S. This will certainly imply that S is a noetherian local ring with maximal ideal J = J(S).

5 Complete discrete valuation rings

619

Let s be a nonunit of S. Since s is integral over R, it satisfies an equation f(s) = 0, where f(X) = xn + a1xn-l +···+an for some ai ER, 1 ::S; i ::S; n. We may assume that s satisfies no monic equation of lower degree. Since an = -s(sn-l + · ·· + an-1) ands is a nonunit, an is a nonunit of R. Thus an E P. We now claim that all the ai are in P. For if not, let ar be the last one not in P: ar f-1 for all g Let

EG

V be an RG-module and let pv : G -+ A t i t ~ ( Vbe ) defined by pv(g)(v) = gv

for all g E G,v E V

Then ( V , p v ) is a representation of G over R to which we refer as the representation of G afforded by V. The representation of G afforded by the regular module V =RG RG is called the regular representation of G over R. In what follows, we write pv pw to indicate that pv is equivalent ~v

to PW.

The following observation demonstrates that the study of representations of G over R is equivalent to the study of RG-modules.

Lemma 1.1. The map V H pv is a bijective correspondence between RG-modules and representations of G over R. Furthermore, for any RGmodules V and W , V W if and only if pv pw. N

Proof. Assume that p : G --+ A t i t ~ ( V is ) a representation of G over R. Then, by Proposition 8.1.1, p extends to a homomorphism p* : RG -+ E n d R ( V ) of R-algebras given by p*(Crgg) = C rgp(g), g E G, rg E R. In this way p determines a representation p* of RG. Conversely, the restriction to G of any representation of RG gives a representation of G over R. Furthermore, p1 p2 if and only if p ; ,., &. The desired conclusion is therefore a consequence of a general property of representations of algebras, namely Proposition 2.4.1. N

Let pv be the representation of G afforded by an RG-module V . We say that pv is irreducible ( indecomposable, completely reducible) if V is simple (indecomposable, semisimple). By Lemma 1.1, if pv is irreducible (indecomposable, completely reducible), then any representation equivalent to pv is also irreducible (indecomposable, completely reducible). If p : G Aut~(v) is a representation of G over R and V is R-free of finite rank n, then n is called the degree of p and is denoted by deg(p). By a matrix representation of G over R , we understand any group homomorphism --f

X : G + GL,(R)

for some n 2 1

1 Deflnitions and elementary properties

625

The integer n 2 1 is called the degree of the representation X and is denoted by deg(X). Assume that V is an R-module which is free of finite rank n. Then any choice of an R-basis of V leads to the identification of A z t t ~ ( Vwith ) the group GLn(R) of all invertible n x n matrices over R. Hence if p : G + A u t ~ ( vis) a representation of G over R, then by using one of such identifications we obtain a matrix representation p* : G

GLn(R)

of G over R to which we refer as a matrix representation of G corresponding to p. Conversely, any such matrix representation p* determines a representation p : G A u ~ R ( Vwhere ) V is the R-module of all n x 1 matrices over R and p ( g ) ( v ) = p*(g)v for all g E G, v E V . We say that p* is irreducible if p is irreducible. Two matrix representations p1 : G + GL,(R) and p2 : G ---+ GLk(R) of G over R are called equivalent , written p1 p2, if n = k and there exists M E GL,(R) such that .--f

N

p&)

= Mpi(g)M-'

for all g E G

Let V be an R-module which is free of finite rank. For any representation p : G -, A u t R ( V ) , denote by { p } the equivalence class of p and by { p * } the equivalence class of a matrix representation p* of G corresponding to p.

Lemma 1.2. The map { p } +i { p * } is a bijective correspondence between the equivalence classes of representations of G over R of the f o r m p : G + AutR(V), where V is a free R-module of finite rank, and the equivalence classes of matrix representations of G over R.

Proof. It clearly suffices to verify that pi p2 if and only if pi p;. Because the latter is an obvious consequence of the definition of equivalences, the result follows. N

N

Let V be an R-module which is free of finite rank and let p :G

--t

Au~R(V)

be a representation of G over R. For each f E E n d R ( V ) , let t r ( f ) denote the trace of the linear transformation f . Then the map xp:G-tR

Generalities

626

given by

XP(d

= tr(p(g))

for all 9 E G

is called the character of p . The character xx of a matrix representation X : G -+ G L n ( R )is defined by x x ( g ) = tr(X(g))where tr(X(g))is the trace of the matrix X(g).

L e m m a 1.3. (i) Equivalent representations of G of the form p : G AutR(V), where V is a free R-module of finite rank, have the same characters. (ii) x p = xP*,where p* is any matrix representation of G corresponding to p . (iiz) For all x , y E G , x P ( x - l y s ) = x p ( y ) . --f

Proof. Direct consequence of the definitions and the fact that similar matrices have the same trace. Let p; : G -+ A u t ~ ( qbe ) a representation of G over R, where J$ is a free R-module of finite rank, 1 5 i 0, we have F G E F F,G, where IF, is the field of p elements. Hence F G is definable over a finite subfield of F , in which case the result follows by virtue of Theorem 14.5.1 (iv) (b). Lemma 1.6. Let p be a matrix representation of G afforded by the FG-module V and let V = v k 3 v k - 1 3 3 V1 I) Vo = 0 be a chain of submodules of V . Then, upon replacing p b y an equivalent representation if necessary, we have P l M

*

0

P2(9)

P(9) =

. * *

*

... * . - ..

0

0

..-

Pk(9

1

I

J

Generalities

628

where p; is a matrix representation of G afforded by K / V , - l , 1 5 i 5 k. Proof. Choose an F-basis for V corresponding to the given chain of submodules of V and take p to be the matrix representation of G with respect t o this basis. W Corollary 1.7. Let x be the character of G afforded by an FG-module V . Then x is the sum of characters of G afforded by the composition factors of

v.

Proof, Apply Lemma 1.6 to a composition series of V. W Corollary 1.8. Let V and W be FG-modules and let xv and x w be the characters of G afforded b y V and W , respectively. (i) If V and W a% simple, then V 2 W i f and only if xv = x w . (ii) If charF = 0, then V 2 W if and only if xv = X W .

Proof. (i) This is a direct consequence of Lemma 1.5. (ii) Let V1,. . . ,V, be all nonisomorphic simple FG-modules and let n; 2 0 (respectively, m; 2 0) be the multiplicity of V , as. a composition factor of V (respectively, W ) . By Maschke’s theorem V and W are semisimple. Hence, by Lemmas 1.1 and 1.3 (i), the result will follow provided we show that xv = xw implies m; = n; for each i E (1,. . . ,r}. By Corollary 1.7, we have xv = C i = l n ; ~and ; x w = C:==lm;xi. Hence, if x v = x w , then C:==l(n; - m;)Xi = 0. Thus, by Lemma 1.5, n; - mi = 0 in F. Since charF = 0, we deduce that ni = m;, as desired. Let x be a character of G over a field F of characteristic 0. By the degree of x, written degO(), we understand the F-dimension of any FGmodule which affords x. By Corollary 1.8, d e g ( X ) is well defined. It is clear that d e g ( x ) = ~ ( 1 ) .

Lemma 1.9. Let F be a field of characteristic p > 0 and let P be a normal p-subgroup of G. Then (i) FG * I ( P ) C J ( F G ) . (ii) P belongs to the kernel of an arbitrary irreducible representation of G over F . (iii) If P is a Sylow p-subgroup of GI then J ( F G ) = F G . I ( P ) .

1 Definitions and elementary properties

629

Proof. (i) By Lemma8.1.21, J(FP) = I ( P ) . Since P a G , F G . I ( P ) = I ( P ) .F G is a nilpotent ideal of FG. Thus FG - I ( P ) C J ( F G ) . (ii) Let V be a simple FG-module which affords a given irreducible representation p of G. By (i), I ( P ) 5 J(FG) and so I ( P ) V = 0. Hence, for each g E P , v E V ,(g - 1). = 0 and therefore P E K e r p . (iii) Owing to Proposition 8.1.7, F G I F G . I ( P ) S F ( G / P ) . Hence, by Maschke's theorem, FGIFG . I ( P ) is semisimple. Now apply (i). Lemma 1.10. Let F be an arbitrary field, let x be the character of G afforded by an FG-module V of F-dimension m and let g E G be of order n. Assume that F contains all n-th roots of unity in the algebraic closure of F , Then x(g) = ~1 . . - E, where each E ; is an n-th root of unity (in particular, E ; E F). Furthermore, if charF n, then there is an F-basis vl, . . . ,v,, of V such that

+ +

Proof. Put H =< g >. We claim that allirreducible F-representations of H are of degree 1; if sustained, it will follow from Corollary 1.7 that x(g) = ~1 i- E, for some n-th roots of unity ~ 1 , .. . ,E,. Furthermore, under the additional assumption that c h a r F ( n, FH is semisimple. Hence it will also follow that V is a direct sum of one-dimensional FH-submodules, as required. To substantiate our claim, we first note that by Lemma 1.9 (ii), we may assume that c h a r F n, in which case F contains a primitive n-th root of unity E . Then g H E ; , 0 5 i 5 n - 1, gives us n irreducible representations of H over F. On the other hand, since FH is semisimple it is a direct product of, say m, finite field extensions of F. Since m is equal to the number of irreducible representations of H and m 5 n = dimFFH, it follows that g H E ' , 0 5 i 5 n - 1 are all irreducible F-representations of H , as we wished to show. H

+

Let g be an element of G and let p be a prime. Then g can be written in a unique way g = gpgpt,where g p is a p-element, gb has order not divisible by

p and gpgpt = gptgp. The elements g p and of g , respectively.

gpt

are called the p and p'-parts

Corollary 1.11. Let x be the character of G afforded by an FG-module V of F-dimension m and let g E G be of order n.

Generalities

630

+

(i) x(g) = ~1 *.. -t E~ where each E; is an n-th root of unity in the algebmic closure of F . (ii) For any integer p , x(gp) = E: +- - .. EL. (iiz) If charF = p > 0, then x(g) = x(gpj).

+

Proof. We may clearly assume that F contains all n-th root of unity in the algebraic closure of F . Hence (i) follows from Lemma 1.10. As we have seen in the proof of Lemma 1.10, all irreducible F-representations of H =< g > are of degree 1. Let p be a matrix representation of H afforded by V . Then, by Lemma 1.6, we may assume that all entries of p(g) below the main diagonal are zero and the elements on the main diagonal are ~ 1 , .. . ,E ~ Then the diagonal entries of p(gp) = p(g)p are E!, . . . ,&k,proving (ii). To prove (iii), note that p ( g ) = p(gpt)p(gp) and p ( g p ) and p(gpj) have zero entries below the main diagonal. Furthermore, by Lemma 1.9 (ii), the diagonal entries of p(gp) are equal to 1. Hence p(g) and p(gpt) have the same diagonal entries, proving that x(g) = x(gpt). H Corollary 1.12. let F be a subfield of (E and let G over F . Then, f o r any g E G, x(g-') = x(s).

x

be a character of

Proof. We may assume that F contains a primitive n-th root of unity where n is the order of any given g E G. Hence, by Lemma 1.10, m

m

i=l

i=l

as desired.

Corollary 1.13. Let F be a subfield of (r: , let x be a character of G over F and let g E G. Then (i) x(g) E IR if and only i . f x ( g ) = x(g-l). (ii) x(g) E Q if and only if x(g) = x(gM) for all p 2 1 coprime t o the order n of g. Proof. (i) This is a direct consequence of Corollary 1.12. (ii) Let E be a primitive n-th root of unity in (E and, for each p 2 1 coprime to n, let ufl E Gal(Q ( E ) / Q ) be defined by u p ( & )= ~ f l .Then, by Corollary 1.11 (ii), x(gp) = o,(x(g)). Since x(g) E Q if and only if for any such p, a,(x(g)) = x(g), the result follows.

.

2 Splitting fields

631

Corollary 1.14. Let F be a field of characteristic p > 0 and let H be a subgroup of G such that each p-regular element of G is conjugate to an element of H . If xv and xw are the characters of G aft'orded by simple F G modules V and W , respectively, such that x v ( h ) = x w ( h ) for all h E H , then V S W . Proof. Given g E G, we have gpt = x-'hx for some x E G, h E H . Hence, by Corollary 1.11 (iii) and Lemma 1.3 (iii), xv(g) = x v ( h ) = x w ( h ) = xw(g). Now apply Corollary 1.8 (i). We close by remarking that the above result is essentially Theorem 1 in a paper of Cossey, Hawkes and Willems (1980).

2

Splitting fields

In what follows, F denotes an arbitrary field and G a finite group. We say that F is a splitting field for G if F is a splitting field for the group algebra FG. Recall that, by definition, the latter means that every simple FG-module is absolutely simple. A representation of G afforded by an absolutely simple FG-module will be called absolutely irreducible . By an absolutely irreducible character of G over F, we understand the character of G afforded by an absolutely simple FG-module. If E / F is any field extension, then we shall frequently identify E G with E F G (see Corollary 8.1.2). When convenient, we shall also treat interchangeably the terms "representation of G" and "matrix representation of G" (see Lemma 1.2). We begin by recording some elementary properties arising from the general theory of splitting fields for algebras contained in Sec.1 of Chapter 11.

Proposition 2.1.

Let F be a splitting field for G, let

pi : G

-+ GL,,(F)

(1 I i

Ir)

be all nonequivalent irreducible matrix representations of G and let pf : F G +. M,,(F)

(1 5 i

5 r)

be given by p b ( C x g g ) = C z g p ; ( g ) , xg E F, g E G. Then each pt is a surjective homomorphism of F-algebras and the map

Generalities

632

is an isomorphism of F-algebras. In particular, r

dirnFJ(FG) = ]GI - c n ’ i=l

Proof. Let V , be an FG-module which affords pi, 1 5 i 5 r. Then, by Lemma 1.1, V,, . ..,Vr are all nonisomorphic simple FG-modules. Furthermore, by the definition of p t , p;* is the matrix representation of FG afforded by V,, 1 5 i 5 r . Thus pi,. . . ,pr all nonequivalent irreducible matrix representations of FG. The desired conclusion is now a consequence of Corollary 11.1.5. Proposition 2.2. for G, then so is E.

Let EIF be a field extension. If F is a splitting field

Proof. This is a direct consequence of Proposition 11.1.6. I Proposition 2.3. Let G1 and G2 be finite groups. If F is a splitting field for G1 and Gz, then F is a splitting field f o r G1 x G2. Proof. By Corollary 8.1.4, the F-algebras F G I ~ F F F Gand ~ F(G1 xG2) are identifiable. Now apply Corollary 11.1.4. Let E/F be a field extension and let V be an EG-module. We say that V is realizable over F if V 2 WE for some FG-module W . This is, of course, a special case of a general definition pertaining to arbitrary algebras over fields (see Sec. 1 of Chapter 11). A matrix representation p : G -+ GL,(E) of G over E is said to be realizable over F if there exists C E GL,(E) such that C-’p(g)C E GL,(F) for all g E G Thus p is realizable over F if and only if the EG-module V which affords p is realizable over F . Proposition 2.4. Let EIF be a field extension such that E is a splitting field for G. Then F is a splitting field for G if and only if each irreducible matrix representation of G over E is realizable over F. Proof. The desired assertion is a special case of Proposition 11.1.11in which A = FG. I

2 Splitting Aelds

633

Proposition 2.5. Let F be a field of characteristic p > 0, let n be the exponent of G and let F contain all n-th roots of unity in the algebraic closure of F . Then F is a splitting field for G. Proof. Let E be the algebraic closure of F and let x be an irreducible character of G over E . By Lemma 1.10, x ( g ) E F for all g E G. Hence, regarding x as an irreducible character of EG, we have x ( a ) E F for all a E F G . Since F G is definable over a finite subfield of F , the result follows by virtue of Corollary 14.4.2.

We next remark that a celebrated theorem of Brauer asserts that the above proposition is also true in case charF = 0. We shall present a proof of this result at a later stage when we acquire more machinery. We next determine all groups for which all irreducible representations over splitting fields are of degree 1. The following result can be obtained as an easy consequence of a powerful theorem of Morita (1951) pertaining to quasiprimary blocks. We offer instead an elementary self-contained argument.

Theorem 2.6. For any given splitting field F of G , the following conditions are equivalent : (i) All irreducible F-representations of G are of degree 1 . (ii) G is abelian or charF = p > 0 and G’ is a p-group. (iii) G is abelian or charF = p > 0 , a Sylow p-subgroup P of G is normal and G I P is abelian. Proof. It is clear that (ii) and (iii) are equivalent. That (iii) implies (i) is a consequence of Proposition 2.1 and Lemma 1.9 (ii). Finally, we assume that (i) holds and prove (ii). If charF = 0, then J ( F G ) = 0 and, by Proposition 2.1, G is abelian. Hence we may assume that charF = p > 0. Let p : G + GL,(F) be the regular matrix representation of G . By assumption and Lemma 1.6, we may assume that for each g E G , each entry of p ( g ) below the main diagonal is zero. We claim that p(G’) is a pgroup. Since p is faithful, it will follow that G‘ is a p-group, hence the result. Since G’ lies in the kernel of every representation of G of degree 1, the diagonal entries of p ( g ) with g E G‘ are all equal to 1. But charF = p > 0 and so each p ( g ) with g E G’ is a p-element, as desired. Corollary 2.7.

Let G be an abelian group of exponent n and let F be

Generalities

634

the field obtained by adjoining to the prime field of characteristic p 2 0 all nth roots of unity. Then F is the smallest splitting field for G of characteristic p 2 0 and all irreducible F-representations of G are of degree 1.

Proof. Let E be the algebraic closure of F . By Theorem 2.6, all irreducible E-representations of G are of degree 1. Since each of them is realizable over F , it follows from Proposition 2.4 that F is a splitting field for G. Since G is abelian of exponent n, it has an element of order n. Thus any splitting field for G of characteristic p 3 0 must contain F .

3

Counting simple modules over splitting fields

Our aim in this section is to characterize the number of nonisomorphic simple FG-modules, where G is a finite group and F is a splitting field for G. We begin by recording some preliminary information concerning finite-dimensional algebras over fields. In what follows, A denotes a finite-dimensional algebra over a field F and [ A , A ]the commutator subspace of A. Assume that F is of prime characteristic p and put

T ( A ) = { a E AlaP" E [ A , A ] for some n 2 1) Note that, by Proposition 8.3.1, T ( A ) is an F-subspace of A containing [A,A]. In fact, we know much more, namely T ( A ) = [A,A] J ( A ) and J ( A ) is the largest left ideal of A contained in T ( A ) (see Theorem 15.5.5). Another significance of T ( A ) comes from the following result.

+

Theorem 3.1. Let A be a finite-dimensional algebra over a field F of characteristic p > 0 and let F be a splitting field for A . Then the number of nonisomorphic simple A-modules is equal to d i m F ( A / T ( A ) ) .

Proof. Put S = [ A , A ] ,T = T ( A ) ,A = A / J ( A ) and T = T / J ( A ) . Let r be the number of nonisomorphic simple A-modules. Since F is a splitting field for A , it follows from Corollary 11.1.4 that

n T

A2

Mni(F)

i=l

for some positive integers n l , . . . ,nT. Now put = ( [ A A] , t J ( A ) ) / J ( A )and A; = M n i ( F ) . Then obviously S = [A,A] and I- = T ( A ) . Let S; = [A;,A;]

3 Counting simple modules over splitting fields

635

and T; = T ( A ; ) ,1 5 i 5 r . Then r

d i m ~ ( A / T=) d i m ~ ( A / T=) C d i p n ~ ( A ; / T ; ) i=l

and thus it suffices to verify that

dimF(A;/T;) = 1

(1 5 i 5 r )

To this end, denote by e,t the n; x n; matrix with (s,t)-th entry equal to 1 and all other entries 0. Then, for any s # t ,

e,t = esjejt - ejtesj E S; and

ess - ett = w e t s - etsest E si Thus S; contains the np - 1 linearly independent elements e,t, 1 5 s, t 5 n, s # t and ess - e l l , 2 5 s 5 n. Because S; T;, this shows that dimF(A;/T;) = dimFA; - d i m F z 5 n: - (n; 2 - 1) = 1 On the other hand, the elements of m5’; are all n; x n; matrices with trace 0. In particular, ell @ S;. However, e:l = ell for all rn 2 1, so ell # Ti. Thus T; # A; and so from d i m ~ ( A ; / T i5) 1 we deduce that dimF(A;/T;)= 1, as required. Let p be a prime or zero. An element g of a finite group G is called p-regular if p = 0 or if p > 0 and the order of g is not divisible by p. We say that a conjugacy class of G is p-regular if all its elements are p-regular. Let p be a prime and let C1 = {1},C2,. . . ,C, be all p-regular classes of G. Then the sets

5’; = { g E Gig,! E C;}

(1 5 i 5 r )

are called p-regular sections of G. It is an immediate consequence of the definitions that G = u;==,S;with S; n Sj = 0 for i # j and each S; is a union of some conjugacy classes of G , while S1 is a union of all conjugacy classes of p-elements of G (including 1).

p

L e m m a 3.2. Let G be a finite group, let F be a field of characteristic 5’1, S2,. . . ,S, be all p-regular sections of G. Then

> 0 and let

T ( F G ) = { c x g g E FGJ

c

gES,

og = 0 for all

i E {1, ...,r } }

Generalities

636

Proof. We first claim that there exists n 2 1 such that m = 0 (mod n ) and all g E G

m

gp = gpt for all

(1)

Indeed, write [GI = pak with ( p , k ) = 1. Since ( p , k ) = 1, there exists a positive integer n such that pn l ( m o d k ) . Replacing n by its multiple, if necessary, we may assume that n 2 a. Let m be divisible by n and let g E G. rn Then gFm = 1 and, since pm E l(modk), we have g;, = gpt. Thus

proving (1). Let z = C g E G z g gE FG. Then, by ( l ) , there exists m 2 1 such that gPm = gpl for all g E G and such that x E T ( F G ) if and only if xpm E [FG,FG].By Proposition 8.3.1,

=

xpm

c

z:mgPm(mod[FG,FG])

sEG

so x E T ( F G ) if and only if CgEGzimgptE [FG,FG]. Invoking Lemma 8.3.3, we infer that 2 E T ( F G ) if and only if

c

xim =

o

for all i E { I , . . . ,r }

gESi

Bearing in mind that

the required assertion follows. H The crucial case of the following result, namely the case where p is a prime dividing the order of G is due t o Brauer (1935).

Theorem 3.3. Let G be a finite group and let F be a splitting field for G . If charF = p 2 0 , then the number of nonisomorphic simple FG-modules is equal to the number of p-regular conjugacy classes of G . Proof. If p = 0, then all conjugacy classes of G are p-regular and by Maschke’s theorem, FG is semisimple. Since F is a splitting field for G , it

4 Brauer's permutation lemma

637

follows that the number of nonisomorphic simple FG-modules is equal to dirnFZ(FG). Hence the case p = 0 follows from Proposition 8.1.9. Now assume that p > 0 and use the notation of Lemma 3.2. Put T = T(FG) and define $:FG +F x F x

... x F

( r copies)

$(c

"gg) = ( P l , * * * , P T )

g€G

where Pi = CgESi zg,1 5 i 5 r. Then $ is obviously a surjective F-linear map and, by Lemma 3.2, K e r $ = T . Thus dirnF(FG/T) = r and the result follows by virtue of Theorem 3.1.

4

Brauer's permutation lemma

The aim of this section is to present a variant of a combinatorial result known as Brauer's permutation lemma (Brauer (1941, Lemma 1)). An extension of this result to nonsplitting fields will be presented in the next section. In what follows, F denotes an arbitrary field and G a finite group. As usual, all FG-modules are assumed to be finitely generated. Given an FGmodule V , we denote by I n v ( V ) the F-space of all G-invariant elements of V , i.e. I n v ( V ) = {v E V l g v = v for all g E G} We say that V is a permutation module if there exists a basis of V on which G acts as a permutation group.

{vl,

. . . ,v,}

Lemma 4.1. Let V be a permutation FG-module and let B = {vl, . . . ,vn} be a basis of V o n which G acts as a permutation group. If B1,.. . ,B, are G-orbits of B and wi = CzEBi x , then 201,. . . ,w, is an F-basis of I n v ( V ) . Proof. It is clear that w1,.. . ,w, are F-linearly independent elements of I n v ( V ) . Furthermore, if V = A;v; E I n v ( V ) , X i E F, and vi,Vj lie in the same orbit, then A; = A j . So the lemma is true. Theorem 4.2. Let F be an arbitrary field, let XI,.. . ,x m be all irreducible F-characters of G and let K1,.. . ,Km be subsets of G which satisfy the following two conditions :

Generalities

638

(a) Each xi is constant on each Kj. (b) For any given g E G, there exists j = j ( g ) E (1,. . . ,m } such that x;(g) = x i ( K j ) for all i E (1,. .. ,m}. Let A be a group acting on { X I , . . . , x m } and ( K 1 , . . . , K m } and let

x;(I_ 1. Let I m ( J ( A )@ R B ) be the canonical image of J ( A ) @ R B in A @ R B. Then

[ I m ( J ( A )@ R B ) r = I m ( J ( A ) n@ R B ) E I m ( J ( R ) A@ R B ) = J(R)(A@RB) Hence I m ( J ( A )@ R B ) / J ( R ) ( A@ R B ) is a nilpotent ideal of

( A @ R B ) / J ( R ) ( A@ R B ) and so

I m ( J ( A )@ R B ) / J ( R ) ( A@JRB )

J ( ( A@ R B ) / J ( R ) ( A @ RB ) ) = J ( A @ R B ) / J ( R ) ( A@ R B )

It follows that I ~ ( J ( A ) @ R BC )J ( A @ R B )and similarly I ~ ( A @ R J ( BC) ) J ( A@RB). Now put C = I m ( A @ R J ( B ) )t I m ( J ( A )@ R B ) . Then, by Proposition 3.2.5, as R-algebras A @ R B 2 ( A@R B ) / C Since we have shown C C_ J ( A @ R B ) , it follows that

( ( A@ R B ) / C ) / ( ( J ( A@ R B ) ) / C )

( A @ R B ) / J ( A@ R B )

= ( ( A@ R B ) / C ) / J ( ( A@ R B ) / C ) g

( A @ R B ) / J ( A@ R B )

Finally, J ( A ) A = J ( R ) B = 0 and so, by Proposition 3.2.10, A@RB2 r?@iiB which proves (i). (ii) If R is a field and A is a separable R-algebra, then by Proposition 11.3.9, J ( A @ R B) = 0. Now apply (i). H For the rest of this section, R denotes a principal ideal domain and a complete local ring and F = R / J ( R ) . We remind the reader that, by

8 Representations of direct products

657

Corollary 11.3.7, all finite-dimensional semisimple algebras over perfect (in particular, finite) fields are separable.

Theorem 8.6. Let A , B be R-algebras which are finitely generated Rmodules and let V ,W be A and B-modules, wspectively, which are R-free of finite mnk. Given an A , B or AQ?RB-moduleU , let Eu be the endomorphism algebm of U and let EU = Eu/J(Ev).Then the following properties hold : ( E V '@F EW)/J(EV'8F E W ) (i) EV@W (ii) Assume that Ev is a separable F-algebra. Then (a) EvBw E v B F Ew (b) V Q R W is an indecomposable A@RB-module if and only if Ev'8~Ew is a division algebm. In particular, if Ev 2 F and W is indecomposable, then V @ R W is indecomposable. (c) If V ,W are indecomposable and the division F-algebms Ev,Ew are central simple of coprime indices, then V ' 8 W~ is indecomposable. (iii) If F is a finite field, then V @ R W is indecomposable if and only if V ,W are indecomposable and ( d i m F E v , dimr;.Ew) = 1.

Proof. (i) By Proposition 7.6, Ev8w = Ev 8 Ew. Furthermore, by Lemma 5.3.3, both Ev and Ew are finitely generated R-modules. Hence (i) follows from Proposition 8.5 (i). (ii) Property (a) is a consequence of Proposition 8.5 (ii). If Ev @IF Ew is a division algebra, then by (a), V @IBW is indecomposable. Conversely, assume that V @ RW is indecomposable. Then, by Theorem 5.3.6 (ii), E v ~ w is a division algebra. Hence, by (a), Ev @IF Ew is a division algebra. Assume that Ev F and W is indecomposable. Then Ev ' 8 Ew ~ E Ew and, by Theorem 5.3.6 (ii), Ew is a division algebra. Hence, by the above, V @R W is indecomposable, proving (b). Assume that V,W are indecomposable (so that, by Theorem 5.3.6 (ii), Ev,Ew are division algebras) and that the division F-algebras Ev,Ew have centre F and coprime indices. Then, by Theorem 12.3.2, E v @ F Ew is a division algebra. Hence, by (b), V '@RW is indecomposable, proving (c). (iii) If V @ R W is indecomposable, then by (b) Ev @IFEw is a division algebra. Hence both Ev and Ew are division algebras and so V ,W are indecomposable. Hence we may assume that V ,W are indecomposable, in which case, by Theorem 5.3.6 (ii), Ev, Ew are division algebras over the finite field F . It follows from Wedderburn's theorem that Ev and Ew are finite fields. By (b), V '@RW is indecomposable if and only if Ev ' 8 Ew ~ is

Generalities

658

a field. The desired conclusion is now a consequence of Corollary 8.4. Using conventions and notation introduced at the begining of this section we now prove the following result, in which R denotes a principal ideal domain and a complete local ring and F = R / J ( R ) .

Theorem 8.7. I n d ( R H ) , then

(Blau (1974)). (i) If Ind(R(G x H ) )

Ind(RG) 8

Ind(RG)8 Ind(RH) = Ind(R(G x H ) )

(ii) If charF IG( and Ind(RG)8 Ind(RH) G Ind(R(G x H ) ) , then Ind( RG) 8 Ind(RH) = Ind( R( G x H ) )

(iii) If E ~ ~ R G ( V ) / J ( E ~ ~ % RF G for ( V )each ) V E Ind(RG), then

(iv) If F is a finite field and V E Ind(RG), W E Ind(RH), then V @ R W is indecomposable if and only if the F-dimensions of

are coprime. Proof. (i) Let V E Ind(RG), W E Ind(RH) and write V 8~ W = @;=1Uj where each Uj E Ind(R(G x H ) ) . By hypothesis, Uj 2 V, @ R Wj for some V;. E Ind(RG), Wi E Ind(RH), 1 5 i 5 n. P u t m = r a n b ~ ( W ) , mj = ranlc~(Wj), t = ranlc~(V), t j = r a n l c ~ ( V , )1, 5 i 5 n. Then

and

(V 8~ W ) R H= t W

@r=ltiWi

The unique decomposition property (Theorem 5.3.6 (iii)) implies V, 2 V and Wj S W for 1 5 i 5 n. Then V 8~ W E n(V 8~ W ) ,so n = 1 and

v 8~W E Ind(R(G X H ) ) .

(ii) Let U E Ind(R(G x H ) ) . Since (G x H : H ) is a unit in R , there exists V E I n d ( R H ) such that U is a direct summand of R(G x H ) @RH V

659

8 Representations of direct products

(by Theorem 10.2.4 (ii), Proposition 4.9.2 (ii) and the unique decomposition property). Now we have the following R(G x H)-isomorphisms

R ( G x H ) @ R H V2 ( R G @ R R H ) @ R H% V R G @ R ( R H @ R H Vg) R G @ R V and we write RG = @;"=lV, where each V, E Ind(RG). Then U is a direct summand of @7=lV,@ R V . Since, by hypothesis, V , @ RV is indecomposable, we deduce that U 2 V,@RVfor some i by the unique decomposition property. (iii) This is a direct consequence of Theorem 8.6 (ii)(b). (iv) This is a direct consequence of Theorem 8.6 (iii). H Corollary 8.8.

Assume that charF;( [GI and that EndRG(V)/J(EndRG(V)) F

for each V E Ind(RG). If { K l i E I } are all nonisomorphic modules in I nd(R G) and {WjIj E J } are all nonisomorphic modules in I n d ( R H ) , then

{K 8~ Wj(i E I,j

E J}

are all nonisomorphic modules in Ind(R(G x H ) )

Proof. By Theorem 8.7 (ii), (iii), we have Ind(RG) I n d ( R H ) = Ind(R(G X H ) ) Now apply Corollary 7.3. H The following result for the case where H is a pgroup is due to Brauer and Feit (1966). Corollary 8.9. Let F be a field of prime characteristic p which is a splitting field for a p'-group G and let H be any finite group. If {Vl,. . . ,Vn} are all nonisomorphic simple FG-modules and {WjIj E J } are all nonisomorphic finitely generated indecomposable FH -modules, then

{V, @ WjlI 5 i 5 n , j E J } are all nonisomorphic finitely generated indecomposable F ( G x H)-modules.

Proof. We first note that, by Maschke's theorem, all modules in I n d ( F G ) are simple. Furthermore, since F is a splitting field for G , EndFG(V) !2 F for all V E Ind(FG). Now apply Corollary 8.8 for R = F . H

Generalities

660

For the rest of this section, F denotes an arbitrary field and all F-algebras are assumed to be finite-dimensional. As an easy application of the preceding results, we now record a number of additional properties. First, we need the following useful observation. Lemma 8.10. Let A , B be F-algebms and let A / J ( A ) be a separable F-algebra. Then (i) J ( A @ F B ) = J ( A )@ F B 4- A @ F J ( B ) (ii) ( A @ F B ) / J ( A@ F B ) Ei ( A / J ( A )@ F ( B / J ( B ) ) Proof. It is clear that N = J ( A )@ F 33 t A @ FJ ( B ) is a nilpotent ideal of A @ F B . Furthermore, by Proposition 3.2.5,

and, by Proposition 11.3.9, the right hand side is a semisimple F-algebra. Hence J ( A @F B ) = N and the result follows. W Corollary 8.11.

Let F be an arbitrary field and let G , H be finite

groups. Then

+

J ( F ( G x H ) ) = FG J ( F H ) FH * J ( F G ) Setting A = FG and B = F H , we may identify A @IF B with F(G x H ) . Under this identification A @F J ( B ) becomes FG . J ( F H ) and J ( A )@IF B becomes J ( F G )- FH = FH - J ( F G ) . Since, by Corollary 11.3.8, A / J ( A ) is separable, the result follows by Lemma 8.10. Proof.

Corollary 8.12.

Let A , B be F-algebras and let A / J ( A ) be a separable

F -algebra. Then (i) For each semisimple A-module V and each semisimple B-module W , the ( A @ F B)-module V @ F W is semisimple. (ii) For each simple A-module V and each simple B-module W , the ( A @ F B)-module V @ F W is simple if and only if EndA(V)@ F EndB(W)

is a division algebra

(iii) Assume that F is either a splitting field for A or for B . If V,, . . . ,V, and W,, . . . ,W,,, are all nonisomorphic simple A and B-modules, respectively, then wjp 5 i 5 71,l 5 j 5 m} {Vi

8 Representations of direct products

661

are all nonisomorphic simple ( A @ F B)-modules.

Proof. (i) Let V be a semisimple A-module and let W be a semisimple B-module. Then J(A)V = 0 and J ( B ) W = 0. Hence, by Lemma 8.10, J ( A @ F B)(V @F W ) = 0 and therefore V @ F W is semisimple. (ii) By (i), V @F W is semisimple. Hence V @ F W is simple if and only if it is indecomposable, i.e. if and only if EndA@B(V@ W ) is a division algebra. Now apply Proposition 7.6. (iii) By hypothesis, EndA(V,) % F for each i or E n d s ( W j ) 2 F for each j. Hence, by (ii), each I$@ F Wj is simple. Furthermore, by Lemma 11.1.12, the modules {q @ F W j } are pairwise nonisomorphic. Finally, every simple (A@FB)-moduleoccurs as a direct summand of ( A / J ( A ) ) @ F ( B / J ( B(by )) virtue of Lemma 8.10 (ii)) and is therefore of the form T/;. @ F Wj for some i andj. H Corollary 8.13. Let F be a n arbitrary field and let G , H be finite groups. (i) For each semisimple FG-module V and each semisimple FH-module W , the F ( G x H)-module V @F W is semisimple. (ii) For each simple FG-module V and each simple FH-module W , the F ( G x H)-module V @F W is simple if and only if E n d F G ( V ) @F E n d F H ( W )

is a division algebra

(iii) Assume that F is a splitting field for G or for H . If V1,. . . ,V, and Wl,. . . ,W, are all nonisomorphic simple F G and FH-modules, respectively, then { v, @ F wj 1 1 5 i 5 n , 1 5 j 5 m } are all nonisomorphic simple F ( G x H)-modules.

Proof. Put A = F G and B = F H . Then, by Corollary 11.3.8, A/ J ( A) is separable and so the result follows by virtue of Corollary 8.12.

Corollary 8.14. Let F be an arbitrary field, let G , H be finite groups and let V,W be simple F G and FH-modules, respectively. If EndFG(V) and E n d F H ( W ) are fields of coprime F-dimension, then V @ F W is a simple F ( G x H)-module. Proof. We know from Corollary 11.3.8 that F G / J ( F G ) and F H / J ( F H )

Generalities

662

are separable F-algebras. Hence, by Theorems 11.3.6 (iii) and 2.1.2 (iii), both field extensions EndFG(V)/F and E n d F , y ( W ) / F are separable. Now apply Lemma 8.3 and Corollary 8.13 (ii). H

Corollary 8.15. Let F be a finite field, let G ,H be finite groups and let V , W be simple F G and FH-modules, respectively. Then V @ F W is a simple F ( G x H)-module if and only if the fields EndFG(V) and E n d F H ( W ) are of coprime F-dimension.

Proof. Apply Corollaries 8.13 (ii) and 8.4. H

9

Changing the characteristic

Let G be a finite group. Our aim is to provide some connections between the representations of G over fields of characteristic 0 and those over fields of prime characteristic. Let p be a prime number. By a p-modular system , we understand a triple ( F ,R, I for all g E G, cp E GV. For every g E G, there is an hi E H and a permutation i H ~ ( i ) of { 1,. . . ,n } such that g i l g = hig.;f.,. Hence n

%cp)

=

csi 8 (gcp)(s;l)

n

=c si 8 cp(g;'g)

i= 1

i=l

n.

n

n

= C S S T ( i ) 8 cp(g;($ i=l

n

= S ( c g T ( i )8 P K ; ) ) ) i=l

= g w ,

proving that 0 is an RG-homomorphism. If CZ1gi 8 cp(gi') = 0, then cp(g;') = 0 for all i E { 1,. . . ,n } . Because { g l ' , ... , g i l } is a right transversal for H in G , for any given g E G we have g = h g r l for some h E H , i E (1,. . . ,n}. Thus cp(g) = h c p ( g r l ) = 0 and so cp = 0, proving that 8 is injective. Finally, the elements g l ' , . . . , g i l constitute a basis of a free left RHmodule RG. It follows that for every set {v; E V ( 1 5 i 5 n}, there is a cp E GV such that cp(g;') = vi, 1 5 i 5 n. Thus 0 is surjective and the result follows. We close by showing that induction commutes with ground field extensions.

Proposition 1.23. Let H be a subgroup of G , let EIF be a field extension and let V be an FH-module. Then

(V G )S~( V E ) ~ as EG-modules Proof. Let {vili E I } be an F-basis of V and let T be a left transversal E T ,i E I } and (t@(l@vi)lt E T , i E I } and Efor H in G. Then {l@(t@vi)lt bases of ( V G )and ~ ( V E ) respectively. ~, Hence the map t9 : ( V G ) , + ( V E ) ~

2 Induction and semisimplicity

681

defined by O ( l @ ( t 63 vj)) = t @ (1 @ vi) is an E-isomorphism. Fix g E G , t E T and write g t = t’h for some t’ E T , h E H . Then, for z = 163 ( t 63 vi), we have

O(gz) = 0(1 @ (gt 63 vi)) = O ( 1 @ (t’ @ hv;)) = t’ @ (1 @ hv;) = t’~h(l63vi)=t’h63(163vi) = g t 63 (1 63 Vi) = gO(z), as desired. H

2

Induction and semisimplicity

This section is based on a work of Knorr (1977) which treats group algebras over fields. In what follows, G denotes a finite group and R a commutative ring. Our aim is to provide various criteria for semisimplicity of induced modules. All the results below are obtained as an easy consequence of some properties exhibited in Sec.1.

Lemma 2.1, Let H be a subgroup of G. Then

J(RG) G RG * J ( R H ) if and only if J(RG) C J ( R H ) RG *

Proof. Apply Lemma 1.17 (iii) for X = J ( R H ) . H Theorem 2.2. Let H be a subgroup of G and let R be semilocal. Then the following conditions are equivalent : (i) For any simple RH-module V , ( V G )is~semisimple. (ii) J ( R H ) RG C RG J ( R H ) (iii) J ( R H ) * RG = RG J ( R H ) In particular, if these conditions are satisfied, then for any given positive integer n, (RG * J(RH))”= RG * J(RH)”

-

-

Proof. (i) j (ii) : By Proposition 8.1.18 (iii), RH is semilocal. Hence V = R H / J ( R H ) is semisimple, so by assumption ( V G )is~semisimple and thus J ( R H ) G a n n ( V G ) . Now J ( R H ) = a n n ( V ) and hence, by Proposition 1.18 (i), a n n ( V G )= Id(RG - J ( R H ) ) . Hence J ( R H ) - RG C RG J ( R H ) . +

Induced Modules

682

(ii) and so

+ (iii)

: It follows from (ii) that RG J ( R H ) * RG +

RG * J ( R H )

c c

C RG - J ( R H )

RG * J ( R H ) RG Id(RG * J ( R H ) ) = I d ( J( R H ) RG) *

by Lemma 1.17 (iii). Hence RG . J ( R H ) E J ( R H )- RG, proving (iii). (iii) + (i) : It is clear from (iii) that RG . J ( R H ) is an ideal of RG. Let V be a simple RH-module and let X = ann(V). Then J ( R H ) X and so RG J ( R H ) C RG * X . Invoking Proposition 1.18 (i), we deduce that

c

J ( R H ) E RG J ( R H ) *

Id(RG * X ) = ann(VG)

Since RH is semilocal, it follows from Proposition 2.2.4 (ii) that ( V G )is~ semisimple.

Theorem 2.3. Let H be a subgroup of G and let R be semilocal. Then the following conditions are equivalent : (a) For any simple RH-module V , V G is semisimple. (ii) J(RG) C RG * J ( R H ) (iii) J ( R G ) C J ( R H )* RG.

Proof. (ii)

@ (iii) : Apply Lemma 2.1. (i) j (ii) : Put W = R H / J ( R H ) . Since RH is semilocal (Proposition 8.1.18 (iii)) and J(RH)W = 0, it follows from Proposition 2.2.4 (ii) that W is semisimple. Hence, by hypothesis, W Gis semisimple and so J ( RG)WG = 0. But, by Proposition 1.16, WG "= RG/RG . J ( R H ) and thus

J(R G ) C RG * J ( R H ) , as required. (ii) + (i) : Let V be a simple RH-module and let W = R H / J ( R H ) . Since V is simple, V Z R.H/M for some maximal left ideal M of R H . Hence V 2 W/U where U = M / J ( R H ) . By Proposition 1.13 (iii), V GE WG/UG and so it suffices t o show that W Gis semisimple. But J ( R G ) E R G . J ( R H ) implies that J(RG)WG= 0. Since RG is semilocal (Proposition 8.1.18 (iii)) it follows from Proposition 2.2.4 (ii) that W Gis semisimple. 4

Lemma 2.4. Let H be a subgroup of G. If for any simple RG-module V , VH is semisimple, then J ( R H ) J(RG). The converse is true under the additional assumption that R is semilocal.

c

3 Induction of dual and contragredient modules

683

Proof. Let I be any maximal left ideal of R G and let V = R G / I . By hypothesis, VH is semisimple. Hence J ( R H ) V = 0 and so J(RH) C I . Thus J ( R H ) J(RG). Conversely, assume that J ( R H ) & J ( R G ) and that R is semilocal. Let V be any simple RG-module. Since J(RG)V = 0, we have J ( R H ) V = 0. Since R H is semilocal (Proposition 8.1.18 (iii)), it follows from Proposition 2.2.4 (ii) that VH is semisimple. W Theorem 2.5. Let H be a subgroup of G and let R be semilocal. Then the following conditions are equivalent : (i) J(RG) = R G . J ( R H ) (ii) J(RG) = J ( R H ) . R G (iii) For any simple RH-module V, V G is semisimple and for any simple RG-module U , UH is semisimple. Moreover, under the stronger assumption that R is artinian, each of the above conditions is equivalent to (iv) For any simple RG-module U , ( U H ) is~ semisimple. Proof. The equivalence of (i), (ii) and (iii) follows froin Theorem 2.3 and Lemma 2.4. Furthermore, it is clear that (iii) implies (iv). Assume that R is artinian and let V be a simple RH-module. Then VG is a finitely generated module over an artinian ring RG, hence VG is artinian, We may thus choose a simple submodule U of VG. By Proposition 1.6 (ii), 0 # H o r n ~ ~ ( U , v ~H)o ~ R H ( U H , V ) .Thus there is an exact sequence UH -, V -+ 0. By Proposit.ion 1.13 (i), the latter gives rise to the exact sequence (UH)' + VG -, 0. Hence (iv) implies that VG is semisimple. Finally, let U be any simple RG-module. Then (iv) implies ( U H )is~ semisimple, hence UH is semisimple by Proposition 1.15. An extension of the above result, in the special case where R is a field, will be provided at a later stage (see Theorem 10.1).

3

Induction of dual and contragredient modules

Throughout, R denotes a commutative ring and G a finite group. As usual, an RG-module means a left RG-module. A detailed treatment of dual modules over arbitrary R-algebras was presented in Sec.9 of Chapter 3. We begin by recalling the following piece of information.

Induced Modules

684

Let V be an RG-module. Then the dual V of V is a right RG-module R) with the action of R G given whose underlying additive group is HOWLR(V, by for all a E RG, v E V, f E V (fa)(v) = f(av)

If V is a right RG-module, then the dual P of V is an RG-module whose ) the action of R G given by underlying group in H o r n ~ ( v , Rwith ( a f ) ( v ) = f(va)

for all a E RG,v E V , f E V

We avoid the previously used notation V* since it will be reserved for contragredient modules. Proposition 3.1. Let H be a subgroup of G and let V be an R H module. Then (VG)^ % (V)G

Proof. Let T be a left transversal for H in G. By Proposition 1.3 (i), a typical element of V G can be uniquely written in the form C t E T t 8 vt with vt E V. Because {t-llt E T } is a right transversal for H in G, each element of ( V ) Gcan be written uniquely in the form CtET $t 8 t-', $t E V . Consider the map 8 : (V)G--t (VG)" defined by +t tET

8

t-')I(Ct 8 v t ) = c $ t ( v t ) tCl'

tET

Then 0 is obviously an R-homomorphism. If O(CtCT$t @ t-') = 0, then +t(vt) = 0 for all t E T,vt E V . Hence each+! ,T = 0 and so 0 is injective. If X E (V')", then X(&Tt@vt) = CtE~Xt(wt), where X t ( v t ) = X ( t 8 v t ) . But each A t is in V , hence 8(&- At @ t - l ) = A and therefore 0 is surjective. By the foregoing, we are left t o verify that for x = $ 8 t - l , E V , t E T , and g E G, 8 ( x g ) = d ( x ) g . To this end, write t - l g = ht,' for some tl E T . Then xg = $ 8 t - l g = $h @ t l 1 and therefore

+

On the other hand, if

t2

E T with

tz # t l , then

3 Induction of dual and contragredient modules

685

Consequently, 8 ( x g ) = 8 ( s ) g and the result follows. Let V be an RG-module. Recall that an element v E V is said to be G-invariant (or simply an invariant element ) if gv = v for all g E G. It is clear that the set InvG(V) of invariant elements in V is the largest R-submodule of V on which G acts trivially. If W is another RG-module, then Horn@, W ) becomes an RG-module via

(gf

= d f( g - l v ) )

( 9 E G, E v,f E HomR(V,W ) )

and H O ~ R G ( V , W =)InWG(HOmR(V,W)). The special case of the RG-module H o m ~ ( vW, ) ,in which W = R and gr = r for all r E R , g E G deserves a separate treatment. In this case, Ho~R(V R), = $' as an R-module and the action of G on H o ~ R ( VR), is given by (. E v,9 E G, f E Q) ( g f ) ( v )= f ( g - l v ) We shall denote this module by V* and refer to V' as the c o n t r a g r e d i e n t of

v.

Proposition 3.2. Let H be a subgroup of G , let V be an RG-module and let W be an RH-module. Then (i) [HomR(W,VH)IG HomR(WG,v )

(ii) [ H o ~ R ( V ' , W ) ] ~H o ~ R ( V , W ~ ) Proof. In what follows, T denotes a left transversal for H in G. (i) A typical element f E [ H o ~ R ( W , V Hcan ) ] be ~ uniquely written in

Consider the map

defined by

8(f)(Ct €3 W t ) = C t f t ( w t ) tET

(wt E

w>

t€T

Then 8 is obviously an R-homomorphism. If e(f) = 0, then 8( f ) ( t €3 WJ) = t f t ( w ) = 0 for all t E T , w E W . Thus each f t = 0 and therefore 8 is injective. If x E H O ~ R ( W ~then , V )X(CtETt €3 wt) = &T tXt(wt), where

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At(wt) = A ( t @ tot). Since each At is in H o ~ R ( W VH , ) we , see that 8(f) = A for f = CtET t @ At. Hence 8 is surjective. Fix g E G and, for any t E T , write g t = t'ht with ht E H and t H t' a permutation of T . Thus, if f = &T t 63 f t , then gf = C t E T t' 63 htft. Furthermore,

and

proving (i). (ii) A typical element the form

f=

f E [ H o m ~ ( V . q , Wcan ) l ~ be uniquely written in

xt

(ft E HomR(vH,W))

@ ft

tET

Consider the map

given by

O(f)(v)

=

@ ft(t-lv) tET

Then 0 is clearly an R-homomorphism. If S(f) = 0, then f t ( t - l v ) = 0 for all t E T , v E V . Therefore f t = 0 for all t E T , so f = 0 and hence 8 is injective. If A E H O ~ R ( V , W ' )then , X ( v ) = CtET 2 8 At(t-'v) for some At E H o ~ R ( V W H ,) . Thus 8(z) = A for z = CtETt @ A t , proving that 8 is

3 Induction of dual and contragredient modules

687

surjective. Fix g E G and, for each t E T , write gt = t'h, with t of T . Then, for all v E V ,

H

t' a permutation

and

proving (ii).

Corollary 3.3. Let H be a subgroup of G and let W be an RH-module. Then (W*)G 2 (WG)*

Proof. Apply Proposition 3.2 t o the case where V = R and G acts trivially on V . Corollary 3.4. The group algebra R G is self-contragredient, that is, (RG)*Z R G as left RG-modules. Proof. Let H = 1 and let W be the RH-module R on which H acts trivially. Then W G= RG @RH R % RG and W* Z W . Hence, by Corollary 3.3, (RG)*2 (WG)*2 ( W * ) G2 RG, as desired.

Whilst on the subject of contragredient modules, we will provide some further properties which will be applied in our subsequent investigations. Most of the results recorded are immediate consequences of corresponding properties of dual modules for arbitrary R-algebras established in Sec.9 of Chapter 3. This remark arises from the fact that the left action of g E G on f E V * is the same as the right action of g-' on f in

v.

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688

@zl @r=l

Lemma 3.5. (i) For any RG-modules V1,. . . ,V,, ( K)* 2 V,.. (ii) For any RG-module V which is a finitely generated projective Rmodule, V is indecomposable if and only if V* is indecomposable. (iii) If R is a field, then for any finitely generated RG-module V , V is simple if and only if V* is simple. (iv) If V is an RG-module which is a finitely generated projective Rmodule, then V* is a finitely generated projective R-module and V Z V**.

Proof. (i) Apply Corollary 3.9.2, (ii) Apply Corollary 3.9.6. (iii) Apply Proposition 3.9.11. (iv) Apply Proposition 3.9.5. H

Lemma 3.6. so is v*.

If V is a finitely generated projective RG-module, then

Proof. By hypothesis, RG @ - - @ RG S V @ W for some RG-module W . Hence, by Corollary 3.4 and Lemma 3.5 (i), V* is isomorphic t o a direct summand of RG @ . - -@ RG, as desired. H Lemma 3.7. Let V be an RG-module which is R-free of finite rank and let I' : G -, GL,(R) be the matrix representation of G afforded by V with respect to a given R-basis of V . Then, with respect to the duaZ basis, the matrix representation I'* of G afforded by V* is given by I'*(g) = T ( g - l ) for all g E G , where tI'(g-l) is the transpose of I'(g-'). In particular, the character x* of r* is given by x*(g) = x(9-l)

for all g E G

Proof. Apply Proposition 3.9.4. Given

f E HomRG(U, V ) ,we define f* E HomRG(V*, U * ) by f * ( a )= a 0 f

(aE V*)

Lemma 3.8. Let U,V and W be RG-modules and let f E HomRG( U , V ) , g E HomRG(V,W ) ,

(i) I f f is a split surjection (injection), then f * is a split injection (surjection).

3 Induction of dual and contragredient modules

V

(ii) If U

5W + 0

is exact, then

0+ is exact. (iii) If 0

+

689

f V3W U --*

0+

+

w*z v*c u* 0 is exact and R is self-injective, then

w*3v*c u* + 0

is exact.

Proof. (i) and (ii) : Apply Lemma 3.9.1. (iii) This is a direct consequence of the assumption that R is an injective R-module. H Let V be an RG-module and let U be a submodule of V . We define UL by

uL = {Ic, E V*lIc,(V)= 0)

Then U' is obviously an RG-submodule of V * . It is also clear that for any submodule U1 of U , U' E U k . Lemma 3.9. Let U be a submodule of an RG-module V and let W be an RG-module. Then (i) For any g E HomRG(V, W ) ,lierg* = (Img)* and ( l i e r g ) l 2 Img'. (ii) I f f : V + V / U is the natural homomorphism, then I m f * = U' and

( V / U ) *2

u'

(iii) If V = U @ W , then V* = UL @ W' and U' S W', W' S U*. (iv) If R is a field and V is a finitely generated RG-module, then (a) V*/u' u* and dimRU da'mRU' = dimRV. (b) The map U H UL yields an inclusion reversing bijection between the submodules of V and those of V * .

+

Proof. Apply Propositions 3.9.3 and 3.9.11. Lemma 3.10. Let V and W be RG-modules. If as an R-module W is finitely generated and projective, then

HomRG(V, W ) 2 HomRG(W*, V * )

as R-modules

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690

Proof. Apply Corollary 3.9.10. Lemma 3.11. Let V and W be RG-modules such that V is a finitely

generated projective R-module. Then

Proof. By Proposition 3.6.14, the map $ : V* @ R W * 3 (V @ R W ) * given by

is an R-isomorphism. Given x E G, we also have

as required. Lemma 3.12. Let V and W be RG-modules such that V is a finitely generated projective R-module. Then as RG-modules. (i) V*@ R W S HomR(V,W ) as RG-modules. (ii) V € 3 W ~ Z HomR(V*,W )

Proof. (i) By Proposition 3.6.12, the map a : V*@RW+ H o r n ~ ( VW , ) given by

.(f

€3 ID)(.)

= f(v)w

(f E v*,w E w,v E V )

is an R-isomorphism. Given g E G , v E V , w E W and f E V * ,we have

as required. (ii) Apply (i) and Lemma 3.5 (iv). H

4 Reciprocity theorems

4

691

Reciprocity theorems

Throughout this section, G denotes a finite group and F a field. All modules over group algebras are assumed t o be finitely generated. Given FG-modules U and V , we denote by i ( U , V ) the intertwining number for U and V which is defined by i(U , V ) = dimFHomFG(U, V ) . Our point of departure is the following result which will be obtained as a consequence of the behaviour of induced modules over arbitrary F-algebras.

Theorem 4.1. Let H be a subgroup of G , let V be a simple FH-module and let W be a simple FG-module. Denote by m the multiplicity of W as a simple constituent of V G / J ( V G )and by n the multiplicity of V as a simple constituent of SOC(W H ) .Then

m . i ( W , W )= n . i ( V , V ) I n particular, if F is a splitting field for G and H , then m = n.

Proof. Apply Theorem 11.2.2 for B = F H and A = F G . Corollary 4.2. (Berman (1958)). Let H be a subgroup of G, let V be a simple FH-module and let W be a simple FG-module. Denote by m the multiplicity of W as a simple constituent of V G and by n the multiplicity of V as a simple constituent of W H .If charF 4 IGl, then m=

n - dimF(EndFH(V)) dimF(EndFG( w ) )

Proof. Since the characteristic of F does not divide IGl, F G is semisimple by Maschke’s theorem. Hence J ( V G ) = 0 and SOC(WH) = W H . Now apply Theorem 4.1. Corollary 4.3. (Frobenius Reciprocity). Let H be a subgroup of G , let F be a splitting field for G and H and let charF 4 IGI. If V and W are simple F H and FG-modules, respectively, then the multiplicity of V as a simple constituent of WH is equal to the multiplicity of W as a simple constituent of vG. Proof. Our hypothesis on charF ensures that J ( V G ) = 0 and that SOC(WH) = W H .The desired conclusion is therefore a consequence of Theorem 4.1.

m

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692

Lemma 4.4. Let F FG-modules. Then

be

a splittingfield for G and let U and W be simple

Proof. If there is a nonzero FG-homomorphism from P ( U ) to W , then

W 2 P ( U ) / J ( F G ) P ( U )2 U Since i ( P ( U ) , U )= i ( U , U ) = 1, the result follows. The following result can be obtained from a work of Nakayama (1938).

Theorem 4.5. (Nakayama Reciprocities). Let H be a subgroup of G, let F be a splitting field for G and H and let V ,W be simple F H and FGmodules, respectively. Then (i) The multiplicity of P ( V ) as a direct summand of P(W)His equal to the multiplicity of W as a simple constituent of V G . (ii) The multiplicity of P ( W ) as a direct summand of P(V)Gis equal to the multiplicity of V as a simple constituent of W H . (iii) The multiplicity of V as a simple constituent of P(W)His equal to the multiplicity of W as a simple constituent of P ( V ) G . Proof. (i) Let m be the multiplicity of P ( V ) as a direct summand of P ( W ) H .Then, by Lemma 4.4, m = i ( P ( w ) ~ , v On ) . the other hand, by ) Since, by Lemma 11.4.7, Proposition 1.6 (ii), ~ ( P ( W ) H ,=Vi(P(W),VG). i ( P ( W ) , V Gis) the multiplicity of W as a simple constituent of V G ,(i) is established. (ii) Let n be the multiplicity of P ( W ) as a direct summand of P(V)G. By Lemma 4.4, n = i(P(V)G,W).On the other hand, by Proposition 1.6 (i), i(P(V)G, W ) = i(P(V),W H ) .Since, by Lemma 11.4.7, i ( P ( V )W , H )is the multiplicity of V as a simple constituent of W H ,property (ii) follows. (iii) Let Ic be the multiplicity of V as a simple constituent of P ( W ) H . Then, by Lemma 11.4.7, k = i ( P ( v ) , P ( w ) ~Hence, ). by Theorem 15.5.1 (vi), Ic = ~ ( P ( W ) H , P ( VThus, ) ) . by Proposiiton 1.6 (ii),

k = i(P(W),P(V)G) Hence, by Lemma 11.4.7, k is equal to the multiplicity of W as a simple . constituent of P ( v ) ~ H

4 Reciprocity theorems

693

Let H be a subgroup of G, let F be a splitting field for G and H and let V and W be simple FH and FG-modules, respectively. Then, owing to Theorem 4.5 (ii), the multiplicity of P ( W )as a direct summand of P(V)' is equal to the multiplicity of V as a simple constituent of W H .It is natural to enquire about the multiplicity of P ( W ) as a direct summand of V G . More precisely, does this multiplicity coincide with the multiplicity of P ( V ) as a direct summand of WH? Our next aim is to provide a positive answer. The proof, as we shall see below, will require much more sophisticated machinery. As a starting point, let us record some properties of an elementary nature pertaining to an arbitrary field F.

Lemma 4.6. Let W be a simple FG-module. Then (i) For any primitive idempotent e of FG, eW

#0

if and only if

FGe 2 P ( W )

(ii) P(W)*E P ( W * ) , whem W* is the contmgredient of W .

Proof. (i) By Lemma 11.4.7, eW # 0 if and only if i ( F Ge , W) # 0. Since the latter is obviously equivalent to F Ge E P ( W ) ,property (i) is established. (ii) Since P ( W ) is projective indecomposable, so is P(W)* by Lemmas 3.5 (ii) and 3.6. Since W* is simple (Lemma 3.5 (iii)), P ( W * ) is indecomposable. Hence, by Theorem 15.5.1 (v), it suffices to show that

Soc( P(W ) * )2 Soc(P( W * ) ) But, by Theorem 15.5.1 (ii), Soc(P(W*))2 W* hence we must show that

W* E Soc(P(W)*) By definition of P ( W ) ,there is a surjective FG-homomorphism f : P ( W )+ W . Hence, by Lemma 3.8 (ii), the map f * : W * P(W)* is an injective FG-homomorphism. But P(W)* is projective indecomposable, hence by Theorem 15.5.1 (ii), W* 2 Soc(P(W)*),as desired. W --f

The following theorem is the key ingredient for the proof of the main result.

Theorem 4.7. (Robinson (1989)). Let F be a field of characteristic p > 0 which is a splitting field for G, let W I ,. . . ,W, be all nonisomorphic

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694

simple FG-modules and let W be an FG-module. Denote by x; the character of G afforded by W; and put z; = CsEGx;(g-l)g,1 2 i 5 r. Then (i) For each i E ( 1 , . . . ,r}, z;W 2 n;W; where n; 2 0 is the multiplicity of P(W;) as a direct summand of W . In particular, z;W # 0 if and only if P( W,) is isomorphic to a direct summand of W . (ia) zjFG S m;W;, where m; = dimFWi, 1 5 i 5 r.

Proof. (i) We may write W = U1 @ .. - @ U t , where each U; is indecomposable, in which case z;W = z;Ul @ . @ z;Ut. Thus it suffices t o prove that if W is indecomposable, then (a) z;W = 0 if W P(W;) (b) z;W 2 W; if W 2 P(W;) We first show that (a) and (b) hold for principal indecomposable FG-modules. Then we show that if zjW # 0, then W is isomorphic t o a principal indecomposable FG-module. This will obviously complete the proof of (i). Let e be a primitive idempotent of F G and let W = FGe. Then z;W # 0 if and only if z;e # 0. Hence, by Theorem 17.6.5 (iii), z;W # 0 if and only if eW, # 0, which by Lemma 4.6 (i) happens if and only if W % P(W;). Furthermore, if W E P(W;), then SOC(W) P(W;)/J(FG)P(Wi) 2 W; by Theorem 15.5.1 (5). But, by Theorem 17.6.5 (iv), z;W E Soc(W) and hence z;W = Soc(W). Thus, if W E P(W;),then ziW 2 W;. Let W be an indecomposable’FG-module. By the foregoing, we are left t o verify that if z;W # 0, then W is isomorphic to a principal indecomposable FG-module. To this end, suppose that z;W # 0. Choose w E W with z;w # 0. Consider the homomorphism of left FG-modules cp : F G -+ W given by ~ ( z = ) z w for all z E F G . Then z;FG Iiercp and so there is a primitive idempotent e E F G with z;FGe Kercp. This implies that ziFGe # 0 and hence, by the foregoing, F G e E P(W;). Moreover, because Soc(FGe) = z;FGe is simple, F G e n Kercp = 0. Hence F G e is isomorphic t o a submodule of W. But FGe is a projective FG-module, hence FGe is also injective (since F G is symmetric). Thus F G e is isomorphic t o a direct summand of W . Because W is indecomposable, we deduce that W FGe, proving (i). (ii) Since F is a splitting field for G , it follows from Proposition 11.4.8 (iv) that F G 2 @S,lmjP(Wj)

4 Reciprocity theorems

695

The desired assertion is therefore a consequence of (i). We are now in a position to undertake the proof of the main result.

Theorem 4.8. (Klyachko (1979), Robinson (1989)). Let H be a subgroup of G , let F be a field of characteristic p > 0 which is a splitting field for G and H and let V and W be simple F H and FG-modules, respectively. Then the multiplicity of P ( W ) as a direct summand of V G is equal to the multiplicity of P ( V ) as a direct summand of W H .

Proof. Let W1,W2,. ..,W, be all nonisomorphic simple FG-modules and let V1,.. . ,I4be all nonisomorphic simple FH-modules. Let XI,. . . ,At be the elements of R e y ( F H ) as the 21,.. . ,t, were for Rey(FG) (see Theorem 17.6.5). Denote by m;j the multiplicity of P ( W ; ) as a direct summand of qGand put kj = d i m F K , 1; = dimFWi

(1 5 j 5 t , 1 5 i 5

T)

Then we have

FGXj

E!

(FHXj)'

2

(kjV,)G

(by Proposition 1-16) (by Theorem 4.7 (ii))

E kjTG

(1)

Observe also that, by Theorem 4.7 (i),

diml;z;FGXj = nijl;

(1 5 i 5 r, 1 5 j 5 t )

(2)

where n;j is the multiplicity of P ( W ; )as a direct summand of FGXj. Owing to (l),kjm;j = n;j and hence, by (2),

dimFziFGXj = kjm;jl;

(1 5 i 5 r, 1 5 j 5 t )

(3)

For any x = C x g g E F G , xg E F , g E G , put x* = C xgg-'. Then the map x H x* is an anti-automorphism of FG. If W: is the contragredient of Wi, then by Lemma 3.5 (iii), (iv), W;, . . . ,W,+are all nonisomorphic simple FGmodules. Similarly, V;, . . . ,V,+are all nonisomorphic simple FH-modules. If x: is the character of G afforded by W:, then by Lemma 3.7 x t ( g ) = x i ( g - ' ) for all g E G. Hence

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This shows that zr performs the role of zi with respect to WT. Similarly, A; performs the role of X j with respect to V;. We now note that dimF(tjFGAj) = dimF(AgFGzr), since x H x* is an anti-automorphism of FG. Again, by Theorem 4.7 (i) and our foregoing observation on A;, we have

dimF(Xj*FGzl) = S j j d i m F V j ’ = kjs;j where sjj is the multiplicity of P(V;)as a direct summand of ( p G z f ) ~It. follows from (3) that m..l.= 8 “ (4) I.? t

1.l

On the other hand, by Theorem 4.7 (ii), FGzr Ei (dimFWT)W;*= ljW;*. Therefore sjj = l;p;j is the multiplicity of P ( V ; ) as a direct summand of ( W ~ ) HInvoking . (4), we deduce that m;j = pjj for all i E (1,. . . ,r } , j E (1,. . . ,t}. But, by Lemma 4.6 (ii), pij is the same as the multiplicity of P ( V j ) as a direct summand of (W ; ) H,hence the result. W Our next result, which is of independent interest, will be obtained with the aid of Theorem 4.8. We remind the reader that the numbers Cjk in the theorem below are Cartan invariants of FG.

Theorem 4.0. (Robinson (1989)). Let H be a subgroup of G , let F be a field of chamcteristic p > 0 which is a splitting field for G and H and let W l , ,. . ,W, and V1,. . . ,V, be all nonisomorphic simple F G and F H modules, respectively. Denote by mij the multiplicity of P(W;) as a direct summand of vj” and by Cjk the multiplicity of wk as a composition factor of P(Wi). Then, for any given i E { 1, . . . ,r } ,

j=1 k=l

with equality if Wj is projective. Moreover, if H is of p‘-index and the equality holds, then W; is projective. Proof. By assumption, tion factor of P ( W ; ) and so

Wk

occurs with multiplicity

Cjk

as a composi-

4 Reciprocity theorems

697

Owing to Theorem 4.8, ( W k ) H contains P ( 4 ) as a direct summand with multiplicity m k j . Because P(V1), . . . ,P ( & ) are nonisomorphic, we see that @f=lmkjP('Cj) is a direct summand of ( W k ) H . Consequently

This proves the required inequality, by applying ( 5 ) and ( 6 ) . Now suppose that Wi is projective. Then so is ( W ~ ) H by, virtue o f Proposition 1.4 (iii). Accordingly,

and, because Cik = 0 for i # k , C j i = 1, the desired equality holds. Conversely, assume that the equality holds and H is of p'-index. Then there is certainly some k E (1,. . . ,r } and some j E (1,. . . ,t } such that C j k m k j # 0. In particular, Cik # 0 and therefore w k is a composition factor of P(Wi). Moreover, because mkj # 0 and cjk # 0, the equality in (5) must hold. Thus (Wk)H @f=,mkjP('Cj) and so ( W k ) H is projective. Because H is of p'-index, it follows from Proposition 1.4 (iii) that Wr, is also projective. But C,k # 0, so wk is a projective composition factor of P(W;).Since by Theorem 15.5.1 (ii) S o c ( P ( W ; ) )2 Wi is simple, we have i = k and Wi is projective. H We next present another result related to Theorem 4.8.

Theorem 4.10. (Robinson (1989)). Let H be a subgroup of G, let F be an arbitmry field and let W1,.. . ,WT and V1,.. . ,V, be all nonisomorphic simple F G and F H -modules, respectively. Denote by m;j the multiplicity of P ( W ; ) QS a direct summand of %G and by C j k the multiplicity of v k QS a composition factor of P(V,), 1 5 i 5 r , 1 5 j , k 5 t . Then, f o r any j E {l,...,t},

with equality if and only if Vj is projective.

Induced Modules

698

Proof. For any given k E { 1 , . . . ,t } , v k occurs tion factor of P ( 4 ) and so

On the other hand, for any such k, P(Wi) occurs summand of VF which implies

Cjk

times as a composi-

m;k

times as a direct

mikdimFP(Wj)

(8)

T

( G : H)dimFVk 2 i=l

The desired inequality now follows by comparing (7) and (8). Assume that V, is projective. Then C j k = 0 for k # j and cjj = 1. On the other hand, by Proposition 1.4 (ii), l$G is projective and so

V;: E eT=lmijP(Wj), proving the desired equality. Conversely, assume that the equality holds. Then there exist i E { 1 , . . . ,r } and k E { 1 , . . .,t } such that cjkm;k # 0. Since C j k # 0, m ; k # 0, the equality in (8) must hold and so VG k

Pi

-

@{=imikP(Wi)

It follows that V f is projective. Hence, by Proposition 1.4 (ii), v k is projective. But Cjk # 0, so v k is a projective composition factor of P ( 6 ) . Since, by Theorem 15.5.1 (ii), SocP(V,) 2 V, is simple, we have k = j and V, is projective. We close by providing some applications of Frobenius reciprocity.

Theorem 4.11. (Janusz (1966)). Let H be a subgroup of G , let F be a splitting field for G and H and let charF # [GI. Let V be a simple FG-module and let e be the block idempotent of F G such that V belongs to the block FGe. Iff is a primitive idempotent of F H such that VH contains F H f with multiplicity 1, then e f is a primitive idempotent of F G such that V Z FGef.

Proof. By Proposition 18.1.16, applied t o X = 0 and Y = F H f , we see that F G f 2 ( F Hf ) G

5 Tensor products

699

Multiplication by e annihilates all the direct summands of F Gf not isomorphic t o V. Hence . F G ef is a direct sum of copies of V. By Corollary 4.3, the number of copies is 1. Thus F G e f V as desired.

Corollary 4.12. Let F be a splitting field for all subgmups of G and let c h a r F 4 [GI. Suppose there is a chain of subgroups

and that V , is a simple FH;-module, 0 5 i 5 r, V = V,, such that V,+1 has multiplicity 1 in ( K ) H , + ~0 , 5 i 5 r, and dimFV, = 1. If ei is the block idempotent of FH; such that V , belongs to the block FH;e;, then e = eOe1 - - - eT is a primitive idempotent of F G such that V Z FGe.

Proof. Since dimFV, = 1, we see that e, is also a primitive idempotent of FH,. The desired conclusion now follows by applying induction and Theorem 4.11. Let V be a simple FG-module where F is a splitting field for G with c h a r F J /GI. If e is the block idempotent of F G such that V belongs t o F G e and V affords the character x, then it is easy t o show (see Lemma 2.7 in the next chapter) that g€G

In view of this, Corollary 4.12 tells that the FG-module V in it affords the representation which can be written in the field generated over the prime subfield of F by the values of the characters xi for 0 5 i 5 r . Of course, our observation is of interest only in the case where c h a r F = 0. We shall pursue this topic in our future study of characters.

5

Tensor products

In what follows, G denotes a finite group and R a commutative ring. Given RG-modules V and W , we write V @ R W for the inner tensor product of V and W . Recall that V @ R Wis an RG-module viag(u@w) = gv@gw, v E V, w E W , g E G. If G1 and G2 are groups and V, is an RGi-module, i = 1,2, we denote by V1#V2 the outer tensor product of V1 and V2. We remind the reader that V1#V2 is an R(G1 xG2)-module via (gl,g2)(~1@02)= g1vl@g~v2,

Induced Modules

700

g; E G;, v; E fi, i = 1,2. A detailed treatment of inner and outer tensor products is given in Sec.7 of Chapter 17.

Theorem 5.1. Let H be a subgroup of G, let U be an RH-module and let V be an RG-module. Then (i) (u @ R V H ) G E U G @ R v (ii) If H d G, then ( V H )E! ~R ( G / H )@ R V where the RG-module structure on R(G/H) is given by g(g1H) = gglH.

Proof. (i) Let T be a left transversal for H in G. Then the map

f : UG x

v + (U@RVH)G

given by

is balanced. Hence f determines a unique R-homomorphism

UG@V (t 8 u) 8 v

r:

(U@VH)G t @ (u 63 t-lv)

where the tensor products are over R. Similarly, the map

( U @ VH)G

--f

uc 63 v,t 8 (u €9 t-'w)

H

(t 63 u) 63 2,

is an R-homomorphism which is inverse to f *. Hence f * is an R-isomorphism.

proving (i). (ii) Let R be the RH-module on which H acts trivially. Then, by (i), applied to U = R and by Lemma 1.19 applied t o I = 0, we have

v~)'

( V H )2~(R € 3 ~ as required. W

RG @ R v 2 R ( G / H )@ R v

5 Tensor products

701

Theorem 5.2. Let H ; be a subgroup of a group Gi, and let V, be an RH-module, i = 1,2. Then ( h # v p o 2

2

v:l#vp

Proof. For convenience, we identify R(H1 x H 2 ) and R(G1 x G2) with RH1 € 4 RH2 ~ and RG1 € 4 RG2, ~ respectively. With these identifications, ( a @ ~ ) ( w ~ ~ w ~ ) = a q @ ~ b v ~EVl,v2EV2,aE forallu~ RH1,bE RH2anda similar formula holds for the action of RG1 € 4 RG2 ~ on #V2G2. To prove the result, we have to establish an isomorphism

Let T;, i = 1,2, be a left transversal for Hi in G;. Then, by Proposition 1.3 (i), we have the following direct decompositions of R-modules :

given by

f"tl €4 $ 2 ) @ 1(.

€4

V2)l

= (tl €4

0 1 ) c3

(t2 @ 0 2 )

is obviously an R-isomorphism, we see that f induces an R-isomorphism

f*

:

v w. --+

Finally, given g l E G I , g2 E G2, write git; = s;hj with s; E

i = 1,2. Then

as required.

I";, h; E H ,

Induced Modules

702

6

Mackey’s theorems

In what follows, R denotes a commutative ring and G a finite group. The process of induction is closely related to the concept of imprimitivity of a module. We say that an RG-module W is imprimitive if W can be written as a direct sum =

w @;Jv;

of R-modules W; with n > 1 such that for each g E G, W; H gWi is a permutation of the set {Wl,. , . ,Wn}. The latter is equivalent to the assumption that G acts on the set {Wl,. . . ,W n }with each g E G sending Wi to gWi. We refer to {Wl,. . . , W n }as a system of imprimitivity for W. The following useful observation ties together imprimitive and induced modules.

Lemma 6.1. (i) Let H be a proper subgmup of G, let V be a n R H module and let T be a left transversal for H in G. Then {t @J V ( t E T } is a system of imprimitivity for VG and the action of G on it is transitive. (ii) Let W be an imprimitive RG-module with { W1,. . . ,W n }a s a system of imprimitivity and let G act transitively on { W l , .. ., Wn}. If H is the stabilizer of any U E {Wl,. . . ,Wn}, then U is an RH-module such that

W 2 UG. Proof. (i) This follows from Proposition 1.3 (i) and the fact that the left multiplication by g E G induces a transitive action of G on the set {tHlt E T } . (ii) It is clear that U is an RH-module. Let T be a left transversal for H in G. Then {Wl, ...,W n } = {tUlt E T } and W = e t E T t U (direct sum of R-modules). Hence, by Proposition 1.3 (i), (iv), the map t 18U -, tU, t @ u H tu induces an R-isomorphism f : U G -, W . Given g E G, t E T , we may write gt = t’h for some t’ E T , h E H . Thus

as desired. W Let H be a subgroup of G, let V be an RH-module and let g E G. We define gV to be the R(gHg-l)-module whose underlying R-module is V and on which the elements z E gHg-’ act according to the rule : z

*

2,

= (g-’zg)w

6 Mackey's theorems

703

Lemma 6.2. Let H be a subgroup of G, let V be a n RH-module and let g E G. Then (i) gV E g 8 V and hV 2 V for all h E H (ii) (")G VG (iii) For all z, y E G, "YV 2 "(Y V ) (iv) V is simple (semisimple, indecomposable) if and only if so is gV (v) If R is a subring of a commutative ring S , then

=

(")s

2 "Vs)

as S(gHg-')-modules

where Vs = S 8~ V . (vi) g ( V * )= (")* Proof. (i) The map f : gV --t g @ V , v H g 8 v is an R-isomorphism, by Proposition 1.3 (iv). Given z = ghg-', h E H , we have

f ( z * V) = f(hv) = 9 @ hv = gh 8 v = ghg-'(g 8 V) = zf(w), proving that gV 2 g 8 V . If h E H , then hV 2 h 8 V = 1 8 V 2 V , as desired. (ii) Choose a left transversal T for H in G with g = t E T . Then the stabilizer o f t 8 V is tHt-'. Hence, by (i) and Lemma 6.1 (ii),

VG E (t @ VI) G2 (")G as asserted. (iii) Invoking (i), we have

"YV 2 (zy) 8 v = z(y 8 V )2 "(y 8 V ) s "(YV) as required. (iv) This follows from the fact that W is an RH-submodule of V if and only if W is an R(gHg-l)-submodule of gV. (v) Observe that (gV)sandg(Vs) coincide as S-modules. Let z E gHg-', s E S and v E V . Then the action of z on s @ v E ~ ( V Sis)given by z

* (s 8 v) = (9-lzg)(s 8 v ) = s 8 9-'zgv

Taking into account that the corresponding action in (gV)sis given by z(s 8 v) = s 8 2

*

2,

= s 8 g-lzgv,

Induced Modules

704

property (v) follows. (vi) By definition, " V ' ) = (gV)' = V* as R-modules. Moreover, for any given 5 E G, the action of 5 on f E "V') is the same as the action of 2 on f E (gV)', as desired. H We refer to the R(gHg-l)-module gV as a conjugate of V . Let H be a subgroup of G and let V be an RH-module. Then, by Lemma 6.2, {g E Nc(H)IgV g V } is a subgroup of G containing H ; we shall in future refer to it as the inertia group of V . In what follows, we put t

VtHt-ln.9

= W)tHt-'rlS

Theorem 6.3. (Mackey Decomposition, Mackey (1951)). Let H and S be subgroups of G, let T be a full set of double coset representatives for (S,H ) in G and let V be an RH-module. Then

( V G ) s2 @ t € T ( t V t H t - l " S )

S

Proof. Let {gl, . . . ,gn} be a left transversal for H in G. Then, by Proposition 1.3 (i), V G= @L1g; @ V

(direct sum of R-modules)

Put X = {gi 63 V ( 1 5 i 5 n}. Then G and, in particular S, acts on X . Furthermore, gi @ V and gj @ V lie in the same S-orbit if and only if gi and gj belong to the same double (S,H)-coset. For each t E T , let Wt be the sum of the gj 63 V for which gj E S t H . Then each Wt is an RS-module and

( V G ) s= @t€TWt Setting V, to be the restriction of tV to R(tHt-' n S), we are therefore left to verify that Wt g (I$)'. Let I ( 1 , . . . ,n} such that S t H = U,€rg;H Then S acts transitively on the set {gi @ Vli E I } and under this action the stabilizer of t @ V is {S

E Slst 8 V = t 8 V } = { s E Slt-'st E H} = tHt-' n S

6 Mackey's theorems

705

Hence, by Lemmas 6.2 (i) and 6.1 (ii), Wt &' (G)' as desired.

Theorem 6.4. (Mackey Tensor Product Theorem, Mackey (1951)). Let H and S be subgroups of G, let V be an RH-module and let W be an RS-module. Then

where the tensor products are over R and T is a full set of double coset representatives for ( S ,H) in G. Proof. Applying Theorems 5.1 (i) and 6.3, we have

V G@ W G % ( ( V G ) s @W ) G BtET[{tKHt-l"S)S

@ WIG

On the other hand, by Theorem 5.1 (i),

The desired assertion now follows by the transitivity of induction. W In what follows, given an RG-module V , we denote by InwG(V) the R-module of all G-invariant elements of V .

Lemma 6.5. Let H be a subgroup of G, let T be a left transversal for H in G and let V be an RH-module. Then the map

{

Invjy(V)

--t

Inwc(VG) XteTtBv

is an R-isomorphism. Proof. It is obvious that if v is H-invariant, then C t E T t8 v is Ginvariant. Because the given map is obviously an injective R-homomorphism, it suffices to show that any G-invariant element of V Gis of the form CtET t@v for some v E InvH(V). So assume that z E V G is G-invariant and write z = CtET t @ vt with each vt E V . Put w = v, for a unique s E T n H. Then the equalities t s = s and hz = z for each t E T , h E H, imply, respectively that each vt = v and hv = w for each h E H , as desired. W

Induced Modules

706

Lemma 6.6. Let F be a field and let V and W be finitely generated FG-modules. Then

Proof. Applying Lemma 3.12 (i), we have

as asserted. 1 We are now ready to prove the following classical result. Theorem 6.7. (Mackey Intertwining Number Theorem, Mackey (1 951)). Let F be an arbitrary field, let H and S be subgroups of G and let V and W be finitely generated F H and FS-modules, respectively. Then

i(vG, wG> = C i ( t ~ H t - l n SWtHt-lnS) , tET

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. We have i(VG,WG) = d i m ~ l n v ~ ( ( V8~W) *G ) (by Lemma 6.6) = dimFlnvG((V*)G8 W G ) (by Corollary 3.3) = dimF~nvG(tV,*H,-lnS8 W t H t - l n S ) G (by Theorem 6.4) t€T

=

d i m ~ l n v ( ~ 8~WtHt-lnS) ~ ~ - 1 ~ (by ~ Lemma 6.5) t€T

=

C i(tVtHt-lns, WtHt-lns)

t€T

where the last equality follows from Lemmas 6.6 and 6.2 (vi). In what follows, F denotes a field and all FG-modules are assumed to be finitely generated. Given FG-modules V and W , we say that they are disjoint if V ,W have no composition factors in common. To provide some applications of Mackey’s theorems, we first record the following lemma.

6 Mackey's theorems

707

Lemma 6.8. Let V and W be FG-modules and let charF [ [GI. Then V and W are disjoint if and only if i(V,W ) = 0.

Proof. By Maschke's theorem, we may write V = @V,, W = where the {Vj} and { W k } are simple FG-modules. Hence

@Wk,

Because v, w k if and only if i(6,w k ) = 0, we see that i(v,w ) = 0 if and only if Vj 9 w k for all j , k. Since the latter is equivalent to the disjointness of V and W , the result follows. Corllary 6.9. Let H and S be subgroups of G , let V and W be F H and FS-modules, respectively, and let charF If IGl. Denote by T a full set of double coset representatives for ( S ,H ) in G. Then V G and W G are disjoint if and only if for all t E T , the F(tHt-' flS)-modules t & H t - l n S and W t H t - l n S are disjoint.

Proof. Apply Lemma 6.8 and Theorem 6.7. Corollary 6.10. Let H be a subgroup of G , let F be an algebraically closed field with charF /' IGl and let V be an FH-module. Denote by T a full set of double coset representatives for ( H ,H ) in G. Then V G is simple if and only if V is simple and for all t E T - H , the F(tHt-' n H)-modules t & H t - l n H and & H t - l n H are disjoint.

Proof. Applying Theorem 6.7 for W = V and S = H , we obtain

) 1. Since F is algebraically closed, V G is simple if and only if i ( V G , V G= Since for t E H , t q H t - l n H K H t - l n H (Lemma 6.2 (i)), it follows that V G is simple if and only if V is simple and for all t E T - H ,

The desired conclusion now follows from Lemma 6.8.

Induced Modules

708

If p is a representation of a subgroup H of G and p is afforded by an FH-module V , then the induced representation pG of G is defined to be the representation afforded by V G .

Corollary 6.11. Let H be a subgroup of G , let F be an algebmically closed field with charF f [GI and let p be a one-dimensional representation of H. Then pG is irreducible if and only if, for each g E G - H, there exists x E gHg-' n H such that p ( z ) # p ( g - l z g ) .

Proof. Let V be an FH-module which affords p. Then, for any g E G, the representation g p of gHg-' given by gp(z) = p ( g - ' s g ) is afforded by gV. Because, for any given g E G, the F(gHg-' n H)-modules gVgHg-lnH and VgHg-lnHare simple, they are disjoint if and only if they are nonisomorphic. In terms of representations, the latter is equivalent to the existence of z E gHg-' n H such that p ( x ) # "(z) = p(x-'gz). Now apply Corollary 6.10 and the result follows. We close by examining induction from normal subgroups.

Theorem 6.12. Let N be a normal subgroup of G , let R be a commutative ring and let V and W be simple RN-modules. Then (i) If gV V for all g E G - N , then V G is simple. (ii) V G 2 W G if and only if V E gW for some g E G.

Proof. (i) Let U of ( V G )and ~

# 0 be a submodule of V G . Then UN is a submodule

(VG)N = $tcTt (8 v where T is a transversal for N in G. Hence ( V G ) (and ~ therefore U N ) is semisimple. Let X be a simple submodule of UN. Then X is a simple submodule of ( V G ) and, ~ by hypothesis and Lemma 6.2 (i), the simple RN-modules t @ V , t E T , are nonisomorphic. Hence, by Lemma 1.4.5 (ii), X = to@V for some to E T . For any g E G, gX C U which implies t@V U for all t E T . Thus V G = U and V G is simple. (ii) If V E gW for some g E G, then by Lemma 6.2 (ii), V G 2 WG. Conversely, suppose that V G E W G .If T is a transversal for N in G, then ( V G )=~& E T ~8 V and ( W G )=~@tcTt (8 W Because ( V G )E~ ( W G )and ~ 1 8 V is an isomorphic copy of V contained

7 Counting induced modules

709

in ( V G ) ~ V , is isomorphic to a submodule, say X of ( W G ) ~Since, . by Lemmas 1.4.5 (i) and 6.2 (i), X is isomorphic to t 8 W S tW for some t E T , the result follows.

Corollary 6.13. Let N be a normal subgroup of G, let F be a splitting field for G and let V be a simple FN-module. Then V G is simple if and only if gV V for all g E G - N . Proof. Let H be the inertia group of V and let V G be simple. By Theorem 6.12 (i), it suffices to show that H = N . Assume by way of contradiction that H # N. Then ( V G )is~a semisimple FN-module such that V is a simple constituent ( V G )with ~ multiplicity n > 1. On the other hand, by Theorem 4.1 applied to H = N , W = V G = V G / J ( V G )we , see that n = I, a contradiction. W

7

Counting induced modules

Throughout, N denotes a normal subgroup of a finite group G and F an arbitrary field of characteristic p 2 0. Our aim is to find the number of nonisomorphic FG-modules induced from the simple FN-modules. As an application, we provide some information on induced modules over group algebras of F'robenius groups. Given an FN-module V and g E G, denote by x and g x the characters of N afforded by V and gV, respectively. Then, by the definition of gV, we have g x ( 4 = x ( g - 1 4 for all x E N We shall say that two F-characters a,@of N are G-conjugate if @ = ga for some g E G. It is clear that G-conjugacy of characters of N is an equivalence relation. L e m m a 7.1. Let V and W be simple FN-modules and let xv and x w be the characters of N afforded by V and W , respectively. Then V G2 W G if and only if x v and xw are G-conjugate.

Proof. By Theorem 6.12 (ii), V G % W G if and only if V 2 gW for some g E G. Now apply Corollary 17.1.8 (i). W L e m m a 7.2.

Let F be a field of characteristic p 2 0 , let N be a normal

710

Induced Modules

subgroup of G and let x be a p-regular element of N . Denote by T, and K X the F-conjugacy classes of G and N , respectively, containing x . Then

Proof. Let n and m be the least common multiples of the orders of p-regular elements of G and N , respectively. Let E be a primitive n-th root of unity over F and denote by I, the multiplicative group consisting of those integers p , taken modulo n, for which E H E P defines an automorphism of F(E) over F. Note that n = mk for some natural number k and 6 = E' is a primitive m-th root of unity over F. Assume that s E T,. Then s = g-'sPg for some g E G and some p E I,. If p = r ( m o d m ) , 0 5 r 5 m - 1, then s = g-lx'g. The automorphism E H E P of F(E) over F induces the automorphism 6 H 6 P = 6' of F(6) over F. Hence r E I , and s = ( g - l ~ g )E~K g - l X g . Since every automorphism of F(6) over F extends to that of F(E) over F, we see that if two elements of N are F-conjugate in N , then they are F-conjugate in G. This proves the opposite inclusion, as desired. H Theorem 7.3. Let N be a normal subgroup of G and let F be an arbitrary field of characteristic p 2 0. Then (i) The number of G-conjugacy classes of irreducible F-characters of N is equal to the number of F-conjugacy classes of p-regular elements of G contained in N . (ii) If charF y IN1 and F is a splitting field for N , then for any g E G the number of irreducible F-characters x of N with g x = x is equal to the number of conjugacy classes I< in N such that 9Kg-l = I m. This completes the proof of the theorem.

As an application, we now prove the following classical result. Theorem 2.5. (Osima (1952), Nakayarna (1938)). Let H be a subgroup of G, let F & a splittingfield for H with charF [ lGl, and let X I , . . . ,xr be all irreducible F-characters of H . Then the number of linearly independent F-characters of G among the induced characters x y , . . . ,x: is equal to the number of conjugacy classes of G which contain an element of H .

.

Proof. Let 91,. . ,g , be the representatives for the conjugacy classes of G containing an element of H and let m be the dimension of the vector space over F spanned by xp, . . ,xp. Replacing each 9; by its conjugate, if necessary, we may assume that g; E H , 1 5 i 5 s. By Proposition 1.2 (i), each X: takes value 0 on each conjugacy class of G which contains no element of H . This shows that m = rank(A), where A is the r x s matrix defined by (1 5 i 5 r, 1 5 j 5 s) A = (XG(9j))

.

Setting B = ( X i ( g T ' ) ) , 1 5 i 5 T , 1 5 j 5 s, it follows from Theorem 2.4 that t A B = diag(nl,n2, ...,n,)

2 Orthogonality relations

747

where ni = ICc(gi)l # 0 in F (since charF [ IGl), 1 5 i 5 s, and t A is the transpose of A . Hence C = t A B is nonsingular and so r a n k ( C ) = s. It follows that s 5 rank(A) = m 5 s Thus m = s and the result is established. W

By the regular F-character of G, we understand the F-character of G afforded by the regular FG-module FGFG.

Lemma 2.6. Let F be a splitting field for G with charF 4 IGI and let p be the regular F-character of G. Then (i) P ( d = 0 ifs # 1 a n d p ( 1 ) = IGl (ii) p = C:=1 xi(l)xi, where xi,. . .,xT are all irreducible F-characters of G. (iii) If x = C x g g E F G , xg E F , g E G , then regarding each xi as a character of F G , 7

zg

= 1q-l CXi(l)Xi(Wl) i= 1

Proof. (i) By taking the elements of G as an F-basis for F G , we see that p ( g ) is just the number of elements x E G with gx = x . (ii) Let V1,. ..,Vr be all nonisomorphic simple FG-modules, let n; = d i r n ~ Kand let xi be the character of G afforded by V;, 1 5 i 5 r . By Maschke's theorem and Proposition 11.4.8 (iv), we have FG S-! $:=ln;K. Hence p = Ci=lnixi = Ci=lxi( 1)xi, as required. (iii) We have zg-l = xg t Ctfgzttg-l. By taking the values of the regular character of both sides, it follows from (i) and (ii) that T

C x i ( l ) X i ( x g - ' ) = XglGI, i=l

as desired. H

Lemma 2.7. Let F be a splittingfield for G with charF not dividing lGl, let e l , . . . ,e, be all block idempotents of F G and let xi be the irreducible F-character of G afforded by a simple FG-module belonging to the block FGei, 1 5 i 5 r . Then ei = 1GI-l

C x;(l)xi(g-l)g SEG

An Invitation to Characters

748

Proof. Write ei = CgEGegigwith egi E F , 1 5 i 5 r . Then, by Lemma 2.6 (iii), we have P

egi

= [GI-'

xk(l)Xk(eig-I) k=l

Since x k ( e i g - l ) = 6 i k x k ( g - l ) , the result follows. I

It is now an easy matter to provide the following generalization of the orthogonality relations exhibited in Theorem 2.3 (i), (ii). Theorem 2.8. (Generalized Orthogonality Relation). Let F be a splitting field for G with charF 4 IGl, let X I , . .. ,x, be all irreducible F-characters ofG and let x E G. Then

Proof. B y Theorem 17.10.4, xi(1) # 0 in F , so the right-hand side is legitimate. Let e l , . . . ,e, be as in Lemma 2.7. Then eiej = 6;jei and we compare the coefficients of the group elements on both sides. The coefficient of x E G on the right-hand side is lGl-1t5ijxi(l)xi(x-1) and that on the left-hand side is

IGl-2xi ( 1> X j( 1)

c

Xi((W

)-I ) X j ( q - l )

s€G

Equating these expressions and substituting x for z-l, the result follows. We next provide an extension of Lemma 2.7 to nonsplitting fields.

Theorem 2.9. Let F be an arbitraryfield with charF \GI, let n be the exponent of G and let E be a primitive n-th root of 1 in the algebraic closure of F . Let E / F be a field extension such that E is a splitting field for G , let XI,.. .,X, be all irwducible E-characters of G and let ei

=

c

xi(l)xi(g-l)g

(15 i 5 r)

g€G

For each u E G a l ( F ( & ) / F and ) x = C x g g E F(&)G,xg E F ( E ) ,g E G, put = C a ( x , ) g . Then the group G a l ( F ( E ) / F )acts on the set { e l , . . . ,e,} as a permutation group and the orbit sums are all block idempotents of F G .

Ox

2 Orthogonality relations

749

Proof. By Lemma 2.7, el ,. ,.,e, are all block idempotents of EG. Adjoining E to E , if necessary, we may assume that F ( E ) E . Then each e; E F ( E )and e l , . . . ,e, are all block idempotents of F ( E ) G .It is clear that for each u E G a l ( F ( & ) / F5, cf is a ring automorphism of F ( E ) Gand that z E FG if and only if u x = z for all u E G a l ( F ( € ) / F ) .This shows that the group G a l ( F ( & ) / Facts ) as a permutation group on { e l , . . . , e r } and that the orbit sums, say u l , . . . ,ut, are central idempotents of F G with 1 = t ui.If U k = u; + u i for some orthogonal central idempotents u i , uz of F G and 'Ilk is the orbit sum of e,, then uke, # 0 or uies # 0. Hence, since u i , u$ E FG, we must uk = u i or uk = uz. Thus u1,. ..,ut are all block idempotents of FG, as required. The following easy application of Theorem 2.9 will be used in the proof of the Witt-Berman's induction theorem.

Corollary 2.10. Let G = AB be a semidirect product of a normal abelian subgroup A and a group B , let F be an arbitraryfield with charF 4 IAl and let E be a primitive n-th moot of 1 over F where n is the exponent of A . Assume that for any b E B , a E A, bab-' = a+' for some p = p(a,b) such that E H E P determines an automorphisrn of F ( E )over F . Then each simple FA-module can be extended to an FG-module.

Proof. The group G acts on F A via (ab)o z = a(b5b-l) for all a E A, b E B , x E F A . In this way F A becomes an FG-module whose restriction to F A is the regular module F A . Since A is abelian and charF does not divide IAI, any simple FA-module is identifiable with a block of A. Hence it suffices to show that beb-' = e for any block idempotent e of F A and any b E B . By Corollary 17.2.7, F ( E )is a splitting field for A and all irreducible F(E)-representations of A are of degree 1. Hence, by Theorem 2.9, e is the xi(a-')a under the action of GuZ(F(E)/F), orbit sum of ei = IAI-' CaEA where x; is some irreducible F(&)-characterof A. Finally, given b E B , we have beib-l = ]A]-' x;(a-')aP = a ( e i ) aEA

where u E G a l ( F ( & ) / F is ) the inverse of result follows.

E

H

E+'.

Thus beb-l = e and the

An Invitation to Characters

750

C. Intertwining numbers and applications Let V and W be FG-modules. Recall that the intertwining number i(V,W ) is defined by

If cp and 11, are F-characters of G, then we put

We next provide the following important application of orthogonality relations. Theorem 2.11. Let F be a n arbitrary field of characteristic 0 a n d let cp and 2/, be F-characters of G afforded by FG-modules V a n d W , respectively. Then < cp,$ >= i(V,W)

Proof. First assume that V and W are simple FG-modules. If V $4 W , then i(V,W ) = 0 and, by Corollary 17.1.8, cp # 11, in which case < cp, >= 0 by Theorem 2.3 (i). Assume that V E W and put D = Endl;.~(V).Then D is a division algebra and, by Theorem 14.4.1,

+

where m(D)is the index of D,each pi is an irreducible E-character of G ( E is the algebraic closure of F ) and r = (F(cp) : F ) . On the other hand, by Theorem 14.3.2,

-w)2 F(cp)

so that r = ( Z ( D ): F ) . Now, by the definition of m(D),

m(D)2 = (D: Z ( D ) ) and, by Theorem 2.3 (ii), < cpi,cpj

>= 6ij. It follows that

< cp,$ > = < p,(p >= r n ( q 2 T = (D: Z ( D ) ) ( Z ( D ): F) = (D : F ) = i(V,W)

2 Orthogonality relations

751

as required. Turning to the general case, let V1,.. . , V, be all nonisomorphic simple FG-modules. Then V @;="=,;K and W S $y==lt;V;.for some integers k;,t; 2 0. If pi is the F-character of G afforded by K, then we must have i(x, =< cp;, cp; > by the special case proved above. Hence

v)

n

i(V,W ) =

C kjtji(V;,K) i=l n

i=1 n

n

as required. W

Corollary 2.12. (Frobenius reciprocity). Let H be a subgroup of G, let F be an arbitrary field of characteristic 0 and let cp and $ be F-characters of H and G, respectively. Then

Proof. Let V and W be F H - and FG-modules affording cp and $, respectively. By Proposition 18.1.6,

Ho~FG(W V ~) 2, HomFH(V,W H ) and so i ( V G , W )= i ( v , W ~ )Hence, . by Theorem 2.11,

< PG,$ >= i ( V G , W )= i(v,wH)=< V , $ H > as required.

Corollary 2.13. Let H be a subgroup of G , let F be an arbitrary field of characteristic 0 and let cp and $ be irreducible F-characters of H and G, respectively. Denote by m the multiplicity of $ as an irreducible constituent of 'pG and by n the multiplicity of 'p as an irreducible constituent of $H. Then m < ?I,$,>= n < cp,cp>

An Invitation to Characters

752

Proof. Let V and W be F H - and FG-modules affording cp and $, respectively. By Theorem 18.4.1, we have

rn.i(W,W) = n . i ( V , V ) Hence, by Theorem 2.11,

m < $,$>= m . i ( W , W ) = n , i ( V , V )= n < cp,y > as required.

Proposition 2.14. Let H be a subgroup of G , let F be an arbitrary field of characteristic 0 and let x be the F-character of H afloded by an FH-module V . Then (i) < x,x > xG,xG I< > with equality if and only if

EndFH(V) S' EndFG(VG)

(as F-algebras)

(ii) If x is irreducible and < x,x >=< xG,xG >, then xG is irreducible. Proof. (i) We first note that the map

{

EndFH(V)

f

-+

EndFG(vG)

l@f

is an injective homomorphism of F-algebras. Hence i(V,V) 5 i(VG,VG) with equality if and only if EndFH(V) 2 EndFc(VG). Since, by Theorem 2.11, < x,x >= i(V,V) and < xG,xG >= i(VG,VG),property (i) follows. (ii) By (i), < x,x >=< xG,xG > implies EndFH(V) 2 E n d F c ( v G ) . Since, by hypothesis, EndFH(V) is a division algebra, we see that EndFG(VG) is also a division algebra. Because FG is semisimple, we conclude that V G is simple. Thus xG is irreducible. I

Proposition 2.15. Let H be a subgroup of G, let F be an arbitrary field of characteristic 0 and let x be the F-character of G a#orded by an FG-module V . Then (i) < x,x >I< X H ,X H > with equality if and only if EndFG(V)

EndFH(VH)

(as F-algebras)

(ii) If x is irreducible and < x,x >=< X H , X H >, then X H is irreducible.

2 Orthogonality relations

753

Proof. (i) It is clear that E n d F c ( V ) is a subalgebra of EndFH(V). Hence i(V,V)5 ~ ( V H , V H with ) equality if and only if

Since, by Theorem 2.11, < x , x >= i(V,V) and < X H , X H >= ~(VH,VH), property (i) is established. (ii) By (i), < x,x >=< X H , X H > implies EndFG(V) 2 EndFH(VH). Hence, since V is simple, EndFH(VH) is a division algebra, Finally, since FH is semisimple, VH must be simple. Thus XH is irreducible. Proposition 2.16. Let F be an arbitrary field of characteristic 0, let H and S be subgmups of G and let T be a full set of double coset represeatatives for ( S ,H ) in G. Then, for any F-characters and 11, of H and S respectively,

< v G , P >=

c
0 and D = E n d F c ( V ) . Since x ( 1 ) = n ( D : F ) = n < x , x > by Theorem 2.11, property (ii) is established. Let X I , . . . ,xr be all irreducible F-characters of G , let ei = e ( x ; ) and let p be the regular representation of G. Then p = CF=lnix; where n; = X i ( l ) / < x ; , ~ >. ; Let x = C x g g E FG, xg E F , 9 E G. Then, by the argument in the proof of Lemma 2.6 (iii),

where each

x; is regarded as a character of FG. Write e; = C g E G e,;g

with

An Invitation to Characters

754

egi E F , 1 5 i 5 r. Then, by (2),

Since Xk(e;g-') = & k x k ( g - ' ) , the result follows.

Corollary 2.18. Let F be an arbitrary field of characteristic 0, let H be a subgroup of G and let cp and $ be irreducible F-characters of H and G , respectiuely . (i) If < $ H , I + ~ H >=< $,$ >, then $H is an irreducible F-character of

H such that FHe($H) 2 FGe($). (ii) If < cpG, cpG >=< cp, cp >, then cpG is an irreducible F-character of G such that FGe(qG) M ( G : H ) ( F H e ( v ) ) Proof. (i) By Proposition 2.15, $H is irreducible and EndFc(V) % EndFH(VH) where V is a simple FG-module affording $. Setting D = EndFG(V) and n = $(1)/ < $,$ >, it follows from Proposition 2.17 (ii) that

FGe($) 2 kfn(D) !2 FHe($H), as required.

(ii) By Proposition 2.14, qG is irreducible and

EndFH(V) % EndFG(VG) where V is a simple FH-module affording cp. Put n = (pG(l)/ < cpG,cpG > and k = cp(l)/ < q,(p >. Then n = (G : H ) L . Setting D = E n d ~ ~ ( vit ~ ) , follows from Porposition 2.17 (ii) that

FGe(cpC) 2 M d D ) 2 q C : H ) ( M k ( q ) EZ M ( G : H ) ( F H ~ ( ( P ) ) as required.

H

Corollary 2.19. Let F be an arbitrary field of characteristic 0 and let K be subgroups of G. Then (i) I f x is an irreducible F-character of G such that

3 Class functions and character rings

755

< X K ,X K >=< x,x > and X K is an irreducible F-character of h'. (ii) If cp is an irreducible F-character of H and < (pG,(pG >=< cp,(p >, then < Q " , Q ~ >=< (P,Q > and Q" is an irreducible F-character of K . then

Proof. (i) Let V be an FG-module affording x. Since

it follows from Proposition 2.15 (i) that E n d ~ ~ (%vEndFH(VH). ) Hence

and so, by Proposition 2.15 (i), < X K , X K >=< x , x >. Therefore, by Proposition 2.15 (ii), x~ is irreducible. (ii) Let V be an FH-module affording 9. Since < cpG, cpG >=< 9, cp >, it follows from Proposition 2.14 (i) that EndFH(V) E EndFc(VG). Since

dimFEndFH( v)5 damFEndFK(VK)5 dimFEndFG(VG) we have

EndFH(V)

EndFK(vK)

Hence, by Proposition 2.14 (j), < pK,pK>=< Q,(P Proposition 2.14 (ii), (p" is irreducible, as desired. W

3

>.

Therefore, by

Class functions and character rings

A. Generalities Let G be a finite group and let R be a commutative ring. We denote by f ( G , R ) the set of all functions from G to R. Define the sum a p, the multiplication crp and the scalar multiple r a for a,p E f(G, R) and r E R by

+

all g E G. Then f (G, R) becomes a commutative R-algebra which is R-free of rank [GI. In fact, as an R-module, f ( G , R ) is identifiable with

for

HowLR(RG,R).

An Invitation to Characters

756

We now introduce a distinguished subalgebra of f(G, R ) , namely the Ralgebra of class functions. A map a : G + R is called a class function if a(%)= a(y-'scy) for all s,y E G The set of all R-valued class functions on G will be denoted by C f ( G , R ) . It is clear that C f ( G , R ) is a subalgebra of f(G,R). Since C f(G,R ) is identifiable with the set of all R-valued functions on the conjugacy classes of G , we see that C f ( G , R ) is a free R-module of rank n , where n is the number of conjugacy classes of G. It is clear that, as an R-module, C f ( G ,R) is identifiable with H o ~ R ( Z ( R GR). ), Denote by m the characteristic of R. Then Z / m Z can be identified with the prime subring of R. By a generalized &-character of G, we understand an element of C f ( G ,R ) of the form

where xl,. ..,xr are &-characters of G. Since for any two R-characters a and p of G, ap is an R-character of G (see Lemma 17.7.1 (iv)), it follows that the set Ch(G,R) of all generalized R-characters of G is a subring of C f ( G ,R). We refer to Ch(G, R ) as the ring of generalized R-characters of G. Assume that H is a subgroup of G and let a E C f ( G , R ) . Then the restriction CWHof cr to H is an R-valued class function on H . We refer to CWH as the restriction of cr to H . As in the case of induced characters, there exists a dual process of induction . Namely, given /3 E C f ( H , R ) ,we define PG : G --$ R by

where T is a left transversal for H in G and, by convention, P ( s ) = 0 for all x E G - H . Since P is a class function, it is clear that PG E C f ( G ,R ) and PG is independent of the choice of T . Of course, if 1HI is a unit of R, then obviously PG(9) = IH1-l P(+4 (9 E G )

c

xEG

It is important to observe that if p is an R-character of G, then PG is the induced character of G (see Proposition 1.2 (i)). For this reason, we refer to PG (for any P E C f ( H , R ) )as the induced class function

.

3 Class functions and character rings

757

Lemma 3.1. For any subgroup H of G, the maps

Resg:Cf(G,R)-+Cf(H,R),cy-a~

-

In& : C f ( H , R )

C f ( G , R ) ,PI- PG

are R-linear and the maps R e s g : Ch(G,R ) + Ch(H,R), a H

In&

: C h ( H , R ) -, Ch(G,R),/3 H PG

are Z / m Z -linear, where m = charR. Proof. This is a direct consequence of the definitions.

Lemma 3.2. Let a E C f (G,R ) , P E C f ( H , R). (i) ( a H P ) G = apG (ii) For any ideal I of C f ( H , R ) ,IndE(1) is an ideal o f C f ( G , R ) . (iii) For any ideal J ofCh(H,R) , Ind$(J) is an ideal of Ch(G,R). Proof. (i) Let T be a left transversal for H in G. Then, for all g E G,

as required. (ii) Given a,P f I , we have by Lemma 3.1 that ac - PC = (a- P)' E In(ac,(I). Let a E C f ( G , R ) and p E I . Then, by (i), ( a ~ p=) aPG ~ f IndZ(1) as required. (iii) The proof is identical to that of (ii). From now on, we concentrate on the case where R = F is a field.

Lemma 3.3. Let F be a field of characteristic p 2 0 and let XI,.. . ,xT be all irreducible F-characters of G. Then Ch(G,F ) is a free Z / p Z -module with basis x i , . . . ,x T . Proof. Let x be an F-character of G. Then, by Corollary 17.1.7, x = & &xi for some A* E Z / p a . Hence each element of Ch(G,F ) can be

An Invitation to Characters

758

written in the form Xixi for some A; E 2% / p Z . Since, by Lemma 17.1.5, X I , . . . ,xTare F-linearly independent, the result follows.

Lemma 3.4. Let F be a field of characteristic p 2 0 , let G o be the set of all p-regular elements of G and let V be the vector space over F of all functions G o -+ F which are constant on the F-conjugacy classes of p-regular elements of G . Then the restrictions X1IG O, .. ,xTJG O of all irreducible F-characters XI,.. .,xT of G form an F-basis of V .

.

Proof. By Corollary 17.1.8 (i) and Theorem 17.5.3, T is precisely the number of F-conjugacy classes of p-regular elements of G. Hence dimFV = r. Furthermore, by Lemma 17.5.5, each x;IG0 belongs to V. Since XI,.. . xr are F-linearly independent (Lemma 17.1.5), it follows from Corollary 17.1.11 (iii) that X1(Go,. ,xrlG0 are also F-linearly independent. So the lemma is true.

..

B, Splitting fields We begin by observing that if F is a splitting field for G such that charF does not divide [GI, then Lemma 3.4 shows that X I , . ,.,xT form an F-basis for Cf (G, F ) . However, in this case, we can prove a much stronger assertion with the aid of the map

{

C f ( G , F )x Cf(G,F) ( m i )

defined by

< f1,h >= 1q-l

-

+

F < fllf2 >

c f*(.)fz(.-l>

XEG

Theorem 3.5. Let F be a splittingfield for G with charF [ /GI. Then the above map defines a symmetric nonsingular bilinear form on C f (G,F ) . Furthermom, the following properties hold : (i) If X I , . . . ,xr are all irreducible F-characters of G, then X I , . . . ,xT form an orthonormal basis for C f (G,F ) . (ii) If C1,. . . ,C, are all conjugacy classes of G , g; E C; and hi = ICil, then the elements

759

3 Class functions and character rings

are all block idempotents of Cf(G, F ) and dimPC f(G, F)ei = 1, 1 5 i 5 r . In particular, e l , . . ,e, is an F-basis for Cf ( G ,F ) and Cf ( G ,F ) is a direct product of r copies of F .

.

Proof. It is clear that the given map defines a symmetric bilinear form. Furthermore, by Theorem 2.3 (i), (ii), we have < xi,xj >= S i j , 1 5 i , j 5 r. Since T coincides with the number of conjugacy classes of G , we are left to verify (ii). By Theorem 2.3 (iii), 0 1

ei(g) =

g $ C; g E C;

if if

Hence e l , . . . ,e, are nonzero mutually orthogonal idempotents of C f ( G ,F ) with 1 = el - e,. It follows that e l , . .. ,e, is an F-basis for Cf (G, F ) , dirnFC f ( G ,F)ei = 1 and

+ +

C f ( G ,F ) = $;==,Cf(G,F)e; F x

-.-x F

( r factors)

as desired.

Theorem 3.6. (Berman (1958)). Let F be a field of characteristic 0, let G be a finite group and let C h ( G , F ) be the ring of generalized F characters of G. Then Ch(G,F ) has no nontrivial idempotents.

Proof. For any field extension E / F , C h ( G , F ) Ch(G,E) and so we may assume that F is a splitting field for G. Let ei be defined as in Theorem 3.5 (ii), 1 5 i 5 r. Then, upon a suitable rearrangement of indices, any given idempotent e 0 in C f ( G , F ) can be written in the form

+

e = el t

-..t e k

(1 5

L 5 r)

-

Hence the coefficient of the identity character of G in e is (hl t -+hk)lGI-'. I f e E Ch(G,F ) , then (h1-k - - - hk)lGI-' E Z . Hence k = r and e = 1, as desired. +

+

Lemma 3.7. Let F be a field of characteristic 0 which is a splitting field for G. Let X I , . ..,X, be all irreducible F-characters of G, let x be an arbitrary F-character of G and let cy be a generalized F-character of G . Then

An Invitation to Characters

760

(i) x = nlxl t - * .t n T x , for some unique integers n, 2 0. (ii) n; =< x , x i > and x is irreducible if and only if < x , x >= 1. (iii) a is an irreducible character ifand only if< a,a >= 1 and a(1)> 0. Proof. (i) It is clear that such integers n; 2 0 exist and their uniqueness follows from Lemma 3.3. (ii) Apply (i) and the fact that, by Theorem 3.5, < x j , x j >= S;j. (iii) If a is an irreducible character, then a ( 1 ) > 0 and, by Theorem 3.5, < a,a >= 1. Conversely, assume that < a,a >= 1 and a(1) > 0. By Lemma 3.3, we may write a = mlxl ... t m,Xr for some mi E ZC; Hence < a,a >= m: = 1 which shows that a = f x i for some i. Since a(1)> 0, we deduce that a = x i , as asserted.

+

.

The following result is a special case of Theorem 2.11 whose proof relies heavily on the theory of Schur index. However, the assumption that F is a splitting field makes a direct proof almost trivial.

Lemma 3.8. Let F be a field of characteristic 0 which is a splitting field for G and let (Y and p be F-characters of G which are aflorded by FG-modules V and W , respectively. Then i ( V , W ) =< a,p > let

Proof. Let Vi, .,.,V , be all nonisomorphic simple FG-modules and xi be the F-character of G afforded by V,, 1 5 i 5 r . Then V

$i==,niV,,W

&lmiV,

(a; 2 0, mi 2 0)

and i(V,W) = Cbl njrni. On the other hand, a = C:=l nixi and /3 = C;'=lmix; so that by Theorem 3.5, < a,p >= & n;m;.

Theorem 3.0. (Frobenius Reciprocity). Let H be a subgroup of G , let F be a field of characteristic 0 which is a splitting field for H and G and let Q E C f ( G ,F ) , /3 E Cf(H,f). Then

< aH,p >=< a,pG > Proof. By Lemma 3.1 and Theorem 3.5, we may harmlessly assume that a and /3 are irreducible F-characters of G and H , respectively. Let W and V be simple FG and FH-modules which afford a and p, respecP is the multiplicity of V as a simple tively. Then, by Lemma 3.8, < ~ H , >

761

3 Class functions and character rings

constituent of W H ,while < a,PG > is the multiplicity of W as a simple constituent of V G .The desired assertion is therefore a consequence of Corollary 18.4.3.

CY

Let F be a field, let H be a subgroup of G and let g E G. Given E C f ( H , F ) , define ga E C f ( g H g - ' , F ) by "a(.) = " ( g - l s g )

for all x E gHg-'

In what follows, we put

Theorem 3.10. Let F be a field of characteristic 0 which is a splitting field for all subgroups of G. Let H and S be subgroups of G, let T be a full set of double coset representatives for ( S , H ) in G and let a E C f ( H , F ) , p E C f ( S ,F ) . Then S (i) ( C Y G ) S= &T(taYtHt-'nS) G (ii) ' PG = &T(t%Ht-lnS 'PtHt-Ins) G G (iii) < ,p >= &T ct % H t - I n S , P t H t - l n S > Proof. We may harmlessly assume that CY and p are F-characters of H and S, respectively. Then (i) and (ii) follow from Propositions 1.10 and 1.11, respectively. Property (iii) follows from Theorem 18.6.7 and Lemma 3.8. H C. Q= -characters

From now on, we concentrate on the case where F = we use the following abbreviations :

(I:.

In this case,

Cf(G) = Cf(G,cC), Ch(G) = Ch(G,cC) Let V be a vector space over (I: . By a hermitian form on V , we understand any map x + (I:, ( x , y ) -< z , y >

v v

such that the following properties hold : (a) < s,y >= < 9,s > (b) < Aizi t A 2 . 2 , ~ >= A1 < 5 1 , >~ +A2 CC)

< S,AlYl + x2y2 >= x 1 < z,y1 > + x 2

~ < z,y2 >

(Ai,A2

E

a)

An Invitation to Characters

762

We say that the hermitian form is positive definite if < I , X >> 0 for all x # 0. For any such form, we define the norm 11111 of I to be

Lemma 3.11. (Cauchy-Schwarz Inequality). Assume that the hermitian form above is positive definite. Then, for all x , y E V ,

I < Z , Y > I s 1141

llYll

Proof. The required assertion follows by expanding

< ax t p y , and setting a =< y , y

t p y

>r0

(.,p

EC )

> and p = - < x,y >.

Consider the following map

defined by xEG

Proposition 3.12. The above map defines a hermitian positive definite form on C f (G)such that (i) The irreducible C -characters X I , . . . ,x,.form an orthonormal basis for C f W (ii) Cr=1k x i , C;'=1p i x i >= Ci=1& P i , & , p i E (iii) (Cauchy-Schwartz Inequalitp]).For any a , p E Cf ( G ) ,

I < a,P > I 5 1141 IlPll *

(iv) ( h b e n i u s Reciprocity). If H is a subgroup of G, CY E Cf(G) and ,B E C f ( H ) ,then

< pG,a >=< P , a H > Proof. (i) That the above map is a hermitian positive definite form on Cf(G)&obvious. If x is a C-character of G, then for all g E G, x ( g - ' ) = x ( g ) by Corollary 17.1.12. Hence, by Theorem 3.5, the irreducible C -characters of G form an orthonormal basis for Cf(G). (ii) This is a direct consequence of (i).

3 Class functions and character rings

763

(iii) Apply Lemma 3.11. (iv) Write Q = Xix; and p = p j + j , where A;, p j E CC and $1,. ..,$t are all irreducible (I: -characters of H . The restriction to Ch(G)of the given hermitian form on Cf(G) coincides with the restriction to Ch(G) of the bilinear form on C f ( G )given prior to Theorem 3.5. Consequently, by Theorem 3.9,

&

Hence, t

T

as desired. H We close by examining the character ring C h ( G ) . It will be convenient to record the following elementary observations.

Lemma 3.13. Let G be an abelian group of exponent n, let H be a c y c k group of order n and let G =< g1 > x -.- x < gk >. For each i E (1,. . . ,k } , choose x; E Hom(G,H ) such that x ; ( g j ) = 1 for j # i and x;I < gi > is injective. Then

Proof. First of all, it is clear that < X I , . . . , x k >=< X I > X X < >. Since the order of x; is equal to the order of < g; >, we see that G E< x1 > x - . . x < x k >. Finally, since ]GI = IHorn(G,H)I, the result *

xk

follows.

Corollary 3.14. For any finite group G , Horn(G, (I: *) E G/G'.

An Invitation to Characters

764

Proof. It is clear that Hom(G, (I: *) 2 Horn(G/G', (I: *). If n is the exponent of GIG' and E~ is a primitive n-th root of 1, then Hom(G/G', (E *) = Hom(G/G', < gn >). The desired conclusion now follows by lemma 3.13. We are now ready to prove the following result.

Theorem 3.15. (Saksonov (1966)). Let U be the unit group of the character ring Ch(G) of G. Then (i) Every element offinite order in U is of the form fx for some x E Hom(G,(C *), (ii) The torsion subgroup of U is isomorphic to (GIG') x Z 2 . Proof. It is clear that (ii) is a consequence of (i) and Corollary 3.14. Let e l , . . . , e r be the idempotents of Cf(G) given by Theorem 3.5 (ii). If cy E U is of finite order m and X I , . ,xr are all irreducible (I2 -characters of G, then by Theorem 3.5 (ii),

..

for some ni E B ,pi E

a = 2 n i X i =ie$

(C

i=l

i= 1

Since am= 1, we have pT = 1 and so \ p i [ = 1 for each i E ( 1 , . . . ,r } . Hence, by Proposition 3.12 (ii) and Theorem 2.3 (iii),

< a,a > = =

r

r

i=l r

i=l

En: = C J ~< ~ ei,eiJ>' r

C < ei,e; >= 1GI-l C hi i=l

i= 1

= 1,

cT==,nf

proving that = 1. Thus cy = f x i for some i E { l , . .. , T } . Since xz" = 1 and the (I: G-module which affords x?" is of (I:-dimension ~ i ( l ) ~ " , we see that xi(1) = 1. Thus xi E Hom(G, (I: *) and the result follows. Corollary 3.16. (Saksonov (1 966)). Let G and H be finite groups such that Ch(G) 2 C h ( H ) , Then GIG'S H / H ' .

Proof. It is a consequence of Theorem 3.15 (ii) that (GIG') x Z 2 2 ( H / H ' ) X ZC, 2

3 Class functions and character rings

765

and hence GIG‘ E Hf HI. 1 Let X I , X ~ .,. . , x r be all irreducible CC -characters of G with for all g E G. Then

xl(g)

= 1

Given

E

T

xixj

=

zijkxk k=l

where { Z + } are uniquely determined nonnegative integers. (1,. . . ,T } , let i’ E (1,. . . ,r } be determined by

For future use, we next exhibit some relations between the structure constants % i j k of the character ring C h ( G )of G.

Proposition 3.17. With the notation above, the following properties hold : (i) Z i j k = Z j i k fOT all i, j , k . (ii) z ; j l = 1 if i = j’ and z j j l = 0 otherwise. (iii) z j j k = z j ’ k j . Proof. (i) This follows from the fact that (ii) By Lemma 3.7 (ii), we have

xjxj

as required. (iii) By Lemma 3.7 (ii), we have

and taking complex conjugates of both sides yields

as desired. 1

=xjxi.

An Invitation to Characters

766

D. Prime and maximal ideals Our point of departure in this subsection is the following standard fact.

Lemma 3.18. Let S be a noetherian integral domain and let I an ideal of S . Then nz==, In = 0.

# S be

nr=lIn,

Proof. Setting V = it follows from Lemma 1.6.17 that V = IV. Next, in the notation of Lemma 1.7.1, put f = identity and x = 1 + s1 - .- t s,. Then x 3 l ( m o d I ) and xV = 0. Since x # 0 and S is an integral domain, we deduce that V = 0, as desired. 1

+

We now describe all prime and maximal ideals of the character ring

Ch(G,F ) where F is any field of characteristic 0. This will be achieved with the aid of Theorem 3.19 below. Let X be a finite set, R a noetherian integral domain of characteristic 0 such that each prime number is a nonunit of R and S a ring of functions :

such that (a) The constant function c : X -, R with c ( x ) = 1 for all x E X belongs to s. (b) S is additively generated by finitely many functions fi, . . . ,fn which are linearly independent over R.

Theorem 3.19. (Banaschewski (1963)). With the notation above, the following properties hold : (i) For any x E X and any maximal ideal M of R such that R I M has prime characteristic, Mz = {fE Sl&> E M } is a maximal ideal of S and each maximal ideal of S is of this type, (ai) Under the stronger assumption that the additive group of R is finitely generated (instead of R being noetherian), for any x E X ,

pz = { f E Slf(.) = 0) is a non-maximal prime ideal of S and every such ideal of S is of this type.

3 Class functions and character rings

767

Proof. (i) It is clear that M , is an ideal of S. Furthermore, since

g! M,, M , is a proper ideal of S. Now if f , g E S are such that fg E M,, then f(s)g(z) E M . Hence either f(s)E M (i.e. f E M,) or g(s) E M (i.e. g E M,).Thus M , is a prime ideal. Now R I M is of nonzero characteristic p and p S E M,. Since SIPS is finite, the integral domain S / M x is also finite. c

Thus S / M x is a field and therefore M , is a maximal ideal. Conversely, let I be any maximal ideal of S. Then the additive group of the field S / I is finitely generated by (b). Hence S / I has nonzero characteristic p since any subgroup of a finitely generated additive group is again finitely generated, whereas the additive group of Q is not. Now p is a nonunit of R and so there exists a maximal ideal M of R containing p . It will be shown below that I = M , for some s E X . Choose 21,. .. ,s, E X such that d = det( fi(sj))# 0, and take m to be a positive integer such that d E M"-' - M". Here, such 21,.. . ,s, exist by the linear independence over R of the f1,. . . ,f, and the choice of m is possible since d # 0 and n g l M * = 0 by Lemma 3.18. Next, consider any f f nXExM,, and put f" = Cc;fi with suitable c; E 22. Then

C c ; f i ( z j )=

fm(sj) = f(Xj)"

and by using the adjugate Ckd

=

(ajk)

E M"

of the matrix

fm(Zj)cYjk

(1 I j 5 n)

(fi(sj)), we

obtain

(1 I k 5 n)

E M"

+

Now, if some C k # M , then there exist p E R and y E M with 1 = P C k y, which leads to d = P C k d yd E M", contradicting the choice of m. Thus, for each k E {I,. . ,n},ck E M n = p z , i.e. Ck = bkp with b k E z . It now follows that f" = pC bJ; E p S . Thus f" E I and f E I since I is prime. This shows that

.

+

z

nzEXMxG

I and again, by the primeness of I, we must have M , I for some s E X . Since M, is maximal, we deduce that I = M,, as desired. (ii) It is clear that each P, is a prime ideal of S and P, is not maximal, since it is not of the type M , (our assumption on R ensures that R is not a field). Conversely, let Q be a non-maximal prime ideal of S. If the integral domain S/Q has prime characteristic, then S/Q is finite, hence a field, and thus Q is maximal, a contradiction. Thus S/Q must have characteristic zero. Note that P, Q for some s E X, since

nrEXPx= 0 C

Q

768

An Invitation to Characters

Now S/ P, is also an integral domain of characteristic zero, finitely generated as an additive group, and so any of its nonzero prime ideals is maximal. Thus the prime ideal Q/P,of SIP, is either maximal or zero. Because maximality here would imply the maximality of Q in S, we have Q = P'. H Theorem 3.20. (Atiyah (1961), Banaschewski (1963)). Let F be an arbitrary field of characteristic 0, let Ch(G,F ) be the ring of all generalized F-characters of G, and let R = ZZ [ E ] where E is a primitive n-th mot of unity (n = exponent of G ) in the algebraic closure ? of F . (i) For any maximal ideal M of R and for any p-regular element g E G , where p is a prime in M ,

is a maximal ideal of Ch(G,F ) and each maximal ideal of C h ( G ,F ) is of this type. (ii) For any g E G ,

is a non-maximal prime ideal ofCh(G,F ) am$ every such ideal of Ch(G,F ) is of this type.

Proof. (i) We wish to apply Theorem 3.19 with X = G and S = Ch(G,F ) . The functions f E Ch(G,F ) map G into R , the principal character c = 1~belongs to S, and the irreducible F-characters XI,.. . ,xn of G are linearly independent over R. The latter is true since XI,...,xn are even Flinearly independent by observing that each xi is a positive integer multiple of an orbit sum of irreducible $-characters under the action of G a l ( F ( & ) / F ) . Thus all conditions of Theorem 3.19 are fulfilled. By the foregoing and Theorem 3.19 (i), it suffices to verify that for any z E G, M, = M, for some pregular element g E G. To this end, write z = g y for some p-regular element g E G and some pelement y of G with g y = y g . Now the restriction of each character xi to the cyclic subgroup < z > of G is an integral linear combination of the irreducible p-characters a l , . . . ,a, of < x >, Since the orders of the roots of unity a j ( y ) are powers of p , one sees that a j ( y ) - 1 E M ;hence

4 Representations of abelian groups

769

and thus x ; ( x ) - x ; ( g ) E M , which extends to X(x) - X(g) E h4 for any X E c h ( G , F ) . It follows that X(z) E M if and only if X(g) E M for any X E ch(G,F ) . Thus M , = Mg,proving (i). (ii) This is a direct consequence of Theorem 3.19 (ii).

4

Representations of abelian groups

In what follows, G denotes a finite abelian group and F a field. I f charF = p > 0 and p divides [GI, then the Sylow p-subgroup P of G belongs to the kernel of every irreducible F-representation of G (see Lemma 17.1.9). For this reason, in studying irreducible F-representations of G, we may harmlessly assume that charF 4 1GI. The following lemma disposes of the trivial case where F is a splitting field for G.

L e m m a 4.1. Let F be a splitting field for G with charF 4 (GI, let n be the exponent of G and write G =< 91 > x . x < gk >. Then (i) F contains a primitive n-th mot of 1, say E , and all irreducible F representations of G are of degree 1 (hence they are precisely the elements of Hom(G, < E >). (ii) G 2 Hom(G,< E >) =< > x - . . x < x k >, where xi E Hom(G, < E >) is defined by x;(gj) = 1 for j # i and x ; ( g ; ) = E ; , where ~i is a primitive I < g; > I-th root of 1.

Proof. Apply Corollary 17.2.7 and Lemma 3.13. The following notation will be used in the rest of this section : G a finite abelian group of exponent n. F an arbitrary field with charF4 IG[. &d a primitive d-th root of 1 over F , dln. G = Hom(G,< t n >). ex = PI-' C g E G x(s-1)9,x E 6 F(x)the field obtained from F by adjoining all values of equivalently, F ( x ) = F ( Q ) where d = ( G : K e r x ) . F" the direct product of m copies of F .

x

E G or,

For any x E G and u E Gal(F(&,)/F),put " x = U O X . It is obvious that ux E G. We shall call two characters a,P E G, F-conjugate if

p=

0

a!

for some u E Gal(F(&,)/F)

An Invitation to Characters

770

It is clear that F-conjugacy of characters is an equivalence relation. For any 2 E F(&,)G and 0 E Gal(F(&,)/F), denote by gz the image of x under the automorphism of F(&,)G induced by u. It is clear that U

ex = ebx

x E G, cr E GuZ(F(E~)/F)

for all

(1)

We remind the reader that, by Lemma 2.7, {exlX E G} is the set of all block idempotents of F(&,)G. Furthermore, it is a consequence of the definition of ex that ge, = x ( g ) e x for all g E G (2) Finally, let x be an irreducible F(&,)-character of G. Then x can be viewed as a homomorphism from G to F ( x ) . For convenience, we shall identify x with its extension to an F-algebra homomorphism from FG to F ( x ) . We are now in a position to undertake the proof of the main result. Theorem'4.2. (Berman (1958), Perlis and Walker (1 950)). Let G be a finite abelian gmup of exponent n, let F be an arbitmryfield with charF 4 [GI and let E, be a primitive n-th root of 1 over F. Denote by ( X I , . ..,xs} a full set of representatives of equivalence classes of F-conjugate irreducible characters of G over F(E,) and, for each i E (1,. , . ,s}, let %i be the sum of all F-conjugates of xi. Then (i) The natural map FG + F(Xj) induced by x; : F G 3 F(Xj) is an isomorphism of F-algebras. Furthermore, 21,...,2s are all irreducible F-characters of G. (ii) FG Z n d l , F(&d)ad, where ad is the number of F-conjugacy classes of G consisting of elements of order d and (F(&d): F ) is the number of elements in each of these classes.

.

Proof. (i) For each i E {l,., , s } , denote by EXi the orbit sum of exi under the action of GuI(F(&)/F).Then, by (1) and Theorem 2.9, {ZxiJ1 5 i 5 s} is the set of all block idempotents of FG. Hence F G = $f==lFGdxi and each FGExi is a field. Furthermore, since by (2), geXi = xi(g)exi,FGZxi affords 2;. It therefore suffices to show that the map

is an isomorphism of F-algebras. We claim that 2Zxi = 0 implies x ; ( x ) = 0; i f sustained, it will follow that the above map is a well defined surjective homomorphism of F-algebras and hence a desired isomorphism. Because,

5 Inductive sources

771

by (2), g e x I = X ; ( g ) e x I g, E G, we have zexi = X i ( z ) e , , . Now e x I is a primitive idempotent of F(E,)G which occurs in the decomposition of Zx, and all the ex, are linearly independent over F(e,). Hence zZXi = 0 implies x;(z) = 0, as required. (ii) Let d(nand let md be the number of elements in a given F-conjugacy class of G consisting of elements of order d . Then md is equal to the number of automorphisms of F(&d)over F induced by those of F(E,) over F . Hence md = (F(&d): F ) . NOW put

Md = ( X E G l x ( G ) =< &d >}, f i d = { e x ( X E Md} XEMd

Every element of the group G a l ( F ( & d ) / F induces ) a ring automorphism of F(Ed)G and in this way Gal(F(&d)/F)acts on the set as a permutation group, Observe that the size of each orbit under this action is equal to md, since for any x E Md there exists g E G such that x ( g - ' ) = E d . Since, by (i), FGE, 2 F ( x ) = F(ed), we see that FGed S F ( ~ d ) ~where d bd is the number of orbits. Let td be the number of cyclic factor groups of order d . Then = lMdl = p ( d ) * td = p ( d ) x (number of cyclic subgroups of order d ) = number of elements of order d in G

mdbd =

lfidl

= ???,dad

and hence b d = ad. Since 1 is the sum of pairwise orthogonal ed's with d ranging over all divisors of n , the result follows.

Corollary 4.3. Let G be a finite abeliian group of exponent n and, for any dln, let ad be the number of cyclic subgroups of order d in G. Then

QG

= nQ

(Ed)"d

4 .

Proof. Note that two elements a , b E G are Q -conjugate if and only if < a >=< b >. Now apply Theorem 4.2 (ii) for F = Q .

5

Inductive sources

Throughout, F denotes an arbitrary field of characteristic 0 and G a finite group. All FG-modules are assumed to be finitely generated. Given an F -

An Invitation to Characters

772

character x of G, we write Irr(x) for the set of all irreducible constituents of x. Let S be a subgroup of G, let g E G and let cp be an F-character of S.For any g E G, we denote by gcp the F-character of gSg-' given by

Recall that the inertia group G(cp)of 9 in G is defined by

Hence S a G(cp). Note that if H is a subgroup of G containing S, then the inertia group H(cp) of p in H satisfies

If cp and II,are F-characters of G, then we put

< cp,$ >= 1GI-l

c

(P(g)+(g-')

s€G

If V and W are FG-modules, then i(V,W)denotes their intertwining number, i.e.

i(v,w )= dimFHOmFG(V, w )

Recall that, by Theorem 2.11, if V and W afford 'p and $, respectively, then

Let X be a set consisting of some F-characters of a subgroup of G. We say that induction to G is an isometry on X if

< p,$ >=< (pG,gG>

for all cp,+ E

x

Let Y be a set of F-characters of G such that cpG E Y for all cp E X . We say that induction to G is an isometric bijection of X onto Y if it is a bijection and an isometry on X .

Theorem 6.1.

(Dade (1985)). Let S

H be subgroups of G and let

cp be an irreducible F-character of S . Then the following conditions are equivalent : (i) Induction to G is an isometry on Ir.(cpH).

5 Inductive sources

773

(ii) Induction to G is an isometric bijection of Irr(cpH) onto Irr(cpG). (iii) < ( g ~ ) g s g - ~ n s , ~ g s g>= - ~ n0 s for all g E G - H . Proof. (i) (j (ii) : By definition, (ii) implies (i). Conversely, assume that induction to G is an isometry on Irr(cpH). If x E Irr(cpH), then obviously xG is a constituent of cpG. By hypothesis, < x,x >=< xG,xG>. Hence, by Proposition 2.14 (ii), xG is irreducible, i.e. xG E Irr(cpG). Assume that cp,+ E Irr(cpH)with cp # $. Then, by Theorem 2.3 (i), < cp,+ >= 0. Hence < cp",+G >= 0 and therefore, by ( l ) , cpG # $ G . Finally, write Irr(cpH)= { X I , , . ,xT}so that cpH = CiZlnixi for some positive integers ni, 1 5 i 5 r . Then x f , . . . ,xp are distinct irreducible F-characters of G such that

.

Thus Irr(cpG) = {xp,. . . ,xp} as required. (iii) : By Proposition 2.16, (i)

where T is a full set of double coset representatives for (S,S) in G. The summands in (3) for which S t S H add up to < cpH,cpH > by Proposition 2.16. Equality (iii) says that the summands in (3) are zero. Since, by ( l ) , all summands in (3) are nonnegative, we deduce that (iii) is equivalent to

< q H , Q H >=< cpG,QG > Since vH =

nixi,

we have

On the other hand, by (2), we also have

< cpG,vG>=

C n;nj < X;G ,xjG > i,j=l

Assume that i # j . Then, by (1) and Theorem 2.3 (i),

ninj

< xiG ,xjG >> 0 = ninj < xi,xj >

(4)

An Invitation to Characters

774

with equality if and only if < xF,xy >= 0. On the other hand, if i = j, then G G 2 xi,xi > n! < xi ,xi >>_ni

x?,x?

by Proposition 2.14 (i), with equality if and only if < >=< x;,x; >. Thus (4) holds if and only if induction to G is an isometry on Irr(cpH),as desired. H

Following Dade (1985), we define an inductive source of G (with respect to F ) to be any irreducible F-character 9 of a subgroup of G such that induction to G is an isometric bijection of

IrT(cpG(q))

onto

rrr((pG)

Corollary 5.2. Let cp be an irreducible F-character of a subgroup S of G . Then the following conditions are equivalent : (i) Q is an inductive source of G. (ii) Induction to G is an isometry on Irr(qG(p)). >= 0 for d l g E G - G(cp). (iii) < (g(P)gSg-lnS,~gsg-ins

Proof. This is a special case of Theorem 5.1 in which H = G(cp). Corollary 5.3. Let N be a normal subgroup of G. Then any irreducible F-character cp of G is an inductive source of G.

Proof. We apply Corollary 5.2 for S = N . Since S Q G, we have g-lSg n S = S. Also, by definition of G(cp), g q # cp for all g E G - G(cp). Hence, by Theorem 2.3 (i), = 0 for all g E G - G(p). Thus cp satisfies condition (iii) of Corollary 5.2 and the result follows. Note that a weaker form of Corollary 5.3, namely the fact that induction to G is a bijection of Irr(cpG("))onto Irr(cpc), is equivalent to Proposition 1.3 (i).

Corollary 5.4. Let cp be an irreducible F-character of a subgroup S of G and let H be a subgroup ofG with S H C G(cp). If induction to G is an isometry on I r r ( p H ) ,then cp is an inductive source of G and H = G(cp).

Proof. By Theorem 5.1, equality (iii) of Theorem 5.1 holds for all g E G- H . On the other hand, no g E G(q) can satisfy condition (iii) of Theorem

5 Inductive sources

775

5.1. Thus G(cp) H and so G(cp) = H . Then, by Corollary 5.2, cp is an inductive source of G, as desired.

CoroIIary 5.5. Let cp be an irreducible F-character of a subgroup S of

G and let < c p , c p >=< cpG,cpG >. Then cp is an inductive source of G and S = G(cp). Proof. Apply Corollary 5.4 for H = S.

m

Corollary 5.6. Let an irreducible F-character cp of a subgroup S of G be an inductive source of G. If H is a subgroup of G with S C H , then cp is an inductive source of H .

Proof. By Corollary 5.2, equality (iii) of Corollary 5.2 holds for all g E G - G(cp). Since H(cp) = G(cpn H it also holds for all 9E H

- H(cp) s G - G(V)

This proves the result, by applying Corollary 5.2.

Theorem 5.7. (Dude (1985)). Let S C H Ii be subgroups of G, let cp be an irreducible F-character of S such that cp is an inductive source of G, and let X E Irr(cpH((")), p E Irr(cpK(9)).Then (i) < A",p" >=< A"('P),p >

(ii) < AH,(PK)H >=< A , P H ( v ) > (iii) Induction to G is an isometric bijection of

Proof. Since A" = (AH)", property (ii) follows from (i) by applying Frobenius reciprocity (Corollary 2.12). Write Irr(XK(p))= {A,,.. . ,A,} so that ~ ~ (= 9 nlX1 ) -t ... nTAT

+

for some positive integers n;. Then

An Invitation to Characters

776

By Corollary 5.6, (o is an inductive source of K which means that induction to Ir' is an isometric bijection of Irr(cp"(q)) onto I r r ( # ) . Because p E Irr(cpK(9))= {XI,.

..,A,}

we have T

=

T

< X"(9),p>,

proving (i). To prove (iii), we first note that, since X E Irr(cpH(9)), we have 1rr(~'(9))

1rr(cp'(9)), ~ r r ( ~ ' > Irr(cpG)

On the other hand, since cp is an inductive source of G, induction to G is an isometric bijection of Irr(y'(9)) onto Irr(cpG). Hence it suffices to verify that p E I ~ T ( X ' ( ~ ) ) if and only if p' E Irr(XG) which amounts to

< XG(9),p >> 0 if and only if

< XG,pG >> 0

Since the latter is a consequence of (i) applied to Ir' = G, the result is established.

Theorem 5.8. (Dade (1985)). Let S H be subgroups of G , let cp be un irreducible F-churucter of S such that cp is an inductive source of G, and let X E Irr(cpH(q)).Then AH is an inductive source of G if and only if X is an inductive source of G(cp) and G(cp)(X) N G ( H ) . In that case we have : (i) X is an inductive source of G. G(XH)(v)= G(cp)(W = G(X) (iii) G(AH)= G(X)H ( i ~ ) G(X) n H = H(cp)

w

Proof. Put 1c, = AH. By Corollary 5.6, cp is an inductive source of H , so II,E Irr(cpH).Assume that tj is an inductive source of G. Then induction to G is an isometric bijection of Irr($'($)) onto Irr(qG).By Corollary 5.6, y is an inductive source of G($). Since GG($) = XG($), it follows from Theorem

5

Inductive sources

777

5.7 (iii) (with G = G(+)) that induction to G(+) is an isometric bijection of Irr(XG(4)(9)) onto I T ~ ( + ~ ( $ Thus ) ) . induction to G is an isometric bijection of Irr(XG(+)(q)) onto Irr(AG). In particular, induction to G is an isometry on ~ r r ( ~ G ( + ) ( q ) ) . Assume that g E G(+)(cp). Then g E N G ( H ) n N G ( H ( ( ~ ) >Hence, . by Proposition 1.9 (i), = XH = "AH) = (")H

+

and S(p('P))

= (sv)W'P) = #('P)

Therefore AH = ( g X ) H and g X E Irr(cpH(")).But cp is an inductive source of G (hence of H by Corollary 5.6) and so A = gX. This shows that g E G(X) and thus G(?,b)(cp) G(X). Hence, by Corollary 5.4 (with S = H(cp), H = G(+)(cp) and cp = A) we have G(A) = G($)(cp). This proves (i) and that the first and last groups in (ii) are equal. Since X is an inductive source of G, it follows from Corollary 5.6 that X is an inductive source of G(cp). The inclusions

imply G ( Y ) ( X )G NG(H). Assume that X is an inductive source of G(cp) and G(cp)(X) N G ( H ) . We now claim that ?,b is an inductive source of G, G(cp)(X) = G(X) and (iii) and (iv) hold; if sustained, this will complete the proof by applying the preceding paragraph. Since A is an inductive source of G(p), induction to G(p) is an isometric bijection of Irr( XG('")(')) onto Irr(XG(q)).By Theorem 5.7 (iii), induction to G is an isometric bijection of Irr(XG("))onto Irr(AG). Hence induction to G is an isometric bijection of Irr(XG('f')('))onto Irr(AG). Since

c

it follows from Corollary 5.4 (with y = A, 5' = H(cp), H = G(cp)(A)) that G(cp)(X) = G(X) and X is an inductive source of G. Since G(X) = G(cp)(X) C_ N G ( H ) , it follows from Proposition 1.9 (i) that

" A H ) = ( g X ) H = AH

c

for all g E G(X)

Hence G(X) G(AH)and so G(X)H G G(XH). Since A is an inductive source of G, it follows from Corollary 5.6 that X is

An Invitation to Characters

778

an inductive source of G(X)H. Hence induction is an isometric bijection of I r r ( X G ( X )onto ) both Irr(XK) and Irr(XG)where Ii = G(X)H. These last two sets are precisely and IT^($^) since XK = = and = +G. Therefore induction is an isometric bijection of XG = onto 1rr(?jG).Hence, by Corollary 5.4, $ is an inductive source of G and G ( X H )= G ( X ) H . Finally, by (ii), we have

and H(V)

c G(XH)(cp)n H = G(X) n H,

thus completing the proof. H

Corollary 5.9. Let S C H E G ( 9 ) be subgroups of G, where cp is an irreducible F-character of S and an inductive source of G , and let X E Irr(cpH). Then A is an inductive source of G if and only if A is an inductive source ofG(cp). In that case G(X) = G((p)(X).

Proof. Since H N G ( H ) and so

G('p),we have H = H(cp). Furthermore, G(X) G((p)(X)

G(X)

NG(H)

The desired conclusion is therefore a consequence of Theorem 5.8.

Corollary 5.10. Let S H be normal subgroups of G, let 'p be an irreducible F-character of S and let X E I r r ( f l ( p ) ) . Then X is an inductive source o f G , G(XH)((p)= G((p)(X)= G(A), G(XH)= G(X)H a n d G ( X ) n H = H(cp).

Proof. By Corollary 5.6, cp is an inductive source of H, so AH E Irr(cpH). Since S G and H Q G, it follows from Corollary 5.3 that cp and AH are inductive sources of G. The required assertions now follow from Theorem 5.8.

Chapter 20

Induction Theorems and Applications A principal objective of this chapter is to establish a number of classical results pertaining to induced characters. These include Brauer’s characterization of characters, a converse of Brauer’s theorem, necessary and sufficient conditions for a class function to be an R-generalized character ( R is a subring of (c ), rational valued characters, etc. The main tool in our investigation will be the Witt-Berman’s induction theorem and its converse discovered by Berman. As one of the many applications, we establish an elegant result due to Turull (1982) which asserts that if V and W are Q G-modules such that (Cv(g))= d i m q (Cw(g))for all g E G , then V 2 W .

dim^

1

The Witt-Berman’s induction theorem

The discussion below will require a considerable amount of notation. Therefore it will be especially useful t o assemble most of it in one place. G is a finite group of exponent n. F is an arbitrary field of characteristic 0. E~ is a primitive m-th root of 1 over F. R is the integral closure of Z in F ( E ~ ) . Im is the multiplicative group consisting of those integers p , taken mod) F. ulo m ,for which E~ H ELdefines an automorphism of F ( E ~over C ~ R ( G , Fis) the ring of all R-linear combinations of F-characters of G (here each F-character of G is regarded as a class function from G to F ( E ~ ) ) . Ch(G,F ) is the ring of all generalized F-characters of G. 779

Induction Theorems and Applications

780

The following lemma due to Solomon (1961a, Lemma 2) will play an important role in the proof of the main result.

Lemma 1.1. Let H =< g > K be a semidirect product of the normal cyclic subgroup < g > of order m and a group K such that for each x E K there exists p E 1, such that xgx-' = g p , Denote by 5' the integral closure of Z in Q ( E ~ ) . Then there exists an S-linear combination X of F-characters o f H such that m if i E Im X(gi) = 0 if i #Im Proof. Let p :< g >+ Z be given by p(gi) = m if i E 1, and p(gi) = 0 if i # I,. Let xo, .. .,xrn-l be all irreducible Q (&,)-characters of < g >. Since p is a class function on < g >, it follows from Theorem 19.3.5 that

c

m-1

p=

cixi

i=O

where

m-1

Then ci E S and each 0 E Gal(F(E,)/F) fixes ci. Since u fixes p and permutes the x;,we see that the F-conjugate characters in (1)appear with the same coefficients. Thus, by Theorem 19.4.2 (i), p is an S-linear combination of irreducible F-characters of < g >, say p = s1a1 t * - .4- s k a k , 3; E s, ai is an irreducible F-character of G. By Corollary 19.2.10, each ai can be extended to an F-character pi of H. Hence X = s& t - .- t $& satisfies the desired property. Let H be a subgroup of G and let p be a prime. We say that H is Felementary with respect to p if the following two conditions hold : (i) H is a semidirect product < g > Ii' of the normal cyclic subgroup < g > of order m coprime to p and a pgroup K . (ii) For any x E K ,there exists p E I , such that zgx-' = g p . We say that G is F-elementary if G is F-elementary with respect to some prime p. To describe a typical situation in which F-elementary subgroups arise, we need the following definition. Let g E G. Then the F-normalizer N of g is defined to be

N = ( 5 E G1zgx-l = g p for some p E Im}

1 The Witt-Berman’s induction theorem

78 1

where m is the order of g .

Lemma 1.2. Let N be the F-normalizer of g E G , let p be a prime and let K be a Sylow p-subgroup of N . Then (i) N is a subgroup of G and < g > d N . (ii) If g is p-regular, then H =< g > K is an F-elementary (with respect to p) subgroup of G. Proof. This is a direct consequence of the definitions. W

Lemma 1.3. Let g E G be an element of order m and let 5’ be an integrd domain containing Z [ E ~ ] .If P is a prime ideal of S containing a prime p , then for any F-character x of G, x ( g ) f X(gpl)(modP ) . where ~k and E; are the p and p’-parts of Proof. Write E~ = ELEL, E ~ Then . E~ = ER(modP) since ~k = l ( m o d P ) . On the other hand, since k gpt = g9 it follows from Corollary 17.1.11 that we may write x ( g ) = EZ and k

for some t i

2 1. Hence x ( g ) = X(gpt)(modP ) as required.

The following result is known as the Witt-Berman’s Induction Theorem.

Theorem 1.4. (Witt (1952), Berman (1956b)). Let G be a finite group and let F be an arbitrary field of characteristic 0. Then any F-character of G can be written in the form

where each z; E group of G.

Z and each x; is a n F-character of an F-elementary sub-

Proof. (Solomon (1961a). For the sake of clarity, we divide the proof into a number of steps. Step 1. Given a subgroup H of G , let [ C ~ R ( H , F )C] C ~ ~R(G,F) consist of all R-linear combinations of F-characters of G induced from F characters of H . Then, by Proposition 19.1.7 (i)? [ChR(H,F)]’ is an ideal

Induction Theorems and Applications

782

of C ~ R ( GF ), . NOW put

where H runs over all F-elementary subgroups of G with respect to distinct F ), . primes dividing the order of G. Then VR(G,F ) is an ideal of C ~ R ( G Similarly, if VZ ( G , F ) is defined as VR(G,F)by replacing R by Z , then V z (G,F) is an ideal of the character ring Ch(G,F). We wish to show that V z (G,F) = Ch(G,F ) which will imply the assertion of the theorem. Denote by 1~ the trivial character of G, i.e. 1G(g) = 1 for all g E G. In this step we claim that it sufices to show that 1~E VR(G,F). Assume that lc E VR(G,P).Then 1~ = El=,T ~ Q ; ,T; E R, a;is induced from an F-character of an elementary subgroup of G . Put r o = 1 and consider the Z5 -module & = Cf=, 22 ~i R. Since the additive group of Ro is torsion-free and finitely generated, Ro is Z -free of finite rank. Moreover, Ro/Z is a torsion-free (hence free) Z -module. Thus we may = TO = 1. Hence choose a Z-basis {PI,.. . , P k } of l& with k

1~= CPjpj

(pj E

vz

(

~

9

~

1

)

j=1

where = 1and P I , . ..,p k are Q -linearly independent elements of R. Now each P j is a Z -linear combination of all irreducible F-characters y1,. . . ,ys of G. Writing p j = Zjmy,, zjm E Z , 1 5 j 5 k, we then have

Because 71,. . . ,ys are F-linearly independent, we obtain

Bearing in mind that

PI,, . . ,P k

m=l

are Q -linearly independent, it follows that

1 The Witt-Berman's induction theorem

783

Since V z ( G ,F ) is an ideal of C h ( G , F ) ,it follows that VZ (G, F) = Ch(G,F), as claimed. Step 2. Let p be a prime and let g E G be pregular. Our aim here is to exhibit $ = & E VR(G,F)for which the following two properties hold : (i) $(z) = 0 if z E G is pregular and 2 is not F-conjugate to g. (ii) $(g) E Z and $(g) = l(modp). To begin with, we denote by N the F-normalizer of g in G. By Lemma 1.2, N is a subgroup of G with < g > d N . Furthermore, if K is a Sylow p-subgroup of N , then H =< g > K is an F-elementary (with respect t o p) subgroup of G with IHI = mt, where m is the order of g and t is the order ) XG E VR(G,F). of I, we have X(y-'zy) = 0. On the then by definition of A, X(y-'zy) = 0. other hand, if y-'zy = g p with p # Im, Since the equality y-lzy = g p for some p E Im implies that z and g are F-conjugate, we deduce that XG is 0 on all p-regular elements of G which are not F-conjugate to g. Finally, since X(y-'gy) # 0 implies y-lgy = g p for some p E I,, i.e. y E N ,we have

Hence by choosing z E 22: with z ( N : K )= l(modp),the function II,= zXG satisfies the required properties. Step 3. Let p be a fixed prime dividing n and let P be a prime ideal of R containing p. We wish to find Q E VR(G,F)such that for all g E G, a(g) l ( m o d P ) . To this end, we first note that, by Lemma 1.3, x ( g ) = x(g,t)(modP) for all g E G and all F-characters x of G. Hence the same congruence holds for all a E C ~ R ( G F ), . Let {gl, . . .,gk} be all representatives for the F-conjugacy classes of pregular elements of G. Then, by Step 2, for each i E (1,. . . ,I c } , there exists $i E VR(G,F)such that

=

+,i(gi)

= l ( m o d P ) ,+;(gj)

= 0 for j

#i

Induction Theorems and Applications

784

Set c t = $ ~ i - . - - + $ % . Thencr(gi)= l ( m o d P ) f o r a l l i ~(1, ...,k}. Finally, given g E G, g,! is F-conjugate to some gi. Hence, by Lemma 17.5.5 a(g)

= cY(gpt)I a(gi) = 1(modP)

as required. Step 4. Completion of the proof. By Step 1, it suffices to show that 1~E VR(G,F).Fix a prime p dividing IGl and write IGl = pks with ( p , s ) = 1. Then it suffices to show that S ~ E G VR(G,F ) . Let cy E VR(G,F ) be as in Step 3. Then

Our construction of cy did not depend on the choice of the prime ideal P containing p and so choosing a sufficiently large t , we have +)pt

= l(modpk~)

for all g E G

Now put 8 = S ( ~ G- a p t ) E C ~ R ( G , F )We . claim that 0 E VR(G,F);if sustained, it will follow that S ~ EG VR(G, F ) (since sap' E VR(G, F)),which will complete the proof. Let 91,. . . ,g, be the representatives for the F-conjugacy classes of G. For each j E (1,. ..,r } , let pj be defined on < gj > as follows : pi($) = mj if i E Im,and pj(g$) = 0 otherwise, where mj is the order of gj. As we have seen in the proof of Lemma 1.1, each pj is an R-linear combination of irreducible F-characters of < gj > and so each py E VR(G,F ) . Furthermore, we know that each py is 0 on all elements of G which are not F-conjugate to gj. An easy calculation shows that

where Nj is the F-normalizer of gj E G. Moreover,

Indeed, since 0 and p?, 1 5 j 5 T , are constant on F-conjugacy classes of G, it suffices to verify (3) for each gi, and this is true by virtue of (2). Finally, we know that e(gj) = s ( 1 - cy(gj)p') E O(modlG1R)

1 The Witt-Berman’s induction theorem

785

Hence all the coefficients in the right-hand side of (3) belong t o R. Thus 8 E V R ( G , F )and the result follows. I

To prove an application of Theorem 1.4, we need the following lemma. Lemma 1.5. Let H be a subgroup of G and let S be the ring of all functions f : G F satisfying the following properties : (i) f is constant on F-conjugacy classes of G. (ii) fH E C h ( H , F ) Then S 2 C h ( G , F ) and, for any ideal I of C h ( H , F ) , I n d s ( 1 ) = {aG\aE I } is an ideal of S . --$

Proof. That S 2 C h ( G , F ) is a consequence of Lemma 17.5.5. Given I , we have aG - PG = (a- P)G E Ind$(I). Let f E S and a E I . Then f E C f ( G ,F ) , a E C f ( H ,F ) and fHa E I . Hence, by Lemma 19.3.2 (i), ( f ~ a= )f a~G E Indg(1) as required. I a,P E

Theorem 1.6. (Berman (1958)). Let f : G F be any map. Then f E C h ( G ,F ) if and only if the following conditions hold : (i) f is constant on F-conjugacy classes o j G . (ii) For any F-elementary subgroup H of G, fH E C h ( H ,F ) . --f

Proof. Let S be the ring of all functions f : G + F satisfying (i) and (ii). Denote by I the 23 -submodule of C h ( G ,F ) generated by F-characters which are induced from F-characters of F-elementary subgroups of G. By Lemma 1.5, S _> Ch(G,F ) and I is an ideal of S. Since, by Theorem 1.4, 1 = C h ( G ,F ) we have S = Ch(G,F ) . W We close by proving a converse of Theorem 1.4.

..,H , be S U ~ ~ ~ OofU Theorem 1.7, (Berman (1958)). Let H I ,Hz,, G such that each F-character of G is an integral linear combination of characters induced from F-characters of the subgroups Hi. Then, for any F elementary subgroup H of G, there exists Hi, i E (1,. . . ,n } which contains a conjugate of H .

Proof. By hypothesis, H is a semidirect product < g > K of the normal cyclic p’-subgroup < g > and a p-group K (where p is a prime).

~ S

Induction Theorems and Applications

786

Furthermore, K is contained in the F-normalizer N of g in G. Choose a Sylow psubgroup P of N containing K . Then < g > P is an F-elementary subgroup of G containing H . It will be shown that there exists Hi which contains a conjugate of E =< g > P, which will complete the proof. Let x i j be an irreducible F-character of H ; . Then, by Proposition 19.1.2,

where 61,. . .,b, are the representatives for the conjugacy classes of Hi contained in the conjugacy class of G containing g and z i j t = ( C ~ ( b t: )C ~ , ( b t ) ) (by convention, each x i j ( b t ) = 0 if there are no such representatives). Let Nit be the F-normalizer of bt in Hi and let Nt be the F-normalizer of bt in G . We claim that zijt = ( N t : N i t ) (5) Indeed, the set of all elements z in G with z-'btz = b r , for a fixed p , forms a coset of G with respect to CG(bt). Therefore lNtl = l C ~ ( b t )where l~ s is the number of elements in < bt > which are F-conjugate to bt. Similarly, IN i t I = ICH,(bt )IS proving (5). By hypothesis, we may write 1 = Ci,ja i j X i j with a;j E Z and so, by (41, zijt f O(modp) for some i , j , t (6) It follows from (5) and (6) that some Nit contains a Sylow p-subgroup Q of Nit C H;.Since bt = z-lgz for some 2 E G, we Nt. Hence < bt > Q deduce that z-'Pz is a Sylow p-subgroup of N t . Thus y - ' ( ~ - ~ P z ) y = Q for some y E Nt. Therefore (sy)-'E(zy) =< bt > Q C Hi and the result follows. m

2

Brauer's theorems

In this section, we derive a number of fundamental results of Brauer as a consequence of the Witt-Berman's induction theorem. Let G be a finite group and let p be a prime. We say that G is pelementary if G is a direct product of a cyclic group and a pgroup. We also say that G is elementary if it is pelementary for some prime p . Observe that elementary groups are nilpotent and hence (by Corollary 18.12.4) M groups. By a linear character , we understand a character of degree 1.

2 Brauer’s theorems

787

Theorem 2.1. (Bmuer’s Characterization of Characters, Brauer (1947), (1953)). (i) Any C -character of G is an integml linear combination of characters induced from linear characters of suitable elementary subgroups of G. (ii) Let a : G 4 C be a class function. Then CY is a generalized character of G if and only if for every elementary subgroup E of G, CYE is a generalized character of E . (iii) Let a : G -, C be a class function. Then CY is an irreducible (I= chamcter i f and only if the following properties hold : (u) For every elementary subgroup E of G, CYEis a generalized character of E . (b) 1GI-l C g E G 4 M g - l ) = 1 (c) a(1)> 0.

Proof. (i) It is clear that a subgroup E of G is elementary if and only if it is (I: -elementary. Now apply Theorem 1.4 and the fact that, by Corollary 18.12.4, elementary groups are M-groups. (ii) This is a special case of Theorem 1.6 in which F = (I: . (iii)Apply (ii) and Lemma 19.3.7. (iii). H Theorem 2.2. ( A converse to Brauer’s Theorem, Green (1955)). Let H I , . ..,H , be subgroups of G such that any (I: -chamcter of G is an integral linear combination of (c -characters induced from the characters of the subgroups Hi,1 5 i 5 n. Then, for any elementary subgroup E of G, there exists i E ( 1 , . . . ,n ) such that Hi contains a conjugate of E .

Proof. This is a special case of Theorem 1.7 in which F = (I:. Theorem 2.3. (Bmuer (1945), (1947)). Let G be a finite group of exponent n and let F be a field over which the polynomial X n- 1 splits into linear factors. Then F is a splitting field for G.

Proof. The case charF = p > 0 is a consequence of Proposition 17.2.5. Assume that charF = 0. Then, by hypothesis, F 2 Q ( E ) where E is a primitive n-th root of 1. Hence, by Proposition 17.2.2, it suffices to show that Q ( E ) is a splitting field for G. Let V be a simple (I: G-module which affords the (I:- character x and let F = Q ( E ) . By Proposition 11.1.11, it suffices to show that V is realizable

Induction Theorems and Applications

788

over F . Since x ( g ) E F for all g E G, we have F ( x ) = F . Hence it suffices is equal to 1. By Theorem 2.1 (i), we to show that the Schur index rn&) may write

x = Cz;xG

zi E

z

(1)

where {xi} are linear characters of elementary subgroups of G. It is clear that each X? is afforded by some FG-module. By Theorem 14.4.1 (i), the multiplicity of x in each character xf is a multiple of r n ~ ( x )On . the other = 1 and the hand, by (l),the multiplicity of x in Czixf is 1. Thus rn&) result follows. H For future use, we need an easy generalization of Theorem 2.1 (ii). Let R be a subring of (c , By an R-generalized character of G, we understand an R-linear combination of irreducible (c -characters of G.

Theorem 2.4. Let LY : G --t (c be a class function and let R be a subring of (c . Then Q is an R-generalized character of G if and only i f for any elementary subgroup E of G, Q E is an R-genemlized character of E .

Proof. Let S be the subring of Cf (G) consisting of all p such that PE is an R-generalized character of E for any elementary subgroup E of G. Denote by K the subring of Cf(G) consisting of all R-generalized characters of G. Then obviously S 2 K . Let I be the R-submodule of K generated by all (I:-characters which are induced from (c -characters of elementary subgroups of G. If X E S and x is a (c -character of E , then X - xG = ( X E X ) ~(Lemma 19.3.2 (i)). Hence I is an ideal of S. Since, by Theorem 2.1 (i), 1~ E I we have S = K , as desired. H To prove our find result, we need the following application of Theorem 2.4.

Lemma 2.5. Let IGl = p"k with ( p , k ) = 1, p-prime, let g be a p regular element of G and let R be the ring of algebraic integers in (Q ( E ) , where E is a primitive k-th root of unity. Define Q : G (I: by --f

if xp1 is conjugate to 9 otherwise Then a is an R-generalized character of G.

3 Rational valued characters

789

Proof. Let E be an elementary subgroup of G. By Theorem 2.4, it suffices to show that C X Eis an R-generalized character of E . Write E = El x E2 where El is a pgroup and E2 is a $-group. If g is not conjugate t o an element of E2, then a~ = 0 which implies the result. Assume that gl, . . . , g r are all distinct elements of E2 which are conjugate to g . Then the elements x E E with xpt conjugate to g are precisely the elements of the cosets E l g i , 1 5 i 5 r . Let x be an irreducible (I: -character of E . Then, by Proposition 17.8.1, x = x1 x x 2 where x i is an irreducible (c -character of Ei, i = 1,2. Hence

cc T

IEI < Q E , X >=

T

4 t g ; ) ; y ( t g i= )

i=l tcE1

If

XI

#l

~then ~

kCXz0 i=l

c

Xl(4

tEEi

CtEEl , x l ( t ) = 0 and so < C Y E , X>= 0. If x i

= l ~then~ ,

T

i=l

Since 1EI = desired.

l E l l JE2l and

lE2l divides k, it follows that < CYE,X> E R , as

Corollary 2.6. Let IGI = pnk with ( p , k ) = 1, p prime, let R be the ring of algebraic integers in Q ( E ) , where E is a primitive (GI-th mot of unity and let P be a prime ideal of R with p E P . Then, f o r any given x,y E G, xp: and yp: are conjugate if and only if for any irreducible (c -character x of G, x ( 4 = X ( Y > (mod PI Proof. Assume x(x) G X(y)(rnodP ) for all irreducible (I: -characters x of G. Let a : G + (I: be defined by a(g)= k if gp: is conjugate to xp: and otherwise a ( g ) = 0. Then, by Lemma 2.5, is an R-generalized character of G and so a ( . ) 3 a ( y ) ( m o d P ) .By definition, a(.) = k. If a ( y )= 0, then k E P which is impossible since p E P and ( p , k ) = 1. Thus a(y) = k and hence ypt is conjugate to z p t . The converse being a consequence of Lemma 1.3, the result follows.

3

Rational valued characters

In this section, we prove a classical result of Artin concerning rational valued complex characters of finite groups and provide some applications. In what

Induction Theorems and Applications

790

follows, G denotes a finite group and all characters are assumed to be C characters. We say that a character x of G is rational valued if

x(g) E Q

for all 9

EG

It is clear that any Q-character is rational valued, but the converse need not be true. Given a subgroup H of G, we write 1~ for the principal character ) 1 for all h E H ) . For any cyclic subgroup H of G , of H (i.e. 1 ~ ( h = a ( H ) : H + Z is defined by a ( H ) ( z )= JHJ if H =< 2 > and cr(H)(z)= 0 otherwise. It is clear that a ( H )is a class function.

Lemma 3.1. Let H be a cyclic subgroup of G. Then (i) a(H) is an integral linear combination of characters of H induced from principal chamcters of cyclic subgroups of H . (ii) For any given g E G, c r ( H ) G ( g ) = ~ N G ( H ) Iif g is conjugate to a generator of H and a ( H ) G ( g ) = 0 , otherwise.

Proof. (i) The case H = 1 being obvious, we argue by induction on / H I . To this end, we first show that a(s)H =

*

1H

SCH

where S ranges all subgroups of

H.For each z E H , we have

where, by convention, a ( S ) ( y ) = 0 for y E H - S. Since a ( S ) ( z )= 0 if S #< 2 > and a ( S ) ( z ) = IS) if S =< z >, we have ‘ & C-H a ( S ) H ( ~ ) = \HI, proving (1). It follows from (1) that

a(H) = I H l l H -

C CK(S)~ SCH

Applying the induction hypothesis to each proper subgroup 5’ of H , the desired assertion follows by the transitivity of induced characters. (ii) Setting a = a ( H ) , we have

3 Rational valued characters

791

where, by convention, a(z-'gz) = 0 if z-lgz # H . Hence, by the definition of a , we have a H ( g )= 0 if g is not conjugate to a generator of H . Assume that g = y - l h y where h is a generator of H and y E G. Since aG is a class function, a G ( g ) = aG(h). On the other hand, z-*hz E H if and only if z E N G ( H ) since H =< h >. Thus, by (2),

as desired. W

Theorem 3.2. (Artin (1924), (1931)). Let x be a rational valued character of G . Then ( G ( xis an integral linear combination of characters of G induced from principal characters of cyclic subgroups of G.

Proof. Let H I , . . . ,H , be all nonconjugate cyclic subgroups of G and let hi be a generator of H i , 1 5 i 5 n. Consider the class function B on G defined by

By Lemma 3.1 (i) and transitivity of the induction, it suffices to show that B(g) = IGlx(g) for all g E G. Now fix g E G and choose a unique j E (1,. ..,n } such that < g > is conjugate to H j . Then, by Lemma 3.1 (ii), a(H;)G(g)= 0 for i # j , and Ct(Hj)G(g)= ING(Hj)l. Thus

But g and hj are Q-conjugate, so x ( g ) = ~ ( h j(by ) the proof of Lemma 17.5.1 (iv)) and therefore B ( g ) = IGlx(g). In what follows, we denote by R(G) the Q -space of all functions from G to Q which are constant on the Q -conjugacy classes of G.

Corollary 3.3. Let H I , . ..,H , be all nonconjugate cyclic subgroups of G. Then (i) G has precisely r irreducible Q -characters, say X I , . . . ,x7. (ii) X I , . . . ,xr and ( l ~ , ). .~ ,, ( 1 ~ , . are ) ~ two Q -bases for R(G).

.

Induction Theorems and Applications

792

By Lemma 19.3.4 (with F = Q ), it suffices to show that each 1 )G,. ~ .~. ,( 1 ~ ~ ) ' .If H is any cyclic subgroup of G, then H is conjugate to some H i , in which case ( 1 ~ =) ( ~ 1 ~ ~ The ) ~ desired . assertion is now a consequence of Theorem 3.2.

Proof.

Q -character

x of G is a Q -linear combination of (

The following lemma due to Turull (1982) will enable us to provide an interesting application of Corollary 3.3 (namely, Corollary 3.7).

For any X E R(G), let A* : G -+ Q be defined by

Lemma 3.4.

X*(g) = 1/1 < g > I

X(x)

for all g E G

x€

Then the map X

I+

A* is an automorphism of the Q -space R(G).

Proof. It is clear that X H A* is a Q -linear map of R ( G ) into R ( G ) . Let G be a counter-example of minimal order. Then there exists X E R(G) and g E G such that X(g) # 0 and A* = 0. For any subgroup H of G, we have AH E R ( H ) and (AH)* = (X*)H = 0. Hence, by the choice of G , AH = 0 for every proper subgroup H of G. Thus G =< g > and X(z) = 0 if x E G and G #< x >. But if x E G and < g >=< z >, then X(z) = X(g), so X*(g) # 0,a contradiction. Let F be a field and let V be an FG-module. Given g E G , we put

Cv(g) = {w

E VJgv =

v}

Then Cv(g) is the maximal F < g >-submodule of V on which < g > acts trivially. In what follows all FG-modules and (Q G-modules are assumed to be finitely generated.

Corollary 3.5. (Turulk (1982)). Let V and W be QG-modules such that dimQ (Cv(g)) = dim^ (Cw(g)) for all g E G Then V Z W .

Proof. Let xv and xw be the characters afforded by V and W, respecHence, tively. Then xv, xw E R(G) and, by hypothesis, (xv)* = (xw)*. by Lemma 3.4, x v = xw and so, by Corollary 17.1.8 (ii), V % W.

3 Rational valued characters

793

Corollary 3.6. (Turu11 (1982)). Let F / Q be a finite Galois field extension and let V,W be FG-modules. If V is simple and

then

V E “W

for some u E Gal(F/(Q )

Proof. Let xv and x w be the characters afforded by V and W ,respectively. Then ~ ( x v=) C “(xv), X(xw) = C “ ( x w ) ,where u ranges over G a l ( F / Q ), are Q -characters of G. Hence X(xv), X(xw) E R(G). By hypothesis, X(xv)* = X(xw)* and so, by Lemma 3.4, X(xv) = X(xw). Since V is simple, we deduce that xv = “xw for some u E Gak(F/(Q),as we wished to show. W

.

Corollary 3.7. (Turu11 (1 982)). Let XI,.. ,X, be the conjugacy classes of cyclic subgmups of G, let Hj E X i and let aij be the number of double cosets HjgH; in G, 1 5 i , j 5 T . Then a;j is independent of the choice of

representatives H ; in X; and

(ajj)

is a nonsingular symmetric matrix.

Proof. Owing to Corollary 3.3 (ii), { ( 1 ~ , ) ~ 1I1i 5 r } is a Q -basis for R(G). By Mackey decomposition (Proposition 19.1.10), for any g E G , ((lH,)G)

= C ( l t H , t - ’ n < g > ) t

where t runs through a set of representatives of the double cosets < g > tH; in G. Hence [ ( l ~ , ) ~ ] * (isg )equal to the number of double cosets < g > t H ; in G. This shows that the a;j are independent of the choice of H ; in Xi. Furthermore, by Lemma 3.4,

is a Q-basis for R(G) and so ( a i j ) is nonsingular. Finally, since obviously aij = aji for each i , j , the result follows. W

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Chapter 2 1

Central, Faithful and Permutation Characters We have collected in this chapter various theorems concerning central, faithful and permutation characters. After introducing central characters and providing a number of their fundamental properties, we establish a classical formula due to Burnside (1911). This formula expresses the structure constants of Z ( F G ) in terms of irreducible F-characters of a finite group G, where F is a field of characteristic 0 which is a splitting field for G. We then confront the following problem : Given a function a! : G --+ F, what are necessary and sufficient conditions for a t o be of the form a! = Xx for some X E F and some irreducible F-character x of G ? A classical result, due to Weil(1938), which provides a complete solution of this problem is presented. The next topic is devoted to a detailed study of character kernels. As one of the applications, we show that if x is an irreducible (C-character of a nilpotent group G, then ~ ( 1 divides ) ~ (G : Z ( x ) ) . Faithful characters, i.e. characters with trivial kernels, are of special importance. One of our results provides a generalization, due to Brauer (1964a), of a classical theorem of Burnside (1911). Namely, we show that if X is a faithful (I: -character of G which has exactly TZ distinct values, then each irreducible (c -character of G is a constituent of one of the characters A', 0 5 i 5 TZ - 1, where A' = la is the principal character. Among other results, we provide necessary and sufficient conditions for a group G t o have a faithful irreducible character. We also investigate in detail 795

Central, Faithful and Permutation Characters

796

faithful characters of nonabelian groups G for which G' E Z ( G ) . As one of the applications, we then completely describe all irreducible characters of extra-special pgroups. Permutation characters play a fundamental role in character theory. We examine these characters in detail. One of the results presented demonstrates that if T is a set of prime numbers and H a Hall n-subgroup of a ?r-separable group G, then H Q G if and only if every irreducible constituent of ( 1 ~ is) of~ #-degree. As an easy application, we prove a theorem due to Fong (1961) which asserts that if G is a psolvable group, then G has a normal abelian Sylow p-subgroup if and only if each irreducible character of G is of p'-degree. It should be pointed out that the corresponding result holds without the assumption that G is p-solvable. However, the proof, due t o Michler (1986), uses the classification theorem of finite simple groups. As a further application of permutation characters, we prove a theorem, due to Saksonov (1987), which asserts that a finite group G is nilpotent if and only if for every Sylow subgroup S of G and every irreducible constituent x of (IdG, 131) = 1.

(XW,

1

Central characters

In what follows, G denotes a finite group and F is a field of characteristic 0 which is a splitting field for G. By an algebraic integer in F , we understand an element in the integral closure of 23 in F . In case F = (I:, we use the term algebraic integer . We begin by recalling the following terminology which was introduced earlier for arbitrary F-algebras. Let V be a simple FG-module. Then, since E n d ~ c ( V= ) F , each element t E Z(FG) acts as an element of E n d r ~ ( Von ) V . The corresponding map $ : Z(FG) + F such that tv

= $(t)o

for all

J

E Z ( F G ) ,o E V

is called the central character of F G afforded by V (or, simply, a centra1 character of F G if V is not pertinent to the discussion). It is clear that 11 is a homomorphism of F-algebras, Furthermore, if x is the character of F G afforded by V , then obviously

1 Central characters

797

In what follows, each irreducible F-character of G is identified with the corresponding irreducible F-character of F G. Also, for any F-character x of G, Q ( x )denotes the field obtained from Q by adjoining all x ( g ) , g E G.

Theorem 1.1. Let X I , . . . ,x, be all irreducible F-characters of G, let C1,. . . ,CT be all conjugacy classes of G and, for each i E ( 1 , . . . ,r } , let c: = &C, 9 and ei = 1 ~ I - l Xi(1)xi(g-')g

C

BEG

Then

(i) The maps w; : Z(FG) -, F given by w ; ( z ) = x ; ( z ) / x i ( l ) are all central characters of F G and all irreducible F-representations of Z(FG), l < a < T .

(ii) w i ( e j ) = 6;j for all i,j E ( 1 , . . . , r } . (iii) For each I E Z(FG), x = C i = l w ; ( z ) e i . (iv) w ; ( C T ) = ICjlx;(gj)/xi(l)is an algebraic integer in Q ( x i ) ,g j E Cj.

(4 c;

=EL

Jw C.lxi(g )e..

Proof. (i) That the wi are all central characters of F G is a consequence of ( 1 ) . Since F is a splitting field for G of characteristic 0, Z(FG) is a direct product of r copies of F. Hence there are precisely T nonequivalent irreducible F-representations of Z( FG) and so the w; are the only irreducible F-representations of Z( FG). (ii) By Lemma 19.2.7, e l , . . . ,e, are all block idempotents of F G (in particular, they constitute an F-basis for Z ( F G ) ) . If I4 is a simple FGmodule which affords x;, then V, belongs to the block FGe; (Lemma 19.2.7). This means that ei acts as the identity on V;. and annihilates each vj for j # i, proving that w;(ej) = 6;j, 1 5 i ,j 5 T . (iii) This follows from (ii) and the fact that e l , . . . ,eT is an F-basis for Z(FG). (iv) If gj E Cj, then x;(CT) = ICjlx;(gj) which proves the formula for anm& with each anmk E 1z . Then w;(CT). Now write C$C$ = T

wi(c,')ui(c$)=

a n m k W i ( C k +) k= 1

Z3 w;(C$) is a subring of F which is a finitely generated 23 Hence module. Therefore, by Lemma 16.1.1, ui(C!) is integral over 23. (v) Apply (iii) and (iv). H

Central, Faithful and Permutation Characters

798

Keeping the notation of Theorem 1.1, let us write r

C:C$ = Ca;jkCk+ k=l

Then aijk is the number of pairs (a,b) E C; x Cj such that where c E Ck

ab = c

The following classical result provides an explicit formula for a;jk in terms of irreducible F-characters X I , . . ,xr of G.

.

Theorem 1.2. (Burnside's Formula, Burnside (1911, p. 916)). With the notation abowe,

Proof. Applying (2), we have

Hence, by Theorem 1.1 (iv),

which implies

Multiplying both sides of the above by Xt(gkl) and summing over t E (1,. ..,TI,it follows from Theorem 19.2.3 (iii) that

as desired. We close by providing the following application of Theorem 1.1.

1 Central characters

799

Theorem 1.3. (Wed (1938, p p . 87-88)). For a n y given function a :

G + F , the following conditions are equivalent : (i) a = Ax for some irreducible F-character x of G and some A E F . (ii) For all x , y E G , a ( z ) a ( y )=

IGI

c

a(x:a-'yz)

(3)

zEG

Proof. (Feit (1967)). If a satisfies (3) and a ( 1 ) = 0, then a = 0. Hence we may harmlessly assume that a ( 1 ) # 0. Note further that any a satisfying (3) is obviously a class function (by taking x = 1). Thus we may also assume that a is a class function. Let Cl, . . . ,C, be all conjugacy classes of G, let g; E C; and let the map w : Z ( F G )+ F

be defined by

Our plan is to show that a satisfies (3) if and only if

(4) are as in (2). Since for any A E F and any irred cible ch racter Ax satisfies (3). Conversely, if a satisfies (3), then by (4)w is a homomorphism of F-algebras; hence by Theorem 1.1 (i), it wiil follow that w = w; for some i E ( 1 , . . . , r } . Therefore a ( g ) / a ( l )= x i ( g ) / x j ( lfor ) all g E G and so a = [ a ( l ) / x ; ( l ) ] xas; ,required. To prove the required assertion above, note that for any g E G, where

aijk

x of G , a = Ax satisfies (4),it will follow that

Central, Faithful and Permutation Characters

800

It follows that

Thus a satisfies (3) if and only if a satisfies (4), as desired.

2

Character kernels and faithful characters

In what follows, G is a finite group and all characters are assumed to be (I:-characters. Given a (I: G-module V , recall that the kernel K e r V of V is defined to be the kernel of the representation of G afforded by V .

A. General properties Let

x be a character of G.

Then the kernel K e r x of

x is defined by

We say that x is faithful if I i e r x = 1. The set of all irreducible characters of G will be denoted by I r r ( G ) . Theorem 2.1. Let x be a character of G. Then (i) K e r x = K e r V where V is a (c G-module which afsords x. (ii) If x = nlxl t t ntXt with n; > 0 and x; E I r r ( G ) , then

I n particular, b y taking have

x

to be a faithful character (e.g. n~Elrr(G)K~= rX

x

is wgular), we

2 Character kernels and faithful characters

801

(iii) If x = XG f o r some character X of a su,bgroup of G , then K e r x = n,€Gg(li'er X)g-]

(iw) If A f I i ' e r x is a n abelian normal subgroup of G f K e r x and irreducible, then x ( 1 ) divides ( G : A ) .

x

is

Proof. (i) Let p be a (C-representation of G afforded by V . If g E K e r p , then obviously x ( g ) = ~ ( 1 ) Conversely, . assume that x ( g ) = n for n = ~ ( 1 )By . Lemma 17.1.10, p(g) is similar to diag(e1,. . . ,E ~ where ) each - - E~ = n. Since I ~ i l= 1, it follows E ; is a root of unity. Hence x ( g ) = that each E ; = 1. Thus g E I i e r p , as asserted. (ii) This is a direct consequence of (i). (iii) Apply (i) and Proposition 18.1.21. (iv) By (i), we may assume that K e r x = 1. Now apply Theorem 17.10.4 and the result follows.

+

Lemma 2.2. Then

Let H be u subgroup of G and let

x

be a character of G.

( GI :H ) < X , X > with equality if and only if x ( g ) = 0 for all g E G - H .

Proof. By definition,

I H I < X H , X H >=

lX(g>I2 = IGl < X, X >

IX(h)lz 5 h W

s€G

because ] x ( g ) I 2 2 0. The equality holds if and only if x(g) = 0 for all g E G - H , as asserted. We next introduce a subgroup of G closely related to K e r x . Namely, given a character x of G , we put

Z(X)= (9 E GI

Ix(s)l= x ( 1 ) )

Property (v) in the following result is due to Burnside (1911). Theorem 2.3. Let x be a character of G . Then (i) Z ( x ) is a normal subgroup of G such that Z ( X ) / K e r x is u cyclic subgroup of Z ( G / I i e r x). Furthermore, if x E I r r ( G ) , then

Z ( X ) / K e r x = Z ( G / I < e rx)

Central, Faithful and Permutation Characters

802

(ii) Z ( G ) = n , , h ( G ) z ( X ) * (iii) I f x E Irr(G), then ~ ( 15 )(G ~ : Z ( x ) ) . The equality occurs if and only i f x ( g ) = 0 for dl g E G - Z ( X ) . (iw) I f x E Irr(G) and G / Z ( x ) is abeliara, then ~ ( 1=)(G ~ : Z(x)). (u) I f x E I T T ( G )and C is a conjugacy class o f G with ( ~ ( lIC() ) , = 1, then for all g E C , either g E Z ( x ) or x ( g ) = 0. Proof. (i) Let p be a (I:-representation of G which affords x and let g E G. By Lemma 17.1.10, p ( g ) is similar to diag(&1,...,E,,.) where each t E ~ it, follows is a root of unity and n = ~ ( 1 ) Since . x ( g ) = ~1 t that 1x(g)1 = n if and only if all ci are equal. Thus g E Z ( x ) if and only if p ( g ) = diag(a,. . , E ) where E is a root of unity. Hence Z ( x ) is a normal subgroup of G such that 2 ( X ) / K e r x is a cyclic subgroup of Z ( G / K e r x). Finally, if x E I r r ( G ) and g K e r x E Z ( G / K e r x ) , then p ( g ) E Z(p(G)). Hence p ( g ) = diag(c,. . . , E ) for some root of unity E and so Ix(g)l = ~ ( l ) ,

.

as required. (ii) It is clear that Z(G) Z ( x )for each x E Irr(G). Conversely, assume that g E Z ( x ) for all x E Irr(G). Then, by (i), g K e r x E Z ( G / I < e r x )and so [g,h]= g-lh'lgh E K e r x for each h E G,x E Irr(G) Hence, by Theorem 2.1 (ii), g commutes with each h E G, i.e. g E Z(G). (iii) Put Z = Z ( x ) . Then, by (i), xz = x(1)X for some linear character X of 2. Hence < x z , xz >= x(1)2 < A, x >= x ( 1 ) 2 It follows from Lemma 2.2 that ~ ( 1 5) ( G ~ :2 ) < x,x

>= ( G :2 )

with equality if and only if x ( g ) = 0 for all g E G - 2. (iv) Assume that g E G- Z ( x ) . By (iii), it suffices to show that x ( g ) = 0. By (i), we have .z = g-'h-'gh # K e r x for some h E G. Since G / Z ( x )is abelian, we also have z E Z(x).Let p be a representation of G which affords x. Then p ( z ) = diag(e,. . . , E ) where E is a root of unity and E # 1 since z $! K e r x . Hence p ( g z ) = p ( g ) p ( z ) = & p ( g ) and therefore x(g.z) = E X ( g ) . But g z = h-'gh, so x ( g z ) = x ( g ) and x ( g ) = E X ( g ) . Since E # 1, we deduce that x ( g ) = 0, as asserted. (v) Fix g E C. Owing to Theorem 1.1 (iv), x(g)lCl/x(l)is an algebraic ,)= 1 , there exist m,n E 1z such that rnx(1) t integer. Because ( ~ ( lI)Cl

2 Character kernels and faithful characters

803

nlCl = 1. Hence

is an algebraic integer. Since mx(g) is also an algebraic integer, we see that X = x(l)-'x(g) is an algebraic integer. Assume that g !$ Z(x). Then Ix(g)l < x(1) and 1x1 < 1. We now proceed to verify that x(g) = 0. Let E be a primitive complex r-th root of unity, where T is the order of g, and let I' = Gal(Q ( E ) / Q ). Because x(g) is a sum of x(1) roots of unity, so is o(x(g)) for all u E I'. Thus Io(x(g))I 5 x(1) and Iu(X)l 5 1 for all u E r. Setting a = nOEra(X), it follows that la1 < 1. But a is fixed under I?, so a E Q . On the other hand, a is an algebraic integer, so CY E Z . Hence (I! = 0 and so X = 0. Thus x(g) = 0 and the result follows. Our next result is closely related to Theorem 2.2 (iv).

Theorem 2.4. Let G. Then

x

be an irreducible chamcter of a nilpotent group

x(1>21(G: Z(X>>

Proof. We may regard x as an irreducible character of Gllr'erx. Since, by Theorem 2.3 (i), Z(G/lierX) = Z(X)/KerX, we may assume that x is faithful in which case Z(x) = Z(G) is cyclic by Theorem 2.3 (i). If G is abelian, then x(1) = 1 and so we may also assume that Z(G) # G. By Proposition 17.8.1, we may further assume that G is a p-group. We now proceed by induction on IGI. Let Z/Z(G) = Z(G/Z(G)). Since G is a p-group, there exists a normal Z and (N : Z(G)) = p . Since N / Z ( G ) subgroup N of G with Z(G) c N is cyclic, we see that N is abelian. Furthermore, we claim that (G : C) = p where C = C G ( N ) .Indeed, since N 2 , we have for y in N - Z(G) and f 2 in G that [z,y] E Z(G). Since the map G + Z(G), g H [g,y] is a homomorphism and since y* E Z(G), we see that the image of G is the subgroup M # 1 of Z(G) generated by some elements of order p . But Z(G) is cyclic, so [MI = p and since K e r f = C,(G : C) = p as claimed. By Clifford's theorem, (Proposition 19.1.11), we may write

for some integer e 1 1 and all distinct G-conjugates XI,. , . ,xr of some irreducible character x1 of N . Since N is abelian, x;(l) = 1 for each i .

Central, Faithful and Permutation Characters

804

Z(G) since x is faithful, which is impossible. Thus If r = 1, then N r > 1. On the other hand, C is contained in the inertia group of x1 and so r = p as (G : C) = p . Now C is the inertia group of X I , so by Clifford's theorem, x = pG for some irreducible character p of C with p~ = ex1. Let K = I i e r p . Then N K / K C Z(C/K)and hence, by induction, ~ ( 1 ) ~ divides (C : N ) . But x(1) = pp(l), while (G : Z(G)) = p2(C : N ) . Thus ~ ( 1 divides ) ~ (G : Z(G)),as asserted.

We next provide the following generalization of a classical theorem of Burnside (1911, Theorem IV of Chapter XV). Theorem 2.5. (Brauer (1964~)). Let X be a faithful character ofG which has exactly n distinct values a1 ,. ..,a,. Then each irreducible character x of G is a constituent of one of the characters X i , 0 5 i 5 n - 1, where Xo = 1~ is the principal character. Proof. Put Xi = { g E G(X(g)= a ; } , 1 5 i 5 n. Assume by way of contradiction that x is not a constituent of X j for j E (0,. ..,n - 1). Then, for any gi E Xi,j E { & I , . . . , n - I}, n

i=l

=

g€Xi

&icxcs, gEX, i=l

The nonvanishing of the Vandermonde determinant implies that for each i E { O , l , ..,,n - I}, CgEXi x ( g ) = 0. But X is fajthful, so X;= (1) €or some i. Thus x(1) = 0, a contradiction.

B. Gaschutz's theorem Our aim here is to provide necessary and sufficient conditions for G to have it faithful irreducible character. As a preliminary, we establish two auxiliary results.

Lemma 2.6. Let N # 1 be a normal subgroup of G and let M I , .. . ,h4t be all normal subgroups of G which aw. maximal with respect to the property of being properly contained in N . Then N is generated by a single conjugacy class of G if and only if N # UfZl M ; .

2 Character kernels and faithful characters

805

Proof. If N is generated by the conjugacy class of G containing g E N and if N = UF=,M;, then g E Mi for some i and hence N = M i , a contradiction. Conversely, assume that N # UI,lM;. I f g E N - (uf=,Mi), let M be the subgroup of G generated by the conjugacy class of G containing Mi for all i, we then have M = N , as desired. H g. Since M Lemma 2.7. (The Exclusion Principle). Let X be a set of N elements and let P(1),P ( 2 ) ,...,P ( n ) be n properties. Denote by N; the number of elements of X with property P ( i ) and, more generally, by Ni,j z...i, the number of elements of X which satisfy each of the properties P(i1))P ( i z ) ,. . . ,P ( i T ) . Then the number N ( 0 ) of elements of X which do not satisfy any of the properties P(l),P( 2), . . . ,P(n ) is given by

Proof. An element x of X which does not possess any of the given properties is counted once in the term N and does not enter any other summand. Otherwise, x satisfies precisely s(1 5 s 5 n ) properties, say P ( j l ) , . . ,P ( j 3 ) .Hence x contributes 1 to the term

.

for any choice of

il,

. . . ,ik

. . . ,j3}) i.e.

E {jl,

for

( )

choices. Thus the

contribution of x to the right-hand side is

Hence the right-hand side counts each element which does not possess any of the given properties exactly once and any other element zero times, as required. H

Corollary 2.8. Let H I , . . . )H , be subgroups of G. Then

Central, Faithful and Permutation Characters

806

Proof. Apply Lemma 2.7 to the case where X = G and P ( i ) is the property that g E H i . H We say that H is a minimal normal subgroup of G if H # 1 , H a G and 1 and H are the only normal subgroups of G contained in H. The socle Soc(G) of G is defined to be the product of all minimal normal subgroups of G. If G = 1, then by definition Soc(G) = 1. Thus G = 1 if and only if Soc(G) = 1.

Theorem 2.9. (Gaschutz (1954)). Any given group G has a faithful irreducible character if and only if Soc(G) is generated by a single conjugacy class of G. Proof. We may clearly assume that Soc(G) # 1. Let X I , ... , x r be all irreducible characters of G and let N 1 , . . . ,Nt be all minimal normal subgroups of G. Put m= Xi(1I2

C

Ker(Xi)=l

where an empty sum is understood to be zero. We claim that

This is clearly proved by a straightforward modification of the exclusion principle. Indeed, let x E { X I , . .. ,x r } . If x is faithful, then x is counted once in the first sum and does not enter any other summand. Now assume that Mj,, ,Mj,, are all minimal normal subgroups of G contained in Ker(X). Then contributes ~ ( 1 to) the ~ term

. .. x

for any choice of contribution of

x

il,.

. . , i k E { j l , . . . ,jn}, i.e. for

to the right-hand side of (1) is

choices. Hence the

2 Character kernels and faithful characters

807

= x(1)2(1 - 1)" = 0

proving (1). Now for any normal subgroup N of G , we have

C

~ i ( 1= ) ~(G : N

)

and therefore, by (l), t

m = IGI t

C(-l)k k=l

( G : Nil . - .N i k )

il c G l n K , it follows that G1 f l Ii' = Z(G). The assumption ]GI > p3 ensures that 11'is not contained in GI. Since Z ( K ) GI, we deduce that K is a nonabelian group, so K' is a nonidentity subgroup of G'. But G' = Z(G) = Z ( K ) is of order p , hence K' = Z( K ) is of order p . Moreover, since

3 Permutation characters

813

we see that l i ' / Z ( K ) is elementary abelian. Thus li' is indeed extra-special. It therefore follows, by induction, that li is the central product of nonabelian subgroups Gi, 2 5 i 5 r , of order p3 and that lKl = p2'-l. Thus G is the central product of G I ,G2,.. .,G , and as G1 fl li' = Z(li), we have [GI = p2'+'. This completes the proof of the theorem. 4 The following theorem completely describes all irreducible characters of extra-special pgroups.

Theorem 2.18. Let G be an extra-special p-group of orderp2'+l. Then G has exactly the following irreducible characters : (i) p2' linear characters. (ii) p - 1faithful irreducible characters xi of degree p', which vanish outside Z ( G ) and satisfy ( x i ) Z ( q = p'A;, where A; is a faithful linear character of z(G).

Proof. The group G satisfies the hypotheses of Theorem 2.14 for n = p . Since IG'l = p , we have JG/G'I = p2'. Hence G has exactly p2' linear characters. Since (G : Z(G))'12= p', it follows from Theorem 2.14 that the faithful irreducible characters of G are given by (ii). Finally, since the sum of the squares of the degrees of characters in (i) and (ii) is p2'

+ ( p - qp2'

= p2'+l = IGI 7

the result follows.

3

Permutation characters

All characters below are assumed to be (I:-characters. Given a subgroup H of a finite group G , we write 1~ for the principal character of H . A character x of G is said to be a p e r m u t a t i o n c h a r a c t e r if x is afforded by a permutation (r: G-module. For convenience, let us introduce the following group- theoretic information. A group G acting on a set X is called r - t r a n s i t i v e on X provided for any two sets ( 2 1 ,... , x T } and {yl, ...,y,.} of r elements in X there exists g E G such that gz; = y; for all i E (1,. . . , r } . Thus G is transitive on X if and only if it is l-transitive on X .

L e m m a 3.1.

Let G act transitively on a finite set X and let H be the

814

Central, Faithful and Permutation Characters

stabilizer of x E X . Then G is ,%transitive on X if and only if H acts transitively on X - {z}.

Proof. I f G is 2-transitive on X, then clearly H acts transitively on X - {x}. Conversely, assume that H acts transitively on X - { x } . Let 51,x2 and y1,y2 be elements of X with 51 # 2 2 , y1 # y2. Then g q = y1 for some g E G. Put y = 9x2. Then y # y1 as X I # 5 2 . Now the subgroup fixing y1 is conjugate to H and hence by our assumptions there exists an element h of G fixing y1 and transforming y into y2 (as y and 5 2 are distinct from yl). The element t = hg then transforms X I into y1 and 5 2 into y2, as desired. Lemma 3.2. Let V be a permutation G G-module, let X be a basis of V on which G acts as a permutation group and let XI,. ..,X , be all G-orbits of X . If Hi is the stabilizer of xi E X; and x is the character of G aflorded by V , then n

x = C(lHJG i=l

Proof. This is a special case of Lemma 18.6.1 (ii). R

Theorem 3.3. Let G act on a finite set X , let (r: X be the corresponding permutation C G-module and let x be the character of G aflorded by G X. Then (i) The number of G-orbits in X is equal to < x,lc >. In particular, G is transitive on X if and only if < x,1~>= 1. (ii) Suppose that G is transitive on X and let H be the stabilizer of 5 E X . Then the number r of the H-orbits on X is equal to both < x,x > and the number of ( H , H)-double cosets in G. (iii) G is 2-transitive on X if and only if x = 1~ t X for some irreducible character X # 1 ~ . Proof. (i) It is clear that < x,1~ > is the dimension of the invariant subspace of (I= X . The desired assertion is therefore a consequence of Lemma 17.4.1. (ii) By Lemma 3.2, we have x = (1~)’. Hence, by Frobenius reciprocity and (i),

< X’X >=< X,(lH)G >=< XH,1H >= T Let

T be a full set of double coset representatives for ( H ,H ) in G. Then, by

4 Irreducible characters of p' and p-power degrees

815

Theorem 19.3.10,

as required. (iii) Let XI = l ~x 2 ,, . . . ,xT be all irreducible characters of G. We may assume that G is transitive on X . Hence, by (i), x = ~G+C;P=~ n;X; for some integers ni 2 0. Then, by (ii), T =< x,x >= 1 C&2n?. Since, by Lemma 3.1, G is 2-transitive on X if and only if r = 2, the result is established.

+

4

Irreducible characters of p' and ppower degrees

Throughout, G denotes a finite group. All characters are assumed to be (c characters and, for any subgroup H of G, 1~ denotes the principal character of H . Let n be a set of prime numbers. A group G is called a n-group if IGl is divisible only by primes in a. A group G is n-separable if each of its composition factors is either a n-group or a n'-group, where n' is the complementary set of prime numbers. In the special case where n = { p } , we use the term p-solvable instead of n-separable. Clearly n-separability and K'separability are equivalent. It is an easy consequence of the definitions that subgroups and homomorphic images of n-separable groups are n-separable. A Hall a-subgroup of a group G is a n-subgroup whose index in G is divisible only by primes in n'. If n = { p } , then a Hall n-subgroup is nothing else but a Sylow p-subgroup of G. It is well known (see Theorem 26.1.6) that if G is n-separable, then G possesses a Hall n-subgroup (in particular, a psolvable group possesses a Hall p'-subgroup). In general, however, Hall n-subgroups need not exist. Again, let n be a set of primes. We say that a character x of G is of ?r-degree if x(1) is divisible only by primes in n. In case ?r = { p } , we say that x is of ppower degree, instead of n-degree. Similarly in case A = { p } , we say that x is of p'-degree instead of n'-degree. As a point of departure,. we prove the following three group-theoretic results. Lemma 4.1. Let f be an automorphism of a group G. I f p is a prime such that p divides the order off but p does not divide the order of G, then f ( C ) # C for some conjugacy class C of G.

816

Central, Faithful and Permutation Characters

Proof. We may clearly assume that the order o f f is p . Let us identify G with its image in S,, n = )GI, via right regular representation of G. Then H =< G, f > is a subgroup of S, with H = G. < f >,Gn < f >= 1 and G d H . Since f-'gf = f ( g ) for all g E G, f does not commute with some go E G. The number of elements in the H-conjugacy class of go is a multiple of p . Since p does not divide the order of G, the H-conjugacy class of go is a disjoint union of p-conjugacy classes of G. The automorphism f must therefore interchange some of the conjugacy classes of G as asserted. H

A subgroup H of G is characteristic in G if f ( H ) = H for all automorphisms f of G. A group with no nontrivial proper characteristic subgroups is called characteristically simple . It is clear that if H is a minimal normal subgroup of G, then H is characteristically simple. Lemma 4.2. (i) A characteristically simple group is the direct product of isomorphic simple groups. (ii) If H is a minimal normal subgroup of G, then either H is an elementary abelian p-group for some prime p or H is the direct product of isomorphic nonabelian simple groups.

Proof. It is clear that (ii) is a consequence of (i). Let G be characteristically simple and let G1 be a nontrivial subgroup of G of least possible order. Denote by H a subgroup of G of maximal order of the form H = G1 x G2 x x Gk, where Gi Q G, Gi S GI, 1 5 i 5 k. Clearly H a G and we shall show that H = G. If f is an automorphism of G, then f(G;) 4 G and f ( G i ) G; 2 GI. Assume that f(G;) H for some automorphism f and some i E { 1,...,k}. Then f(G;) n H c f(Gi), whence If(G;) r l HI < IG1l. But f(Gi) n H a G, so f(Gi) n H = 1 by our minimal choice of GI. Hence H f(G;) = H x f(Gi) satisfies the same conditions as H , but has larger order, contrary t o the definition of H . Thus j ( G ; ) H for all i E (1,. . ,k} and all automorphisms f of G and so f ( H ) = H for all such f. Hence H is characteristic in G and, since G is characteristically simple, we deduce that G = H = GI x . . x G,. Finally, any normal subgroup of GI is a normal subgroup of G, so that G1 must be simple by our minimal choice of GI, as desired.

.

Lemma 4.3. A minimal normal subgroup of a n-separable group is either a n-group or a r'-group.

4 Irreducible characters of p’ and p-power degrees

817

Proof. Let N be a minimal normal subgroup of the n-separable group

G. Then, by Lemma 4.2 (ii), N is the direct product of isomorphic simple groups N ; , 1 5 i 5 k. Since N d G, there exists a composition series of G in which N1 is a composition factor of G. Since G is n-separable, N1 (and hence each N ; ) is either a n-group or a n’-group, as required.

Lemma 4.4. Assume that G has a Hall n-subgroup H such that each irreducible constituent of ( 1 ~ is) of ~ a’-degree. Then the same property holds for N and G I N , where N is any normal subgroup of G. Proof. Let N be any normal subgroup of G. Then H n N is a Hall n-subgroup of N and H N / N is a Hall n-subgroup of G I N . For each character x of N , x is a constituent of ( x G ) ~Hence . X’nN is a constituent of ( x ~ ) H , - , N . Furthermore, if x is a constituent of (lHnN)N, then 1HnN is a constituent of X H n N . Thus, by Frobenius reciprocity,

< x,“lH)GIN > = < X G , (ldC >=< ( X G ) H , 1H > = < ( X G ) H n N , 1HnN > ( H : ( H n N ) ) - ’ f . 0

Hence for each irreducible constituent x of (lHnp/)N, there is an irreducible constituent X of ( 1 ~ for) which ~ x is a constituent of AN and thus for which x(1) divides A(1). By hypothesis, X is of a’-degree and so x is also of ddegree. Thus N satisfies the required property with respect t o H n N . Since 1 H N is a constituent of ( I H ) ~ (1HNIG ~ , is a constituent of ( I H ) ~ . If x is an irreducible character of G with N C K e r x and x is a constituent of ( ~ H N )then ~ , x is a constituent of ( 1 ~ ) By ~ .hypothesis, x is of n’-degree and so GIN satisfies the required property with respect t o H N I N .

Theorem 4.5. (Gallagher (1962 b)). Let n be a set of prime numbers and let H be a Hall n-subgroup of a n-separable group G. Then H a G if and only if every irreducible constituent of ( 1 ~ is) of~ a’-degree. Proof. If H a G, then by Theorem 2.1 (iii), K e r ( 1 ~ = ) ~H . Hence each irreducible constituent of ( 1 ~is )an~irreducible character of G / H and so is of n‘-degree by Theorem 17.10.4. Conversely, assume that every irreducible constituent of ( 1 ~ is) of~ n’degree. Denote by M a minimal normal subgroup of G. Then, by Lemma 4.3, M is either a n-group or a n’-group. We now argue by induction on

818

Central, Faithful and Permutation Characters

IGl. If G = M , then there is nothing to prove. Hence we may assume that G # M . Then G / M is a n-separable group, which by Lemma 4.4, satisfies the same property as G. Thus, by induction, G/M has a normal Hall nsubgroup N / M . If M is a n-group, then N is a normal Hall n-subgroup of G (hence H = N 4 G). If N # G, then N is n-separable and, by Lemma 4.4, N satisfies the same property as G. Hence, by induction, N has a normal Hall n-subgroup, which is then a normal Hall n-subgroup of G. Finally, assume that M is a n‘-group and N = G . Then G = M - H and M n H = 1. Hence “lH)GIM = X(0X

c X

where x ranges over the irreducible characters of M. Let x be an irreducible character of M. Then there is an irreducible constituent X of ( 1 ~such ) ~ that x is a constituent of AM. Let K be the inertia group of x. Then, by Clifford’s theorem, (G : K) divides X(1). By the hypothesis, X is of d-degree and hence (G : K )is divisible only by the primes in n’. But (G : #) divides ]HI,so (G : K ) = 1 and K = G. Thus each irreducible character of M is G-invariant. It follows from Brauer’s permutation lemma that H fixes all conjugacy classes of M . Hence, by Lemma 4.1, H centralizes M . Thus G = M x H and so H Q G, as desired.

Corollary 4.6. Let p be a prime and let P be a Sylow p-subgroup of Q p-solvable group G. Then P 4 G if and only if every irreducible constituent of (1p)’ is ofp’-degree. Proof. This is a special case of Theorem 4.5 in which n = ( p } and H=P.. An alternative proof of the above corollary is contained in a work of Saksonov (1987).

Corollary 4.7. (Fong (1961)). Let p be Q prime and let G be a p solvable gmup. Then G has a normal abelian Sylow p-subgroup if and only if each irreducible character of G is of p’-degree.

Proof. The sufficiency is a consequence of Theorem 17.10.4. Conversely, assume that the degree of every irreducible character of G is relatively prime to p . Then, by Corollary 4.6, G has a normal Sylow p-subgroup

4 Irreducible characters of p' and p-power degrees

819

P. If P is nonabelian, then P has an irreducible character X with plX(1). But then, by taking an irreducible constituent x of A', it follows from Frobenius reciprocity and Clifford's theorem that x(1) is divisible by p, a desired contradiction. W Note that the result above holds without the assumption that G is psolvable. The corresponding proof, due to Michler (1986, Theorem 2.3), uses the classification theorem of finite simple groups.

Corollary 4.8. Let p be a prime and let G be a p-solvable group. If H is a subgroup of G and x is an irreducible character of H with p l x ( l ) , then plX(1) for some irreducible character X of G. Proof. Assume by way of contradiciton that each irreducible character of G is of $-degree. Then, by Corollary 4.7, G has a normal abelian Sylow p-subgroup P . Hence P n H is a normal abelian Sylow p-subgroup of H , contrary to Corollary 4.7. Note that in view of the Michler's result mentioned above, Corollary 4.8 is true without the assumption that G is p-solvable.

Theorem 4.9. (Berkovich (1989)). Let p be a prime, let P be a Sylow p-szlbglurup of G and let N be the intersection of the kernels of all nonlinear irreducible characters of G of p'-degree (put N = G if there are no such characters). Then N n G ' n P P' Proof. Put Q = N n G' n P and assume by way of contradiction that Q g PI. Then there exists a linear character x of P such that Q K e r X . The induced character xG has degree (G : P ) f O(modp). Let X be an irreducible constituent of xG. Then, since Q K e r x , it follows from Frobenius reciprocity that Q K e r A. Hence p(X(1) for all nonlinear , xG has irreducible constituents X of xG. But p does not divide x G ( l ) hence a constituent p of degree 1. Then p p = x and so

Thus Q

g Kerp.

Since G'

K e r p , we deduce that

NnG'nP=QgG',

820

Central, Faithful and Permutation Characters

which is a contradiction. We say that G has a normal p-complement N if N is a normal subgroup of G such that G = N P and N f l P = 1, where P is a Sylow p-subgroup of G.

Theorem 4.10. (Isaacs and Passman (1964)). Every nonlinear irreducibke character x of G has p-power degree if and only if G has a nornzal a belian p-complemen t. Proof. The sufficiency is a consequence of Theorem 17.10.4. Conversely, assume that every nonlinear irreducible character x of G has a p-power degree. If N is any normal subgroup of G, then by Frobenius reciprocity and Clifford’s theorem, X(1) is a power of p for any nonlinear irreducible character X of N . Invoking induction on IGl, it therefore suffices t o find a normal subgroup N of G of index p, because the normal abelian pcomplement for N will be one for G. We may clearly assume that G is nonabelian. Let X I , . ..,Xn be all nonlinear irreducible characters of G. Then

Since the right-hand side is divisible by p , we see that pl(G : G’). Thus G has a normal subgroup of index p , as desired. Our final aim is to provide an application of Corollary 4.6. We need the following result which is of independent interest.

Theorem 4.11. (Gatlagher (1964)). For any finite group G, the following conditions are equivalent : (i) For any elements 91,.. . ,gn E G of pairwise relatively prime orders, g1 , . .g, = 1 implies g1 = g 2 = - = gn = 1. (ii) If the restriction of an irreducible character x of G to any Sylow subgroup S of G contains ls, then x = 1 ~ .

-

Proof. Let P I , . . . ,pn be all distinct primes dividing [GI, let S; be a Sylow pi-subgroup of G and let cyi = Let XI = 1 G , ~ 2 , ... , xr be all

4 Irreducible characters of p' and p-power degrees

821

irreducible characters of G and write T

j=l

Next put

Then t 2 1 and t = 1 if and only if fly==,cy;(gi) = 0 for (91,. . . ,gn) # (1,.. , 1). Because ai is 0 on &elements, we may assume that in the definition o f t , g; ranges over the pi-elements of G. Bearing in mind that cw;(g;) > 0 for any pi-elements gi, we infer that t = 1 is equivalent to condition (i). Given an n-tuple (91,. ..,gn), put h; = 91. .gi for i E { 1 , . ..,n - 1). Then

.

-

h l ,...,hn-1

Applying Theorem 19.2.8 n - 1 times, we obtain n

t

1 = cjc ni=l zi')Xj(l)n-2 n

n

n?=,

Since, by Frobenius reciprocity, zil = 1, we see that t = 1 if and only if for j # 1 there exists i with z;j = 0. This shows that t = 1 is equivalent t o (ii), thus completing the proof.

Remark 4.12. It was conjectured by P. Hall (1937) that property (i) in Theorem 4.9 characterizes solvable groups. The truth of this conjecture was established many years later by investigation of minimal simple groups

822

Central, Faithful and Permutation Characters

classified by Thompson (see Gorenstein (1968, Chapter 16, Sec.5) or Feit (1967, p. 51)).

Corollary 4.13. (Saksonou (1987)). A finite group G is nilpotent if and only if for every Sylow subgroup S of G and every irreducible constituent x of (Is)', (X(l),lSl>= 1.

Proof. If G is nilpotent, then G obviously satisfies the required property. Conversely, assume that for every Sylow subgroup S of G and every ) , = 1. If x is an irreducible irreducible constituent x of (Is)', ( ~ ( lISl) constituent of (1s)' where S ranges over all Sylow subgroups of G, then x(1) = 1 since (x(l),IG() = 1. But then obviously x = l a . Thus, by Frobenius reciprocity, G satisfies property (ii) in Theorem 4.11. Hence, by Remark 4.12, G is solvable. Applying Corollary 4.6, we deduce therefore that each Sylow subgroup of G is normal. Hence G is nilpotent and the result is established. I

Chapter 22

Character Tables This chapter contains a discussion of some of the questions about finite groups G which can b e answered if the characters of G are known. Namely, we are concerned with group-theoretic information which can be extracted from a character table. We also discuss in detail the effect of Galois actions on character tables. More precisely, let E be a primitive n-th root of 1, where n is the exponent of a group G and let F be a subfield of a . Then the group G a l ( F ( & ) / F ) acts on the rows and columns of the character table of G in an obvious manner. It is natural to enquire whether the permutation group induced by GaZ(F(&)/F)on the rows of the character table of G is permutation isomorphic to the permutation group induced by GaZ(F(&)/F)on the columns of the character table. A positive answer is provided in case GaE(F(&)/F) is cyclic. In general, however, we show that the answer is negative. The corresponding result, due to Thompson (1970a), is derived as a consequence of a nonduality theorem discovered by Thompson (1970a).

1

Group information

In this section, we discuss some of the questions about groups G which can be answered if the characters of G are known. All characters below are assumed to be (L: -characters and all groups are finite. We begin by fixing the following notation : XI,.. . ,xr are all irreducible characters of G. C1,. ..,C, are all conjugacy classes of G ga E c; l, let 8i be the restriction of wi to Z ( F G ) . If z E Z ( F G ) and O;(zfi) = 0, then uj(zfj) = wj(zej) = 0 , so nej = 0. Thus, for each u E G a l ( F ( & ) / F ) , u ( z e ; ) = 0 = z a ( e j ) and so z f j = 0. Since O;( f j ) = 0 for all j # i, we obtain Z ( F G ) f ; E F;, as desired. W

Corollary 2.4. Let F be a subfield of (c and let E be a primitive n-th root of unity, where n is the exponent of G . If the permutation group induced by G a l ( F ( & ) / F on ) the rows of the character table of G is permutation isomorphic to the permutation group induced by G a l ( F ( & ) / F on ) the columns of the character table, then

Proof. We keep the notation of Theorem 2.3. By hypothesis, the permutation group induced by Gal(F(E)/F)on the irreducble (c -characters of G is permutation isomorphic to the permutation group induced by G a l ( F ( & ) / F ) on the conjugacy classes of G . Thus, we may assume that for each i E (1,. ., ,s}, xj and Ci have the same fixed group H i . Hence 4 = Lj is the fixed field of H; and the result follows by virtue of Theorem 2.3. W Corollary 2.5. (Thompson (1970 a)). If p is an odd prime and G is a p-group, then for any subfield F of (c ,

C ~ F ( G2) Z( F G ) Proof. Let E be a primitive n-th root of 1, where n is the exponent of G. Then Gal(F(E)/F)is cyclic. Now apply Corollaries 2.2 and 2.4. H Theorem 2.6. (Thompson (1970 a)). There exist finite 2-groups G of arbitrarily large derived length such that

Proof. Let S be a free group on n 2 2 generators. Put S = SO and let

Sm+l=< x212 E

s, >

(rn = 0,1,2,. * .)

Character Tables

832

Define G by G = G m = Gm(n) = S/Sm. Then G is a finite 2-group and GIG' is the direct product of n cyclic groups of order 2m. Furthermore, the exponent of G is 2m. Now, if H = Sl/S,,, and h E H , then h2m-1 = 1. On the other hand, if g E G - H, then since H is the intersection of all subgroups of G , gG' is not a square in G/G', so gG' has order 2m in GIG'. In particular, g has order 2m. For each g in G, let K ( g ) be the field generated over Q by all x ( g ) where x ranges over I r r ( G ) . For each integer t 2 1, let Q t be the field obtained from Q by adjoining a primitive t-th root of unity. Then, by the previous remarks,

The last equality holds since as x ranges over the linear characters of G, x ( g ) ranges over all 2m-th roots of unity. Finally, choose m 2 3. Because the dihedral group of order 2m-' is a homomorphic image of G, G has an irreducible character which determines the real subfield of Q 2m. But this field is neither Q 2m nor is contained in Q 2m-1. Thus, by Theorem 2.3, C ~ (QG ) Z ( Q G ) . Since the Sylow 2-subgroups of the general linear group of degree n over the field of two elements are homomorphic images of G,(n), we see that the derived length of G n ( n ) goes to infinity with n. So the theorem is true.

Corollary 2.7. (Thompson (1970 a)). There exists a finite 2-group G such that if E is a primitive n-th root of 1, n = exponent of G , then the permutation group induced by Gal(Q ( E ) / Q ) on the rows of the character table of G is not permutation isomorphic to the permutation group induced by Gal(Q ( & ) / Q) on the columns of the character table. Proof. Apply Theorem 2.6 and Corollary 2.4.

3

Character tables for A5 and S5

We next provide the character table for the simple nonabelian group of smallest order, namely the alternating group A5 of degree 5 .

Proposition 3.1. If G = A5, then 1, (12)(34), (123), (12345) and (13524) are representatives of the conjugacy chases of G and the character

3 Character tables for A5 and S5

833

Proof. The assertion concerning the conjugacy classes of G is obvious. Let us also note that the number of elements in the conjugacy classes of G are 1, 15, 20, 12 and 12, respectively. The character x1 is of course the principal character of G. Now G is 2transitive on { 1,2,3,4,5} and we denote by x the corresponding permutation character of G. Then, by Theorem 21.3.3, x2 = x - x1 is an irreducible character of G of degree 4. By counting the fixed points of (12)(34), (123), (12345) and (13524), we see that the values of x 2 are given by the second row. The group H =< (12345) > is a Sylow 5-subgroup of G and N G ( H ) =< (12345),(14)(23) >, (G : N G ( H ) )= 6 The group G acts on the set X of left cosets of N G ( H ) with respect to G. Since H acts transitively on the 5 nontrivial cosets in X, it follows that G is 2-transitive on X. Let x be the corresponding permutation character of G. Then, by Theorem 21.3.3, x3 = x - x1 is an irreducible character of G of degree 5. An easy calculation shows that the values of x3 are given by the third row. Counting the character degrees n;, we have 5

60=Cn:= 1 2 + 4 2 + 5 2 + n : + n i i= 1

Hence n? t ni = 18 which implies n4 = Theorem 19.2.3 (iii),

715

= 3. If g = (123), then by

5

3 = ICG(g)I =

C Ixi(g)12 i=l

= 3 -tIX4(S)l2

+ lX5(S)l2

Character Tables

834

Thus x4(g) = x5(9) = 0. Let p; be the representation of G which affords x;,i = 4,5, and let g = (12)(34). Then d e t p i ( g ) = 1 and the eigenvalues of p ; ( g ) are 1,-1,-1. Thus x4(g) = x5(g) = -1. Finally, let g = (12345) and h = (14)(23). Since h-lgh = g-l, we have

and hence x ; ( g ) is real. Similarly, if g = (13524), then x ; ( g ) is real. Now fix i E {4,5} and put X = xi((12345)) and p = xi((13524)) Then, by orthogonality relations, we have 0 = IGI < Xi,Xl >=

c

Xi(4

xEG

= 3 - 15

and 60 = [GI < x ; , x i

-+ 12X + 1

2 ~

>= 9 +- 15 -t 12X2+- 12p2

Thus we must have X -t p = 1, X2 -t p2 = 3 which implies X = (1/2)(1 t &), p = (1/2)(1 p = (1/2)(1 -t 6). This concludes the proof.

- 6) or X = (1/2)(1 - A),

We close by providing the character table of the symmetric group S5 of degree 5. Proposition 3.2. If G = S5, then 1, (12), (123), (l2)(84), (123)(45), (1234) and (12345) are representatives for the conjugacy classes of G and the character table of G is given by 1 x1 x2

: 1 : 1

x 3 : 4 x4

:

4

5 x6 -' 5 x r : 6 x5

*.

(12) 1 -1 2 -2 -1

(128) 1 1 1 1

(1234) 1

(12345) 1

-1

1 -1 -1 0

1

-1 -1

0 0 1 -1

0

0

0

0

1

3 Character tables for As and SS

835

Proof. The statement concerning the conjugacy classes of G is obvious. Observe also that the number of elements in the conjugacy classes of G are 1, 10, 20, 15, 20, 30 and 24, respectively. The character x1 is the principal character of G, while x2 is a faithful linear character of &/As. It is clear that if x is an irreducible character of G , then so is x x 2 . The group G is 2-transitive on {1,2,3,4,5}. If x is the corresponding permutation character of G, then by Theorem 21.3.3, x 3 = x - x1 is an irreducible character of G of degree 4. The values of x3 are clearly given by the third row. Setting x4 = ~ 3 x we 2 obtain another irreducible character of G of degree 4 whose values are given by the fourth row. The group H =< (12345) > is a Sylow 5-subgroup of G and

NG(H)=< (12345), (1243) >, (G : N G ( H ) )= 6 By the argument in the proof of Proposition 3.1, G is 2-transitive on the left cosets of N G ( H ) in G. This provides us with the irreducible character x 5 given by the above table. The character x6 is given by x6 = x S x 2 . Since 120 = ~ ; ( 1= ) ~84 4-~ 7 ( 1 ) we ~ , have x7(1) = 6. By looking at the character degrees of A5 and applying Clifford's theorem, we see that x7 = xG where x is any irreducible character of As of degree 3. Thus the values of x7 are given by the last row and the result follows.

xi'=,

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Chapter 23

Zeros of Characters This chapter is devoted entirely to the study of zeros of C -characters x of a finite group G, i.e. to the study of those elements g E G for which x(g) = 0. Our first result, due to Gallagher (1966), asserts that if H is a subgroup of G and x is an irreducible character of G with x(g) # 0 for all g E G - H , then XH is irreducible and, for any g E G - H,x(g) is a root of unity. As an immediate consequence, we derive a classical theorem of Burnside (1903), which shows that any nonlinear irreducible character of G has at least one zero. Let H be a subgroup of G and let x be an irreducible character of G such that x ( g ) = 0 for all g E G not conjugate to elements of H . What information can be obtained about such a character ? One of the results presented (Ferguson and Isaacs (1989)) shows that X H is reducible and that there exists a positive integer n and a generalized character X of H such that X~ = nx. A separate section is devoted to counting the number n(x) of zeros of a given irreducible character x. For example, it is shown that n(x) 2 ( ~ ( 1 -) ~l)lZ(x)l, and n(x) 2 x ( l ) z ~ Z ( x )unless l 1x1 takes only values 0, 1 and ~ ( 1 ) .We also present a powerful condition for a character value to be zero. Namely, we establish a classical result, due to Brauer and Nesbitt (1941), which asserts that if the degree of an irreducible character x of G is divisible by the order of a Sylow p-subgroup of G, then x(g) = 0 for each g E G whose order is divisible by p . A separate section is devoted to some recent results due to Zmud. In particular, we prove an elegant theorem, due to Zmud (1977, 1990), which asserts that if G is a nonabelian group, then G is generated by the zeros of 837

Zeros of Characters

838

a suitable irreducible

(I: -character

of G.

1 Burnside’s theorem Throughout this section, G denotes a finite group and all characters are assumed to be (I:-characters. Here we demonstrate that if x is an irreducible nonlinear character of G, then x ( g ) = 0 for some g E G. This will be derived as a consequence of a more general result which will be needed later.

Lemma 1.1. Let a1,. . . ,an be the complex roots of a monic polynomial f ( X ) E Z [XIof degree n 2 1 and let Ia;I = 1 for each i E (1,. ,n } . Then each cri is a root of unity.

..

Proof. For each k 2 1, put fh(X)= nbl(X- at). Then b ( X ) E Z [XIsince, up to the sign, the coefficients of fk(X)are the elementary symmetric functions of at,. .. ,a! and each of these functions is a polynomial

.

of the elementary symmetric functions of a1 ,.. ,on. Since ~ ( Y=~1I for each i E {l,...,n } , we see that the coefficients of various fk(X) are uniformly bounded. Hence there are only finitely many such polynomials. Therefore there are only finitely many roots of these polynomials. Hence some two distinct positive powers of a; must coincide, as desired.

Lemma 1.2. Let (Y # 0 be an algebraic integer with all conjugates a1, . . . ,a, (each repeated the same number of times). Then T

i=l

with equality if and only if

(Y

is a root of unity.

Proof. We know that the arithmetic mean of positive real numbers is at least the geometric mem and that the equality occurs if and only if these numbers are equal. In particular, l-

In:=’=,

with equality if and only if all lail are equal. But a;[is a positive rational integer, so (1) holds with equality if and only if all lail are equal

1 Burnside’s theorem

839

nr==,

and I ail = 1. Since the latter is equivalent to all [ail= 1, the result follows by virtue of Lemma 1.1.

Lemma 1.3. Let g E G be an element of order rn, let

c = {s”I(P,m) = 1) and let

x be a character of G with x(g) # 0.

c

Then

lx2(x)l 2 ICl

x€C

with equality if and only if x(g) is a root of unity.

Proof. Let E be a primitive rn-th root of unity. Then for any p 2 1 with ( p , r n ) = 1, we have

x2(sP> = q.t(x2(s>> where up E G a l ( Q ( E ) / Q ) is defined by up(&)= E P . Hence {x2(x)1x E C} are all conjugates of the aIgebraic integer x2(g) # 0 (each repeated the same number of times). Since x2(g) is a root of unity if and only if x(g) is a root of unity, the result follows from Lemma 1.2. 4

Theorem 1.4. (Gallagher (1966)). Let H be Q subgroup of G and let x be a n irreducible character of G such that x(g) # 0 for all g E G - H . Then X H is irreducible and, for any g E G - H , x(g) is a root of unity. Proof. We have < x , x >= 1 and < X H , X H >2 1 with equality if and only if XH is irreducible. The equivalence relation x-y,=,y>

on G

partitions G-H into disjoint subsets, say G-H = UfZ1Cj. Then, by Lemma 1.3, IGI =

c 1x2(s)l c =

g€G

hEH

2 IHI

cc t

Ix2(h>> i-

Ix”g>l

i = l gEC,

+ IG - HI = /GI

Hence C h E H 1x2(h)l = IHI and &C, 1x2(g)1= ICiI,1 I i I t. Thus XH is irreducible and, by Lemma 1.3, x(g) is a root of unity for all g E G - H . H

Zeros of Characters

840

Corollary 1.5.

(Burnside (1903)). Let

x

be an irreducible character

of G with x(1) # 1. Then x(g) = 0 for some g E G .

Proof. Apply Theorem 1.4 for H = 1. W

2

Characters vanishing off subgroups

Let H be a subgroup of G and let x be an irreducible character of G such that x ( g ) = 0 for all g E G not conjugate to elements of H. What conclusion can be drawn about such a character x? In this section we show that XH is reducible and that there exists a positive integer n and a generalized character X of H such that XG = nx. The proof of the following result is due t o D. Gluck (see Isaacs (1983)).

Theorem 2.1. Let H be a subgroup of G and let x be an irreducible character ofG. Assume that x(g) = 0 for all g E G not conjugate to elements of H . Then XH is reducible.

Proof. By Proposition 19.1.7 (ii), (xH)' = ( 1 ~ x. ) ~Furthermore, 2 1 whenever x(g) # 0 (by hypothesis) and

(lH)'(g)

( 1 ~ )1)~=( (G : H ) > 1 Applying Frobenius reciprocity, we then have

as asserted.

Our next aim is to demonstrate that if x is an arbitrary character of G satisfying the hypothesis of Theorem 2.1, then there exists a generalized character X of H and a positive integer n such that nx = XG. The following lemma due to Ferguson and Isaacs (1989) will clear our path.

2 Characters vanishing off subgroups

841

Lemma 2.2. Let a be a generalized character of G such that a(g) # 0 for all g E G and let s = o ( g ) , where g runs over a set of representatives for the conjugacy classes of G. Then s E Z and s/a,defined by ( s / a ) ( g )= s a ( g ) - ' , g E G , is a generalized character of G.

n,

Proof. Our hypothesis on a ensures that Q has an inverse in the ( c space c f ( G ) of class functions of G. Hence the map r : c f ( G ) + c f ( G ) defined by I'(cp) = ya is a nonsingular linear transformation of cf(G). The matrix A of r with respect to the basis Irr(G),consisting of all irreducible characters of G , has entries in Z . On the other hand, the characteristic functions of the conjugacy classes of G constitute another basis for c f ( G ) . With respect to this basis, the matrix of I? is diagonal, with entries equal to the various a(g). Thus s = det(I') = d e t ( A ) E Z . Let A* denote the adjoint of A so that A-' = s-'A*. Then A* has integer entries since so does A . Hence sA-' has integer entries and therefore S r - ' ( l G ) is a Z -linear combination of I r r ( G ) . Because S r - ' ( l G > = s / o , the result follows. H

Theorem 2.3. (Ferguson and Isaacs (1989)). Let x be a character of G and let H be Q subgroup of G such that x(g) = 0 for all g E G not conjugate to elements of H . Then there exists Q positive integer n and a generalized character X of H such that XG = nx. ) ~see, that a ( h ) # 0 for all h E H . Now Proof. Setting a = ( l ~ we define the class function P on G by

Then both Pxa and x are zero on elements g E G which are not conjugate to elements of H and P ( g ) a ( g ) = 1 for the remaining g E G. Thus

x

= =

Pxa = ( P x ) ( W G ( ( P x ) H ~ H ) ~ (by Lemma 19.3.2

(i))

= (PHXH)~ On the other hand, by Lemma 2.2, n P H is a generalized character of H for some positive integer n. Thus = n & x H is a generalized character of H such that XG = nx, as desired.

Zeros of Characters

842

3

Gallagher's theorems

The results of this section will require a considerable amount of notation. Therefore it will be especially useful to assemble most of it in one place. In what follows G denotes a finite group and all characters are assumed to be (I:-characters. Moreover the following notation will be used : Irr(G) is the set of all irreducible characters of G. n(x) is the number of elements g E G such that x(g) = 0, x E Irr(G). n(g) is the number of x E Irr(G) such that x(g) = 0, g E G. Z(X) = (9 E GI Ix(s>l= X ( W K(g) is the smallest normal subgroup of G containing {[g,z]Iz E G} r ( G ) is the number of conjugacy classes of G.

Theorem 3.1. (Gallagher (1962 a)). With the notation above, we have (i) For each x E Irr(G), n(x) L ( ~ ( 1 )-' 1)lZ(x)l. (ii) For each x E Irr(G),n(x) 2 x(l)'lZ(x)l unless 1x1 takes only values 0, 1 andx(1). (iii) For each g E G, n(g) 2 r(G) - ICG(g)l (iv) For each g E G, n(g) = r(G)- IcG(g)l if and only af K(g) = G' and Ix(g)l = 0 or 1 for all x E Irr(G).

Proof. (i) and (ii). Given x E Irr(G), it follows from Lemma 1.3 that

with equality only if Ix(g)l takes only the values 0, 1 and ~ ( 1 )Because . n(x) is a multiple of IZ(x)l, inequality implies n(x) 2 x(l)'~Z(x)l, as desired. (iii) and (iv) For each g E G, we have

where the first sum is over those x with Ix(g)l = x( 1) and the second sum is over the others. Note that g E Z(x) if and only if x is a character of G/K(g). Hence the first sum is IG/IC(g)l and the number of characters in the first

3 Gallagher's theorems

sum is r(G/K(g)).Thus, if X I , . . then t = r ( G )- r ( G / K ( g ) )and

843

. ,xt are the characters in the second sum,

Let E be a primitive m-th root of unity, where m is the exponent of G. Then the group Gal(Q ( e ) / Q ) acts on the set { X I , . . . ,x t } via

If S1,.

..

)

Sk are the orbits of { X I , . . . , x t } , then by Lemma 1.2,

with equality only if Ix;(g)l takes only the values 0 and 1. Hence, by (l),

with equality only if Ix(g)l takes only the values 0, 1 and x ( 1 ) for each x E I r r ( G ) . Because IG/K(S)I 2 r ( G / K ( g ) ) with equality if and only if K ( g ) = G', it follows that

with equality if and only if K ( g ) = G' and Ix(g)l = 0 , l or x ( 1 ) for each x E I r r ( G ) . But 1x(g)1 = x(1) only for the characters of G / K ( g ) ,which is only for x(1)= 1 in this case. This completes the proof of the theorem. H The following assertion is a part of Clifford theory which is a subject in itself and will be examined at a later stage. However, since the result is of a very elementary nature, its verification can be achieved by an ad hoc argument which we present below.

Lemma 3.2. Let N be. a normal subgroup of G such that GIN is cyclic of order n. If x is a G-invariant irreducible character of N , then there exist precisely n irreducible characters of G extending x and xG is their sum.

Zeros of Characters

844

Proof. By Frobenius reciprocity, the number of irreducible characters of G extending x cannot exceed n . On the other hand, if X is a character of G extending x , then A is irreducible; hence so are Ap1,. . .,Apn (where p 1 , . . . ,pn are all linear characters of GIN) since each Api extends x . Furthermore, each Ap, is a constituent of xG by Frobenius reciprocity and ~ ~ (= 1n ( )X p j ) ( l ) ,so xG = C7=lXpj. Thus it suffices to show that x can be extended to a character of G. Let 'l : N + GL,((L:) be an irreducible matrix representation of N which affords x and let g E G be such that G I N =< gN >. Since g x = x and g x is afforded by the matrix representation gI' given by ( T ) ( n )= I?(g-lng) for all n E N , there is a matrix p ( g ) E GL,((C ) such that p ( g ) - l r ( n ) p ( g )= r ( g - l n g ) for all n E N . Then

.

(for all n E N, i E (1,. . ,n)) P(g)-ir(n)P(s)' = Replacing p ( g ) by its scalar multiple, if necessary, we may assume that p ( g ) n = r(n0) where g n = no E N . Defining I?* : G + GL,(C) by r * ( g j , ) = p(g)jr(n)

it follows that

(0

5 j 5 n - 1, n E N )

I'* is a representation of G which extends I?, as desired.

Theorem 3.3. (Gallagher (1966)). Let N be a normal subgroup o f G with N # G and let G j i x each conjugacy class of N. Then, f o r any nonlinear irreducible character x of G, there exists g E G - N such that x ( g ) = 0 .

Proof. Assume by way of contradiction that x is a nonlinear irreducible character of G with x ( g ) # 0 for all g E G - N . Put H =< g l N > where g1 E G - N . Then H / N is cyclic, H fixes each conjugacy class of N and x ( g ) # 0 for all g E G - H . Hence, by Theorem 1.4, X H is irreducible and so we may assume that GIN is cyclic of order n. Let XI,. . . ,xr be all irreducible characters of N . By Corollary 17.4.3, each xi is G-invariant since G fixes all conjugacy classes of G. Hence, by Lemma 3.2, each xi extends in exactly n ways to characters x j j , 1 5 j 5 n, of G and each irreducible character of G is precisely one of the xij, 1 5 i 5 r , 15jSn. By Theorem 1.4, x ( g ) is a root of unity for all g E G - N . Hence the character x X - 1 vanishes off N (here x ( g ) = x ( g ) for all g E G). Thus < X X - 1, x;j > depends only on i and so

4 Applications and related results

045

for suitable integers ai. By Corollary 1.5, x ( g ) = 0 for some g E G. Hence g E N and evaluating (2) at g, we get

naiX;(g)= 0 (mod n),

-1 = i

a contradiction.

A strengthened version of the result below is given by Theorem 5.11 to be proved in Sec. 5 . Corollary 3.4. Each nonabelian group G is generated b y the set of all zeros of all of its irreducible characters.

Proof. Let N be the subgroup of G generated by all of the zeros of all irreducible characters of G. By Theorem 1.4, the restriction of each irreducible character of G to N is irreducible. It follows that G fixes the characters of N and hence G fixes the conjugacy classes of N . Because no irreducible character of G has zeros off N ,it follows from Theorem 3.3 that N = G, as required.

4

Applications and related results

We first present the following remarkable consequence of Corollary 1.5 discovered by Brauer (1954) and Wielandt.

Theorem 4.1. Let C1, ...,CT be all conjugacy classes of G and let

CT = Cscc, g , 1 5 j 5 r . Then G = G' if and only if r

t

Proof. Let x1 = l ~~ 2, , .. . ,xT be all irreducible characters of G and let w1,.. . ,wT be the central characters of C G given by Theorem 21.1.1 (i). Then, by Theorem 21.1.1 (iv) and the orthogonality relation

j=1

j=1

Zeros of Characters

846

Let m E Q be defined by 21.1.1 (iv),

'&lCjl

= m n j Z 1ICjl. Then, by Theorem

n

w i ( C c?)= m j=1 lCjl= wl(mjn c:) =1 r

r

t

j=1

(3)

Applying (2) and (3), we see that (1) holds if and only if ui(& CT) = 0 for i # 1, or, equivalently, for any i # 1 there exists zi E G with xi(zi) = 0. But xj(zi) = 0 implies xi(1) # 1; hence by Corollary 1.5, (1) is equivalent to the fact that x1 is the only linear character of G, or equivalently G = G'. This completes the proof of the theorem. Our next result is extracted from a classical paper of Feit and Thompson (1963, Lemma 4.3).

Proposition 4.2. Let N be a normal subgroup of G and let x be an irreducible character ofG with KerX 2 N . If g E G is such that CG(g)nN = 1, then x(g) = 0.

Proof. Let XI, X2,. . . ,At be all irreducible characters of GIN and let XI,~ 2 , . .. ,xs be the remaining irreducible characters of G. If C G ( g ) n N = 1, then the natural homomorphism G GIN induces an injective homomorphism cG(fJ) --+ C,,N(gN). Thus, by Theorem 19.2.3 (iii), --f

t

i=l

j=1

.

Hence xj(g) = 0 for all j E { 1,.. ,s}, as required.

As a preliminary to the next result, let us record the following elementary fact. Lemma 4.3. Let R be a commutative ring of characteristic 0 and let n,m be coprime positive integers. If r E R is such that n divides mr, then n divides r.

Proof. By hypothesis, mr E Rn and 1 = an Hence T = Tan rpm E Rn, as desired.

+

+ pm for some a ,p E 23 .

5

Zmud’s theorems

847

We close by providing the following powerful condition for a character value to be zero. Theorem 4.4. (Brauer and Nesbitt (1941)). Let x be a n irreducible character of G and suppose that p is a prime such that x(1) is divisible by the order of a Sylow p-subgroup of G. Then x(g) = 0 for each g E G whose order is divisible by p .

Proof. (Gallagher (1966)). Denote by a the class function on G given by

{

4 7 ) =

;(g)

if p divides the order of g otherwise

We must show that a = x. Because 0 < < a,a >I< x,x >= 1, with equality if and only if a = x, it suffices to show that a is a generalized character of G. By Theorem 20.2.1 (ii), it therefore suffices to prove that < ~ E , X > E Z for each elementary subgroup E of G and each irreducible character X of E . For each such E , we have E = P x Q, where P is a p-group and Q is a group of order prime to p . Hence a(.) = 0 for all 5 E E - Q and QQ = XQ. Therefore, for each irreducible character X of E ,

< a E , x >= I P I - ’


(4)

For z E Q, we have Cc(.) 2 P, so (G : CG(Z))divides (G : P ) . On the (in the other hand, by Theorem 21.1.1 (iv), x(1) divides x(s)(G : CG(S)) ring of algebraic integers of Q (x)). Hence x( 1) divides x(z)(G : P) and so (P(x(1)divides IGlx(z). From the hypothesis that x(1) is divisible by the order of a Sylow psubgroup of G it now follows from Lemma 4.3 that ]PI divides ~ ( s for ) all x E Q. Hence IPI divides IQI < XQ,XQ > and so IPI divides < XQ, XQ >. Thus, by (4), < a ~A , > is an integer as required. W

5

Zmud’s theorems

Throughout, G denotes a finite group. Our aim is to obtain sharpened versions of Theorem 1.4 and Corollary 3.4. One of the main results (Theorem 5.11) asserts that if G is nonabelian, then G is generated by the zeros of a suitable irreducible (c -character of G. If G is nonabelian and nilpotent, then it is shown that G is generated by the zeros of each irreducible nonlinear (c character of G.

Zeros of Characters

a48

Lemma 5.1. Let p be a prime and let g E G be of order pn with n > 0. Then, for any C -character a of G and any x E C G ( g ) ,

a(gz)P" where R = Z [ E ] and

E

=a(q"

(modpR)

is a primitive complex pn-th root of unity.

Proof. Put H = C G ( g ) and write CIH = a1 t - - t . aT,where each ai is an irreducible (I: -character of H . Since g E Z ( H ) , we have a i ( g ) = E ; ( Y ; ( ~ ) where E i is a p"-th root of unity. Since x E H , it follows that q ( g x ) = E ~ o ~ ( z )(1 5 i 5 r ) We deduce therefore that t

a(gz)P"

t

= ( c a i ( g z ) ) p " = (CE;ai(z))P" i=l

i=l

as required.

In what foIlows, the subgroup generated by an empty set is understood to be 1.

Lemma 5.2. Let p be a prime, let x be an irreducible (I: -character of G and let N be the (necessarily normal) subgroup ofG generated by the zeros of x. (i) If G has a p-element g with Ix(g)l = 1, then x(1)= f l ( m o d p ) . (ii) If G # N and p divides [ G I N / ,then x ( 1 ) 1 f l ( m o d p ) . (iii) ( ~ ( l) )G,/ N ) = ) 1.

Proof, (i) We may assume that the order of g is pn with n a = xx and 5 = 1 it follows from Lemma 5.1 that a(g)P"

> 0.

Setting

= a(1)P" = a( 1 ) ( m o d p R )

where the last congruence holds since for any integer a, up" ~ ( m o d p ) . Now a ( g ) = lx(g)I2 = 1, a(1) = ~ ( 1and ) ~so ~ ( 1 E) 1 ~( m o d p R ) . Hence

5 Zmud's theorems

849

~ ( 1E)1~ ( m o d p ) and therefore x(1) = f l ( m o d p ) , proving (i). (ii) Let P be a Sylow p-subgroup of G. Since p divides \GIN\, we have P N. Hence we may choose g E P - N . Then, by Theorem 1.4 (applied to H = N ) , Ix(g)l = 1. The desired assertion is now a consequence of (i). then x(1) (iii) We may assume that G # N . If p divides [GIN[, f l (rnodp) by virtue of (ii). But then p [ ~ ( l as ) ,required. H

=

Lemma 5.3. Let H be a proper subgroup of G and let x be a n irreducible (1: -character of G such that x = +G for some character of H and such that the zeros of x generate a proper subgroup N of G. Then (i) G = N H and H n g-lHg c N n H for all g E G - H . (ii) nseGg-lHg E N n H .

+

Proof. (i) Since x(1) = (G : H)+(l), it follows from Lemma 5.2 (iii) that ((G: H),(G : N ) ) = 1 Since (G: N H )divides both (G: H )and (G : N ) , we deduce that G = N H . Hence GIN E H / H o , where HO = H n N , and therefore, since G # N , we have Ho # H . Now put m = (G: H ) so that

mlHol= IN1

(1)

Next put S = G - N and let g E S. Then, since x ( g ) = ]HI-'

+(tgt-')

(+(z) = 0

for z E G - H )

tfG

and since x(g) # 0, we have g E t - * H t for some t E G. Hence, setting H t = t - l H t , we have

s c utecHt

Since S is a G-invariant subset of G,we have

s = S n(

u ~ ~ H = ugl(s ~ ' ) nP

i )

= u"a=1 (S n H ) ~ ;

where t l , . . . ,t , is a right transversal for H in G.Applying (1) and the fact that S n H = H - Ho, we deduce that

Zeros of Characters

850

= m(lHI - IHol) = IGl -IN1 = IS1

whence I UEl (S n = Czl I(S n H)ttl. Hence the sets (S n H ) t l , 1 5 i 5 m, are mutually disjoint. We may, of course, assume that t l = 1. Then, setting X = S n H , we have X n = 8 for all i > 1. Finally, fix g E G - H . Then g E H t i for some i > 1. Since X is H-invariant, we have X n Xg = X n X t l = 8. Taking into account that Ho n X9 = X n Hog = 8, we deduce that H)tlI

Xtl

H n H g = (Houx)n(H;uXg) = (Ho n H i ) u ( H n~ xg)u ( X n H;) u (Xn x = H ~ ~ H : c_ Ho,

g )

as required.

(ii) This is a direct consequence of (i). H

Lemma 5.4. Let x be an irreducible (I: -character of G such that the zeros of x generate a proper subgroup N of G and let M be a normal subgroup of G which is not a proper subgroup of N. Then X M is irreducible and, under the stronger assumption that A4 N, the zeros of X M generate a proper subgroup of M .

Proof. If M = G, then the assertion is trivial and if M = N then the assertion is a consequence of Theorem 1.4. We may therefore assume that M N and M # G. By Clifford's theorem (Proposition 19.1.12), X M = e(A1+ * .+A,) where XI = X is irreducible, A 1 , . . . ,A, are all distinct conjugates of A, e is the ramification index and n = (G : H ) where H is the inertia group of A. Assume that n > 1. Then H # G and by Clifford's theorem, x = t,hG for some character $ of H . Hence, by Lemma 5.3 (ii), a

a contradiction. Thus n = 1 and so X M = eX. P u t S = G - N . S i n c e M g N , w e h a v e M n S + @ . F i x g E M n S . Since x(g) = e$(g) and, by Theorem 1.4, x(g) is a root of unity, then e divides 1

5 Zmud's theorems

851

in the ring of algebraic integers. Hence e = 1 and so X M = A is irreducible. Finally, put T, = ( 9 E GIX(g) = 0 } , Tx{g E GIX(g) = 0) and let Ir' be the subgroup of M generated by Tx. Because Tx = M n T,, we have

Thus K

#M

and the result follows.

We are now ready to prove the following theorem which is a strengthened version of the first assertion of Theorem 1.4. (Zrnud (1990)). Let x be an irreducible a! -character of x genemte a proper subgroup N of G and let M be a subnormal subgroup of G which is not a proper subgroup of N . Then X M is an irreducible character of M and, under the stronger assumption that M N , the zeros of X M generate a proper subgroup of M .

Theorem 5.5.

G such that the zeros of

-

Proof. Let M = Gk a - - 4 GI a G be a part of a composition series for G passing through M . As in the proof of Lemma 5.4,we may assume that M # G a n d M g N. S i n c e M g N , w e h a v e G j g N f o r a l l i E ( 1 , . . . , I c } . In particular, since G1 d G and G1 g N , it follows from Lemma 5.4 that X G is ~ an irreducible

character of G1 whose zeros generate a proper subgroup of G I . Since G2 d G1 and G2 K where Ir' is generated by the zeros of X G ~ it, follows similarly that xc2 is an irreducible character of G2 whose zeros generate a proper subgroup of G2. Continuing in this way, we finally conclude that X M is an irreducible character of G whose zeros generate a proper subgroup of M. This completes the proof of the theorem.

Corollary 5.6. Let x be an irreducible (c -character of G such that the zeros of x generate a proper subgroup N of G and let M be a subnormal subgroup of G with M g N . Then X M ~ Nis an irreducible character of MnN. Proof. By Theorem 5.5, X = X M is an irreducible character of M whose zeros generate a proper subgroup K of M . Since M n N is a subnormal subgroup of M which is not a proper subgroup of K , it follows from Theorem 5.5 that X M ~ N is an irreducible character of M n N. Finally, since

Zeros of Characters

852

the desired assertion follows.

Let x be an irreducible U2 -character of G whose zeros generate a proper subgroup of G. If M is a maximal normal subgroup of G , then X M is irreducible. Corollary 5.7.

Proof. This is a direct consequence of Theorem 5.5. H Corollary 5.8. Let x be an irreducible nonlinear (c -character of G whose zeros generate a proper subgroup N of G and let M be a normal subgroup of G which does not contain zeros of x. Then M C N .

Proof. Assume by way of contradiction that that M is not a proper subgroup of N . Then, by Lemma 5.4, X M is irreducible. Since x(1) # 1, x ( g ) = 0 for some g E M (by Corollary 1.5). Since this is contrary to the assumption that A4 does not contain zeros of x, the result follows.

Let x be an irreducible nonlinear CI: -character of a nilpotent group G. Then G is generated by the zeros of x, Corollary 5.9.

Proof. Since G is an M-group (Corollary 18.12.4), x is induced from a linear character of a proper subgroup. Hence x is induced from a character of a maximal subgroup M of G which must be normal since G is nilpotent. Then X M is reducible and therefore the desired assertion follows from Corollary 5.7. W

As a preliminary to our next theorem, we record the following result. Lemma 5.10. Let x be an irreducible C -character of G and let x(x) = 0 for some x E G . Then CG(X) c N where N is the subgrozsp of G generated by the zeros of x. Proof. We first show that CG(X) C N. Assume by way of contradiction that &(x) N. Then N does not contain at least one Sylow subgroup of CG(Z).Hence there exists a prime p and a Sylow p-subgroup P of CG(Z) such that P N . Let g E P - N . Then sg fZ N and so, by Theorem 1.4, x ( z g ) is a root of unity. Let o(g) = p". Setting cr = xx, it follows from

5

Zmud's theorems

a53

Lemma 5.1 that a(gz)P"

= cr(5)P"

(modpR)

Since x ( z g ) = x ( g s ) is a root of unity, a ( g s ) = lx(gz)12 = 1. On the other hand, a(%)= lx(z)12 = 0 and so 1 = O(modpR), a contradiciton. Thus CG(Z) N . Assume by way of contradiction that C G ( Z )= N . Then s E Z ( N ) . Since, by Theorem 1.4, X N is irreducible, we have z E Z ( x ) . But then x(z) # 0, a desired contradiction. H We have done most of the work to prove our second theorem.

Theorem 5.11. @mud (1977,1990)). Let G be a nonabelian group. Then G is generated by the zeros of a suitable irreducible (c -character of G. Proof. Assume by way of contradiction that the zeros of each irreducible (c -character of G generate a proper subgroup of G. Let M be any maximal normal subgroup of G. Then, by Corollary 5.7, X M is irreducible for any irreducible CC -character x of G. Hence each irreducible C -character of M is extendible to G which implies, by Theorem 18.7.3 (ii), that each M-conjugacy class is a G-conjugacy class. Hence

Now choose x to be a nonlinear irreducible (c -character of G. Let N be the subgroup of G generated by the zeros of x and let M be a maximal normal subgroup of G containing N . By Corollary 5.7, X M is irreducible and, since x ~ ( 1 #) 1, there exists g E M with x(g) = 0 (see Corollary 1.5). Hence, by Lemma 5.10, c G ( g ) C N C M . But then c G ( g ) = c M ( g ) , contrary to ICG(g)I > I C M ( g ) l . This completes the proof of the theorem. H

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Chapter 24

Characters, Conjugate

Elements and Commutators Let G be a finite group. Our aim is to examine the following two problems. Suppose that z,y E G and y is a product of conjugates of 2 . How many conjugates of 2 will suffice to express y as their product ? The second problem concerns relations between the irreducible characters of a group and the product of commutators of that group. In particular, we investigate the length of the shortest expression of an element g E G' as a product of commutators. Most of the results recorded are due to Robinson (1984) and Gallagher (1962a). One of the theorems presented asserts that if 1 # g E GI and x(1) assumes Ic distinct values as x ranges through irreducible (I= characters of G with x(g) # 0, then g can be written as a product of k - 1 commutators. As an easy consequence, we prove that if G is a p-group which has an abelian subgroup of index p t , then every element of GI is a product of t commutators.

1

Products of conjugate elements

We continue to assume that G is a finite group and that all characters are (I: -characters. Suppose that 5,y E G and y is a product of conjugates of z. How many conjugates of 5 will suffice to express y as their product ? In this section, we shall obtain some answers in terms of irreducible characters. The main tool will be Corollary 22.1.5 which is a consequence of the formula for computing the product of several class sums within Z((I:G) from the character table (see Theorem 22.1.4). 855

Characters, Conjugate Elements and Commutators

856

Theorem 1.1. (Robinson (1984)). Let x , y be nonidentity elements of G such that y is a product of conjugates of x and let k be the number of distinct values of x(x)/x(l) - as x mnges through those irreducible characters of G for which x(z)x(y) # 0 . Then (i) y can be written as a product of k or fewer conjugates of x. (ii) If y cannot be written as a product of k - 1 or fewer conjugates of 5, then there exists an irreducible character x of G such that x(z) = 0 and X(Y) 0.

z

Proof. (i) Let = 1 , . . . ,X k be the distinct values of x(z)/x( 1) as x ranges through the set X of irreducible characters of G for which x(x)x(y) # 0. Denote by Xi the subset of X consisting of all x for which x(z) = Xix(l), 1 5 i 5 k, and put XEX,

Assume by way of contradiction that y is not a product of k or fewer conjugates of z. If XI,...,xrare all irreducible characters of G then, by Corollary 22.1.5. forall 8€{1,

=O

...,k}

j=1

Thus, by the definition of pi, we have k

EXspi = 0

for all s E {1,

...,k}

i= 1

Hence AB = 0 where A is the k x k matrix whose ( & j )entry is X i and B is the k x 1 matrix whose (i, 1) entry is pi. Because XI,. ..,Xk are distinct and nonzero, d e t ( A ) # 0 and so B = 0. Hence

(note that XI# 8 since it contains the principal character of G). Now, by Theorem 21.2.1, X1 consists of all x; for which x E K e r x ; . Hence X1 is precisely the set of irreducible characters of G which contain

< g-'zgJg

E

G>

in their kernels. Thus we cannot have y E < g-lxglg E G >, since if that were the case, then would be a positive integer, contrary to (1). This

1 Products of conjugate elements

857

gives us a desired contradiction. (ii) Assume that y cannot be written as a product of k - 1 or fewer conjugates of 2. Then, as in (i), we have

Assume by way of contradiction that x;(s) = 0 implies x;(y) = 0 for all i E (1,. . . ,r } . Then whenever x;(z)xi(y) = 0, we must have x;(y) = 0, so that k

T

Cpi =

C xj(l)xj(y)

i=l

j=1

=0

(3)

Applying (2) and (3), it follows that C B = 0, where C is the k x k matrix But then d e t ( C ) # 0, so B = 0 leading to the whose ( i,j ) entry is A:-'. same contradiction as in (i).

Theorem 1.2. (Robinson (1984)). Let H be a subgrovp of G with H # G, let y # 1 be an element of < g-'Hg(g E G > and let k be the number of distinct valves of < XH, 1~ > /x(1) -as x ranges through irreducible characters of G such that < XH, 1~ > x( y) # 0. Then y = h 1 - a . h k where each h; is conjugate to an element of H .

Proof. Deny the statement. Then, since 1 E H , y cannot be written in the form hl -. ht,where each h; is conjugate to an element of H , for any t 5 k. Let XI,. . . ,xrbe all irreducible characters of G. Then, by Theorem

-

22.1.4,

for any t 5 k and any t-tuple (hl,. . . ,ht) of elements of H . Now x = xj x x xj ( t times) is an irreducible character of H x .. x H ( t times) and < X H X - ~ H , ~ H ~ . . . X H >=< (Xj)H,1~ > t

..-

Hence, summing (4) over all possible t-tuples, where t _< k is fixed, we obtain

Characters, Conjugate Elements and Commutators

858

.

Thus, for all t E (1,. . ,k}, we have

Applying the argument similar to that used in Theorem 1.1, we deduce that

where x runs through irreducible characters of G such that < X H , 1~ >= ~ ( 1 )It. follows that y $< g-lHglg E G >, a desired contradiction. We close by providing some applications of Theorem 1.1.

Corollary 1.3. (Robinson (1 984)). Let x, y be nonidentity elements of G such that y is a product of conjugates of x. Assume that there are t irreducible characters x of G for which x(y) # 0 . Then y can be written 0s a pruduct o f t - 1 or fewer Conjugates of x. Proof. Assume that the corollary is false. Then, by Theorem 1.1 (i), there exist precisely t irreducible characters x of G for which x(z)x( y) # 0. Furthermore, by Theorem 1.1 (ii), x(x) = 0, x(y) 0 for some irreducible character x of G. Since these two properties cannot hold simultaneously, the result follows. H

+

x

Lemma 1.4. Let g E G and let t be the number of irreducible chamcters of G for which x ( g ) # 0 . Then

.

Proof. Let XI,. .,xt be all irreducible characters 0. Then, by Theorem 19.2.3 (ii),

x of G with x ( g ) #

Write {XI,... ,xt} = XIU . . . U X , as a disjoint union of Galois conjugates, choose A; E X i and put a; = X ; ( g ) l , 15 i 5 s. Then {x(g)21xE X;} are all

1 Products of conjugate elements

859

conjugates of a;each repreated the same number of times. Setting d; = IXil, it follows from (5) and Lemma 23.1.2 that s

i=l XEX, 9

as desired. W

Corollary 1.5. (Robinson (1 984)). Let x,y be nonidentity elements of G such that y is a product of conjugates of x. Then y can be written as a product of at most min(lCG(z)l,IcG(Y)I

- 1)

conjugates of x.

Proof. Let t be the number of irreducible characters x of G for which x(y) # 0. Then, by Lemma 1.4, t 5 [ C ~ ( y ) lSimilarly, . there are at most [ C G ( Zirreducible )~ characters x of G for which x ( z ) # 0. Now apply Corollary 1.3 and Theorem 1.1. W Corollary 1.6. (Robinson (1984)). Let G be simple and let s be the minimum over all 1 # g E G of the number of irreducible characters x of G with x ( g ) # 0 . Let n be the minimum of IcG(g)l over all 1 # g E G and let m be the size of the smallest nontrivial conjugacy class of G. Then (i) For a n y nonidentity z , y E G, y can be written as a product of s2 - s or fewer conjugates of x . (ii) [GI 5 1 m m2 + - - .+- mn2+

+- +

Proof. (i) Choose 1 # go E G with the property that there are precisely x of G such that x ( g 0 ) # 0. Because G is simple, y E < g-lgoglg E G >. Hence, by Theorem 1.1 (i), y is a product of s or fewer conjugates of go. Owing to Corollary 1.3, each of these conjugates of go can be written as a product of s - 1 or fewer conjugates of 2 since go E< g-'xglg E G >. Hence y can be written as a product of s2 - s or fewer conjugates of x. (ii) By Lemma 1.4, s 5 n. Let C be the smallest nontrivial conjugacy class of G and let z E C . By (i), each 1 # y can be written as a product of s irreducible characters

Characters, Conjugate Elements and Commutators

860

n2 - n or fewer conjugates of x . Hence, if Ct is the class sum in Z(C G) of C, then y occurs with nonzero coefficient in (C+)' for some i E { 1 , . . . ,n2 - n } . Since ICl = rn, it follows that /GI 5 1 t m im2

+ - +- rnnz-,

as desired. H

2

Characters and commutators

The object of this section is to give some relations between the irreducible characters of a group and the product of commutators of that group. We also study the length of the shortest expression of an element of g E G' as a product of commutators. Throughout, G denotes a finite group and all characters are assumed to be CC -characters. Given x , y E G, their commutator [z,y] is defined by [x,y] = x-ly-lzy. If x is an irreducible character of G, then denotes the irreducible character of G defined by x ( g ) = x(g) for all g E G. Lemma 2.1.

Let

x

be an irreducible character of G. Then

c

x(11-l =

Z([Z,Yl)

X,?lEC

Proof. By Theorem 21.1.3, we have

and therefore, by Theorem 19.2.3 (ii),

as required. H Theorem 2.2. (Gallagher (1965a)). Let X I , .. . ,xTbe all irreducible characters of G and let n 2 1. Define A n : G -, (c! by An(g) being the number of 2n-tuples ( 2 1 , yl,x2,y2,. . ,x,, yn) of elements of G such that

.

= ( 2 1 , YI] * * * [zn, ~

n ]

2 Characters and commutators

861

Then T

i= I

Proof. Since

A1

:G

-, (C

is a class function, we have r A1

= ccixi i=l

where

Hence

proving the case n = 1. For n > 1, we have

Since, by Theorem 19.2.8,

the desired conclusion follows by induction on n.

Corollary 2.3. (Gallagher (1962~)).Let XI,. . .,xI be all irreducible characters of G and let g E G. Then g is a product of n 2 1 commutators if and only if

Proof. Let X,(g) be as in Theorem 2.2. Then g is a product of n commutators if and only if X,(g) # 0. Now apply Theorem 2.2. H

Characters, Conjugate Elements and Commutators

862

Theorem 2.4. (Gallagher (1962~)). Let XI,.. . ,x3 be all nonlinear irreducible characters of the nonabelian group G. If n >_ 1 is such that 3

i=l

then each element of G' can be written as a product of n commutators.

.

Proof. Assume that g E G' and let XI,. . ,Xi, be all linear characters of G. Then Xj(g) = 1, 1 5 j 5 k, and l x i ( g ) l 5 x i ( l ) , 1 5 i 5 s. If g cannot be written as a product of n commutators, then

a contradiction.

Corollary 2.5. (Gallagher (1962~)).Assume that n 2 1 is such that 4n 2 IG'I. Then each element of G' can be written as a product of n commututors.

Proof. Let x l , . . . ,x3be all nonlinear irreducible characters of G. Then 3

C ~ i ( 1=) ~ ]GI - (G : GI) i=l

= (G : G')(lG'l

- 1)

and therefore

= 4-n(G : G')([G'l - 1) < (G:G') The desired conclusion is therefore a consequence of Theorem 2.4. W The following theorem is an improvement of the above result.

2 Characters and commutators

863

Theorem 2.6. (Gallagher ( 1 9 6 5 ~ ) ) .Let n 2 1 be an integer. Then (i) If ( n 2)!n!> 21G’l - 2, then each element of G‘ is a product of n commutators. (ii) If G is a p-group, with JG’J= p a , and if n ( n 1 ) > a, then each element of G’ is a product of n commutators.

+

+

Proof. For the sake of clarity, we divide the proof into three steps. Step 1. Here we fix some notation and make preliminary observations. Let I P ( G ) denote the set of all elements of G which can be written as a product of n commutators. Then, by Corollary 2.3, for each rn, 0 I m I n, and g $! li”(G), we have

.

Denote by do, d l , . , the distinct degrees of the irreducible characters of G arranged in increasing order and let X i = dr2. Let uj = a ; ( g ) be the coefficient of AT in ( 1 ) . Then for g E G’ - K”(G),

We may arrange the irreducible characters X I , . .. ,xr of G such that XI,. are all nodinear ones. Then, for alI g E

e,

Hence, for all g E G’

..,xs

- li”(G), we have

Denote by A, = A”(X1, Xz, . . .) the greatest lower bound of &1 IuiJ,where the a* are now subject only to (2) with the normalization a0 = 1. By (3), we then have

P ( G ) = G‘ provided A , > IG’J - 1

(4)

Characters, Conjugate Elements and Commutators

864

Step 2.

Our aim here is to prove that

i=1

We first note that condition (2) together with the normalization a0 = 1 are equivalent to the condition that for any polynomial f ( X ) of degree n,

Now put n

f(X) = n(x- X i ) i=l

5 n and

Then f ( X j ) = 0 for 1 5 j

If(Xj)l

n

. Xn for j

_< XI..

> 71, SO by (6),

n

which proves ( 5 ) . Step 3. Completion of the proof. Applying ( 5 ) , we obtain n

n

i=l

i=l

Y

If G is a p-group, the di are powers of p and therefore n

Tl

i=l

i=l

If IG'] = pa, then (1/2)pn(nt1) > IG'l- 1 is equivalent to n(n

+ 1 ) > a. This proves (i) and (ii), by applying (4).

Theorem 2.7. (Robinson (1984)). Let 1 # g f GI and suppose that x ( 1 ) assumes k distinct values as x ranges through irreducible characters of G for which x ( g ) # 0 . Then g can be written as a product of k - 1 corn mu 2 at om.

2 Characters and commutators

865

Proof. Deny the statement. Since 1 E G is a commutator, g cannot be written as a product of t commutators for all t E (1,.. . ,k - 1). Hence, by Corollary 2.3, T

CX;(l)'-2tX;(g) = 0 i= 1

.

for all t E (1,.. . ,k - l}, where XI,. . ,xT are all irreducible characters of G. Let XI,. . .,Xk be the distinct values of x( 1) as x ranges through irreducible characters of G with x ( g ) # 0. Let X;,1 5 i 5 k, be the set of all irreducible characters x of G such that x(g) # 0 and x(1) = A;. Setting

we then have

k

CP;=O

(since g

#

1)

i= 1

Also, for 1 5 j 5 k - 1, C;"=, AT2'p; = 0. Because A T 2 , A i 2 , . . . , X i 2 are distinct, it follows that each pi = 0. In particular, if c q , ...,a, are all linear characters of G, then m

i=l

which is impossible since g E G'.

Corollary 2.8. (Robinson (1984)). Let p be a prime and let G be a p-group. Then (i) If (G : Z ( G ) ) = p k , then each element of G' is a product of [k/2] commutators (here [C/2] is the largest integer n with n 5 k/2). (ii) If G has an abelian subgroup of index p t , then every element of G' is a product o f t commutators. Proof. (i) Owing to Theorem 21.2.4, ~ ( 1 divides ) ~ p' for each irreducible character x of G. Hence each irreducible character of G has degree at most p [ k / 2 ] .Therefore there are at most [k/2]+ 1 possible degrees of irreducible characters of G. Since the identity element of G is a commutator, the desired assertion follows by virtue of Theorem 2.7. (ii) Let A be an abelian subgroup of G of index p t . Then, for any irreducible character x of G, < XA,XA > L x(1) and so, by Lemma 21.2.2,

IGI 2

< X A , X A >>_IAIX(1)

866

Characters, Conjugate Elements and Commutators

Thus x(1) 5 (G : A ) = pt and the result follows by Theorem 2.7. H

Chapter 25

The Frobenius-Schur Indicator This chapter examines some classical aspects of character theory. Namely, given an irreducible (c -character x of a finite group G, we wish to determine necessary and sufficient conditions for x to be realizable over IR and also for x to be real valued, but not realizable over IR..A full answer is given by the so called Frobenius-Schur indicator v(x) of x. We show that v ( x ) = 0 if x is not real valued, Y ( X ) = -1 if x is real valued, but realizable over R , and v(x) = 1 if x is realizable over R . A significance of the F’robenius-Schur indicator is also demonstrated by the fact that if XI,...,xr are all irreducible a-characters of G , then the number t of elements of G of order 2 is given by t = -1 EL1 V ( X i ) X i ( l ) . The chapter ends with an alternative interpretation of v ( x ) . Namely, we show that if x is an irreducible (I:- character of G afforded by a (I: G-module V , then v(x) = 0 if and only if there is no nonzero G-invariant bilinear form on V ;v(x) = 1 if and only if there exists a G-invariant nonsingular symmetric bilinear form on V ; v(x) = -1 if and only if there exists a Ginvariant nonsingular skew symmetric bilinear form on V .

+

1

Unitary matrices

This section contains some preparatory material which will be used in the study of the Frobenius-Schur indicator and linear groups. Throughout, G denotes a finite group. 867

The F’robenius-Schur Indicator

868

Let V be an n-dimensional vector space over

(I:

and let

f:VxV+(C be a positive definite hermitian form. Upon identification of V with C n , a typical example of such a form is n

f((21,...,2,),(yl,...,yn))=C5i~i i=l

( z i , y i ~a )

The unitary group U ( V )= U j ( V )of V is defined by

A unitary basis of V is a basis

v1,

. . . ,wn

such that

The existence of a unitary basis can be deduced as follows. Choose v1 E V such that f ( w 1 , vl) = 1. Next take any vi E V with w: 4 (c v1. Now choose X E (c such that f(v1,Xwl t vk) = 0 and put 02 = p ( X v 1 t wi), where p2 E (I: is such that f ( w 2 , w z ) = 1. Continuing in this way, we obtain a desired unitary basis. A matrix M E GL,((c ) is said to be unitary if

M.f&j=] where t&f is the transpose of A?, &j is obtained from M by taking complex conjugates of the entries of M and I is the identity n x n matrix. We denote by Un((c ) the group of dl n x n unitary matrices.

Lemma 1.1. Let v1,. . . ,w, be a unitary basis of V , let Q E G L ( V ) and let M be the matrix of a with respect to v1,. . ,v,,, Then (i) a E U ( V ) if and only i f a ( v l ) , ...,a(vn) is a unitary basis of V. (ii) Q E U ( V ) if and only if M is a unitary matrix (in particular, we have U ( V )2 Un(C)).

.

Proof. This is a direct consequence of the definitions. The following theorem collects together some of the basic properties of unitary matrices.

1 Unitary matrices

869

The following properties hold : (i) Every permutation matrix is unitary. (ii) A triangular unitary matrix is diagonal. (iii) For any A E M,(C ), there exists U E U,(C ) such that U-lAU is triangular. (iv) For any A E U,,((c), there exists U E U , ( C ) such that U-lAU is diagonal. (v) For any A E U,(C ), all eigenvalues of A have absolute value 1. (vi) For any commuting unitary matrices Al,. . . ,A,, there exists a unitary matrix U such that each U" A,U is diagonal. Theorem 1.2.

Proof. (i) and (ii) : Obvious. (iii) Let Q be a linear transformation of V such that A is the matrix of a! with respect to some unitary basis of V . By Lemma 1.1 (i), the replacement of A by U-'AU with U E U,(C ) is equivalent to a change of unitary basis of V . Choose an eigenvector 0 # v E V of Q and choose v1 = pv for some p E (c such that f ( v 1 , v l ) = 1. If W = (3 E V ( f ( z , v l )= 0 } , then W has a unitary basis v2,. ..,v,. Then v1,v2,. ..,v, is a unitary basis of V and with respect to this basis, the matrix of a becomes

wheer c E C is defined by Q ( V ) = cv. The desired conclusion now follows by induction on n. (iv) Apply (ii) and (iii). (v) By (iv), we may choose U E U , ( C ) such that D = U"AU is diagonal. Then A has the same eigenvalues as D. Since D is unitary, I = D and the eigenvalues of D have absolute value 1. (vi) By induction on n , we may assume that A1,. . . ,A,-1 are diagonal matrices. By (i), we may assume that A l , . . . ,A,-1 are of the form

-

A; = diag(Xj1I1, ...,X;,I,)

(1 5 i

5 n - 1)

where Ij is the n x n identity matrix and if s # t then Xi, Because A, commutes with A l , , ,A,,-1, we must have

..

# Xit

for some i.

A, = diag(B1,. . . ,B,) for some unitary nj x nj matrices Bj. Choosing, by (iv), unitary Uj's with UJ-*BjUj daigonal, it follows that U = diag(U1,. . , U,) is a required matrix. So the theorem is true. H

.

The Frobenius-Schur Indicator

870

Let F be an arbitrary field. A bilinear form f on an FG-module V is said to be G-invariant if

In what follows, all (CG-modules are assumed to be finitely generated. Theorem 1.3. Let V be a (c G-module. Then (i) There exists Q G-invariant positive definite hermitian form f on v. (ii) If V is Q simple (c G-module, then the form f in (i) is uniquely determined, up to multiplication b y a nonzero constant in (I=.

Proof. (i) Let 1c, be any positive definite hermitian form on V . Put f(Vl,V2)

=

c

(%%

?b(mzv2)

EV)

xEG

Then f : V x V

-, (c

is a positive definite hermitian form on V . Since

the desired assertion follows. (ii) Let f' be another form on V as in (i), and let v 1 , . . . ,w, be a a!basis of V such that f(v;,vj) = 6ij. Then each v € V can be written in the form w = f(v, vj)wj and the map V -+ V , w H f'(w, q)wj is (c linear, Denote by wo an eigenvector of this map and by p the corresponding eigenvalue. Then

xjnzl

cjn=l

n

n

j=1

j=1

which implies that f'(v0,vj) = pf(wo, wj) for all j E (1,. . .) r } . Thus f'(v0, v) = pf(vo, v) for all w E V . Now put

W = {w E Vlf'(w,x) = pf(v,z) for all z E V } Then W # 0 and W is a submodule of V . Because V is simple, we have W = V and the desired conclusion follows.

A unitary representation of G is any homomorphism G + V,(C ) for some n 2 1. Two matrix representations p; : G --t GL,((C ), i = 1 , 2 , of G

1 Unitary matrices

over that

(I:

a71

are said to be unitary-equivalent if there exists U E Un(C) such p 2 ( g ) = U-'pl(g)U

for all g E G

Theorem 1.4. (i) Any matrix representation ofG over (I: is equivalent to a unitary representation. (ii) Two irreducible unitary representations of G over (I: are equivalent if and only if they are unitary-equivalent. Proof. (i) Apply Theorem 1.3 and Lemma 1.1 (ii). (ii) If two matrix representations of G are unitary-equivalent, then they are obviously equivalent. Conversely, assume that p1 and p2 are equivalent irreducible unitary representations of G, say pi : G U,((I:), i = 1,2. By hypothesis, there exists A E GLn((I:) such that --f

for all 9 E G

p 2 ( g ) = A-'Pl(g)A

Then p I ( g ) A = Ap2(g) and so P2(d

-1

t -

( A) =

t-

-t

P 2 W A = AP2(d -t = p1(g)A = t A-t- P l ( g ) = t A P l ( g ) - '

It follows that P1 ( g ) W A ) P l ( g - ' )

= AP2(S)P2(g-l

=A V )

whence p l ( g ) A ( t A )= A ( t A ) p l ( g )for all g E G . Since p1 is irreducible, we deduce that A ( t A ) = X I for some X E (I: . Write A = ( a j j ) so that the ( i , i)-th entry of A ( t A ) is n

n

k=l

k=l

Then X is a positive real number and so X = ZZ for some z E (I:. Setting U = z - l A , we have U ( t o ) = I and therefore U is unitary. Since, for all 9 E G, U - l P l ( g ) u = A-'Pl(g)A = P 2 ( 9 ) , the result follows. As a preliminary to our final result, we now record the following observation.

The Frobenius-Schur Indicator

872

Lemma 1.5. Let A be a unitary matrix and let B be a unitary diagonal matrix such that A B = BA. Then there exists a unitary diagonal matrix C such that C 2 =B and A C = C A

Proof. Choose a permutation matrix P such that

where all X i are distinct. By Theorem 1.2 (i), P is unitary and, since AB = B A , we also have (P-'AP)(P-'BP) = (P-'BP)(P-'AP). Thus P-'AP = diag(A1,. , A T ) ,where each A; is a unitary matrix. Now choose p1, ...,p, with = Xi and put Co = diag(p1I,...,p,.I). Then Ci = P-'BP and P-'APCo = CoP-lAP. Because all [ p i ] = 1, Co is unitary. Finally, put C = PC0P-l. Then C is a unitary diagonal matrix commuting with A and C2 = PciP-' = PP-lBPP-' = B ,

..

as desired. W

Theorem 1.6. Let p : G U,(C ) be an irreducible unitary representation of G over (I: with character x = Then (i) Them exists Q unitary matrix U such that tU = f U and --f

x.

-

U-'p(g)U = p ( g )

for all g E G

if and only if p is equivalent to an R -representation of G. Proof. (i) The representation p : G Un(C), p ( g ) = p ( g ) , g E G, affords the character k. By hypothesis, x = so p and p are equivalent. (ii) t U = U

--$

Thus, by Theorem 1.4 (ii), there exists a unitary matrix U satisfying ( 1 ) . Since U - l = 'U and U-' = ' U , U = (")-', it follows from (1) that

-

p(g) = U - ' p O U = tUp(g)(tU)-l = tUU-'p(g)U(tU)-l Thus t U U - l commutes with all p(g) and so tUU-' = X I for some X E Hence tU = XU and so

u = " ( " U )= yxu) = A(")

= X2U

(c.

1 Unitary matrices

873

Thus X2 = 1 and X = f l , proving (i). (ii) Assume that p is equivalent to an IR-representation of G. Then there exists A E GLn((I: ) such that

A-'p(g)A E GLn(IR)

for all g E G

Hence, by (l),

which implies UAA-' = X I for some X E

(1:.

Thus

and so, if tU = - U , we would have

a contradiction. Conversely, suppose that tU = U. By Theorem 1.2 (iv), we may choose a unitary matrix A such that A-'UA = B is diagonal. Then

and so ( t A ) A B = B ( t A ) A . Hence, by Lemma 1.5, there exists a unitary diagonal matrix C such that C2 = B and ( t A ) A C = C ( t A ) A . Now put D = ACA-I so that D 2 = ABA-' = U. We then have

and SO D-' = = D. Finally, put p l ( g ) = D-'p(g)D, g E G. Then, for each g E G, p1(g)

=

D p o D = Du-'p(g)uD-'

= DD-2p(g)D2D-l = p 1 ( g ) ,

thus completing the proof. W

The Frobenius-Schur Indicator

a74

2

The Frobenius-Schur indicator

In what follows, G denotes a finite group. Given g E G, we put ,/j= {z E G(x2= g } Let

x be an irreducible (c -character of G. Then the number v ( x ) given by

is called the Frobenius-Schur indicator of x. We say that over IR if x is the character of an R-representation of G.

x is realizable

Theorem 2.1. ( h b e n i t t s and Schur (1906b)). Let x be an irreducible (c -character of G. Then (i) ~ ( x=) 0 if x is not real valued. (ii) Y ( X ) = -1 i f x is rtal valued, but not realizable over EL. (iii) v ( x )= 1 i f x is Galizable over IR . (iv) For all g E G, r

M I = &(Xi)Xi(S) i=l

where XI,. ..,x,. are all irreducible (c -characters of G. (v) The number t of elements of G of order 2 is given by T

t = -1 t C v ( x i ) x i ( 1 ) i= 1

(vi) t 5 -1 t field f o r G.

x i ( 1 ) with equality if and only if

IR is a splitting

Proof. We first make some preliminary observations. Let p : G + GL,(C ) be a representation of G which affords x. In what follows, we write p(g) = (a;j(g)) for all g E G. By Theorem 1.4 (i), we may assume that each p ( g ) is a unitary matrix. Moreover, we claim that if x is realizable over IR, then we may assume that each q ( g ) E I€€ for all g E G. Indeed, x is afforded by an IRG-module W. We may clearly choose a G-invariant symmetric bilinear form f : W x W -+ 1R which extends to a G-invariant hermitian positive form f : V x V + ([: , where V = (c @'JR W. By choosing an lR.-basis v1, . . . ,v, of V with j ( v i , vj) = 6 ; j , 01,. . . ,v, becomes a unitary

2 The Frobenius-Schur indicator

875

basis of V , which clearly implies the claim. By definition, we have

v(x) =

PI-'

c

t r [ p ( g ) p ( g ) l = IGI-'

g€G

= 1GI-l

c

t-

W g ) P(S

'11

g€G

C Eaij(g)aij(g-l)

(1)

g G i,j

x.

Since X is afforded by the Assume that x is not real valued, i.e. x # representation p : g H p ( g ) , it follows that p and p are not equivalent. Hence, by Lemma 19.2.2 (i) and (l),v ( x ) = 0, proving (i). Assume that x = X and x is not realizable over IR. Then, by Theorem 1.6, -there exists a unitary matrix U such that t U = -U and U - ' p ( g ) U = p ( g ) for a l l g E G . Then U-' = = -U and

p ( g - 1 ) = U-'p(g-')u = - U p ( g - l ) U

876

The Frobenius-Schur Indicator

proving (iii) . Let X : G -+ 6: be defined by X(g) = Ifit. Then X is a class function, so there exist z1,. . . ,t, in (c such that X = Cr=l 2 ; ~ ; Choose . 91,.. . ,gm in G such that G = Ug,& (disjoint union). Then

proving (iv). Property (v) follows from (iv), by taking g = 1, while the inequality in (vi) holds since v ( x i ) 5 1. The equality in (vi) holds if and only if each u ( x ; ) = 1. Since, by (iii), the latter is equivalent to the requirement that 1R is a splitting field for G, the result follows. a Our next aim is to characterize the Frobenius-Schur indicator in terms of bilinear forms.

Lemma 2.2. Let V # 0 be a finite-dimensional IRG-module. Then there exists a positive definite symmetric G-invariant bilinear form on V .

Proof. Let f : V x V + lR.be any positive definite symmetric bilinear form on V . Define II,: V x V -+ IR,by

Then II,is a symmetric G-invariant bilinear form on V . For v have $(v, v ) 2 f(v, v ) > 0 , so II,is positive definite.

# 0 in V , we

Let V be a finite-dimensional vector space over a field F . A bilinear form f : V + V is called skew symmetric if

Lemma 2.3. Let V be a finite-dimensional vector space over a field F with C ~ U T#F2, let B ( V ) be the F-space of all F-bilinear maps V x V F --f

2 The F’robenius-Schur indicator

877

and let the subspaces S ( V ) and A ( V ) of B ( V ) be defined by S ( V ) = { f E B ( V ) (f A ( V ) = {f E B(V)Jf

is symmetric} is skew symmetric}

Then B ( V ) = S ( V )@ A ( V )

Proof. Given f E B ( V ) , define V . Then

z,y E

f

= (1/2)(f

f E B ( V ) by f ( x , y ) = f ( y , z )

for all

t f)+ (1/2)(f - f)

where the first summand is in S ( V ) and the second in A ( V ) . If f E S ( V ) n A ( V ) , then for all z , y E V ,

which implies f = 0 since charF

# 2.

Lemma 2.4. Let F be an arbitrary field, let V be a finite-dimensional FG-module and let V* be the contmgredient of V . Denote b y B G ( V ) the F-space of all G-invariant bilinear forms on V . Then (i) For each f E BG(V),the map f’ : V + V * defined by f’(vl)(v2)= f (v1,v2), vi E V , is an FG-homomorphism. (ii) The map B G ( V ) + HomFG(V,V*),f c1 f’ is an F-isomorphism * )cp’ E & ( V ) given by cp’(v1,212) = whose inverse sends cp E H o ~ F G ( V , V to c p ( S ) ( V Z ) , vi E v. (iii) If V is simple and x is the chamcter of G afforded by V , then the following conditions are equivalent : (a) V 2 V * (6) x(s) = x(s-’) for all 9 E G (c) There is a nonzero G-invariant bilinear form on V . (d) There is a G-invariant nonsingular bilinear form on V . Moreover, if F is a splitting field for G , then the forms in (c) and (d) are unique up to multiplication by an element of F . Proof. (i) The map f’ is clearly F-linear. Since, for all g E G, v1,v2 E

v, f’(sv1)(.2)

= f($Vl,fJ2) = f ( w - 1 v 2 ) = f‘(vl)(g-lvz) = [gf’(.1)1(.2)

The F'robenius-Schur Indicator

a78

the desired assertion follows. (ii) The map f c+ f' is clearly F-linear. If cp = f', then

proving that cp' = f . On the other hand, if f = 'p', then

and so f' = cp, proving (ii). (ii) Since V is simple, so is V* by virtue of Lemma 18.3.5 (iii). Hence, by (ii), (a) is equivalent to (c). By Lemma 18.3.7, the character x* of G afforded by V' is given by Xr(g) = x ( g - ' ) for all g E G. Hence, by Corollary 17.1.8 (i), (a) and (b) are equivalent. It is clear that (d) implies (c). Next assume that (c) holds and let 0 # f E BG(V). By (ii), the map f' : V --$ V* given by f ' ( q ) ( v 2 ) = f ( q , v 2 ) , vi E V , is an FG-isomorphism. In particular, f' is an F-isomorphism and hence, by Proposition 3.9.13, f is nonsingular. This proves (d) and thus (a), (b), (c) and (d) are equivalent. Finally, assume that F is a splitting field for G. If (c) holds, then by (ii) and (a), & ( V ) 2 H O ~ F G ( V , 2 EndFG(V)

v*)

and so d i m F B c ( V ) = 1, as desired. H

We are now ready to prove the following classical result. Theorem 2.5. (Fmbenius and Schur (1906b)). Let x be the irreducible of G aflorded by a (c: G-module V and let Y ( X ) be the FmbeniusSchur indicator of x. (i) v ( x )= 0 if and only if there is no nonzero G-invariant bilinear form on V . (ii) u ( x ) = 1 if and only if there exists a G- invariant nonsingular symmetric bilinear form on V (such a form is necessarily unique, up to multiplication by an element of (I:). (iii) u(x) = -1 if and only if there exists a G-invariant nonsingular skew symmetric bilinear form on V (such a form is necessarily unique, up to multiplication by an element of a!). (I: -character

Proof. (i) The character x is not r e d valued if and only if x ( g ) x ( g - l ) for some g E G. Now apply Lemma 2.4 (iii).

#

2

The Frobenius-Schur indicator

879

(ii) Here we follow the argument presented in Serre (1977, p. 107). Assume that Y ( X ) = 1. Then V E C! @IR W for some IRG-module W . By Lemma 2.2, there exists a positive definite symmetric G-invariant bilinear form fo on W . By scalar extension, fo defines a G-invariant bilinear form f which is symmetric and nonsingular. Furthermore, by Lemma 2.4, f is unique, up to multiplication by an element of (r: . Conversely, assume that f : V x V + (I: is a G-invariant nonsingular symmetric bilinear form on V . By Theorem 1.3, there exists a G-invariant positive definite hermitian form $ : V x V -+ C . For each x E V , there exists a unique element cp(x) E V such that

Then the map cp : V V is an additive bijection such that cp(Az) = xcp(z) for all X E (I: x E V . Hence cp2 = cp o cp is a (I: -automorphism of V . For all x,y E V , we have --f

This means that

(p2

is a hermitian operator. Furthermore, since

the operator cp2 is positive definite. We know that if is a hermitian positive definite operator, then u = v2 for a unique hermitian positive definite operator v and v can be written in the form P(u),where P is a polynomial with real coefficients (if the eigenvalues of u are XI 7 . . . ,A, choose P so that P(Xi) = for all i ) . Hence we may apply this fact to u = cp2, and we put a = cpv-l. Because v = P(cp2),cp and v commute and so a2 = cp2vd2 = 1. Let V = Vo @ V1 be the decomposition of V with respect t o the eigenvalues 1 and -1 of a. Since a(Az) = Xx for all X E a , 2 E V , multiplication by i maps VOonto V,. This shows that V = Vo CB iV0. Finally, since f and $ are G-invariant, cp, v , and a commute with all linear transformations v H gv is a G-invariant R-subspace of V , of V ( v E V,g G). It follows that which proves that x is realizable over IR.. (iii) By Lemma 2.4 (iii), x is real valued if and only if there is a Ginvariant nonsingular bilinear form f on V (which is necessarily unique up

The Frobenius-Schur Indicator

880

to multiplication by an element of CC ). By (ii), it suffices to verify that f is either symmetric or skew symmetric. By Lemma 2.3, we may write f uniquely in the form f = fi f2 with f1 symmetric and f2 skew symmetric. Then f i and fz are G-invariant. Since f is unique, we have either jz = 0 (and f is symmetric) or f1 = 0 (and f is skew symmetric), as desired.

+

Chapter 26

Characters and Hall Subgroups Let N be a Hall subgroup of a finite group G. We wish to examine charactertheoretic conditions under which H has a normal complement in G. It is shown that if H is solvable, then G contains a normal complement to H if and only if each irreducible (I: -character of H can be extended to a character of G. By using the classification of finite simple groups together with the inspection of the character table of some of them, Ferguson (1989) proved the corresponding result without the restriction that H is solvable. The second part of the chapter contains a proof of a classical result due to Blichfeldt (1904) which asserts that if x is a faithful character of G and ?r = { p l p prime, x(1) < p - 13, then G has an abelian Hall ?r-subgroup, and moreover all the Hall ?r-subgroups of G are conjugate in G. While on the subject concerning degrees of faithful characters, we also establish a theorem due to Ito (1953) which asserts that if P is a Sylow p-subgroup of a solvable group G and x is a faithful (c -character of G with x( 1) < p - 1, then P is an abelian normal subgroup of G.

1

An excursion into group theory

In order not to interrupt the main flow of the subsequent exposition of character theory, we present here a number of group-theoretic results. In what follows, unless explicitly stated otherwise, all groups are assumed t o be finite. If H is a subgroup of a group G, a subgroup I( is called a 881

Characters and Hall Subgroups

882

complement of H in G if

A subgroup H of a group G is called a Hall subgroup of G if the order of H is coprime with its index in G. Let n be a set of primes. In Sec.4 of Chapter 21 we introduced the notions of n-groups, n-separable groups and Hall n-subgroups. The normal n-subgroups of a group G play a special role. If K and H are a-subgroups of G and K a G,then H K is obviously a r-group. Thus the subgroup generated by all the normal n-subgroups of G is a n-group. We denote it by O,(G). It is clear that O,(G) is the unique maximal normal ?r-subgroup of G.

Lemma 1.1. (i) If H is a subnorrnaZa-su6gmup of G , then H (ii) O,(G) is the intersection of all maximal n-subgroups of G. (iii) O,(G) is contained in every Hall n-subgmup of G .

O,(G).

Proof. (i) By assumption, there is a series H = &a H I a * ..a Hk = G. If k 5 1, then H a G and H C O,(G) by definition. We now assume that k > 1 and argue by induction on k. Then, by induction hypothesis, H E O , ( H k - 1 ) . Since the latter subgroup is characteristic in H k - 1 , it must be normal in G. Thus O , ( H k - 1 ) C O,(G) and H O,(G), as desired. (ii) Let A4 be a maximal a-subgroup of G. Then O,(G)M is a a-group and so O,(G) C M. On the other hand, the intersection of all maximal r-subgroups is obviously normal in G, so it is contained in O,(G). (iii) Since every Hall n-subgroup is a maximal n-subgroup, the desired assertion follows by (ii).

Lemma 1.2. ( h t t i n i ' s argument). Let H be a normal subgroup of G and let P be a Sylow p-subgroup of H . Then G = N G ( P ) H Proof. If g E G , then g-'Pg E H and g-*Pg is a Sylow p-subgroup of H . Hence g-lPg = h-'Ph for some h E H by Sylow's theorem. Thus gh-' E NG(P) and g E N G ( P ) H ,as desired. f x + I be an exact splitting sequence Lemma 1.3. Let 1 t N iG + of gmups. Then u subgroup K of G is a complement of N an G if and only if there exists a splitting homomorphism (Y : X + G such that a ( X ) = K .

1 An excursion into group theory

883

Proof. If a : X ---t G is a splitting homomorphism, then a ( X ) is obviously a complement of N in G. Conversely, assume that G = N K and N n K = 1. Let Q : X -, G be any splitting homomorphism. Define p : a ( X ) -+ Ii by ~ ( Q ( z )=) x' where x' E Ii is uniquely determined by a ( $ )= nx' for some n E N . Then PQ : X -+ G is a splitting homomorphism with P Q ( X ) = K ,as required. Theorem 1.4. (The Schur-Zassenhaus Theorem). Let N be a normal Hall subgroup of a group G. Then complements of N in G exist and any two of them are conjugate.

Proof. For the sake of clarity, we divide the proof into five steps. Step 1. First assume that N = A is abelian and put X = G / A . Let H ' ( X , A ) and H 2 ( X ,A ) be defined with respect to the action of X on A via conjugation. Then, by Proposition 9.2.5 (vi), H ' ( X , A ) = H 2 ( X , A ) = 1. Since H 2 ( X , A )= 1, it follows from Theorem 9.8.2 (iif) that the extension 1 A + G -+ X + 1 splits. Since H 1 ( X , A ) = 1, the desired assertion follows from Theorem 9.8.1 (ii) and Lemma 1.3. Step 2. Existence in the general case. We argue by induction on IGI. Let p be a prime dividing I N [ ,let P be a Sylow p-subgroup of N and let Z = Z(P). Put K = N , ( P ) and M = N G ( Z ) . Since Z is a characteristic subgroup of P,we have K M . By Lemma 1.2, G = K N and so G = M N . Setting N1 = N n M , we see that N1 0 M and (A4: N1) - (G : N ) = m, say. We may apply the induciton hypothesis to the group M / Z (since Z # 1). If H / Z is a subgroup of M / Z of order m, then M = HN1 and H n N1 = 2. Because IH/Zl = m is relatively prime to 121, it follows from Step 1 that H has a subgroup of order m. Step 3. Conjugacy in the case GIN is solvable. Let A be the set of all prime divisors of m = IG/NI and put S = O,(G). Let H and K be two complements of N in G. Then, by Lemma 1.1 (iii), S C H n K since H and Ii are Hall r-subgroups of G. By passing to the group G/S and noting that O,(G/S) = 1, we may assume that S = 1. We may also assume that m > 1 so that N # G. Let M / N be a minimal normal subgroup of G I N . Since GIN is solvable, it follows from Lemma 21.4.2 (ii) that M / N is an elementary abelian pgroup for some prime p in R . Since H n M 2 ( H n M ) N / N M / N and --f

( M : ( H n M ) )= ( H M : H ) = IN[, we see that H n M is a Sylow p-subgroup of M . Since the same is true for

Characters and Hall Subgroups

884

K n M , it follows that H n M = g-'(fr'n M > g = g-%g

nM

for some g E G. Setting R = H n M , we deduce that RQ < H , g - ' K g >= T , say. We claim that T # G; if sustained it will follow by induction on \GI that H and g - l K g are conjugate in T and so H and K are G-conjugate. To substantiate our claim, assume by way of contradiction that T = G, so that R Q G. Then, since R is a n-group, R C O,(G) = 1. Hence M is a p'-group which is impossible since M / N is a pgroup. Step 4. Conjugacy in the case N is solvable. Let H and Ir' be two complements of N in G . Then HN'/N' and K N ' / N axe conjugate by Step 1. Thus g - l H g C KN' for some g E G. Invoking induction on the derived length of N , we infer that g - ' H g and K are conjugate (being subgroups of order m = IG/NI in the group K N ' ) . Thus H and K are conjugate. Step 6. Conjugacy in the general case. Since ( I N ] ,IG/NI) = 1, either IN1 or IG/NJ is odd. Hence, by the Feit-Thompson theorem (1963), either N or G I N is solvable. The desired conclusion is therefore a consequence of Steps 3 and 4.

Corollary 1.5. Let N be a normal Hall subgroup of G, let rn = IG/NI and let t be a divisor of m . Then any subgroup of G of order t is contained in a subgroup of order m.

H1

Proof. By Theorem 1.4, there is a subgroup H of G of order m. Let be any subgroup of G of order t . Then G = H N and

HIN = HIN n H N = (HIN n H ) N Thus

l H I N n HI = IHIN/NI= lHll = t By Theorem 1.4, we deduce that

H1 = g - ' ( ( H I N ) n H ) g & g - l H g

for some g E G

Since /g-'Hgl = m, the result follows.

Theorem 1.8. Let ?r be a set of primes and let G be a n-separable group (e.g. G is solvable). Then

1 An excursion into group theory

885

(i) G has a Hall n-subgroup. (ii) Every n-subgroup of G is contained in a Hall n-subgroup of G. (iii) Any two Hall r-subgroups of G are conjugate. Proof. Because each n-subgroup of G is contained in a maximal nsubgroup of G, it suffices to show that any maximal n-subgroup P of G is a Hall n-subgroup and that all such subgroups are conjugate. We may assume that IGl # 1 and we argue by induction on [GI. Put S = O,(G) and assume that S # 1. Then 5 ' G P and by induction PIS is a Hall n-subgroup of G/S. Hence P is a Hall n-subgroup of G. If Q is another Hall n-subgroup of G, then PIS and Q / S are conjugate and so P and Q are conjugate. Finally, assume that S = 1. Because G is n-separable and G # 1, it follows that R = O,,(G) # 1. Now P R / R is a n-group and by induction it is contained in a Hall n-subgroup Q / R of G/R. By Corollary 1.5, the nsubgroup P is contained in a Hall n-subgroup Q1 of Q. But P is a maximal Ir-subgroup, so P = Q1 and P is a Hall n-subgroup of Q. Thus P is a Hall subgroup of G. Let Pl be any other Hall n-subgroup of G. Then P R / R and P1R/R are conjugate (since they are Hall n-subgroups of G/R). Hence g-'Plg G P R for some g E G. Applying Theorem 1.4 to P R , we see that g-'Plg and P are conjugate, as desired. Let p be a prime. Recall that a group G is said to be p-solvable if each composition factor of G is either a p-group or a p'-group. Thus G is p-solvable if and only if G is n-separable for n = (p} (or, equivalently, G is #-separable for r = {p}).

Corollary 1.7. Let p be a prime and let G be a p-solvable group. Then (i) G has a Hall p'-subgroup. (ii) Every p'-subgroup of G is contained in a Hall p'-subgroup of G. (iii) Any two Hall p'-subgroups of G are conjugate. Proof. Let n be the set of all prime numbers distinct from p. Then a subgroup is a n-subgroup if and only if it is a p'-subgroup. Since G is Ir-separable, the desired assertions follow from Theorem 1.6. H Let G be a group, possibly infinite, and let H be a subgroup of finite

Characters and Hall Subgroups

886

index n. If { t l , . . . ,t n } is a left transversal for ' H in G, then for any g E G,

gtiH = t g ( q H where the i H g ( i ) is a permutation of the set {1,2,. . . ,n}. Assume that f : H + A is a homomorphism from H to an abelian group A. Then the t r a n s f e r of f , written T j , is the mapping

given by n

Our assumption that A is abelian guarantees that the order of the factors in the product is irrelevant. The most important case of the transfer occurs when f : H H I H ' is the natural homomorphism. In this case, we write TG,Hinstead of Tj and refer to TG,Has the t r a n s f e r of G into H . Thus, by definition, the map --f

is given by n

TG,H(g) = n ( t i 6 ) g t ; ) H '

for all g E G

i=l

If P is a Sylow p-subgroup of a finite group G, then we define G'(p) to be the intersection of all normal subgroups N of G such that GIN is an abelian pgroup. We refer to G'(p) as the p - c o m m u t a t o r subgroup of G. It is clear that G/G'(p)is isomorphic to the Sylow p-subgroup of GIG'. T h e o r e m 1.8. (i) The map Tj : G + A is a homomorphism which does not depend on the choice of the transversal. (ii) If g E G and the < g >-orbits of the set of left cosets of H in G are

then

1 An excursion into group theory

887

(iii) If H is a central subgroup of G , then

(iv) If P is a Sylow p-subgroup of a finite group G, then G'(p) is the ~ P n GI is the kernel of the restriction of T G , to ~ P. I n kernel of T G , and particular, Im(Tc,p)FZ P / ( P n GI)

Proof. (i) We first show that Tf is independent of the choice of the transversal. Let { u l , . . . ,u n } be another left transversal for H in G and suppose that tiH = u;H, 1 5 i 5 n. Then ui = tihi with hi E H and for any 9 E G, 21-1 dt)gu; = hg(i)(ti;)gt;)h; Hence, since A is abelian

The second factor in (2) is trivial, since as i runs over the set ( 1 , . . . ,n } , so does g ( i ) . Thus Tj is independent of the choice of the transversal. Given x , y E G , we have

n

proving (i). (ii) The elements {gjs;ll 5 i 5 k, 1 5 j 5 mi - 1 ) form a left transversal for H in G. Using this transversal, we calculate T f ( g ) . Because gmisiH = siH, the contribution of orbit (1) to T j ( g )is 1 mi

f ((gSi)-l(gsi) (g"~-'s;)-l(g"~-lsi)(s; g * *

8;))

Characters and Hall Subgroups

888

which reduces t o f(srlgmisi), proving (ii). (iii) Observe that s;’gnis; E H , so g m * E H and thus sflgmisi = gmi. Hence, by (ii), k

T f ( 9 )=

n

grn’ = gn,

i=l

as desired. (iv) Put IC = KeT(TG,p). Since G / K is an abelian pgroup, we have G’(p) C K . Decompose the set of left cosets of P into < g >-orbits as in (ii). Then, by (ii),

(n k

TG,P(g)=

silgm’si)P’

i=l

Since G = PG’(p), we may choose the si t o be in G’(p). Hence

where n = (G : P ) and z E G’(p). Thus g E K implies that g n E P’G’(p) = G’(p). It follows that K/G’(p) is a p’-group which forces K = G’(p). Finally, PnKer(TG,p) = PnG‘(p) = PnG’ since G’(p)/G’ is ap‘-group. Since I r n ( T ~ , p2) G/G’(p) 2 PG‘/G’ 2 P / ( P n GI), the result follows. H

Theorem 14. (Griin’s First Theorem, Griin (1935)). Let P be a Sylow p-subgroup of a finite group G , let N = N G ( P ) and let P* be defined by P* =< P n N ’ , P n g-lp’glg E G >

Then P n G’ = P’

and I r n ( T ~ , p2) P/P*

Proof. By Theorem 1.8 (iv), it suffices to show that P n G’ = P’. It is clear that P’ P n G’ and that P’ -'z(yz) E H . Then u = y-lzy E H and z-luz E H . Hence u-'~-~uz E H n N = 1, since N d G and H n N = 1. Thus y - ' s y = ( y z ) - l z ( y z ) ,as required. W

Characters and Hall Subgroups

892

As a preliminary to our next result, recall that, by Theorem 1.6, any n-separable group has at least one Hall n-subgroup.

Theorem 1.13. (Sah (1962)). Let n be a set of primes and let G be a n-separable group. Then the following conditions are equivalent : (i) G contains a normal Hall x'-subgroup of G . (ii) Each Hal2 n-subgroup of G is c-closed. (iii) At least one Hall n-subgroup of G is c-closed.

+ (ii) : This is 'a direct consequence of Lemma 1.12. (ii) j (iii) : Obvious. (iii) =+ (i) : We argue by induction on \GI. Let H be a Hall n-subgroup of G such that H is c-closed in G. Let M be a minimal normal subgroup of G. Then, by Lemma 21.4.3, M is either a n-group or a d-group. Assume that M is a n'-group. Then, by Lemma 1.11 (iv) and induction, G I M has a normal Hall A'-subgroup N I M . Consequently N is a normal Hall n'-subgroup of G. Assume that M is a n-group. By Lemma 1.1 (iii), M E H . Hence, by Lemma 1.11 (iii) and induction, GIM has a normal Hall n'-subgroup L I M . Then M is a normal subgroup of L. Hence, applying Theorem 1.10 (with M = H , L = G , A the automorphism group of L induced by conjugations by elements of H ) , we see that L contains a normal complement N to M . Then N is a normal Hall d-subgroup of G , as required. Proof. (i)

As a further application of Theorem 1.9, we shall establish a classical theorem of Burnside. The following preliminary observation will clear our path. Lemma 1.14. Let P be a Sylow p-subgroup of G . Then there exists a normal subgroup I< of G such that G / K is isomorphic to P / ( P n G').

Proof. Since PG'/G' is a Sylow p-subgroup of GIG', we may write GIG' = PG'/G' x K/G' for some normal subgroup I< of G . Hence G I K 2 PG'/G' "= P / ( P n GI), as desired.

Theorem 1.15. (Burnside). If a Sylow p-subgroup of a group G lies in the centre of its normalizer in G , then G has a normalp-complement.

1 An excursion into group theory

a93

Proof. Let P be a Sylow p-subgroup of G and let N = NG( P). Then, by hypothesis, P C Z ( N ) and, in particular, P is abelian. Hence P n g-lP'g = 1 for all g E G. Since P is a normal Hall subgroup of N , it has a complement H in N by virtue of Theorem 1.4. Thus N = P H and P n H = 1. However, P C Z ( N ) , so P centralizes H and hence N = P x H. Accordingly N' = ( P x H)' = H' and thus P n N' = P n H' = 1. Hence, by Theorem 1.9, P n G' = 1. It follows from Lemma 1.14 that G possesses a normal subgroup K such that G / K E P. Thus K is a desired p-complement.

Corollary 1.16. Let P be a Sylow p-subgroup of a group G. If P is abelian and q does not divide JAut(P)I for any prime q # p dividing [GI, then G has a normal p-complement. Proof. Let g E N c ( P ) and let < g >=< g1 > x . - - x < gn > be the decomposition of < g > into the direct product of primary components. By Theorem 1.15, it suffices to show that g E Cc(P). We may clearly assume that p does not divide I < g > I. Since conjugation by gi induces a homomorphism < g; >+ Aut(P), the assumption that q does not divide JAut(P)I ensures that the image of g; is the identity automorphism. Thus g E &(P) and the result follows. Corollary 1.17. If p is the smallest prime divisor of the order of G and G has cyclic Sylow p-subgroups, then G has a normal p-complement. Proof. Let P be a Sylow p-subgroup of G and let [PI = pn, n 2 1. Then JAut(P)J= p(pn) = pn-l(p - 1) Hence, if Q # p is a prime dividing IGl, then q does not divide IAut(P)I. Now apply Corollary 1.16. We close by providing the following useful observation.

Proposition 1.18. Let P be a normal abelian Sylow p-subgroup of a group G and let Z = P n Z ( G ) . Then there exists a normal subgroup Q of G such that P = Z x Q .

Characters and Hall Subgroups

894

Proof. Let T = (91,.. . , g n } be a transversal for P in G. Define

i=l

Because P a G and P is abelian, this product is independent of the order of the factors. Moreover, since P is abelian, f is independent of the choice of T . It therefore follows that f(glg2) = f(gl)f(gl) for all g1,g2 E P,so f is a homomorphism. Put Q = K e r f. If 2,y E P and y = g-l zg for some g E G, then gg; runs through a transversal for P in G as g; does, so f(z) = f ( y ) . Thus Q d G. Similarly, gig runs through a transversal, so g-'f(s)g = f ( z ) , f ( P ) 2. Finally, let 5 E 2 n Q. Then 1 = f ( 5 ) = zn and ( n , p ) = 1 forces 5 = 1. Thus 2 n Q = 1 and so 2 x Q E P. Consequently,

and therefore P = Z x Q, as desired.

2

Characters and Hall subgroups

All groups considered below are assumed t o be finite and all characters are (I:-characters. Let G be a group. Given a Hall subgroup H of G, when is there a normal complement of H in G ? Our principal goal is to provide a character-theoretic criterion in case H is solvable. By using the classification of finite simple groups together with the inspection of the character table of some of them, Ferguson (1989) established the corresponding result without the restriction that H is solvable. Our second goal is to prove a related result concerning Hall subgroups of complementary orders. Our approach relies heavily on two group- theoretic facts proved in the previous section, namely Theorems 1.10 and 1.13. Lemma 2.1. Let H be a subgroup of G such that each irreducible character of H can be extended to a character of G. Then (i) H is c-closed in G.

2 Characters and Hall subgroups

895

(iz) For each N d H , there exists Ii d G such that K n H = N and each irreducible character of H K / K may be extended to a character of G / K . Proof. (i) Assume that h l , ha E H are G-conjugate and let X I , . . . ,xr be all irreducible characters of H . Since each xi extends to a character of G, we have xi(h1) = Xj(hz), 1 5 i 5 r. Hence, by Theorem 19.2.3 (iii), hl and h2 are H-conjugate. (ii) For each irreducible character x of H with N 2 l i e r x , choose a character X of G extending x. Then KerX _> N and we denote by Ii the intersection of all such KerX. Then li a G and Ii fl H = N . Each irreducible character of H K / K gives rise t o an irreducible character of H/(H nK ) = H/N

via the inclusion map H Hli. By the choice of Ii,this character of H / N can be extended to a character p of G with K C I i e r p , i.e. to a character of G / K . --f

Theorem 2.2. (Sah (1962)). Let H be a solvable Hall subgroup of G. Then the following conditions are equivalent : (i) G contains a normal complement to H . (ii) Each irreducible character of H can be extended to a character of G.

Proof. (i)

j (ii) : This is obvious. (ii) (i) : Assume that each irreducible character of H # 1 can be extended t o a character of G. Let N be a minimal normal subgroup of H . Since H is solvable, N is a p-group by virtue of Lemma 21.4.2 (ii). By Lemma 2.1 (ii), there exists K a G such that K f l H = N and each irreducible character of H K / K can be extended t o a character of G/K. Now IC # 1, so by induction G/K contains a normal complement L / K t o H K / K . Then L 0. If X Q is irreducible, then since ( G : Q ) = p it follows from Clifford's theorem and Lemma 23.3.2 that p l x ( l ) , a contradiction. Thus XQ is irreducible and, by Theorem 21.2.18, x( 1) = q'. By hypothesis, G has

(G : N G ( P ) )= (PQ : PQ') = q2' Sylow p-subgroups. By Sylow's theorem, q2' ( q r - l)(q' -t 1) = q2' - 1, proving that

a desired contradiction. Step 4. By Step 3, we are left to verify that

= l(modp).

Thus p divides

Q is extra-special and N , ( P ) =

3 Degrees of faithful characters

899

PQ’. Assume that Qo is any proper subgroup of Q which is normal in G. Z ( G ) . Indeed, PQo is a proper subgroup of G and, We claim that Qo by induction, P a PQo. Thus PQo = P x Q o , P C Cc(Q0) # G and so P d Cc(Q0). Since P is characteristic in Cc(QO), we have P a G ; hence Cc(Qo)= G and Qo E Z ( G ) . Now let S/Q‘ be a proper subgroup of Q/Q’ such that S/Q’a G/Q’. Then S d G , S is a proper subgroup of Q and so, by the preceding paragraph, S c Z ( G ) , S/Q’ c Z(G/Q’). If Z(G/Q’) = G/Q’, then G‘ = Q’ contrary t o the fact that GIG’ is of order p by Step 2. Hence Z(G/Q’) = T/Q’ with S T Q . Owing t o Proposition 1.18, Q/Q’ = T/Q’ x L/Q‘ for some L a G. By the previous paragraph, T Z ( G ) ,L E Z ( G ) , Q Z ( G ) ,a

c c

c

c

contradiction. By the previous paragraph, Q/Q‘ is a minimal normal subgroup of G/Q‘ which forces Z ( Q ) = Q’, Q/Q’ is elementary abelian. By the first paragraph, Q’ = Z ( G ) and therefore N c ( P ) 2 PQ’. Assume that a proper subgroup H 2 Q’ of Q normalizes P . Then H a PH so PH = P x H and H a G. By the first paragraph, we have H Z ( G ) ,H = Z ( Q ) = Q’ and so N c ( P ) = PQ’. Finally, since Q‘ & Z ( G ) ,X Q I = x( 1)p for a linear character p of Q‘, XQ’ is faithful, Q‘ is cyclic. Choose a,b E Q with Q’ =< [a,b] >. Since Q/Q’ is elementary abelian, we have [a,b]q = [aq,b]= 1. Therefore IQ’I = q and Q is extra-special, thus completing the proof. H

Corollary 3.2. Assume that G has a normalp-complement N f o r some prime p and that G has a faithful character x with x ( 1 ) < p - 1. Then G = P x N , where P is an abelian Sylow p-subgroup of G .

Proof. Let q be a prime divisor of I N ] . Then every Sylow q-subgroup of N is a Sylow q-subgroup of G. Hence all Sylow q-subgroups of G lie in N . Let r be the number of Sylow q-subgroups of G and let Q be one of them. Then rI IN1 and r = (G : N c ( Q ) ) . Hence p 1 r and so there exists a Sylow q-subgroup S of G such that P N c ( S ) . Therefore P S is a solvable group and, by Theorem 3.1, P is abelian and S c N c ( P ) . Hence IN1 divides INc(P)I and so P a G , as desired. H

c

Lemma 3.3. Let H be a subgroup of a group G. If x is a character of G such that x ( h ) = 0 for all 1 # h E H , then /HI divides ~ ( 1 ) .

Characters and Hall Subgroups

900

Proof. We may assume that G = H . Then

) ,desired. Hence ]GI divides ~ ( l as

We next prove the following classical result.

Theorem 3.4. (Burnside (1911)). Suppose that G has a faithful irreducible character x of prime degree p. If p 2 divides IGl, then Z(G) # 1. Proof. Let P be a Sylow psubgroup of G. If P is not abelian, then the degrees of nonlinear irreducible characters are divisible by p. Hence x p is irreducible and, by Schur's lemma, Z ( P ) Z(G). Thus Z(G) # 1. Assume that P is abelian. If 1 # g E P and C is the conjugacy class of G with g E C, then

Thus, by Theorem 21.2.3 (v), either g E Z ( x ) = Z(G) or x ( g ) = 0 . But if x ( g ) = 0 for all 1 # g E P, then by Lemma 3.3, /PIdivides x ( 1 ) = p . Since p 2 divides IGl, we deduce that g E Z(G) for some 1 # g E P .

Corollary 3.5. Suppose that a simple group G has an irreducible character x with x(1) = p, where p is a prime. If P is a Sylow p-subgroup of G, then lPl = p (in particular, by Corollary 1.1 7, p # 2). Proof.

This is a direct consequence of Theorem 3.4.

As a preliminary to the proof of the next theorem, we establish a number of auxiliary results.

Lemma 3.6. Let H and S be nilpotent Hall subgroups of the same order n > 1 in a group G. Then H is conjugate to S in G. Proof. We argue by induction on IGI and choose a prime divisor p of n. Let P and Q be Sylow p-subgroups of H and S, respectively. Then, since H and S are Hall subgroups of G, P and Q must be Sylow p-subgroups of

3 Degrees of faithful characters

901

G. Hence g-'Qg = P for some g E G. Now put L =< H,g-'Sg >. Because H and g-'Sg are nilpotent, P is normal in both H and g-'Sg. On the other hand, HIP and g - ' S g / P are nilpotent Hall subgroups of L I P . Thus, by induction, they are conjugate in L I P . Therefore H and g-'Sg are conjugate in L , which shows that H and S are conjugate in G. H Given a primitive complex n-th root of unity

E,

we put

Qn = Q (E) Lemma 3.7.

I,, m , n > 0 are integers with ( m ,n ) = d, then QmnQn=Q,

Proof. Let t be the least common multiple of m and n. Then, by Galois theory,

Gal(Q t / ( Q

m

n Q n)) 2 G Q ~ ( Q t/Q

m)

x Gal($

t / Q n)

Accordingly,

which shows that

P(~)/((Q m n Q n> : Q ) = ~ + ) ~ / q ( m ) ~ ( n > However, by the definition o f t and d , we have cp(t)cp(d)= cp(rn)cp(n).Thus

Because (Q

d

((Qm Q n ) : Q = d d )= (Q d : 9 c Q n Q n , we deduce that Q d = Q n Q n .

H

We say that an algebraic number a requires n-th roots of unity if a € Q n and

a $ Q m for any m < n

The following lemma is due t o Blichfeldt (1904) and we present a proof due to Brauer (1964a). Lemma 3.8. Let x be a chamcter of a group G. Assume that P I , . . . , p r are distinct primes and that there exist g; E G such that x ( g i ) requires p;'-th

Characters and Hall Subgroups

902

roots of unity for some

ti

> 0 , i = 1,2,. . . ,r . Then G contains an element

of order p y p ? . . . p F .

Proof. Assume that g E G is of order m. Then x ( g ) E Q m . Hence, by Lemma 3.7, if x ( g ) requires n-th roots of unity, then

X(d

E

Q m

nQn

=Qd

where d = ( m , n )

This shows that nlm. Hence pti divides the order of g i , 1 5 i 5 r. Let pmi be the largest power of pi dividing the order n of G. Then Hence t; . Because the A-subgroups of G are abelian, Lemma 3.6 tells us that all Hall A- subgroups of K are conjugate to H in K. Therefore there are precisely m = (IS' : N h . ( H ) ) , such groups. Of course, since q E A , q does not divide m. But g permutes the set of all Hall A-subgroups of Ii by conjugation, so for some Hall n-subgroup M , say, of Ii we have g - l M g = M . Hence < M , g > is a required Hall n-subgroup. Step 2. Reduction to the case where x is irreducible. We argue by in. that x is reducible and write x = x1 x 2 . duction on n = ~ ( 1 ) Suppose Let Ni = K e r x ; so that GIN; is isomorphic to a linear group of degree x i ( 1 ) < n . By the induction hypothesis, GIN; has a Hall n-subgroup H i / N i . If Hi C G, then the inductive hypothesis yields a Hall A-subgroup H of H;. Since no prime in A divides I(G/N;) : (H;/Ni)l = IG : Hil, it follows

+

Characters and

904

Hall Subgroups

that H is a Hall n-subgroup of G and we are done if Hi C G. We may therefore assume that H ; = G and G/Ni is a n-group for i = 1,2. However, N1 n Nz = KerXl fl ICerX2 = K e r x = 1 and hence G is isomorphic to a subgroup of the r-group (GIN*)x (GINZ). Thus we may assume that x is irreducible. Step 3. Here we complete the proof by treating the case G = G'. By Step 2, we may regard G as an irreducible subgroup of GL,((C ), n = x( 1). Since G' = G and det(s-'y-'xy) = 1 for all z,y E G,we have G E S L n ( C ) . Hence if g E Z ( G ) ,then g = E * I and det(g) = E~ = 1. This shows that if 1 # g is a n-element, then CG(g) # G. By Lemma 3.9, if p ; E n and pjl IGl, then for some g; E G, x(g;) requires pi-th roots of unity. Hence if pl,. . .,p, are the distinct primes in A dividing IGl, then G has an element g of order pl - - -pr by Lemma 3.8. By the preceding paragraph, C G ( g ) # G and so by induction C G ( g ) has a Hall a-subgroup H . In what follows, we shall prove that H is a Hall n-subgroup of G. For each p i , 1 5 i 5 r , pi divides the order of g and hence of C G ( g ) . Let P; be a Sylow pi-subgroup of G such that Pi n H # 1 is a Sylow pi-subgroup of H . Now < Pj,H &(Pi n H ) # G, and hence by induction H is a Hall n-subgroup of Cc(P;n H ) . Thus (Pi1divides ( H I . Because this is true for all i E { 1,. . . ,r } , we infer that H is a Hall n-subgroup of G.

>c

The following consequence of Theorem 3.10 can be found in Isaacs (1976).

Corollary 3.1 1. Assume that for every nonlinear irreducible character Then G is solvable.

x of a group G, x(1) is prime.

Proof. The hypothesis is inherited by factor groups and, in view of Clifford's theorem, also by normal subgroups. It therefore suffices to derive a contradiction from the assumption that G is nonabelian simple. Fix an irreducible character x of G with minimal x( 1) # 1. Owing to Corollary 3.5, x(1) = p # 2. Put A = {qlq is prime, q > p t 1) Now G is simple, so x is faithful and therefore Theorem 3.10 implies the existence of an abelian Hall n-subgroup A of G. Given 1 # a E A , we have A C Cc(a). Also the only prime divisors of (G : CG(U)) are 5 p + 1 and so 5 p. Thus, if X is an irreducible character of G with A( 1) # 1, p then (X(1),

(G: Cc(a>)= 1

3 Degrees of faithful characters

905

Invoking Theorem 21.2.3 (v), we deduce that either A(a) = 0 or a E Z(X). But G is simple, so Z(X) = 1 and therefore A(a) = 0. Setting XI,.. .,x,., x,.+l,.. . ,xs t o be all irreducible characters of G with xi(1) = p for i E ( 1 , . . ,T } and x j ( 1 ) # p for the remaining i’s, we deduce that

.

3

i=l

T

i=l

But then l / p is an algebraic integer, a contradiction.

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Chapter 27

Extensions of Characters In this chapter we tackle the following two problems. First, we examine conditions under which an irreducible (I: -character of a normal subgroup of a finite group can be extended to a character of the group. Second, suppose that H is a subgroup of G. What are necessary and sufficient conditions for each irreducible (I: -character of H to be extendible to a (I:-character of G ? Our first result, related to the first problem, asserts that if G = N . H , where N is a normal subgroup of G and H a subgroup of G with N n H N’, then any irreducible G-invariant (E -character x of N with ( ~ ( lIHI) ) , = 1, extends to a (I:-character of G. As an immediate consequence, we derive some classical results of Gallagher (1962b) and Mackey (1951). We then provide some further results, due to Gallagher (1962b), which involve determinants of characters. As one of the applications of the results presented, we prove a theorem, due to Thompson (1970b), which asserts that if the degree of each irreducible (c -character of a group G is divisible by a prime p , then G has a normal pcomplement. Concerning the second problem, we present two results due to Isaacs (1986b). Namely, we show that if R is a set of primes and H is a maximal solvable n-subgroup of G such that each irreducible (I: -character of H is extendible to a (I: -character of G, then G has a normal n-complement and H is a Hall n-subgroup of G. We also prove that if P is a Sylow p-subgroup of G and each irreducible (I:-character of N c ( P )is extendible to a (I: -character of G, then N G ( P )has a normal complement in G. 907

Extensions of Characters

908

1

A general criterion

In what follows, G denotes a finite group and all characters are assumed to be (I: -characters. A map p : G + GL,((I: ) is a projective representation of G if p(1) is the identity matrix and, for each x , y E G , there exists a ( z , y ) E ( c * such that p ( x ) p ( y ) = a ( x , y ) p ( z y ) . Of course if a ( x , y ) = 1 for all x , y E G , then p becomes an ordinary representation. If N is a normal subgroup of G, then m y character of GIN will be identified with a character of G whose kernel contains N . The proof of the following result is based on a work of Thompson (1970 b) (see also Karpilovsky (1984)). Theorem 1.1. Let N be a normal subgroup of G, let x be a G-invariant irreducible character of N and let be a matrix representation of N which aflords x. Then (i) There exists a projective representation p of G such that

r ( n )= p ( n ) and p ( g ) ~ ( g= ) I for all n E N , g E G (here o(g) denotes the order of g and I is the identity matrix). (ii) If G = N - H for some subgroup H of G and if plH is an ordinary representation of H , then x can be extended to a character of G.

Proof. (i) Let g be a fixed element of G. Because x is G-invariant, the representations and gI' of N , where T ( n ) = r ( g - l n g ) , n E N , g E G , are equivalent. Thus there is a matrix e ( g ) such that

e(g>-'r(n)e(g)= r(g-'ng)

( g E G,n E N )

(1)

and we may assume that e(n)=r ( n )

for all n E N

To see that 0 is a projective representation of G, fix have

(2) g1,g2

E G. Then we

1 A general criterion

909

Since 1 E N , O( 1) = r(1)is the identity matrix, proving that 0 is a projective representation of G. If g E N , then by (2), O(g)"(g) = I. Suppose now that g E G - N and let n = o(g). Because e(g") = 0(l) = I , there exists A(g) E cC* such that 0(g), = X ( g ) I . Let p ( g ) 6 C * be such that p(g)" = X ( g ) - l . Setting p ( n ) = 1 for all n E N , it follows that the representation p defined by p ( g ) = p(g)O(g) satisfies the required property. (ii) Let T be a transversal for N n H in H containing 1. Then each element g in G has a unique representation g = tn with t E T , n E N . If g1 = t l n l is another element of G, write t t l = t2n2 with t2 E T , nz E N fl H . Define $ ( g ) = p ( t ) p ( n ) . Then, applying (l),(2) which holds for p and the assumption that p ( x y ) = p ( z ) p ( y ) for all z,y E H , we have

Thus $ is an ordinary representation of G. Since, for all n E N , $(n) = p ( n ) = I'(n), the character afforded by extends x. This completes the proof of the theorem.

+

Corollary 1.2. Let G = N . H where N is a normal subgroup of G and H is a subgroup of G such that N fl H C N'. If x is an irreducible Ginvariant character of N such that ( ~ ( l )[ ,H I ) = 1, then x can be extended to a character of G.

Proof. Put m = x(1) and n = IHI. Keeping the notation of Theorem 1.1, assume that g E N n H . Then p ( g ) = r ( g ) , so d e t p ( g ) = 1 since N f l H C N'. If g E H - N , then by Theorem 1.1 (i), d e t p ( g ) = E , where E~ = 1 and k = o(g). Since ( m , n )= 1, we have ( m , k ) = 1 and so there is a k-th root of unity 6 such that det 6p(g) = 1. Thus we may assume that d e t p ( g ) = 1for all g E H . Calculating the determinants in (3) (with 0 = p), we then have 4g1,gz)" = 1 for all g1,gz E H (4) Now consider the central subgroup 2 =< a ( g l , g z ) I l g l , g z E H > of the group K =< p(g)lg E H >. Then the mapping H + K / Z , h H p ( h ) Z is a surjective homomorphism. By (4) and the assumption that ( m , n )= 1, the orders of 2 and K / Z are coprime. Thus Ir' = 2 x S for some subgroup S

Extensions of Characters

910

of I=< P,r > This is true since by Corollary 2.2, ,Bx(H) and y x ( H ) are equal only for p=y.

Let x be a character of G and let p : G 4 GL,((C ) be a representation of G which affords x. Define

det(X) : G

-, (E

by

[det(x)](g)= detp(g) for all g E G Then det(X) is a linear character of G which does not depend upon a choice of p. Following Isaacs (1976), by the determinantal order of x , written o ( x ) , we understand the order of det(X) in the group Hom(G, (I2 *). Thus o ( x ) = (G : Ker(det(X))) and so o ( x ) divides the order of G.

Theorem 2.5. (Gallagher (19666)). Let N be a normal subgmup of G and let x be a G-invariant irreducible chamcter of N such that

( X ( l ) , ( G: N ) ) = 1

Then (i) x extends to a chamcter of G if and only if so does det(x). (ii) If det(x) eztends to a character X of G, then there is a unique character p of G such that p~ = x and det(p) = A. Proof. I f x extends to a character of G, then obviously so does det(x). Hence it suffices to prove (ii). Assume that det(x) extends to a character X of G. To prove the uniqueness, suppose that x extends to a character 8 of G. Then, by Corollary 2.2, all extensions of x are given by a0,where a ranges over all linear characters of G I N . Since Xdet(B)-l is a linear character of G I N , and x( 1) is prime to (G : N ) , there is a unique linear character ,B of G I N for which P X ( l ) = Adet(O)-l, i.e. for which

det(p8) = p X ( l ) d e t ( e )= x

Extensions of Characters

914

This proves the uniqueness. To establish the existence, assume first that GIN is supersolvable. If G = N ,then put p = x. If G # N ,let M / N be a normal cyclic subgroup of GIN. Since x is M-invariant, Lemma 23.3.2 shows that x has an extension to a character of M . By the preceding paragraph, x has a unique extension 7 for which det(7) = AM. Now for each g E G , 9 7 is an extension of x to M for which det(gy) = AM. Thus 9 7 = 7 and 7 is G-invariant. Hence, by induction, 7 extends to a character of G. Turning to the general case, we first observe that elementary subgroups are supersolvable. Therefore there are unique extensions x ( H ) of x to the intermediate groups for which H / N is elementary and d e t ( X ( H ) ) = AH. Applying uniqueness of these extensions together with the fact that X is G-invariant, we see that they satisfy conditions (i) and (ii) of Theorem 2.4. Thus x can be extended to a character of G and the result follows.

Corollary 2.6. Let N be a normal subgroup of G and let invariant irreducible character of N such that (x(l),(G : N ) ) = 1 and

x

be a G-

det(;y) = 1~

Then there is a unique character p of G such that p~ = x and det(p) = 1 ~ . Proof. The principal character 1~ of N extends to the principal character 1~of G. The desired conclusion is therefore a consequence of Theorem 2.5 (ii).

For an alternative proof of a slightly different version of Corollary 2.6, we refer t o a work of Glauberman (1968). In view of Theorem 2.5, it is important to discover conditions for extending linear characters. The following lemma contained in Isaacs (1976) will enable us to take full advantage of Theorem 2.5.

Lemma 2.7. Let N a G, let x be a G-invariant linear character of N , and for each p J o ( x ) ,let H p E G be such that H , / N is a Sylow p-subgroup of GIN. (i) If x is extendible to a character of each H p , then x is extendible to a chamcter of G. (ii) If ((G : N),o(x)) = 1, then x extends to a unique chardcter X of G for which ((G : N),o(A)) = 1 (in fact o ( x ) = .(A)).

3 Two results of Thompson

915

Proof. (i) Put n = o ( x ) and, for each pln, choose a power x p of x such that x = x p and o(xp) is a power of p . It suffices to show that xp extends to a character a, of G, for then X = napis an extension of x. Since xp is a power of x,which is extendible to H,,we see that xp extends to a character of H p . Thus we may assume that n is a power of p . Let p be a character of H p with p~ = x. Since p does not divide p G ( l), there exists an irreducible constituent Q of pG with p not dividing ~ ( 1 )Now . < Q H , , ~># 0 and so < a ~ , ># x 0. Because x is G-invariant, we have C ~ N= rnx where m = ( ~ ( 1 ) .Thus ( d e t a ) ~ = xm. Since ( p , r n ) = 1 and n is a power of p , we may choose t E Z such that rnt = l(rnodn). Then [(det a ! ) t ] N = xmt= x, proving (i). (ii) By (i), x extends to a character p of G. Put n = o ( x ) and choose m E Z with m(G : N ) f l ( m o d n ) . Setting X = p3, s = m(G : N),we then have AN = xs = x. In particular, o(A) 2 n. On the other hand, ( p " ) ~= 1~ so that pn is a character of GIN and A" = psn = lc, proving that .(A) = n. To establish uniqueness, assume that a is a character of G with ( Y N = x and ((G : N ) , o ( a ) ) = 1. Then (o(X&), (G : N ) ) = 1 and (X&)N = IN so that A& is a character of GIN. Thus XO = 1~ and a = A, as asserted. W

n

Corollary 2.8. Let N be a normal subgroup of G and let x be a Ginvariant irreducible character of N such that ((G : N ) , o(X)x(l)) = 1. Then x extends to a unique chamcter X of G such that ((G : N ) , .(A)) = 1 (in fact, .(A) = o ( x ) ) .

Proof. By Lemma 2.7, d e t x has a unique extension to a character X of G for which ((G : N ) , .(A)) = 1 (in fact, .(A) = o ( x ) ) . Hence, by Theorem 2.5, x extends to a unique character p of G with det p = A. Then o ( p ) = o(N = 4 x ) . Assume that x extends to a character Q of G with ((G : N ) , .(a)) = 1 and put ,b' = detcr. Then ((G : N ) , @)) = 1 and hence p = X by the uniqueness of A. Hence a! = p by the uniqueness of p, as desired.

3

Two results of Thompson

All groups below are finite and all characters are (c -characters. We pursue our goal to discover conditions for extending characters from normal subgroups. Given a prime p , we write O p ( G ) for the subgroup of G generated by all elements of G of order prime to p . If n is a positive integer, then Gn

916

Extensions of Characters

denotes the subgroup of G generated by all elements of the form gn with g E G. We denote by Q , the field obtained by adjoining a primitive n-th root of unity to Q , and if x is a character of G, Q (x)denotes the field generated over Q by all x ( g ) with g E G.

Theorem 3.1. (Thompson 1970b)). Let p be a prime, let P be a Sylow p-subgroup of G and let K be a subgroup of Op(G) containing P n Op(G). Assume that P N G ( K )and that x is a nonprincipal linear character of K such that OP(K)[P,P n Op(G)] C KeTX Then Op(G) has a G-invariant nonlinear irreducible character X of p'-degree such that (i) < X K , > ~ is prime to p. (ii) The Sylow p-subgroup o f G a l ( Q ,,/Q ( x ) )stabilizes A, n = ]GI. (iii) X can be extended to a character of G. (iv) x can be extended to a character of P K with the same order as x. Proof. Put H = OP(G) and Q = P n H . Because K = QOP(K) and Op(K)[P, Q] K e r x, we see that P stabilizes x. Hence xH is stabilized by PH = G. Since the Sylow p-subgroup A of G a l ( Q ,/Q ( x ) )stabilizes x , A stabilizes xH. Let XI,. ..,A, be all irreducible characters of H . Since "('Ax;) = "Xi) for all u E A , g E G, the p-group (G/H) x A acts on {XI,. . . ,A,} via ( g H , o ) : X i H "('Xi). Let 01,,..,01 be the orbits of {XI,. , , A T } under (G/H) x A and let a j be the sum of all elements in O j , 1 I. j 5 t . Since A and G stabilize xH,we have xH = Cs=lnjoj for some integers nj 2 0. Since ~ ~ (= 1 ( H): K ) f O(modp), there exists j such that njcrj(1) f O(modp). Since aj(1) = lOjlX(l), where X E Oj, and (G/H) x A is a p-group, we have lOjl = 1, i.e. O j = {A]. Thus A is an irreducible G-invariant character of H of p'-degree and A is stabilized by A , proving (ii). By Frobenius reciprocity, < XK,X >=< X,xH >, so (i) holds. If X(1) = 1, then Q C_ K e r X , since H = Op(G). Since (i) holds, we obtain XK = x so Q C K e r x . Since K = QOP(K), we have x = 1~ contrary to our assumption. Thus A is nonlinear. Since, by Corollary 1.3 (iii), (iii) holds, we are left to verify (iv). By (iii), X extends to a character (Y of G. Let M = {PI,. . . ,p s } be all linear characters of G/H. Then, by Corollary 2.2, X = { p l a , . .., p S a } is the set of all characters of G which extend A. For each cp E X , let v = ~ ( ( p )= d e t p . Then v(pp;) = p?(')v(p), 1 5 i 5 s. Since M is a Sylow

.

3 Two results of Thompson

917

p-subgroup of Hom(G,(r: *) and (p, X(1)) = 1, { ~ ( c p p ; )1, 2 i 5 s} is a coset of M in Hom(G,CE *). Thus, there is exactly one element 90 E X such that ~ ( 9 0 is ) a #-element of Hom(G,CE *), Since X is stabilized by A , X admits A , so the uniqueness of cpo implies that 90is stabilized by A . This is so since each element of A may be extended to an automorphism of C! to get that ~("9) = "~(9) for all u E A. Put L = Plr' and y = ( 9 0 ) ~ .Then y~ = XK = 6, say. Now write y = yo 71, where yo is a linear combination of irreducible characters 8 of L for which p ( < O K , x >, and y1 is a linear combination of the remaining irreducible characters of L. Since Ir' 4 L and since L stabilizes x, it follows that for each irreducible character 8 of L , 8 K is either orthogonal to x or 8 K = 8 ( l ) x . Hence, if A is the set of all irreducible characters of L of p'-degree whose restriction to Ir' is a multiple of x,then y1 is a linear combination of the elements of A and A admits A. If 8 E A, then O P ( 1 i ) C K e r 8 . Because L/OP(Ir') is a p-group, we see that 8(l) = 1 and so A C Hom(L, (r: *). Let A 1 , . . . ,A, be the A-orbits of A and let 6; be the sum of all elements in A;, 1 5 i 5 t . Because A stabilizes yl, we see that y1 is a linear combination of 61,. . ., S t . Because < 6, x > is prime to p , so is < y ~x , > and thus < ( y l ) Q , x> is prime to p . Since ( n )=~y1(1)x, it follows that p is prime to y1( 1). Thus p i s prime to Si(1) for some i E { 1,.. . ,t } . But A is a pgroup, hence A; = {hi}. Because O P ( K ) C K e r 6;, the order of 6; is a power of p. Since ( 6 ; ) ~= x , we have Q(6;) 2 Q ( x ) . Because A stabilizes hi, we have a)(&) C F , where F is the fixed field of A. Thus ( Q n : F ) = IAl. But A is the Sylow p-subgroup of Gal(Q n / Q (x)),so ( F : Q (x))is prime to p. Because ( Q ( S i ) : Q (x))is a power of p, we have Q ( x ) = Q(6i). Finally, since x # l ~x and , 6i have the same order, thus completing the proof. m

+

As a preliminary to our final result, we record the following lemma. Lemma 3.2. Let m be a natural number, let Q be a subgmup of G of index prime to m and let Z be a subgroup of Q n Z ( G ) . If x is a linear character of Q and x, xz have the same order t with tlm, then xz can be extended to a linear character of G of order t .

Proof. Let A = GaZ(Q ,/Q( x ) ) ,n = [GI, and let p = det(XG). Then A stabilizes x, so stabilizes xGand thus stabilizes p. Hence Q ( p ) C Q (x). Because pz = (xz)",where s = ( G : and ( m , s ) = 1 , we have Q ( X Z ) C

a),

Extensions of Characters

918

Q ( p ) . Since x and xz have the same order, we have Q ( x ) = Q(xz) = Q ( p ) . Hence the order of p is t or 2t. Assume that the order of p is t. Because ( m , s ) = 1, there exists k such that sk f l ( m o d t ) . Setting X = pk, we see that X has order t and Az = xz. Assume that the order of p is 2t. Since Q ( p ) = Q (xz),it follows that t is

odd. Because ( s , t ) = 1, there exists r such that r p2r has order t and extends xz,as required.

a 2 9 f

l ( m o d t ) . Hence

Theorem 3.3. (Thompson (19706)). Let m be a natural number, let H be a subgroup of G of index prime to m and let N be a normal subgroup of G such that G = NH. If S = N n H and x is a linear character of S n N"[G, N ] with S*[H, S]C_ K e r x then x may be extended to a character of H of the same order.

Proof. We may assume that x has order p e , e 2 1, where p is a prime. Let m = T S , where T is a power of p and (p,s) = 1. Note that

is an abelian group and

contains the Sylow psubgroup of (Sn N T [ GN, ] ) / S r n [ HS,] . Hence x can be extended to a linear character p of SnN'[G, N ] of order p e . Since ( p , s) = 1, we have S'[H, S ] C Ker p. Replacing m by T , we may therefore assume that m is a power of p , in which case

O P ( N )C N m

Nrn[G,N]

Let P be a Sylow psubgroup of H , so that P is a Sylow psubgroup of G. Put PI = P fl N , Pz = P fl O P ( N ) . Then N = PlOP(N) and N"[G, N ] = Pr[G,P1]OP(N). Taking into account that G = N H , we obtain

and so

S n N m [ G ,N] = P,"[H,P~](s

nO P ( N ) )

4 Thompson's theorems

919

Because PY[H,Pl] Sm[H,S], we obtain P?[H,Pl]C K e r x . Let x1 be the restriction of x to S n OP(N). Then x and x1 have the same order. By Theorem 3.1, x 1 may be extended to a linear character a of IC = P ( S n O P ( N ) ) of the same order as x. Because H stabilizes x, H must also stabilize X I . Thus

Finally, put Go = H I D , 20 = ( S n O p ( N ) ) / D , PO = K / D . Then a is a linear character of Po which extends xl, where we view a and x1 as linear characters of PO and 2 0 , respectively. Owing to Lemma 3.2, x1 may be extended to a linear character p of Go of the same order as X I , and we regard p as a character of H . Consequently,

Sm[S,H] c Iierp and therefore p is an extension of

4

x,as desired.

Thompson's theorems

In this section, by applying Theorem 3.1, we prove two theorems of Thompson. All characters below are assumed to be (C-characters and all groups are finite.

Theorem 4.1. (Thompson (1970 b)). Let p be a prime and let p divide the degree of each nonlinear irreducible character of a group G. Then G has a normal p-complement.

Proof. Let P be a Sylow p-subgroup of G and let Op(G)be the subgroup of G generated by all elements of G of order prime to p . Since Op(G)a G and G/Op(G)is ap-group, it suffices to show that PnOp(G) = 1. Assume by way of contradiction that P n W (G )# 1. Setting IC = P n Op(G),it follows that P C N G ( K ) and

oP(K)[P,P n OP(G)]= [P, IE Z for all x E I T T ( G ) .

Proof. (i) The desired assertion is equivalent to the requirement that

. .,X, be all Q -conjugacy classes of G and let gi E Xi 1 5 i 5 n . Let XI,. If g E G is of order m, E a primitive m-th root of 1 in 03 and p is a positive integer coprime to m, then for any x E I r r ( G ) , where up E Gal(Q (&)lo ) is defined by a P ( &= ) &EXi X ( d , we have

EP.

Hence, setting X i =

1 Characters and flxed-point spaces

941

which shows that

Xi E Z 5 Finally, given

for d l i E {l,...,n]

(2)

x E Irr(G),it follows from (2) that n

proving (1). (ii) Let z1,22,. ..,irbe representatives for the conjugacy classes C1,. .,,C, of G. Then for any x E Irr(G),

x(l)-llGl < 0,X > =

c x(1>-lwX(z> c

xEG r

=

x(l)-'a(~i)x(~i)lcil

i= 1

and this is an algebraic integer since x(l)-lx(zi)lC~lis an algebraic integer by Theorem 21.1.1 (iv). On the other hand, by (1)

x(l)-*lGl < a,x >E fQ and so (ii) is established. is clearly constant on Q Let x be any (C -character of G. Then conjugacy classes of G. Hence, by Theorem 1.1 (i), lGli is a generalized character of G. Our next aim is to demonstrate that a multiplier m smaller than IGl is sufficient to make rn2 a generalized character. Let cr be any class function of G and let n be a positive integer. We define a new function dra) :G + C by d n ) ( g ) = a(gn)

for all g E G

It is clear that dn)is a class function of G such that for any positive integer m, (,(n))(4 = ,(nd (3)

Fixed-Point Spaces and Powers of Characters

942

If x is a (I:-character of G, then it is not, in general, true that ~ ( is ~ a character; however, as we shall see below, x(") is always a generalized character.

Lemma 1.2. If x is a character (or a generalized character) of G, then for any positive integer n, ~ ( is ~ a generalized 1 character. Proof. For any characters XI,.. .,x, of G and any integers t l , . .., t , , we have (tlX1 * * * -t t r x p = 21$) 4- * * * trx!")

+

+

Hence we may assume that x is a character. Furthermore, by (3) we may also assume that n = p is a prime. Let V be a C G-module which affords x and let W = V 8 - @ V ( p times). Define the linear map f : W -, W by

-

Let

E

be a primitive p t h root of 1 in

(I:

and let

u = {w E WI f ( w ) = &W} We claim that U is a (CG-submodule of W which affords the character ( l / p ) ( x P - @ ) ) ; if sustained, it will follow that x ( P ) is a generalized character, as desired. To substantiate our claim, fix a basis v1,.. . ,v, of V and put

Given z E X , let

Now the group < f > permutes X into orbits of size p . Let 21,...,z, be a set of representatives of these orbits. Then each 5i E U and it is clear that 51,.. . ,g,,, is a basis for U . An easy calculation now shows that U is a (I: G-submodule of W which affords the desired character. H Given a function function

Q

:G

--f

(I:

and a positive integer n, we define the : G + CC

1

1 Characters and Axed-point spaces

943

by the formula x€G,x"=g

Lemma 1.3. If x is a generalized character of G and a is a positive integer, then x(l/") is a generalized character.

Proof. Given X E I r r ( G ) , we have

X E G yEG,yn=x

Y€G

=

< x,X(n) >

Since, by Lemma 1.2, A(") is a generalized character, we have < x,A(") > E Z which implies that x ( ' / ~is) a generalized character.

Lemma 1.4. Let N be a normal subgroup of G such that G I N is cyclic and let x be a generalized character of G such that

x ( z ) = 0 for all z E G - N Then x = aG for some genemlited character

(Y

of N .

Proof. We may view S = I r r ( G / N ) as a group of linear characters of G , in which case S acts on I r r ( G ) by multiplication. Given X E S, we have x = xX, since x ( s ) = 0 whenever X(s) # 1, for all x E G. Thus x is a Z -linear combination of sums of orbits of the action of S on Irr(G). It therefore suffices to show that each orbit sum is induced from some character of N . Let p E Irr(G) and let 8 be the sum of the S-orbit containing p . Fixing an irreducible constituent a of p ~ it ,will be shown that tJ = a', which will complete the proof. First of all, tJ and aG agree on G - N . Indeed, if x E G - N , we can choose X E S such that X(z) # 1. Because tJ is an S-orbit sum, we have OX = tJ and so e(z) = 0. On the other hand,).('a = 0 and so tJ and aG do agree on G - N .

944

Fixed-Point Spaces and Powers of Characters

Next note that each irreducible constituent of 8 is of the form p,L3 for some Because PN = l ~we, have ( ~ P ) N= p ~ Hence . 8 N is a multiple of p~ which is a multiple of the sum of the G-orbit of a. On the other hand, (&)N is a multiple of the G-orbit sum of Q. Consequently to prove that 8 and Q' agree on N (and therefore complete the proof), it suffices to compare degrees and verify that 8(l) = a'(1). Let H C S be the stabilizer of p , Then

P E s.

e(i) = (s : H ) ~ ( I ) If T G is the inertia group of a , then p = 7G for some y E I r r ( T ) and 7~ is a multiple of a. Because TIN is cyclic, we have Y N = a by Lemma 23.3.2. Thus p(1) = (G : T ) a ( l )and 8(l) = (S : H)(G : T ) a ( l ) . Since a c ( l ) = (G : N ) a ( l ) ,we must therefore show that (G : N ) = (G : T ) ( S : H ) . Thus it suffices to verify that (T : N ) = (S : H ) . We claim that H = I r r ( G / T ) ;if sustained, it will follow that ( S : H ) = ISl/lHI = (G: N ) / ( G : T ) = (T : N), as desired. To substantiate our claim, let

P E H . Then

and ( ~ / ? T ) N= Y N = a. But y is the unique character of T lying over Q which induces p , and hence y p = ~ 7 . Since 7 N is irreducible and N c lier(PT), it follows from Corollary 27.2.2 that PT = 1 ~ Thus . E I r r ( G / T ) which shows that H Irr(G/T). Conversely, assume that /3 E Irr(G/T). Then

and so /? E H , thus completing the proof. We have now accumulated all the information necessary to prove the following result.

Theorem 1.5. (Isaacs (1 989)). Let x be a character of a group G and let X be a generalized chamcter of G . Assume that m is any positive integer such that X ( g ) = 0 for all g E G

with g m

#

1

(4)

1 Characters and fixed-point spaces

945

In particular, e i is Q generalized character of G , where e is the exponent of the group G.

Proof. The condition of (4)is vacuously satisfied if m = e is the exponent of G and X € Irr(G). Hence the assertion concerning e i follows from the first statement. For the sake of clarity, we divide the rest of the proof into a number of steps. Step 1. Reduction to the case where G is nilpotent. Assume that ( 5 ) holds in case G is nilpotent. By Brauer's theorem on induced characters (Theorem 20.2.1 (i)), the principal character 1~can be written as a sum of certain cry where the aj are generalized characters of certain nilpotent subgroups Hi G. Then, by Lemma 19.3.2 (i),

Consequently, by Theorem 19.3.9,

and this is a sum of integers by ( 5 ) , because the XH,CY; are generalized characters which vanish on elements h € Hi such that h" # 1. Step 2. By Step 1, we may assume that G is nilpotent. Because the map x H 2 is additive, we may also assume that x is an irreducible character. We proceed by induction on x( 1) and first establish the case where x( 1) = 1. So assume that x is linear. Then, setting Ir' = K e r x , we have R(z) = 0 for z $ K and i ( z ) = 1 for z E K. Now put

so that N is the inverse image in G of the set of all elements in G / K whose order divide m. Because G / K is cyclic, N is a subgroup of G such that ( N : I

4,are G-conjugate.

AH

and this lies in 22 by the inductive hypothesis since +(1) < ~ ( 1 ) . To evaluate SZ, put p(y) = C z E G - H , x p = y X(s). Then

zEG-H YE H

=

Iff1

Hence mlG(-1S2 = ( m / p )< $ , p > and we are left to verify that this lies in 24. Let x E G - H and zp = y E H . Then p divides the order of 2. We distinguish two cases, Suppose first that p does not divide m. Then x m # 1 and so A(x) = 0 by hypothesis. Hence p is identically zero and there is nothing to prove in this case. = 1. Hence, Finally, assume that plm. If x m = 1 and x p = y, then if y(mlp) # 1, then X(z) = 0 for all x with z p = y and so p ( y ) = 0. Now x E G ,xP=y

xEH,zP=y

so p = (X@/P))H - (XH)('/~) and this is a generalized character by Lemma 1.3. Since, by the above, p ( y ) = 0 if Y ( ~ / P )# 1, the inductive hypothesis implies that ( m / p )< $ , p >E Z , as desired.

2

Powers of characters

In what follows, G denotes a finite group and all characters are assumed to be CC -characters. As before, given a positive integer n and a class function

Fixed-Point Spaces and Powers of Characters

948

a :G

+

CC , we denote by

the class function

dn): G

--f

(L:

given by

) the set of irreducible Also, for any character x of G, we write I ~ T ( xfor constituents of x. Our aim is to provide necessary and sufficient conditions for a power of an irreducible character of G to have exactly one irreducible constituent. We also determine necessary and sufficient conditions for a power of a conjugacy class of G to be a single conjugacy class.

x

Lemma 2.1. Let be a character of G. Then, for any positive integer n, there exist characters XI, Xz of G such that

and Irr(Xj)

I T T ( x ~ ) ( i = 1,2)

Proof. Since = 2x - x, the result is trivial for n = 1. We now argue by induction on n. Assume that n > 1. We may then write n = mp, where p is a prime and m is a positive integer. Since m < n, it follows by induction that x(") = p1 - pz, where pj is a character of G with I r r ( p j ) C- ITT(x"), i = 1,2. By the proof of Lemma 1.2, we have

where pi is a character of G afforded by a submodule of a module affording pp, i = 1,2. Therefore Irr(i&) G Irr(pY). We now show that

are the required characters. Indeed, X I and Xz are characters of G such that ~ ( =~X I 1- Xz. Bearing in mind that Irr(X1 4- Xz) E I T T ( ~U I~r )r ( p ; ) , we are left to verify that

2 Powers of characters

949

Since I r r ( p ; ) 2 I r r ( x " ) , we have k;x" = p; t ai for some positive integer ki, and some character a; of G', i = 1,2. Then

for some character follows.

p; of G, i = 1,2. Thus Jrr(pp) C Irr(xn) and the result

In what follows, n(n) denotes the set of all prime divisors of the positive integer n.

Theorem 2.2. (Blau and Chillag (1986)). (i) Assume that x and $ are irreducible characters of G such that xn = k$ for some positive integers n, k with n 2 2. Then x(g) = 0 for all g E G - Z(x), $ = ~ ( " 1k, = x(I)"-' and IGln(n)divides IZ(x)l. (ii) Assume that x is an irreducible character of G such that x ( g ) = 0 for all g E G - Z ( x ) . If n is any positive integer such that IGIA(n)divides IZ(x)l (e.g. (n,IGl) = l), then X" = kzl, for some positive integer k and some irreducible character II, of G (namely, Ic = x( l)"-' and $ = ~ ( " 1 ) . Proof. (i) Owing to Lemma 2.1, we have x(") = XI - A2 for some characters X * , X 2 of G with I r r ( X j ) 2 lrr(xn) = {zl,}. Hence X i = ki$ for some positive integer k;, i = 1,2, and therefore ~ ( = ~tzl, 1for t = Icl - k2. Because x(")(1) = x( 1) = t$( 1) we see that t > 0. Hence x(")is a character of G. Now xn = kzl, and x(")= tzl,, so that X n ( g ) = ( k / t ) X ( n ) ( gfor ) all g E G. Substituting g = 1, we have k / t = ~ ( 1 ) which " ~ ~shows that

Invoking ( l ) , we see that for any given g E G, 1x(g)1 = x(1) if and only i f Ix(g")l = ~ ( 1 )This . implies that g E Z(x) i f and only if gn E Z(x)

(2)

It will next be shown that x ( g ) = 0 for all g E G - Z(x)

(3)

Fixed-Point Spaces and Powers of Characters

950

Assume by way of contradiction that x ( z ) # 0 for some 5 E G - Z(x). By repeated application of (2), 5"' E G - Z(x) for all integers i 2 0 which < x(1) for each such i. Also, since x(z) # 0, repeated means that Jx(zn')J ) 0 for all i 2 0. Moreover, since n 1 2, application of (1) shows that ~ ( x " ' # it follows from (1) that

IX(Z"'>l = lx(l)/x(~n')ln-lIxk > Ix(xn'+')l

nr+l)

1

But then the set (Ix(zni)li 2 0) is infinite, which is a contradiction. Thus (3) is established. Invoking (2) and (3), we have Ix(g)l = Ix(g")l = 1x(")(g)1for all g E G. Hence < x ( ~ ) , x ( " >=< ) x , x >= 1 which shows that x(n)is irreducible and equals $. Since x" = k+, we also have k = x(l)"-'. Finally, let T = ~ ( nand ) assume by way of contradiction that lGln does not divide IZ(x)l. Then IG/Z(x)lr # 1 and so we may choose a nonidentity n-element gZ(x) of G/Z(x). Hence gnJ E Z(x) for some positive integer j. But then, by repeated application of (2), we have g E Z(x),a contradiction. (ii) Suppose that x is an irreducible character of G which vanishes off Z(x) and n is a positive integer such that IGIT(n)divides IZ(x)l. If g E G is such that gn E Z(x), then g E Z(x) since otherwise G/Z(x) would contain a n(n)-element. This shows that, for any given g E G, g E Z(x) if and only if gn E Z(x). Because x vanishes off Z(x), we deduce that

Xn(g) = X(l)n-lX(n)(g) = o for all g E G - Z(X) (4) On the other hand, X Z ( ~ )= x ( l ) p for some linear character p of Z ( x ) . Hence X n M = X(l)nP(g)n = X(l)"-lX(n)(g)

for all 9 E Z(X)

(5)

Taking into account (4) and ( 5 ) , it follows that xn = x(l)"-lx("). Now \x(g)l = \X(")(g)\for all g E G, since x vanishes off Z(x) and g E Z(x) if ~ ) 1. Since x(,)(l) > 0 and, and only if gn E Z(x). Thus < x ( ~ ) , x ( >= by Lemma 1.2, x(") is a generalized character, we deduce that ~ ( is~ an 1 irreducible character. This completes the proof of the theorem. We close by examining the dual situation, i.e. the one in which characters are replaced by conjugacy classes. Let C1, . ..,C, be some conjugacy classes of G. Then their product C1 Cz * C, is defined as follows :

-

+

-

C1.C2...Cn={g1g2 ...gnlgi ~ C ; , l ~ i ~ n )

2 Powers of characters

951

If C' = c, = * . * = C, = C, then this product will be denoted by C". Our next result provides necessary and sufficient conditions for a power of a conjugacy class of G to be a single conjugacy class of G.

Theorem 2.3. (Blau and Chillag (1986)). (i) Assume that C # (1) is a conjugacy class o f G such that C" is a conjugacy class of G for some integer n 2 2. Then the map C + C", x H x n is Q bijection, and C = g N forsome N a G a n d s o m e g E G - N . (ii) Assume that C is a conjugacy class of G with C = gN for some N d G and some g E G - N . If n is a positive integer such that the map C -+ C", x H x n is injective, then gnN is a conjuga.cy class of G and C" = g n N .

Proof. (i) Fix g E C and write C = (9,992,. . . , g g k } for some g 2 , . . . , g k in G. Put N = (1 = g l , g Z , . . . , g k } so that C = g N . We first show that the map C C", x I-+ 2" is a bijection. To this end, note that --f

g"-l(gg;) = gngi E

C"

(1 5 2 5 k)

which shows that C" 2 g"N. Because CG(g) S C G ( g n ) , g E C and gn E C", we have IC"I 5 ICI = IN1 = 1g"NI. Thus C" = gnN and ICI = IC"I. Because {t"lt E C} is a conjugacy class, namely C", we deduce that the map C + C", x H x" is a bijection. By the foregoing, we are left to verify that N is a normal subgroup of G and g $! N . By hypothesis, n 2 2 and so gn(g-'gig)gj = g"-'(ggi)(ggj) E C" = gnN which shows that g"(g-'gig)gj g-'giggj E N

(1 I i , j

5 k)

= gngt for some 1 5 t 5 Ic. Accordingly,

for all i , j E {1,...,k}

(6)

In particular, by taking j = 1, we see that g-lNg = N . Hence any gr in N equals g-'gig for some i. Therefore, by (6), grgj E N for all r , j and thus N is a subgroup. Given y E G, we have y - l g y = gh for some h E N and gN = C = y-'Cy = (y-lgy)(y-'Ny) = gh(y-'Ny) Hence y-'Ny = h-'N = N , proving that N is normal. Finally, if g E N then C = gN = N and so C contains 1, a contradiction.

952

Fixed-Point Spaces and Powers of Characters

(ii) Suppose that C is a conjugacy class of G with C = g N for some N 4 G and some g E G - N. Assume further that n is a positive integer such that the map C -+ C", 2 H x n is injective. Then C1 = {g"lg E C} is a conjugacy class of G such that

Because (C1(= IgN( = Ig"N( by hypothesis, we see that Cl = C" = gnN, as required.

Chapter 30

Det errninant s of Characters This chapter provides a thorough investigation of determinants of characters. All the results presented are due to Gallagher (1965b) and Yoshida (1978). One of the main theorems gives an explicit formula for the determinant of the induced character xG in terms of the determinants of x and ( l ~ ) ' ,where x is a character of a subgroup H of the finite group G. A number of related results of a technical nature is also presented. Using properties of determinants of characters, we then introduce a character-theoretic tranfer. We also provide all necessary properties of this transfer which will be required for future group-theoretic applications. Especially important will be the Yoshida transfer theorem which generalizes a classical Wielandt's theorem. This theorem of Wielandt is equivalent to the assertion that if p is a prime and P is a Sylow p-subgroup of G then the p-structure of H'(G, (I: *) coincides with that of H 1 ( N ~ ( P ()I:,*). As we shall see later, this fact will be closely related to the local control of the Schur multiplier.

1

Determinants of characters

The object of this section is to obtain relations between the determinants of characters and the transfer homomorphism. The main tool is the formula for determinants of induced characters which was originally discovered by Schur (1904) in the case of central subgroups. A number of applications is also provided. Among one of them, we derive a character-theoretic interpretation of the principal ideal theorem for transfer. Another application provides conditions under which an irreducible character x of G restricts irreducibly to

953

Determinants of Characters

954

the kernel of det(x). The section ends with examining the order of det(x)(g), g E G , in terms of ~ ( l the ) , order of g and the number of conjugates of g . In what follows, G denotes a finite group and all characters are assumed to be 02 -characters. Given two square matrices A and B over a commutative ring, we write A 8 B for their tensor product. The direct sum A 63 B of A and B is defined by

Lemma 1.1. Let A and B be respectively n x n and rn x m matrices over a commutative ring R. Then (i) det(A @ B ) = (det A)"' (det B)" (ii) det(A @ B ) = (det A)(det B).

Proof. Property (ii) is obvious. For each k 2 1, let Ik be the identity k x k matrix over R. Then

and so

det(A 6jI B ) = det(A @I Im)det(In6jI B ) = det(A$...$A)det(B$...$B) = (det A)'"(det B)", as required.

Lemma 1.2. Let A be a permutation n x n matrix over a commutative ring R with charR # 2. Then detA = fl and det A = 1 if and only if A affords an even permutation.

Proof. We may clearly assume that n 2 2. Let G be the group of all n x n permutation matrices over R. Since for any A E G, A is a monomial matrix with nonzero entries equal to 1, we have det A = fl. The set N = { A E GldetA = 1 ) is a normal subgroup of G of index 2. Since G is identifiable with the symmetric group of degree n , the result follows. Recall that, for any character x of G, the determinant of det(X),is the linear character of G defined by

det(x)(g)= det(p(g))

( 9 E G)

x, written

1 Determinants of characters

955

where p is a matrix representation of G which affords x. When convenient, we shall treat detO() as a character of GIG’.

Proof. This is a direct consequence of Lemma 1.1. Let 0 E S, the symmetric group of degree n. Then the sign of 0 , written sgn(a), is defined by s g n ( a ) = 1 if o is even and s g n ( a ) = -1 if cr is odd. L e m m a 1.4. Let x be a permutation character afforded by a permutation (I: G-module with a permutation basis B . Then, for any given g E G, (i) d e t ( x ) ( g ) is the sign of the permutation o f B induced by g . (ii) det(X)(g) = 1 if gG’ is of odd order.

Proof. (i) This is a direct consequence of Lemma 1.2. (ii) The map GIG‘ -+ (I: *, xG’ H d e t ( x ) ( s )is a group homomorphism. Hence d e t ( x ) ( g ) cannot be of order 2 if gG”is of odd order. Now apply (i) and the result follows. The special case of Lemma 1.4 (i), in which status of a corollary.

x

= ( 1 ~ deserves ) ~ the

Corollary 1.5. Let H be a subgroup of G and let t l H , . ..,tnH be all left cosets of H in G. Then, for any given g E G, d e t ( ( l H ) G ) ( g )is the sign of the permutation of {tl H , ,t,H} induced by left multiplication by g .

. ..

x

Proof. The desired assertion follows by applying Lemma 1.4 (i) for = ( 1 ~ and ) ~B = { t l H , . . ., t n H } .

Let H be a subgroup of G. If { t l , . . . , t n ) is G, then for any g E G,

a left transversal for

gti = tg(i)H where the i

c)

g ( i ) is a permutation of the set {1,2,.

H in (1)

. . ,n}.

Recall from

Determinants of Characters

956

Chapter 26 that the transfer homomorphism

TG,H; G + N/H' is defined by

The following result for the case where H C Z(G) was established by Schur (1904, p. 25).

let

Theorem 1.8. (Gallagher (1965 b)). Let H be a subgroup of G and and X be characters of H and G , respectively. T h e n (a) det(XG)= [ d e t ( ( l ~ ) ~ ) ] X ( ' ) [ do eTG,H] t(~) (ii) det(X)(G'H)= det(XH)o TG,H.

x

Proof. (i) (Isaacs (1986a)). Let p be a matrix representation of H which affords x and let T be a left transversal for H in G. Then, by Proposition 19.1.1, xG is afforded by the "block monomial" representation pG with block positions indexed by pairs ( t , s ) E T x T . For g E G, the (t,s)-block of p G ( g ) is zero unless gt E s H in which case the block equals p(s-'gt) (here we change the notation in Proposition 19.1.1 in order to conform to ( 1 ) ) . To compute detpG(g), let us permute the rows of p G ( g ) with the aim to rearrange the blocks and put the matrix into block diagonal form. Then there exists a sign function E ( g ) = f l such that E(s)det(PG(s)) =

n

det &-%)

t€T

where in each factor on the right, sH. Thus

3

= s ( t ) in uniquely determined by gt E

det(xG)(g) = 47)det ( P ( n s - W )

(x)(n teT

= &(g)det

s-'gtP)

tET

= &(g)det(X)(TG,H(g))

(by (2))

It remains to compute E(g). If x ( 1 ) is even, the blocks in the matrix pG(g) have even size and there is no sign penalty for their rearrangement. If x ( 1 )

1 Determinants of characters

957

is odd, then E ( g ) is the sign of the permutation of { t H l t E 2') induced by g. By Corollary 1.5, this equals det((lH)')(g>. Thus in all cases 4 7 ) = [ d e t ( ( kG )

>I

x(1) (g),

proving (i). (ii) We first note that, by Proposition 19.1.7 (ii),

(1H)G * A = (AH)G By applying this fact and Lemma 1.3, we obtain, after substituting AH for x in (9,

[ d e t ( l ~ ) ~ ] ~ ( ' ) d e t ( X=) ( ~ ' ~ ) = [ d e t ( ( l H > G ) ~ ~ ( ~ o" TdG~, ,t~(]~, ~ ) By cancelling the first factors, the desired assertion follows.

Proof. We have d e t [ ( l ~= ) ~det((1K)') ] ~ by transitivity of induction. Applying Theorem 1.6 (i) to obtain the desired assertion. H

x

= ( l ~ we ) ~ ,

We now proceed to refine these results in case H is a normal subgroup of G. Theorem 1.8. (Gallagher (1965 b)). Assume that N is a normal subgroup of G. (i) d e t ( ( l N ) G )= 1~ unless the Sylow 2-subgroups of GIN are cyclic and nontrivia 1. (ii) If the Sylow 2-subgroups of G I N are cyclic and nontrivial, then d e t ( ( l N ) G )is the only linear character of G of order 2. Proof. (i) We apply Corollary 1.5 with H = N . Let g E G be such that the coset g N has order n in S = G / N . In the permutation action of G

Determinants of Characters

958

on the cosets G I N , the element g is therefore represented by a product of disjoint cycles, each of length n. Thus

det((lN)G)(g)= ( - 1 ) ( n - 1 ) ( l W n ) = (-1)l’I-l’Vn Thus d e t ( ( l N ) G ) ( g )= 1 unless IS[ is even and ISl/n is odd. The latter case can occur only if a Sylow 2-subgroup li;of S is contained in < g N >. Hence d e t ( ( l N ) G )= 1~ unless the Sylow 2-subgroups of GIN are cyclic and nontrivial, and in that case det((lN)G)# 1 ~ . (ii) We claim that S has a normal 2-complement, which will easily imply the required assertion. To substantiate our claim, denote by L the kernel of d e t ( ( l N ) G )in S. Then L C S and, by induction, L has a normal 2complement M . Since M is characteristic in L , we have A4 4 S. Thus M is the required normal 2-complement. W The proof of the following result uses some elementary facts concerning projective representations of groups. These properties (in a much more general form!) will be established in our future study of projective representations of groups. It would be undesirable to postpone the proof of the theorem below since it will interrupt the flow of exposition of the topic under consideration.

Theorem 1.9. (Gallagher (1965 b)). Let N be a normal subgroup ofG, let x be an irreducible chamcter of N and let A be an irreducible constituent of xG. Then e = (G : N ) x ( l ) / X ( l )E 21; and

(det(X)Ie= [d.t((lN)G)le~(’)[det(~)0 TG,N]

Proof. By F’robenius reciprocity, x is an irreducible constituent of AN. Hence, by Clifford’s theorem, there exists an irreducible character p of the inertia group S of x such that pG = A

and p~ = t x

for some positive integer t . The integer t is the degree of an irreducible projective representation of S I N . By a theorem of Schur, t divides (5’ : N ) and therefore (G : N ) x ( l ) --(S : N ) €22 t = tX(l)(G :S) From now on, for convenience, we abbreviate d e t ( ( l H ) G )by E G , H . Suppose first that there is a proper subgroup H 2 N and an irreducible

1 Determinants of characters

959

character (Y of H such that aG = X and ( Y N . Then, by Theorem 1.6 (i), det(X)

x is an irreducible constituent

= &$$det(a)

0

of

TC,H)

Hence using induction on (G : N ) and the fact that e = (H : N ) x ( l ) / a ( l ) , we have

Hence, in this case, it suffices to show that

In proving (3), we may assume that e is odd. Then a(1) has the same parity as (H : N)x(l). Thus (3) follows from Corollary 1.7. Next assume that there is a proper normal subgroup K 2 N such that G / K is cyclic. If EG,N # l ~we, take the normal 2-complement of GIN as K I N . If the previous case does not hold, then /3 = XK is irreducible by Clifford's theorem. Owing to Theorem 1.6 (ii), we have (det(X))(G:") = det(P) o TG,K

Hence, using induction on (G : N ) and putting f = (K : N)x(l)/X(l), we have (det(X))" = d e t ( @ ) jo TG," =

(~2:) T G J )(det(X) TG,N) o

a

Therefore, in this case, it suffices to show that

If EG,N # l ~then , (G : K ) = e / f is even and ( K : N ) is odd, by our choice , of K . Hence, E ; , ~ = 1~ and EK,N = l ~ so, (4) holds. If EG,N = l ~then by Corollary 1.7, (K:N) &H',N0 TG,K = &G,K

Hence, in proving (4), we may assume that ( K : N ) is odd. Then EK,N = 1~ and (4) holds. Finally, assume that neither of the two previous cases hold. Let L be the

960

Determinants of Characters

kernel of the quotient of the two sides of the equation in the statement of the theorem. Then G / L N is cyclic, so G = L N . Hence (G : L ) = ( N : ( N n L ) ) and so it suffices to prove the equation on N . Because the first case does not hold, = t x , where e = (G : N ) / t . In particular, x is G-invariant, so detO() is a character of N / [ G ,N ] . For g E N , T c , ~ ( g=) g(G:N)mod[G,N] by Theorem 26.1.8 (iii). Thus

(det(X))"(g) = det(x>""9) = det( x)(s ( ' : ~ ) )

= d e t ( x ) 0 TC,N(9), as desired.

As a preliminary to the next result, we remind the reader a standard fact, known as the Principal Ideal Theorem, that for all finite groups G, TG,Qis trivial (for a proof, refer to Zassenhaus (1949, p. 147)).

Corollary 1.10. (Gallagher (1965 b)). Let x be an irreducible character of GI, let X be an irreducible constituent of xG and let e = (G : G')x( l ) / X ( 1 ) Then (5) (det(X))e = det((lcl)G)eX(') Conversely, (5) for all linear characters x of GI implies that TG,G~ is trivial.

Proof. Put N = GI. Then TG,Nis trivial, so ( 5 ) follows from Theorem 1.9. Conversely, if ( 5 ) holds for all linear characters x of GI, i.e. if det(X)" = det((lct)G)",then by Theorem 1.9, T G ,is~trivial.

Corollary 1.11. (Gallagher (1965 b)). Let x be a n irreducible character of GI, let X be an irreducible constituent of xG and let K ( X ) be the kernel of det(X). (i) X ( l ) / x ( l )is a divisor of 2 ( K ( X ) : G') (ii) If K(X)/G' is a Hall subgmup of GIG', then XK(X) is either irreducible or the sum of two distinct irreducible characters. Proof. (i) Squaring both sides of ( 5 ) yields det(X)2e= 1. On the other hand, the order of det(X) is (G : K(A)). Thus (G : Ir'(X)) divides 2e. Since 2e = 2(G : K ( X ) ) ( K ( X ) : G')x(l)/A(l),

1 Determinants of characters

961

X(l)/x(l) divides 2(K(X): G'). (ii) Let a be an irreducible constituent of such that x is a constituent of QG'. Then A( l ) / a ( 1) divides A( 1)/x(1) and therefore divides 2(K(X): G') by virtue of (i). On the other hand, since G/K(A) is cyclic, it follows from Clifford's theorem and the fact that the irreducible projective representations of cyclic groups are of degree 1, that XK(X) is the sum of all distinct conjugates of a. Thus X(l)/a(l) divides (G : K ( X ) ) . Since, by hypothesis, ( ( K ( X ) : G'),(G : K ( X ) ) = 1, the desired assertion follows. We close by examining further conditions under which an irreducible character X of G will restrict irreducibly to the kernel of det(A).

Theorem 1.12. (Gallagher (1965 6)). Let A be an irreducible character of G and let K ( X ) be the kernel ofdet(X). If G splits over K ( X ) , then XK(A) is irreducible unless (G : K(A))/2 is an odd integer, in which case AK(x) may be the sum of two distinct irreducible characters.

Proof. Let a be an irreducible constituent of XK(x) and let H be the inertia group of a. As was mentioned in the proof of Corollary 1.11, AK(A) is the sum of all distinct conjugates of a. Thus it suffices t o show that H = G unless (G : K(A))/2 is an odd integer, in which case (G : H ) = 1 or 2. By Clifford's theorem, X is induced by an extension of a to H . Therefore X ( g ) = 0 for all g E G- H . By hypothesis, G = K ( A ) . S for some subgroup S of G with S n K ( A ) = 1. If S = 1, then there is nothing to prove; otherwise, we may choose a prime divisor p of IS(. Let g be a generator of the Sylow p-subgroup of S(S E G/K(X) is cyclic), and let pa be the order of G. Then det(X)(g)is a primitive pa-th root of unity. If p divides (G : H ) , then g 6 H and so X(g) = 0. Let p be a faithful linear character of < g >. Then A = f(P) and, by Lemma 1.3, (det(A)) = @'(*), where f ( X ) is a polynomial with nonnegative rational integer coefficients. If pJ(G : H ) , then X(g) = f(P(g)) = 0, so f ( X ) is divisible by the pa-th cyclotomic polynomial @(X)given by

@(X)= 1 +

+

.

XZP+' + . . + X ( P - ' ) P " - l

Because @ ( X )is monic, the quotient h ( X ) has also integer coefficients. Ac-

Determinants of Characters

962

cordingly,

=

O(m0dp)

unless pa = 2. Because det(X)(g) is a primitive pa-th root of unity, pa = 2. Hence H = G unless (G : K ( X ) ) / 2is an odd integer, in which case we have (G : H )= 1 or 2. So the theorem is true.

To prove our final result, we need some information of field-theoretic nature. Let F / K be a finite Galois field extension of degree n and let 6 1 , . . . ,un be all elements of GaZ(F/K). For any z E F, the norm N ~ / h ' ( zof ) z is defined by

nw n

NF/K(4=

i=l

It is clear that N F ~ K (E~I< ) and that N F i ~ ( z y = ) N F / K ( Z ) N F / K ( Y )for all s , y E F. Our interest lies in the special case where K = Q and F = Q n, where Q = Q ( E ) and E is a primitive n-th root of unity. The reader may easily verify that z E Z [ E ] is a unit if and only if NQ n , (3) ~ = f l . In what follows, we write an(X)for the n-th cyclotomic polynomial, i.e. c

where E runs over all primitive n-th roots of unity, We are now ready to prove the following standard fact. Lemma 1.13. Let E be a primitive n-th root of unity over Q and let N = N Q , / Q , Then if n = pm , p prime if n is not a prime power In particular, for any composite n, 1 - E is a unit of Z [ E ] . (ii) If n = pm, m 1 1, then them exists a unit u of Z [ E ] such that

(i) N ( 1 - E ) =

(1 - & ) P m - l ( P - l ) = pu

Proof. (i) We first remark that N(1-

E)

= a n ( l ) . Since

QP(X)= 1 -t x + * * * 4- x p - 1

1 Determinants of characters

963

and = QP(XP")

@pm+l(x)

we have @,(1) = p = @ , m ( l ) , so the prime power case is established. Turning to the composite case, we assume by induction that the lemma is true for all proper divisors of n. Since

n

X" - 1

dln,d>l

@d(X)=--

x-1

-

1+x+...+xn-'

ndln,d>l

it follows that n = @ d ( l ) . Applying the induction hypothesis, we deduce that n = en(1)n d @d(l),where d > 1 ranges over all prime divisors - p r S be the canonical decomposition of n. Then the of n. Let n = pr'p;' prime power case dictates that

--

Consequently, n = @a(1) - py'p? * * .p;d = @n(l)n. Thus @n(1) = 1, which proves (i). (ii) Let 6 and 7 = bk be two primitive p"-th roots of unity. Then 1- 7 = 1- bh = (1 - S)(1+ 6 t . * *

+ 6"l)

and so 1- 6 divides 1- y. Similarly 1- y divides 1- 6 and so 1- y = r(1- 6) where T is a unit of Z [ E ] . Since, by (i))

where 6 runs over all primitive pn-th roots of unity, the result follows.

Lemma 1.14. Let n 2 1 and let Then

E

be an n-th root of unity over

if

0.

&=l

if & # 1 Proof. The case which case n 2 2 and

E

= 1 being trivial, we may assume that n- 1

i=O

1-&

-0

E

#

1 in

Determinants of Characters

964

Hence n-1

n-1

n-1

i=O

i= 1

i=I

n-1

= -n, as desired. Returning to determinants of characters, assume that x is an irreducible character of G . We wish to investigate the order of det(X)(g),g E G , in ) order of g and the number of conjugates of g . terms of ~ ( 1 )the

Theorem 1.15. (Gallagher (1965 b)). Let g E G be of order n, let m be the number of conjugates of g and let x be an irreducible character of G. Then the order o f d e t ( x ) ( g ) diwides mn

(n,X(1)) unless n is even and (NG< g >: CG < g >) is odd, in which case the order of det(X)(g)divides

2mn

Proof. For each g E G, let h ( g ) denote the number of conjugates of g and put w,(g) = x ( g ) W / x ( 1 ) Then, by Theorem 21.1.1 (iv), wx(g) is an algebraic integer. Let X be a faithful linear character of H =< g >. Then

and [ d e t ( x ) ]=~ Ad where n- 1

n-1

i=O

i=O

1 Determinants of characters

965

On the other hand, by Lemma 1.14, for each j E {0,1,. . , ,n - l},

c

n- 1

n-1

(CiX”(gj)

=

i=O

i[X(gj)]i

i=O

n(n-1)

if j = O if j # O

Hence, by (6), we have

Taking into account that

we deduce that

Hence crdh(g)/x( 1) is an algebraic integer, if cr is a common denominator for

Now observe that, by Lemma 1.13,

is a common denominator for (7). If 21. and 2((CG(gni2) : cG(g)), then (y

= 21/2

rI

pl/(P-l)

Pb,P>2 is a common denominator for (7). If 21. 2 [ (NG < g >: CG(g)), because

cG(g)

and 2

/’ (CG(gn/2)

c N G < > L CG(gn’’)

: CG(g)), then

Determinants of Characters

966

By raising a d h ( g ) / x ( l ) to a suitable power, we see that d h ( g ) / x ( l ) is an integer, unless n is even and ( N G < g >: C G ( g ) ) is odd, in which case 2 d h ( g ) / x ( l ) is an integer. Thus, in the two cases, ( [ d e t ( x ) ] ~ ) ~and (~)([det(x)]~)'~(~) are powers of

proving the theorem. W

Corollary 1.16. (Gallagher (1965 b)). Let p be a prime and let A be a Sylow p-subgroup of G of exponent e . If A is abelian, then for any irreducible character x ofG the order of [ d e t ( x ) ]divides ~ e / ( e , x ( l ) ) , unless p = 2 , in which case the order divides 2 e / ( e , x ( l ) ) . Proof. The order of [ d e t ( x ) ] is ~ equal to the order of d e t ( x ) ( g ) for some g E A . If m is the number of conjugates of g in G, then ( m , p ) = 1. Since o(g)le, the result follows by Theorem 1.15. H

2

C haracter-theoret ic transfer

In this section, we introduce a character-theoretic transfer discovered by Yoshida (1978). Some important applications will be provided at a later stage. As a point of departure, let us fix some notation and conventions. Throughout, we shall be dealing with a finite group G, the set I w ( G ) of irreducible (r: -characters of G and the ring Ch(G)of generalized characters of G. Some of the properties of the ring Ch(G)were presented in Chapter 19. Recall, in particular, that the additive group of Ch(G) is a free ZI, -module freely generated by Irr(G). We write G* = Hom(G,(I: *) for the group of all linear characters of G. For a character x of G, we write deto() for the determinant of x . Put

Since the additive group of C h ( G ) is freely generated by I r r ( G ) , the map

2 Character-theoretic transfer

967

is a homomorphism of the additive group of C h ( G ) into the multiplicative group G*. By Lemma 1.3 (i), the above map is an extension of the determinant map det : char(G) --+ G* where char(G) is the set of all characters of G. Given a subgroup H of G and 8 E C h ( H ) ,we put

This gives us a map

T$ : C h ( H )+ G* If there is no danger of confusion, we shall use the same symbol Tg for the restriction of T; to subgroups of H* C h ( H ) . Note that, by (2),

for all 8 E C h ( H ) . We also remind the reader that

[ d e t ( ( l H ) G ) ]= 2 1G

(4)

by Corollary 1.5. In what follows,

denotes the transfer homomorphism. As usual, any linear character of H is regarded as a character of H I H ' .

Lemma 2.1. Let H be a subgroup of G. Then (i) T;(a) = ac o TG,Hfor all a E H * . (ii) T; : H * -+ G* is a group homomorphism.

Proof. (i) Given a E H * , we have

= (ao T ~ , ~ ) [ d e t ( ( l(by ~ ) Theorem ~ ) ] ~ 1.6 (i))

Determinants of Characters

968

as desired. (ii) Given a,@E H * , we have

as required. Following Yoshida (1978), we refer to the homomorphism

T: : H'

--+

G*

as the character-theoretic transfer . Our next aim is to investigate the extended map T: : C h ( H ) + G* in detail.

E Ch(G) and let d; = ei(l), i = 1,2. Then

Lemma 2.2. Let

det(B102) = det(81)d2det(82)d1

Proof. If 81 and 02 are characters of G , then the required assertion is Lemma 1.3 (ii). Turning to the general case, write

el = C nix;, e2 = C mjxj i

as asserted. H

j

(ni,mj

E

z, x i , x j E I T T ( G ) )

2 Character-theoretic transfer

969

Theorem 2.3. (Yoshida (1978)). Let H be a subgroup of G and let

e, el, e2 E c ~ ( H ) .

(i) z-g(e)= T g ( d e t ( e ) ) (ii) Tg(eIe2)= T g ( e , ) d z z - g ( e 2 ) d lwhere , di = ei(i), i = i , 2 . (iii) I f H K C G, then @(T&(8))= T$(e) (iv)

I f x E Ch(G), then

(G:H)

G a X H ) = [det(x)l

Proof. (i) The map T i is a composition of the additive homomorphism C hB( H ) eG C h( Ge(i)(iH)G )

{

and the homomorphism

det : Ch(G)-, G*

of the additive group of C h ( G )into the multiplicative group G*. Thus ' 2' is a homomorphism of the additive group of Ch(G) into the multiplicative group of G*. Assume that (i) is true for any character x of H . Writing

0 =xnixi

(n;E

z ,xi E I T T ( H ) )

I

it follows that

Thus we may assume that 0 is a character of H . Then, setting d e t ( ( l H ) G ) ,we have

T:(@> = det(BG),,,H 4 1 )

EG,H

=

(by (3))

= [det(e)o T G , H ] E $ , ~ c(by G ~Theorem ~) 1.6 (i))

Determinants of Characters

970

as desired. (ii) We have

= THG(det(81)d2det(82)d1) (by Lemma 2.2)

thus completing the proof. W

2 Character-theoretic transfer

971

We close by recording some elementary properties of the character-theoretic transfer which will be used in our subsequent investigations. All the results below are due to Yoshida (1978).

Lemma 2.4. (Mackey Decomposition). Let H and S be subgroups of G, let R be a full set of double coset representatives for ( S ,H ) in G and let x be a linear character of H . Then

[THG(x)Is

=

n

TfHg-lnS(gxgHg-lnS)

gER

Proof. By the Mackey decomposition for characters (Proposition 19.1.10), we have

and

Hence

as desired. H

Proposition 2.5. Let H be a subgroup of GI let x E G and let R be a full set of double coset representatives for (< x >, H ) in G. Let x be a linear character of H and, for each g E R, let eg = (< 2

>:< 2 > ngHg-l)

Then

THG(x>(4=

n

x(g-lxegg)

g€R

Proof. Put S =< x > and fix g E R. Since S is cyclic, S has a character A, which extends gXgHS-lnS. Thus S

TgHg-1nS('X~Hg-lnS)=

Determinants of Characters

972

by virtue of Theorem 2.3 (iv). Since x e g E gHg-l nS, it follows from Lemma 2.4 that

as desired. W

Let H be a subgroup of G. By the normal closure of H in G, we understand the intersection of all normal subgroups of G containing H . Clearly, the normal closure of H in G is the subgroup of G generated by all subgroups g - l H g with g in G . Corollary 2.6. Let H be a subgroup of G, let x E G and let S be the normal closure of < x > in G . If x is a linear character of H such that T H ( x > ( x ># 1, then S n H Kerx

Proof. Assume that S r l H 2.5, for all g E R, g - 1 x eg g E H n g - l
g c H n s c KerX

Hence, by Lemma 2.5, T t ( x ) ( s )= 1, a contradiction.

To prove our next result, we need to record some information arising from the duality theory for finite abelian groups. First of all, let us recall that by Corollary 19.3.14,

G* E GIG' For any subgroup H of G, we put

H* = {x E G * J x ( h = ) 1 for all h E H } Then H I is the kernel of the restriction map

Res : G'

+

H*

(5)

2 Character-theoretic transfer

Since x E H I if and only if HG'

973

E I i e r x, we see that

H I 2 (G/HG')* For any subgroup I' of G*, we put

In what follows, we write L(G/G') for the lattice of all subgroups of G containing G' and L(G*) for the lattice of all subgroups of G*.

Lemma 2.7. (i) The map G -, G**, g +, g*, where g*(x) = x ( g ) , g E G , x E G*, is a surjective homomorphism with kernel G'. In particular, G/G' G"*. (ii) The maps

are inverse bijections which reverse products and intersections. (iii) For any H E L(G/G') and I? E L(G*),

(iv) IfG* = rl x

rZlthen GIG' = ( r t / G ' ) x (l?i/G').

Proof. (i) The given map is obviously a homomorphism whose kernel is G'. Since, by ( 5 ) , G** % (G/G')* 2 G/G', the desired assertion follows. (ii) Given H l , H 2 E L(G/G'), we obviously have ( H 1 H 2 ) l = H f n H k and ( H I n Hz)' 2 H f H ; . By ( 5 ) and (6), we have IH'I = (G : H ) for all H E L(G/G'). Hence \(HInH2)'J = J H f H ; ) and so ( H I nH2)' = H f H ; . Let rl,r2E L(G*). Then, by definition, (rlr2)l= I?: n I'i and (rl n r2)*2 l?trk. By (i), for any I' E L(G*) the map G + I'*, g I+ g; is a surjective homomorphism, since the restriction map G** I?* is surjective. Since, by definition, I'l is the kernel of g H g;, it follows that G/r' E I?*. Thus = IGl/lrl since r 2 r* by ( 5 ) . Hence I(r1n I ' Z ) l [ = Ir:I'i( and SO (rln r2)l= rfrk. Let H E L(G/G') and I' E L(G'). Then, by definition, .--f

Determinants of Characters

974

Since ( H I 1 = (G : H ) and lrll = lGl/lI'l, we see that IHI = I H l l l and = lr"l which implies that H = HLL and I' = rl*, as desired. (iii) Since H I is the kernel of the restriction map G* + H * and the latter is surjective, we have G*/H* & ! ' H*. As we have seen in (iv), we also have G / r L E r, as required. (iv) By (ii), we have

G= and

= (rln

r2)l= rfr;

G' = (rlr2)l= r; n r;

as we wished to show. We now end this digression and return to our study of character-theoretic transfer. Given a prime p , we write Gi for the Sylow p-subgroup of G* and G'(p) for the pcommutator subgroup of G. Thus

GIG' = (sG'/G') x G ' ( ~ ) / G ' where S is a Sylow p-subgroup of G. Finally, GP denotes the subgroup of G generated by all g P with g E G.

Theorem 2.8. (Yoshida (1978)). Let p be a prime and let H be a subgroup of G of p'-index. Then

(i) The composition Gi Res + H;

Tz

+

G; is an isomorphism and

H i = I m ( R e s ) x ( K e r T') Furthermow, ( H n G'(p))/H'(p)is a direct factor of H / H ' ( p ) . (ii) Let L be a subgroup of HnG'(p) containing H'(p). Then the following conditions are equivalent : (u) L = H n G'(p) (b) There is no element x E H* such that x is of order p , T$(x) = lg and L C K e r x . (iii) Let P be a p-subgroup of H and assume that

P n H P H ' c K C PnGPG' Then there exists x E H* such that Ir' C K e r x,x is of order p , T$(x)= 1~ a n d X P # 1P.

2 Character-theoretic transfer

975

Proof. (i) Put T = TE and R = Res. By Theorem 2.3 (iv),

T ( R ( X ) )= x ( G : H )

for all

x

E G;

Since ( G : H ) is prime to p , we deduce that T OR : GI: 4 Gi is an isomorphism. Because T OR is an isomorphism, we see that R is injective, T is surjective and ImR n K e r T = 1. Thus ImR 2 Gi 2 H,*/Ir'erT and

Hp* = I m R x IIer T . To prove the second assertion, observe that H; = {x E H*IH'@) G Ir'er x} and so we may identify Hp" with ( H / H ' ( p ) ) * . With this identification, we have

( I m W = ( H n G'(P))/H'(P) Hence, by Lemma 2.7 (iv) applied to G = H / H ' ( p ) , we see that ( H n G'(P))/H'(P) is a direct factor of H / H ' ( p ) . (ii) We abandon the identification of H; with ( H / H ' ( p ) ) * ,in which case we have ( I m R ) ' -= H n G'(p). It follows from Lemma 2.7 that (a) is equivalent to L' = I m R and that

ImR = ( H n G'(p))' C L' Since Hp' = ImRxKer T , we see that L' = ImR if and only if L'nKer T = 1. This shows that (a) is equivalent to (b). Let = {A E G * I A P = 1). By hypothesis, HPH'Ir' H and we put

rl = (HpH'K)' Then we have

rz

and

rz = R ( r )

and so

We must show that there exists x E rl n K e r T with x p way of contradiction that no such x exists, i.e. that P Then

P n GPG' E =

(rln IieT T ) n~r; [(rln IierT)r2]'

#

lp. Assume by (TI

n ICerT)'.

(by Lemma 2.7 (ii))

Determinants of Characters

976

=

rf = H P H ' K

Thus P n GPG' C P n HPH'K = A', a contradiction.

(by Lemma 2.7 (ii))

Chapter 31

Tensor Induction of

Characters In this chapter, we study characters of tensor induced modules. A generalized version of the construction of these modules was introduced by Evens (1961, p, 231). Subsequently Evens (1963) applied tensor induced modules (which he called “monomial G-modules”) to provide a useful generalization of the transfer map in the cohomology of groups. Since then tensor induction has been discovered independently by some authors, and has been used by Dade (1967), Dress (1971), Berger (1977), Isaacs (1982), Knorr (1984) and others to study various problems arising in group representation theory. Our treatment of tensor induced modules is direct without appealing to the wreath product construction. After introducing tensor products of n 2 2 modules over a commutative ring, we provide a definition and some basic properties of tensor induced modules. In particular, we establish transitivity, Mackey decomposition, preservation of tensor products, etc. Turning to tensor induced characters, an explicit description of them is provided. Subsequently, we examine induction determinant property and prove a number of related results. Special attention is drawn to a detailed investigation of tensor induced class functions. Among other results, we prove their transitivity, Mackey decomposition and multiplicativity. We also exhibit a deviation of tensor induced characters from being additive and from preserving multiplication by a scalar. This is done by taking the simplest nontrivial case, where H is a normal subgroup of G of prime index. Given a generalized character Q of a subgroup H of G, it is natural to 977

Tensor Induction of Characters

978

enquire whether the tensor induced character aOG is a generalized character of G. A positive answer is provided (Knorr (1984), Gluck and Isaacs (1983)). We then approach a similar problem in case (I! is a generalized permutation is a generalized permutation character of H, and demonstrate that character of G (Gluck and Isaacs (1983)). It is also proved that aBGis a Q -generalized character of G whenever a is a Q -generalized character of H (Gluck and Isaacs (1983)). The chapter ends with some applications of tensor induced characters to the study of characters of central products.

1

Tensor products of n 2 2 modules

In what follows, R denotes a commutative ring. Our aim is to introduce the tensor product of n 2 2 R-modules which is a generalization of the construction given earlier for the case n = 2. Let Vl, Vz, . . . ,Vn, W be R-modules. A function

f

:

v, x v2 x - - .x v, + w

which is linear in each variable and such that f(v1,.

. .,T V i , . . . ,v,)

= Tf(V1,.

. . ,vi, . .. ,v,)

(T

E R,Vi E K )

is said to be an n-multilinear map , Let F be a free R-module freely generated by the set V1 x x V, and let N be the submodule of F generated by all elements of the following types : a a

-. T V j , . *

(~1,.

3

9

-

~ n) ~ ( ~ 1 , ss 9 ~ i ,

v,)

( ~ i ,V:

-

E K, T E R)

The factor module FIN is called the tensor product of the R-modules Vi, . . ,Vn and is denoted by V1 @ R Vz @ R * - @IRV,. It is also denoted by

.

if no confusion can arise. The image of (VI, . . . ,v,) under the natural homomorphism F + F / N is denoted by v1 @I 02 @ * @ v, or by @Y=ivi. Observe that the R-module

--

1 Tensor products of

?a

2 2 modules

@ Z l K is generated by all the relations :

@r=lwi

with w; E

979

and these generators satisfy

It follows that the function

n.t +@Zlr/;. n

?r:

i=l

defined by

. . ,?In)= w 1 8 -..@ 21, The R-module @ Z l K enjoys the following universal

7r(v1,.

is an n-multilinear map. property.

Lemma 1.1. For any n-multilinear map

there is a unique R-module homomorphism

such that

Proof. Since the elements 01 @ - - -8w,, wi E V,, generate @y=lV, there cannot be more than one such f . If F and N are as in the definition of @L1r/;., then $ extends to an R-homomorphism g : F -+ W with N E K e r g . Hence the induced homomorphism f : F / N -, W satisfies the required property. So the lemma is true. W Lemma 1.2. Let V1,. ..,Vn be R-modules, n 2 2. (i) The map @,blK ---i (@y.:vl:) @ R V,, @?=lwi ct (@;i:wi) @ wn is an R-isomorphism (by convention, @ri:K = V1 and @r.:vv; = v1 if n = 2).

Tensor Induction of Characters

980

(ii) If V, is a free R-module freely generated by the set is a free R-module freely generated by (vi E

my=lvi

Xi,then @r=lVi

Xi)

Proof. (i) This is an easy consequence of Lemma 1.1. (ii) Apply (i), the case n = 2 (Corollary 3.1.11) and induction on n.

2

Tensor induced modules

In what follows, H denotes a subgroup of a finite group G and R an arbitrary commutative ring. All modules below are assumed to be left modules. We begin by constructing tensor induced modules as “multiplicative versions” of induced modules. Let {tl,. . . ,tn} be a left transversal for H in G. Then, for any g E G, gt: = tg(i)h;

(hi = k ( g ) E H )

(1)

where i H g ( i ) is a permutation of the set {1,2,. .. ,n}. If V is an RHmodule, then by Proposition 18.1.3 the R-structure of the induced module V Gis given by V G = $?==,ti@ v and, for each i E { 1,. , .,n}, the map

v + ti 8 v,

2, H

t ; 63 v

is an R-isomorphism. The action of g E G on ti 8 v is defined by g(4

m ).

= gti c3

= tg(;)8 h;v

where h; is given by (1). The construction of tensor induced modules is similar to that of induced modules, namely we replace @ by 8 in the R-structure of V Gand let G act diagonally on the resulting R-module. Since, unlike in the situation of direct sums, the order of tensor factors is relevant, we shall adjust our action of G by suitably permuting the tensor factors and using an equivalent form of (l),namely gt,-i (i) = tihg-i (i) More precisely, we form the tensor product

v@’ = B:=l(ti

V)

over R

2

Tensor induced modules

981

and let G act on VBGby the formula

Proposition 2.1. Let H be a subgroup of G and let V be an R H module. (i) V@' with the action of G given by (2) is an RG-module. (ii) The isomorphism class of VmG depends neither on the choice of the transversal { t l , . . . ,t n } nor on the ordering of elements in this transversal. (iii) If V is R-free with an R-basis X,then VBGis R-free with an R-basis {@:=i(ti 8 vi)lvi E X } In particular, the R-rank of V g G is [XIn.

Proof. (i) For any given g E G , the map n fg

: n ( t ;8 V )

+

@ Z l ( t i€3

V)

i=l

given by

fg(t1 €3 ~

-

1 , . ,tn

8 vn) = @:=1(ti 8 hg-l(i)vg-l(i))

is n-multilinear. Hence, by (2) and Lemma 1.1, the map @:=l(ti

8~

i H)

g(@;=l(ti8 v i ) )

determines a unique R-linear transformation of V@'. Of course, the identity element of G acts as the identity transformation. For each g E G, write gt; = tg(;)h;(g) for some hi(g) E H . Then, for all X,Y f G , ~ = - l ( ; ) ( ~ ) h y - l = - l (= ; ) (+Ys)- 1 ( ; ) ( 4 (3) Hence

Tensor Induction of Characters

982

as required. (ii) Let {sl,.. .,sn} be any other left transversal for H in G arbitrarily ordered. Then there exists a permutation K of { 1,...,n} such that s; = t,(;)si for some z; E H , 1 _< i 5 n, Hence, by (l), -1

gsi = S,-lga(i)2=-igrr(;)~*(j)zi

(4)

Let Wac be defined with respect to the new transversal with the given ordering. Then the map f : W@' + V g G given by

as desired.

(iv) This is a direct consequence of Lemma 1.2 (ii). W

Corollary 2.2. Let H be a subgroup of G and let V be a permutation RH-module. Then VaG is a permutation RG-module.

Proof. By hypothesis, V has an R-basis X on which H acts as a permutation group. By Proposition 2.1 (iii),

Y = {@?=*(ti

8 vi)JviE X }

is an R-basis of V B c . Since, by (2), G acts as a permutation group of Y , the result follows.

Proposition 2.3. (Transitivity of Tensor Induction). Let H 2 Ir' be subgroups of G and let V be an RH-module. Then

2 Tensor induced modules

983

Proof. Let 21,. . . ,x, be a left transversal for IC in G and let y l , . . . ,ym be a left transversal for H in Ii'. Then the elements s;yj (ordered lexicographically) give a left transversal for H in G. The map

{

VBG

(v@K)@G

@i,j(xiYj @ vij)

H

@?=l(xi @

(@?=1(yj @ vij)))

(vij E

V)

is obviously an R-isomorphism. Fix g E G , r E (1,. ..,n},t E (1,. .. , m } and write 9x3 = X r a , ayk = ytb for some a E Ii', b E H , s E ( 1 , . . . , n } and k E (1,. . . , m } . Then gZ,yk = s,ytb and so the (r,t)-th tensor factor of g(@i,j(Ziyj@ vij)) is s r y t @ bv,k. Hence the (r,t)-th tensor factor of f(g(@i,j(siyj@vij))is sr@(yt@6v,k). On the other hand, the (r,t)-th tensor factor of g(@?=l((zi@ (@T=i(yj @ vij))) is 2,. @ zt, where zt is the t-th factor of a(@y=l(yj@ vsj)). Since zt = yt @ bv,k, the result follows. Let V I ,... ,V, be RG-modules. Then g(vi @

@ v n ) = gv1 @

@F.lK is an RG-module via 8 gv, (9 E G,vi E K )

For n = 2, this module is simply the inner tensor product of V1 and Vz defined in Sec.7 of Chapter 17. It is clear that the isomorphism class of @?=lK is independent of the ordering of the tensor factors V,. For this reason, if I = { 1 , 2 , . . . ,n}, we may write unambiguously @iErV, instead of @r=lK. The formulation of the next result uses this convention for the indexing set T .

Proposition 2.4. (Mackey Decomposition). Let S and H be subgroups of GI and let T be a full set of double coset representatives for (S,H ) in G. Then, for any RH-module V,

Proof. Let {gl, . . . ,gn} be a left transversal for H in G. Then VBG = @:=l(gi €4 V)

(5)

Owing to Proposition 2.1 (ii), we may arbitrarily reorder the tensor factors in ( 5 ) . Then we may write

VBG = @tETWt

Tensor Induction of Characters

984

where Wt is the tensor product of all g; 8 V (arbitrarily ordered) for which g; E StH. Since the action of s E S on VBG induces an action on each W t , it is clear that Wt is an RS-module and

Since the R(tHt-')-modules tV and 1 @ V are isomorphic (Lemma 18.6.2 (i)), we are left to verify that for any fixed t E T,

where t C3 V is regarded as an R(tHt-' n 5')-module by restriction. To prove (6), let s1,.. .,Sk be a left transversal for tHt-' nS in S and let gj = s i t , 1 5 i 5 Ic. Then StH = Uik,lg;H is a decomposition into disjoint left cosets and so Wt = @f=l(gj C3 V). By definition,

( t @ v)@'= @t=I(siB ( t

v>>

and so the map --t

@f&i 8 >i.

(t@V)@'S 8:=1(si C3 ( t 8 %)>

is at least an R-isomorphism. Since it obviously preserves the action of s E S,the result follows. H

So far the behaviour of tensor induced modules is similar to that of induced modules. However, in contrast to induction, tensor induction does not preserve direct sums, i.e. in general

Indeed, even in the case where both V and W are R-free, the left and righthand sides need not have the same R-rank by Proposition 2.1 (iii). On the other hand, as we shall see below, tensor induction preserves inner tensor products, and these are not preserved by induction.

Proposition 2.5. Let H be a subgroup of G and let V,W be R H modules. Then (v @ R W ) @ G% V@' @ p R WBG

3 Tensor induced characters

985

Proof. Let g l , . . . , g , be a left transversal for H in G. Then the map

I

(V@RW)@G @Y=l(gi @ (vi @ w ) )

i +-t

V@'G@'RW*G [@Y=l(gi 8 vi)] 8 [@pZI(gi €9

w)]

is obviously an R-isomorphism. Let a = €9;=1(gi 63 (vi €9 wi))and b = [@;=l(gi €3 v;)]€9 [ @ b l ( g i €9 ~ j ) ] .Fix g E G,k E (1,...,n } and write ggs = g k h for some s E (1,..., n}, h E H . Then the k-th tensor factor of g a is gk €9 h(vs €9 ws) = gk €9 (hv, 8 hws)

Hence the k-th tensor factor of f ( g a ) is c

= (gk @ hvs) €3 (gk @ hws)

On the other hand, the k-th tensor factor of g f ( a ) is also c , as desired.

3

Tensor induced characters

Throughout, H denotes a subgroup of a finite group G and R a commutative ring. All RH-modules are assumed to be R-free of finite rank. Let V be an RH-module. Then, by Proposition 2.1 (iii), the tensor induced module V@' is R-free of finite rank. Hence we may consider the character of G afforded by PG. If x is the character of H afforded by V , then we write X@G for the character of G afforded by V@'. We refer to xBG as the character of G tensor induced by the character x. Our first task is to compute xBG in terms of x. As a preliminary to the result below, let us make the following grouptheoretic remark. Assume that H is a subgroup of G and let t , g E G. Then there exists m 2 1 such that

< g > t H = U;;'g"H is a decomposition into disjoint left cosets of H . Here m is characterized as the smallest integer k 2 1 with g k t H = t H , i.e. with t - ' g k t E H . It is clear that l , g , . .. , g m - l is a left transversal for tHt-'fl < g > in < g >. Thus

m = I < g > tHIIH1-l = (< g = (< t - l g t >: (Hn < t - l g t

>: ( t H t - l n < g >)) >>>

Tensor Induction of Characters

986

Proposition 3.1. Let x be the character of H afloded by an R H module V . Fax g E G, let G = uL1 < g > tiH be a decomposition into disjoint (< g >, H)-double cosets and, for each i E ( 1 , . .,k}, put

.

m; = min{rnlt;'gmt; E H }

Then m; = (< trlgti >: (< tf'gti

> n H ) ) = I c g > t;HIIHI-l and k

1 m. X@G(s) = n x ( t T 9 'ti)

i=1

Proof. The equalities regarding m; follow from the preceding remark. Fix t E {tl,. ..,tk}, put S =< g > and denote by W the restriction of the R(tHt-')-module t @ V to tHtel n S. Let X be the character of S afforded by W@'. Then, by Proposition 2.4, it suffices to verify that

where m = IStHl 1HI-l. By the definition of m, l,g, . , , ,gm-' is a left transversal for tHt-' n S in S and gm E tHt-l fl S. By the definition of WmS,we then have W@S=

m-1 @*=O

i

(9 @

(t @ V ) )

Let X be an R-basis of V. Then, by Proposition 2.1 (iii),

Y = {@:;l(gi

€3 ( t @ Vi))lO 5 i 5 rn

- 1, w; E X}

is an R-basis of Was. On the other hand, for any g(@2;1(9i €3 ( t €3

Vi)))

WO, 211,.

..,vrn-l

E X,

= (1 @ g"(t €3 % - l ) @ (@;=Im-1 9i @ (t @ W i - 1 ) ) (2) = (1 €3 ( t @ t-lgrntv,-1)) €3 (€3Z11gi 8 ( t €3 wi-1))

Let p be the matrix representation of H afforded by V with respect to any given ordering 31,. ,z, of X. Put wj = @ z i l ( g i @ ( t @ xj)), 1 5 j 5 n, and let us list ~ 1 , . ,w n as the first n elements of Y , while the remaining elements of Y are taken in an arbitrary order. Then, by (2), the matrix of g with respect to the given ordering of Y is of the form

.. ..

3 Tensor induced characters

987

where t r ( A ) = trp(t-'gmt) = X(t-'g"t) and t r ( D ) = 0. Thus X(g) = x(t-'gmt), proving (1) and hence the result. H

Corollary 3.2. Let x be the character of H afforded by an RH-module V which is R-free of rank 1 (hence x is a character of HIHI). Then, for each g E G , X@'(g) = X ( T G , H ( g ) ) where TG,H: G t H / H ' is the tmnsfer homomorphism.

Proof. Apply Theorem 26.1.8 (ii) and Proposition 3.1. Proposition 3.3. Let H be a subgroup ofG and let x be a n R-chamcter of H . Then

is a permutation character, then so is x@'. (ii) If H is contained in a subgroup K of G, then

(i) If

x

x B G = ( x@ K 1€3' (iii) If X is any other R-chamcter of H , then

Proof. Apply Corollary 2.2 and Propositions 2.3 and 2.5.

Proposition 3.4. (Mackey Decomposition). Let S and H be subgroups of G and let T be a full set of double coset repmsentatives for ( S ,H ) in G. Then, for any R-chamcter x of H ,

(x@')s = n ( t X t H t - ' " S ) * s tET

Proof. Apply Proposition 2.4. H We remind the reader that since tensor induction does not preserve direct sums, the same is true for tensor induced characters, i.e. in general (Xl

+ X2)@'

# XPG t X f G

From now on, all characters are assumed to be (C-characters. The next property, contained in a work of Isaacs (1986a), provides a link between determinants of characters, induced characters and tensor induced characters.

Tensor Induction of Characters

988

Proposition 3.6. of H . Then

Let H be a subgroup of G and let

x

be a character

det(XG)= [det((l ~ ) ~ ) ] ~ ( ' ) ( d e t ( x ) ) @ ~ Proof. Apply Theorem 30.1.6 (i) and Corollary 3.2. Following Isaacs (1986a), by a sign character ,we mean a linear character with values fl. For example, if x is a permutation character, then d e t ( x ) is a sign character(see Lemma 30.1.4 (i)). In particular, by Proposition 3.5,

det(XG)= &(det(X))BG

(3)

where E = [ d e t ( ( l ~ ) ~ ) ]isX a( ~sign ) character. We now investigate the possibility to eliminate the sign character E in ( 3 ) by multiplying x by an appropriate sign character of H before inducing to G. It will be convenient to introduce the following terminology due to Isaacs (1986a). We say that a sign character 6 of the subgroup H of G has the induction determinant property in G if

det((6x)') = det(X)@'

(4)

for all characters x of H . We warn the reader that there does not always exist a sign character of H having the induction determinant property nor is it necessarily unique if it does exist. The next two results are extracted from a work of Isaacs (1986 a). Recall that for a character x of G , o ( x ) denotes the order of det(x).

Proposition 3.6. Let H be a subgroup of G and let x be a character of H . Then o(xG) divides 20(x) and if 6 is a sign chamcter of H having the

induction determinant propertg, then o( ( 6 ~ ) ~divides ) o( x )

Proof. Put X = det()o and n = o ( x ) = .(A). Then A(hn) = 1 for all h E H. By Corollary 3.2 and Proposition 3.5, we have det(xG)= [ d e t ( ( l ~ ) ~ ] ~ o( TG,H) ')(X It follows from ( 5 ) that for all g E G,

det(xG)(s") = f V G , H ( g n ) ) = fX(TG,H(g)n) = f l

(5)

4 Tensor induced class functions

989

and therefore

det(xG)(g") = 1 Thus o(xG)divides 2n. Let 6 be a sign character of H having the induction determinant property. Then, by (4)and Corollary 3.2,

det((6~)~)(= 9 "W)G , H ( S " ) )

=1

thus completing the proof. We close by producing sign characters which have the induction determinan t property.

Proposition 3.7. Let H be a subgroup of G of odd index and let

6 = det(( 1 ~ ) ~ ) Then SH has the induction determinant property for H an G.

Proof. Let x be any character of H . By Proposition 19.1.7 (i), ( 6xG. Hence, by Lemma 30.1.3 (ii),

W(6HX)

1 - bXG(')dei(XG)

~ H x= ) ~

(6)

Because (G : H ) is odd, we have

Thus, by (6) and Proposition 3.5,

as required.

4

Tensor induced class functions

Throughout, G denotes a finite group and all class functions of G are assumed to be 02 -valued. All characters below are (c -characters.

Tensor Induction of Characters

990

In analogy with induced class functions, it is natural t o introduce tensor induced class functions and, in particular, tensor induced generalized characters. Since the tensor induction map is decidedly nonlinear, it would be unreasonable to anticipate that many properties of tensor induced characters can be carried over to generalized tensor induced characters. Nevertheless, surprising as it may be, it will be demonstrated that if x is a generalized character or a generalized permutation character of a subgroup H of G, then so is x @ ~In. case H d G and (G: H ) is prime, we shall exhibit an extent of deviation of tensor induction from linearity. Our treatment of tensor induced class functions is based on a work of Knorr (1984). Let H be a subgroup of G and let cr be a class function of H . Given g E G, let G = Uik,, < g > t i H be a decomposition into disjoint (< g >,H)-double cosets. For each i E (1, * ' ' ,h ) , Put m; = m i n { r n l t ; l g m t ; E H } Recall from Sec. 3 that

> nH>>= I < g > t i ~1l~ I - l

m i= (< t r 1 g t i >: (< t ; l g t i

Following Knorr (1984), we define the map

cr@'G:G-r(i= by

n k

cr@G(g) =

cr(t;lgmit;)

i=l

By Proposition 3.1, if a is a character of H , then is the tensor induced character of G. The following lemma shows that cr H crBG is a map from the class functions of H into the class functions of G. This map is called tensor induction

.

Proposition 4.1. For any class function cr of H , aBGis a well defined class function of G . Proof. Let 5 1 , . . . ,x k be another set of double coset representatives for (< g >, H ) in G. We may assume that x i = aitahi for some h; E H and aj E < g >, 15 i 5 k . Hence " ; l g m i z i = h;'t;1a;lgmia;t;h;

= h;'(t;lg"'ti)hi

4 Tensor induced class functions

991

and so a ( z f l g m l q )= a ( t f ' g m i t ; )since a is a class function of H . This shows that the value of a@'(g) does not depend on the choice of the ti's. Let g1 = ygy-' for some y E G. Then

where the unions are disjoint. It follows that {yt;Il 5 i 5 k} is a full set of (< gl >, H)-double coset representatives. Since

the mi's are the same for

g1

and g. Finally, since

( y t ; )-1 g1m '. ( y t ; )= tf'gm'tj

for all i E {I,. . . ,I C )

we deduce that a@'(gl) = aBG(g).Thus aBGis a class function. H The proof of the next result requires the following group-theoretic property.

Lemma 4.2.

Let H C S be su6groups of G , let g E G and let

be a (< g >,S)-double coset decomposition of G. If n; = (S1-l 1 < g > yjSI, = yi l g n t y ; and S = U j < U ; > V i j H is an (< ~i >,H)-double coset decomposition of S , then Uj

is a (< g

>, H)-double coset decomposition of G.

Proof. By the remark preceding the proof of Proposition 3.1, u; E S and yi-1 g k y; # S for all 1 5 k < ni. Hence, if y i l g t y i E S for some integer

t 2 0 , then yi-1 g t yi E< u; >. If z E G, then 5 E< g > yjS for some i. Hence 2 E < g > y; < u; > v;jH for some j . But

Tensor Induction of Characters

992

and so 3 E< g

> u;v;jH. Assume that

Yk €< g

> YiVijHv;:

C< g > y;S

Thus k = i. Hence yivis = g'y;vijh for some t 2 0, h E H . It follows that y;-1 gt yi = v;,h-'v;' E S and so yi-1 gt y; E < u >. Thus

and so s = j , as required. H

Proposition 4.3. Let S, H be subgroups of G and let a , p be class functions of H. Then (i) (Transitivity). If H C S 5 G, then (a8S)@'= a@' (ii) (Mackey Decomposition). If T is a full set of double coset representatives for (S,H ) in G, then =

JJt"tHt-1nS)

@S

tCl'

(iii) (Multiplicativity) (ap)@' = (ff@G)(P@'G). Proof. (i) Let G = U; < g position. Put n; = 15'1-' Next, let S = Uj and let

> y;S

1 < g > yiSJ

be a (< g >,S)-double coset decom-

and

ui = y;'gniy;

< u; > vijH be an (< u; >, H)-double coset decomposition

< U; > VijHl Then, by Lemma 4.2, G = U;j < g > y;vijIT is a (< g >,IT)-double coset m;j

decomposition. Moreover,

= [Hl-lI

4 Tensor induced class functions

993

Thus

as required. (ii) Let G = U;SgiH be a double coset decomposition. Take s E S and let s = u j < s > y;j(giHgr' n S ) be a double coset decomposition. Then

is a double coset decomposition and

Denoting by

n;j

the above number, we have

as required.

(iii) This is a direct consequence of the definition. Our next aim is t o exhibit a deviation of crBG from being additive and from preserving multiplication by a scalar. We take the simplest nontrivial case, where H is a normal subgroup of prime index.

Tensor Induction of Characters

994

Let H be a normal subgroup of G of prime index p. Then G acts on the set

x =W

l g E GI

Denote by P ( X ) the power set of X, i.e. the set of all subsets of X. Then G acts on P ( X ) via

gY = g Y

for all g E G,Y E P ( X )

Since G is transitive on X, there are precisely two fixed points in P ( X ) , namely 0 and X. Since H is in the kernel of the action, it follows that 2 P - 2 nontrivial subsets of X form orbits of length p = IG/HI under the action of G. Let {XjljE J } be representatives for these orbits, i.e.

where GX,denotes the orbit of Xj. Here of course IJI = p-'(2P - 2). Now let Q and p be class functions of H. For each I E X,let the class function (a,@),of H be defined by

where "a =

gct

if x = g H . Finally, put

Proposition 4.4. Let H be a normal subgroup of G of prime index p and let a , ,kl be class functions of H . Then, with the notation above,

Proof. First we show that the equality holds on G - H . To this end, fix g E G - H. Then G =< g > H and I < g > HI 1HI-l = p . Hence

4 Tensor induced class functions

995

Since [ ( a , p ) j I G ( 9 ) = 0 for all j E J , the equality holds on G - H . Since H is normal in G , it follows from Mackey decomposition (Proposition 4.3 (ii)) that

+

=

[(a P ) 9 H

n P) c b,Ph I(a t

XEX

=

ICP(X)

= ( 0 ) x t (.,P)0t

cc

(a,P)r

i E J IEGX,

Let T be a transversal for H in G. Because

Proposition 4.5. and let X E (I:. Then

Let H be a normal

(A * l H ) @ G = A

*

S U ~ ~ T OofV G ~

1G t p-'(AP

- X)(lH)G

Proof. By the definitions, we have

and

which clearly implies the desired equality. W

If R is a subring of

(I:

and X E R,then in general, p-'(Xp

- A)

@R

of prime index p

Tensor Induction of Characters

996

(e.g. take R = 23 [i], p = 2 and X = i). Consequently, by Proposition 4.5, tensor induction of an R-generalized character (i.e. of an R-linear combination of characters) does not always yield an R-generalized character. The following result which treats the case R = 23 is therefore rather unexpected.

Theorem 4.6. (h'norr (1984)). Let H be a subgroup of G. If Q is a is a generalized character of G. generalized character of H , then Proof. By Brauer's characterization of characters (Theorem 19.6.1 (ii)), it suffices to show that (cY@")E is a generalized character for each elementary subgroup E of G. Let T be a full set of double coset representatives for ( E , H ) in G. Then, by Mackey decomposition (Proposition 4.3 (ii)), we have

Hence it suffices to show that c y B E is a generalized character if cy is a generalized character of a subgroup S of the elementary group E . We argue by induction on ( E : S). If ( E : S) = 1, then there is nothing to prove. Assume that S c S1 c E for some subgroup S1 of E . Then 5'1 is elementary and (Sl: S) < ( E : S ) , ( E : Sl)< ( E : S). Hence, by induction and Proposition 4.3 (i), Q@'I and ( Q @ ' ~ ) @ '= ~ are generalized characters. By the foregoing, we may assume that S is maximal in E . Because E is nilpotent, S a E and ( E : S) = p is a prime. Let x, II, be characters of S such that Q = x - $. Then, by Proposition 4.4 (i)

(x,

The +)j are (up to a sign) products of characters of S, so Cj[(x,- $ ) j l E is a generalized character of E . Thus we are left to verify (-$)aE is a generalized character of E . Bearing in mind that (-@E

= [( - 1s)?)PE = ( - ~ s ) @ ~ I I , @ (by ~ Proposition 4.3 (iii))

= {-1~ t ~ - ' [ ( - lt) ~ l](ls)E}$@'E (by Proposition 4.5)

4 Tensor induced class functions

and that p divides

(-1)P

997

t 1, the result follows. H

We shall provide a n alternative proof of the result above at the end of the section. By a generalized permutation character , we understand an integral linear combination of permutation characters. Our next goal is to prove that if a is a generalized permutation character of a subgroup H of G, then ' @ a is a generalized permutation character of G. In what follows, we write P(G) for the ring of generalized permutation characters of G.

Lemma 4.7. g E G , a(g) E

Proof.

Let a be a generalized chamcter of G such that for each a ( g ) is divisible by /GI*. Then Q E P( G) .

Z and

For any irreducible character

< a , x >= 1GI-l

x of G, we have

a(g)x(g-')E Z g€G

Since each a ( g ) = O(m0dlG1~), it follows that IGl divides < a,x >. Hence IGl-*a is a generalized character. Therefore we can write IGI-'a uniquely in the form x1 -x2 where each xi is a character of G or zero and < X I , x 2 >= 0. Since x1 and x2 are uniquely determined by (GI-la and (G(-'a is a rational valued, we see that x1 and x 2 are fixed by all field automorphisms over Q . Thus x1 and x2 are rational valued. Hence, by Theorem 20.3.2, lGlx1 and lGlx2 lie in P(G). Therefore Q

= IGI(X1 - x2) E P(G)

as required.

Theorem 4.8. (Gluck and Isaacs (1983)). Let H be subgroup f .. is a G and let a be a generalized permutation character of H . Then generalized permutation character of G. Proof. Since a E P ( H ) , we may write a = x1 - x 2 where each permutation character of H . Put

P Then

= x1

t (IG12- 1)xz

p is a permutation character of H such that ,f3(h)E a(h)modlGIZ

for all

hEH

x; is a

Tensor Induction of Characters

998

Hence, by Proposition 3.3 (i) and Theorem 4.6, /3BG is a permutation character of G and &'' is at least a generalized character of G which is integer valued. Furthermore, we have P@'(g)

= a@G(g)modlG12for all

g EG

Hence a@G- /3gG is an integer valued generalized character of G whose values are divisible by lG12. Thus, by Lemma 4.7, a*G -pBG E P ( G ) . Since PBG E P ( G ) , the result is established. For the rest of the section, R denotes the ring of algebraic integers in Given a subring S of (c, by an S-generalized character of G, we understand an S-linear combination of (L: -characters of G . (c.

Lemma 4.9. Let H be a subgroup of G and let a be a generalized character of H . Then a B G is an R-generalized character of G .

Proof. Choose characters x1 and Put Then

x2

of H such that

(I!

= x1 - x2 and

P = x1 t (PI- 11x2

4 is a character of H

such that

a ( h ) = ,B(h)modJGJR for all h E H Hence, by the definition of tensor induction, a*'(g)

= /3@'(g)mod(GIR

for all g E G

It follows that

< a@' - / 3 @ G , ~ >€ R for all irreducible characters x of G, since x has values in R. This shows that a@G - /3@' is an R-generalized character of G . Since ,OBG is a character of G , the result follows. H We remark that, as we have seen earlier, it is not sufficient to assume that a is an R-generalized character in order to infer that aBGis an Rgeneralized character . As a preliminary to our final result, we shall provide a necessary and sufficient condition for a class function of G to be a Q -generalized character.

4 Tensor induced class functions

999

Lemma 4.10. Let E be a primitive n-th root of 1, where n is the exponent of G and, for each a € G a l ( C / Q ) , write a(&)= E ~ ( for ~ ) some uniquely determined integer m(a) with 1 5 m(a) 5 n. If cr : G + C is a class function of G, then a is a Q -generalized character of G i f and only i f g E G, o E Gal(C /Q)

a(cr(g)) = cr(gm(a)) for all

Proof. Assume that a is a Q -generalized character of G. If x is an irreducible character of G, then a ( x ( g ) ) = X ( g m ( u ) ) by Lemma 17.5.1 (i). Since ct is a Q -linear combination of all such x, it follows that a((r(g))= a(gm(")) for all g E G, u E Gal((L:/ Q ). Conversely, assume that a is a class function which satisfies the condition. We may write a = Ex z x x where x ranges over irreducible characters of G and the zx are uniquely determined complex numbers. Then for any g E G, a E Gal(C / Q ), we have

c +,)4x(g))

= fJ(a(s>> =. a(grn(") 1

X

=

c

rxX($m(u))

X

=

Ezx4x(!d) X

From the uniqueness of the zx,we infer that zx = a(+) for all a E Gal((I:/ Q ). Thus all of the zx lie in Q and the result follows. W

Theorem 4.11. (Gluck and Isaacs (1983)). Let H be a subgroup of G is a Q -generalized and let cr be a Q -9enemlazed chamcter of H . Then character of G. Proof. We shall demonstrate that (rBG satisfies the condition of Lemma 4.10 which will prove the result. Let a E Gal(C / Q ), let m = m ( a ) and let g E G. Let G = Uf=l < g > t i H be a decomposition into disjoint (< g >,If)-double cosets and let mi = I < g > tiIfl IHI-'. Then k

u(a@'G(g))= a ( n c t ( t ; l g Y ; ) ) i=l

n k

=

i=l

a(ct(tf1g"lti))

Tensor Induction of Characters

1000

k

CY(t71g""'t;)

= i=l

-

c P G ( g m ) (since

=)

as desired. H We close by recording : An alternative proof of T h e o r e m 4.6 (Gluck and Isaacs (1983)). Given is a generalized a generalized character cr of H , we must show that character of G. It is clear that CY is a Q -generalized character of H . Hence, by Theorem 4.11, a @ ' is a Q -generalized character of G. On the other hand, by Lemma 4.9, a@'G is an R-generalized character of G. Since R n Q = Z , it follows that a@Gis a generalized character of G , as we wished to show. H

5

An application to characters of central products

All groups below are assumed to be finite and their characters to be C characters. A group G is said to be a c e n t r a l product of s u b g r o u p s G I , . ,G, amalgamating D if r > 1 and the following properties hold :

..

( a ) G = < G 1 ,...,G, > (b) Gin < GjIj # i >= D (c) [Gj,Gj]= 1 for i # j

We shall often omit reference to

D if it is not pertinent to the discussion.

Lemma 5.1. Let G be a central product of GI,. .. ,G, amalgamating D . Then (i) Gj d G for each i E (1,.. . , r } (ii) G = G I ** GT (iii) D C Z(G), D = nir,lGi and G / D = G1/D x x GT/D (iv) The map

{

G1 x . - . x G T (gl?**.?gV')

1, G 91*.*gT

is a surjective homomorphism and, furthermore, K e r f is a central subgroup of G I x * - - x G,. Proof.

Properties (i), (ii) and (iii) follow from the definition of central

5 An application to characters of central products

1001

product. Since [Gi,Gj] = 1 for i # j , the map f is a homomorphism; its surjectivity is a consequence of (ii). Finally, if gl . . .gT = 1, then

and the result follows. In what follows, we put

and denote by I r r ( G ) the set of all irreducible (c -characters of G. If xi is a character of Gi,then x1 x - - - x xr denotes the character of G given by

We remind t h e reader that, by Proposition 17.8.1, all irreducible characters of 6' are precisely of the form x1 x *.-x Xr with xi E Irr(Gi). Lemma 5.2. Let G be a central product of G I , . . . ,G, and let Irr(G). Denote by xi an irreducible constituent of X G , and let

Then (i) The xi are uniquely determined by 2 . (ii) 2(z) = x ( j ( z ) )for all 5 E G , where f is given by Lemma 5.1 (dii) K e r f C K e r 2.

x

E

(i.1,

Proof. Since G = G;CG(G;), each xi is G-invariant and, by Clifford's theorem, we can write XG, = mixi for positive integers mi. This certainly implies (i). Because f is surjective, there exists $ E I r r ( G ) with $ ( E ) = x ( f(z)) for all 2 E G. We can write $ = $1 x ... x GT with +i E ITT(G;).Hence, to prove (ii) and (iii), we are left to verify that $i = xi, 1 5 i 5 T . Put ni = $j(l). Then for g E Gj we have

nj+*

Tensor Induction of Characters

1002

Since $i, xi E Irr(G;),we conclude by linear independence that required.

$i

= xi, as

Lemma 5.3. Let N & H be subgroups of G with N a G and let x be a character of H . Then, for all g E N ,

X@G(s> =

JJ x ( t - W tET

where T is a left transversal for H in G. Proof. Given g E N ,write G = (< g >, H)-double cosets. Since

< g > tiH

= t;(tf'

lJtl< g > tiH as a disjoint union of < g > ti)H = tiH

we see that T = { t l , . . . ,tk} is a left transversal for H in G . Since

1 < 9 > t;HI ]HI-'

=1

the assertion follows from the definition of tensor induced characters. W We have now accumulated all the information necessary to prove the following result by the technique of tensor induction.

Theorem 5.4. (Isaacs (1982)). Let N be a central product of subgroups N1,... ,Nr and let a group H act on N and permute the N; transitively. Let G = N H be the corresponding semidirect product and let x be an irreducible G-invariant chamcter of N. Denote by 21 the irreducible constituent of X N ~ . If x 1 is eztendible to a character of N ~ N H ( N ~then ) , x is eztendible to a character of G.

..-x Nr and let f : fi -+ N be given by f(g1,-- - ,gr) = 91 - - .gr (9;E N i )

Proof. Put # = N1 x

By hypothesis, for each h E H , i E (1,. (1,. . . , r } such that h

Let H act on h

. . ,T } , there exists a unique h ( i ) E

Ni = N h ( i )

fi by h

( g l , - - * , g r= ) ( gh-1(1),...,

h

gh-l(r))

(hE H)

5 An application to characters of central products

1003

It is straightforward to verify that h defines an automorphism of

f ( h z )= h j ( z )

for ail h E H , z E

8 and that

rir

Let G = f i H be the corresponding semidirect product and let a : G -+ G be defined by a ( 2 h ) = f ( z ) h for all z E fi, h E H . Then a is a homomorphism which extends f and maps G onto G. By Lemma 5.2, it suffices to extend = x1 x x Xr to G, where xi is the irreducible constituent of X N ~ l i i < T . Denote by W the natural image of N1 in 8 and let K be the product of the remaining Ni's. Then N H ( N ~=)N H ( W )and we have

&NH(N1)= NG(W) = IL'WNH(N1) and Ii n (wiV~(N1)) = 1. Now the subgroup W N H ( N ~is)isomorphic to ) Thus viewing x1 E I T T ( W )we , deduce that the subgroup N ~ N H ( Nof~ G. is extendible to W N H ( N ~Hence ). there exists II, E I r r ( f i N ~ ( N 1 )with ) 'h Ker 1c, and ?,hw= XI. We claim that 8 = $@G is the desired extension of 2 to G. Let T be a left transversal for N H ( N ~in) H. Then T also represents the cosets of f i N ~ ( N 1in ) G. Hence, by Lemma 5.3, for all 5 E fi, 6(z) =

n

?b(t-'zt)

tCT

Since 1c,((gi,. . . ,gr)) = xi(g1) for gi E N ; , it follows that

Since x is invariant under H , we have

Thus

r

as required. W

,

This Page Intentionally Left Blank

Chapter 32

Knijrr’s Generalized Character The problem which will be tackled by the generalized character in the title can be formulated as follows : Problem . Let A be an elementary abelian p-group on which a p’-group G acts faithfully and irreducibly. Is it true that the number of conjugacy classes of the corresponding semidirect product AG is bounded by the order ofA? As has been proved by Nagm (1962) an affirmative answer is equivalent to the famous Brauer conjecture for psolvable groups (see Brauer (1955)). This conjecture, which will be examined in detail in our future study of block theory, asserts that, for any finite group G, the number of ordinary irreducible characters of G in a p-block B is bounded by the order of a defect group of block B , The aim of this chapter is to provide an important tool discovered by Knorr (1984) which will enable us to tackle the problem. The results presented are of independent interest from the viewpoint of character theory. Using these results it will be shown in future that the Brauer conjecture holds in the case where G is solvable with a supersolvable Hall p’-subgroup. A key role in our investigations will be played by a generalized character 6 of G which measures how far an element g E G is from acting trivially on A. The main result asserts that if F is a finite field of characteristic p, G is a p’-group acting faithfully on the FG-module V and if CG(D) is abelian for some D E V , then the number of conjugacy classes of V G is bounded by the order of V .

1005

Kngrr's Generalized Character

1006

1

Definition

In what follows, C h ( G ) denotes the ring of generalized characters of a finite group G. Recall that, by definition, each element of Ch(G) is a Z -linear combination of (C-characters of G. As we know, Ch(G) is a subring of the ring C f (G) of all class functions from G t o (I: . An element cr E C f ( G ) is invertible if and only if a(g) # 0 for all g E G. As usual, we write (Y-' for the inverse of a. Thus, by definition, if a is invertible, then -1

( 9 ) = &)-l

for all g E G

Note that if x is a generalized character of G and x is an invertible element of C f (G), then x-l need not be a generalized character of G.

Proposition 1.1. (h'norr (1984)). Let x be a generalized character of G, let 91,. . .,gn be representataves of the conjugacy classes of G and let

Then

(i) z E Z and, if z # 0 , then zx-' is a generajited character of G. (ii) If 0 # z is relatively prime to [GI and if II, is a generalized character of G such that x ( g ) - l $ ( g ) is an algebraic integer for alE g E G , then x-'II, is a genemlzzed character of G. Proof. (i) The map fx : cf(G) --$ cf(G), f,(X) = Ax is C-linear. Consider the basis ~ 1 , ...,En of c f ( G )given by E ; ( g j ) = 6;j. Then fX(E;) = x ( g i ) E i and so d e t ( f x ) = z. On the other hand, we may take irreducible characters X I , ...,xn of G as a basis. Let M be the matrix of fx with respect to this basis, SO M = (mij) where mij =< xjx,x; > E 23 for all i ,j E (1,. . . ,n}. Hence z = det( f x ) = d e t ( M ) E

Z

Assume that z # 0. Then x is an invertible element of C f ( G ) and we put 9 = 2x-l. Set N = ( n ; j )where nij =< xjv,xi >. Then N is the matrix of f, with respect to X I , . . . ,xn. Hence N M = z I , where I is the identity

1 Definition

1007

matrix. Therefore N = zM-' is the adjugate of M and so N has entries in 21;. In particular, < p , x ; > E Z for all i E {1,...,n } and thus p i s a generalized character of G. (ii) Assume that z # 0, z is relatively prime to JGJand that 11, is a generalized character of G such that x ( g ) - ' + ( g ) is an algebraic integer for all g E G. We have to show that < x - l + , x i >E Z for all i E (1,. . . , n } . By hypothesis, 1 = a t t blGl for some a,6 E Z . Then

< x-'11,,xi > =

< zx-'+,xi > G I G [ < x-'11,,x; > = a < zx-'+,x; > t b x-'(s>+(s)xi(g-') a

c

g€G

is an algebraic integer. Indeed X-'(g)11,(g) is an algebraic integer by assumption, x i ( g - ' ) is an algebraic integer as character value and < zx-'$,x; >E Z by (i). On the other hand,

< x-'+,x; >= z-' < z x - ' + , x ;

>E Q

and thus < X-'11,,xi > E Z ,as desired. H From now on, we fix a prime p and denote by F a finite field of characteristic p . All FG-modules are assumed to be finitely generated. Given an FG-module V and g E G , we put C v ( g ) = {v E Vlgv = u }

Corollary 1.2. Let G be a p'-group and let V be an FG-module. Then (I= given by

the map S : G

--$

is a generalized character of G and, for each g E G,

S ( g ) = pn(g) for some integer n(g) 2

o

Proof. The group G acts as a permutation group of the set V . The corresponding complex character x of G is given by x ( g ) = lCv(g)1 for all g E G. Since 0 E C v ( g ) , we have x ( g ) # 0 for all g E G. Hence, if z is as in Proposition 1.1, then z # 0. Moreover, since C v ( g ) is a subspace of V , x ( g ) is a nonnegative power of p and x ( g ) divides IVl. This shows that S ( g ) = p n ( 9 ) for some integer n ( g ) 2 0 and that z is a power of p . Since G is

Kn6rr's Generalized Character

1008

a $-group, we have (z,[GI) = 1. Finally, define 1c, by 11, = IVl - la. Then, for all g E G,

x(g)-l%W

= 6(g) E

Hence, by Proposition 1.1 (ii), S = x-l+ is a generalized character of G, as desired. H We shall refer to 6 = 6(G,V ) given by Corollary 1.2 as Knorr's generalized character of G with respect to V .

2

Reduction to C&)

In what follows, given a finite group G, we write k(G) for the number of conjugacy classes of G. Our aim is to prove an important result which will indicate relevance of the generalized Knorr's character to the problem stated in the introduction.

Lemma 2.1. Let a : G + C be a class function of G , let 91,. . . ,gn be representatives of the conjugacy classes of G and let X I , . . . ,xn be all irreducible (I: -characters of G. Then

Proof. We have n

n

n

n

i=l

by virtue of Theorem 19.2.3 (iii) and the assumption that a is a class function. So the lemma is true. W

2 Reduction to C c ( t ~ )

1009

Theorem 2.2. (h'norr (1984)). Let F be a finite field of characteristic p , let G be a p'-group and let V be a finite-dimensional FG-module. Form the semidirect product V G corresponding to the action o f G on V and denote by 6 the Know's generalized character of G with respect to V . Assume that there is a v E V such that for any generalized character x of C = &(v) with x( 1) f Omodp,

< X b , X >2 k ( C ) Then V G ) I IVI Proof. For convenience, V will be regarded as a multiplicative group. The action of G on V within V G is given by g v = gvg-' for all g E G, v E V . For the sake of clarity, we divide the proof into a number of steps. Step 1. Here we treat the case where v = 1, hence C = G . Given g E G , let gVG and gG denote the conjugacy classes of g in V G and G , respectively. We claim that

gVGnG=gc

(1)

It is clear that gG C gVG n G. Conversely, assume that u E V , 2 E G are such that y = uzgz-lu-' E G. Then u(zgz-') = yu = (yuy-')y and so y = zgz-' E g G , proving (I). By (l),there are precisely k ( G ) conjugacy classes of V G which intersect G nontrivially. The induced character GVG is zero on all other conjugacy classes. On the other hand, setting S(z) = 0 for z E V G - G , we have for all g E G ,

where r ( g ) = ICv(g)l. Let I r r ( V G ) be the set of all irreducible characters of V G . Then for any

KnBrr's Generalized Character

1010

x

E I r r ( V G ) , x(1) divides \GI by Theorem 17.10.4. Hence x(1) f Omodp and so, by hypothesis, < X G ~ , X G>2 WG) (3)

Thus, if {xi} is a set of representatives of the conjugacy classes of VG, then

=

(by Lemma2.1)

xElrr(V G )

=

C

XG~,XG

x EIrr (VG )

(by Lemma 19.3.2 and Theorem 19.3.9)

= k(VG)k(G)

Dividing by L(G) gives the desired result. Step 2. From now on, we assume that A=

#

1 and put

and N = NG(A)

Then CG(W) = C d N , IAl = p and & ( a ) = C for all 1 # a E A . Hence N / C operates fixed-point-freely on A and, in particular, ( N : C) divides p - 1. Note also that the direct product A x C is a subgroup of V G . Let {c;li = 1,2,. ,k(C)} be a set of representatives for the conjugacy classes of C and let {ajlj = 1,2,. , . ,( N : C)-'(p - 1)) be a set of representatives for the N-conjugacy classes of A - (1). The aim of this step is to show that { a j c ; } is a set of representatives for the conjugacy classes of V G which intersect ( A - (1)) x C nontrivially. In particular, it will follow that there are k ( C ) ( N :C)-'(p - 1) such classes. Let ac E AC with a # 1. Then nan-' = aj for some n E N and some j . Since ncn-' E C, there exist d E C and i such that dncn-ld-' = C i s Then

..

dn(ac)(dn)-l = (dagd-')c; = ajc;

On the other hand, if (ug)(ajc;)(ug)-' = a , ~ for , some g E G , u E V , then (ug)ci(ug)-l = c, and (ug)aj(ug)-' = a, since (ICl,IAI) = 1 and ca = ac

2 Reduction t o C G (V )

1011

for all a E A , c E C. Since a, = (ug)aj(ug)-' = gajg-' and A =< aj >, we have g E N . Hence j = s and g E C. Because (ug)c;(ug)-l E G implies u E CV(gcig-'), this means c, = gcig-' and so r = i, as desired. From the above proof, it also follows that for 1 # a E A , c E C , g E G and u E V the following holds : (ug)(ac)(ug)-' E A x C if and only if g E N and 21 E C v ( g c g - l ) . Step 3. Now define a on A x C by a = (PlA - p) x 6c,where p is the regular character of A. Thus, by definition, a(ac)=

if

a=l

Here we prove the result under the additional assumption that

< ~ X A ~ C , X A>~2C (P - l)k(C)

(4)

for all x E Irr(VG). First of all, aVG vanishes on all conjugacy classes of V G which intersect ( A - { 1)) x C trivially. On the other hand, for all 1 # a E A, c E C , we have aVG(ac) =

p1-l I c y

c c

a((ug)ac(ug)-l)

U E v,9E G

= P-lIcI-l

a((9a9-1)(sc9-1))

UG'v(gcg-'),gEN

=

p-'Icy

c

p~(gcg-1)7r(gcg-1)

s€N

= 1CI-l

c

IVI

g€N

= (N:C)IVI Hence, if then

{xi}

is a set of representatives for the conjugacy classes of VG,

( p - l)k(C)[VI = ( k ( C ) ( N :

=

c

c)-'(p- 1))(N : C)lVl

(by Lemma 2.1)

xEZrr(VG)

=

C

< ~XAXC,XAXC >

xEZrr(VG)

(by Lemma 19.3.2 and Theorem 19.3.9)

Kniirr's Generalized Character

1012

c

>

-

(P- 1 ) W )

(by (4))

XEIrs(VG)

=

k(VG)(p- l)b(C)

Thus cancelling ( p - l)lc(C), the desired conclusion follows. Step 4, By Step 3, it remains t o verify (4). To begin with, we write Y

where 7 runs through I? = I T T ( A )(recall that character of C or xr = 0. Then

I'( = [ A [ = p ) and x-, is a

Taking into account that

we deduce that

< ~ X A X C , X A ~ >C =

PC

=

c


Y,W@ ( X Y - X,)JC,(XY

-XP)

>

7# 0.

Knijrr's Generalized Character

1016

as desired. (ii) Put C = c~(X,a). If Ii E C, then (X,a) = (X,a)h' = (XaYK,aK)

so a~ = 1 and a K = a. Hence, for any J E B , a K + J = a ~ ( a = ~ ) ~ ( Y J , Now C is a subgroup of B and we let B = u8(c t J s ) be a coset decomposition. Then Ycr

= nc1 a€

I

( -l)'A-+J'IaK+

=

Js

s KEG

=

C(- l ) l J J a J sc (- 1)'K' KEC

8

IK i-JsI 3 IICl t lJslmod2. If [Itr( Omod2 for all li E C, then CKEc(-l)lKl = ICI. Assume there is IC E C with IKI = lmod2. Then the Ii's of even cardinality form a subgroup of index 2 in C and therefore CKEc(-l)K = 0. In either case, (C(-'y, is a generalized character and hence so is (C-'(Xy,. Hence all the (iii) Owing to (i), pyp = fXy, for all (p,/3) E summands in the sum are equal. If C = C,(X,a), then there are ( B : C) =

because

21rllCI-1 summands. Thus 2-14

c

X9PYp

>2

< X,XY,

=

1q-l

=

< x,XY, >< x,IcI-'XYa >

>2

(P,P)E(WB

which is an integer by virtue of (ii) and which is obviously positive provided < x,Xy, ># 0. This concludes the proof of the lemma. H

Lemma 3.3. Let 0 be given by (1). Then

Proof. We argue by induction on n. If n = 0, then both sides are equal to l ~by, convention. Now let n > 0, N' = N - { n } and 0' = f l i E N , O i .

3 The abelian case

1017

we have

Accordingly,

Corollary 3.4.

For any generalized character

< xe,x >= Proof. Since < xy,?,, virtue of Lemma 3.3.

c

C 2-14 ICN

x of G,

< X,XYff

>2

(A,a)ETxAI

x >= Cxer < x,Xy,

>2,

the result follows by

Proposition 3.5. Assume that n;="=,H;= 1 and let 8 be given by (1). Then, f o r nonzero genemlized character x of G,

< x4,x >> IGI Proof. For each X E r, let M A consist of those subsets I of N for which there exists p f rz with < x , X p ># 0. Note that N f M A so, in particular, M A # 8. Indeed, since x # 0, there exists u E r such that 0 #
=< x,X(X-lv)

>

and

X-lv

E

r=

Knijrr's Generalized Character

1018

(since Ir'errN = fl?==,H; = 1) and so N E MA. Let m(A) E M A be an element of minimal cardinality. Then we have a map n : r -+ P ( N ) such that for all X E I', the following two properties hold : (i) < x,Ap ># 0 for some p E rrn(A) (ii) < x,Xp >= 0 for all p E rJ if J C m(A) We now partition r into disjoint union

r =UI~N~-'(I) where m-'(I) = {A E I'lm(A) = I } is the inverse image of I under rn. Fix I C_ N and a coset c p r I E I'. Let us show next that the result will follow provided we prove that

Indeed, we have

=

C I uqE[r:rIlm-V)n @II C lm-l(I)I = = [GI,

I cN

=

IcN

as required. By the foregoing, we are left to verify (2). If m-'(I) n 'pri = 8, then there is nothing t o prove. So assume that 61,. . .,ut are distinct elements of l'I such that m - ' ( I ) fl = { V a l , * . * 9 'pat) Then m(cpu,) = I , 1 5 s 5 t , so if J C I and ,f3 E I'J, then < x,'pu,,f3>= 0. Given a E A I , we have ')'a = C(-l)IJlaJ

cprr

J cI

3 The abelian case

1019

and therefore

< X,' @asym > = x(-l)'J' < x,YfJsaJ > JC I

= (-1)lZl

< x,cpasaz >

Now the map AI + I'r, a H a1 is a surjective homomorphism and, by definition of I , there exists p E A1 such that < x,cpulp~># 0. Furthermore, we can choose a, E AI such that (&,)I = a;lal E I'I. Put PS = asp E A I ; then (cpo,,P,) E c p r I x AI for all s E ( I , . . . ,t } . Our next aim is to show that (a) < x,cpa37p,># 0 for all s. (b) The (cpas,Ps)'sbelong to different orbits under the action of B =

P(0 Once this is accomplished, (2) will follow by virtue of Lemma 3.2 (iii). As we have seen before,

< X,cp~S7Os> =

< X,(PG(Ps)I > = (-l)'z' < x,cpas(a,)rPz> = (-1)lIl < X,cpalPI > (-1)'Il

# 0, proving (a). To prove (b), let J be a subset of I such that for s,k E (1,. (90s

0 s ) = (pak Pk)J 9

= ( V b k ( P k )J 9 ( P k )

J,

. ., t } ,

Then as

J -

J J

= P s = ( P k ) - akP

(3)

= ak(Pk)J = a k ( a k ) J P J

(4)

asP

Applying (3), we obtain as"alPr

= (%)PZ = (asP)z J J = (akP )I - (ai)I(DJ)I

= =

( a k ) l [ ( a k )J ] - 2 P I ( P J ) - 2

>

0;' a1[(ak J ] ' 2 P I ( P J ) - 2

and therefore 6s

= ak[(at ) JI2( p J ) 2

Comparing with (4),it follows that as we wished to show. H

( C Y ~ ) J P=J 1. Thus 0,

= Uk and

s

= k,

Kn6rr’s Generalized Character

1020

4

The main result

All groups below are finite and all characters are assumed to be CC -characters.If a and p are class functions of a group G, we say cr 5 p if for all g E G, a ( g ) , p ( g ) and p ( g ) - a(g) are nonnegative real numbers.

Proposition 4.1. (Knorr (1984)). Let F be a finite field of characteristic p and let G be an abelian p’-group acting faithfully on a finitedimensional FG-module V . Let 6 = 6(G,V) be the Know’s genemlized character of G with respect to V and let x be a nonzero generalized character of G . Then

< x4x > L PI Proof. For any submodule W of V, let I i e r W be the kernel of the representation of G afforded by W . Since G acts faithfully on V, we have be a direct decomposition into simple FGI i e r V = 1. Let V = modules and let S; = 6(G,W;).Since ICv(g)l = ICw,(g)l for all g E G and ( V (= (W;l,we have S = nZ16;. Next observe that Cwi(g) is an FG-submodule of W;,since G is abelian. Because W; is simple, we deduce that either Cw,(g) = W; (hence g E H ; = I i e r Wi) or Cw,(g)= 0. Thus

nr==,

nr==,

and therefore

) ~>= , 1, this implies in Because 6i is a generalized character and < ( l ~ , 1~ particular that (G : Hi)divides lWil - 1, say

IWi] = k i ( G : H i ) t 1

(1 5 i 5 n )

for some integer ki 2 1. Let 0, be the generalized character of G defined in the previous section. Then bi

= (k;(G: H i ) t 1 ) l -~ k ; ( l ~ , ) ~ = ((G : Hi) t 1 ) l -~( I H , ) ~ (h- 1)[(G : H i ) k - ( I H , ) ~ ] 2 ( ( G : Hi) 1)lG = 6;

+

+

4

The main result

since Thus

Ici

1021

2 1 and (G : H;)lc - ( 1 ~takes ~ )only ~ nonnegative real values. n

n

i=l

i= 1

Since ny=(=,Hi = K e r V = 1, it follows from Proposition 3.5 that (GI. Thus < xhx > P < X ~ >XP IGI

< xB,x >2

as desired. The following result is the goal we have been striving for throughout this section,

Theorem 4.2. (Knorr (1984)). Let F be a finite field of characteristic p and let G be Q pf-group acting faithfully on a finite-dimensional FG-module V . If there exists v E V such that CG(V) is abelian, then the number of conjugacy classes in the semidirect product V G is bounded by the order of V. Proof. Let C = CG(V)and regard V as an FC-module by restriction. Put 6 = 6(G,V) and note that 6c = S(C,V). Since C is abelian, it follows from Proposition 4.1 that

for any nonzero generalized character x of C. The desired conclusion is therefore a consequence of Theorem 2.2.

This Page Intentionally Left Blank

Chapter 33

Characters of Centralizer Rings Let H be a subgroup of a finite group G and let F be a field of characteristic 0 which is a splitting field for H and G. Given an idempotent e # 0 of F H and the character 1c, of H afforded by FHe, we wish to provide connections between the irreducible constituents of llGand the irreducible characters of the F-algebra eFGe. It is shown that if is irreducible, then the irreducible constituents of the induced character ?+hG determine (by restriction) all distinct irreducible characters of eFGe, and the degrees of these characters of eFGe are precisely the multiplicities of the irreducible constituents of $.' As an application, we provide some information concerning the degrees of irreducible characters (Curtis and Fossum (1968)). It is natural to enquire what happens if we replace the process of induction by the dual process of restriction. The second part of the chapter provides an investigation of this phenomenon. It turns out that the role of eFGe is played by the centralizer CFG(H)of H in FG.

+

1

Characters of eFGe

Throughout this section, H denotes a subgroup of a finite group G and F any field of characteristic 0 which is a splitting field for H and G. The Fcharacters of G and H will be identified with the corresponding characters of FG and F H . For convenience, we divide this section into three subsections.

1023

Characters of Centralizer Rings

1024

A. C h a r a c t e r s of centralizer rings Let e # 0 be an idempotent of FH and let $ be the character of H afforded by FHe. Our aim is to tie together the irreducible constituents of $G and the irreducible characters of the F-algebra eFGe. Recall that e is the identity element of eFGe and that, by Proposition 1.9.3, eFGe is isomorphic to (EndFG(FGe))O. Lemma 1.1. Let e # 0 be an idempotent of F H , let $ be the character of H aflorded by FHe and let x be an irreducible F-character of G afforded by an FG-module V . Then

(i) FGe affords the induced chamcter .roc. (ii) < + G , >= ~ x ( e ) = dimFeV. (iii) eFGe is semisimple and an idempotent u E eFGe is primitive in eFGe if and only if it is primitive in FG. (iv) The field F is a splitting field for eFGe. Proof. (i) This is a direct cosequence of the fact that, by Proposition 18.1.16, (FHe)GE! FGe. (ii) We have V = eV @ (1 - e)V as F-spaces. Since e acts as identity on eV and annihilates (1 - e)V, we see that x ( e ) = dimFeV. On the other hand, by Proposition 1.9.2 (i),

eV

E'

HomFG(FGe,V) as F-spaces

Since, by (i) and Lemma 19.3.8,

property (ii) follows. (iii) By Maschke's theorem and Proposition 1.5.29 (ii),

J(eFGe) = eJ( FG)e = 0, proving that eFGe is semisimple. Since both F G and eFGe are semisimple and uFGu = u(eFGe)u,the desired assertion follows. (iv) Let u be a primitive idempotent of eFGe. It suffices to show that u(eFGe)u = Fu. But, by (iii) and the assumption that F is a splitting field for G , uFGu = Fu. This proves (iv) upon noting that ue = u.

1 Characters of eFGe

1025

The following result is contained in works of Curtis and Fossum (1968) and Ree (unpublished).

Theorem 1.2. Let e # 0 be an idempotent of FH, let t,b be the character of H aflorded by F H e and let $' = nlX1

+

* * *

-k n T X T

where each ni is a positive integer and the x; 's distinct irreducible characters of G. Denote by a; the character of eFGe which is the restriction of x; toeFGe. Then (i) al, . . . ,ar are all distinct irreducible characters of eFGe and deg(a;) = ni, 1 5 i 5 r . In particular, eFGe is commutative if and only if $G is the sum of distinct irreducible chamcters of G. (ii) If x E { X I , . ,xr} and x E Z ( FG) is the sum of all conjugates of g E G, then

..

x ( g ) = IcG(g)lx(eze){

X(ez-le)X(e~e))-' x G

Proof. (i) By hypothesis, < $ G , ~ i>= n; # 0. Hence, by Lemma 1.1 (ii), eT/;: # 0, where V;. is an FG-module affording xi. We may view eV, as an eFGe-module by restriction. Now let z E eFGe. Then zV, C eVs: so that the trace of z on V;. is equal to the trace of z on eV,:. Therefore et;: affords a!i. If v # 0 and v E eT/;:,then eFGev = eV,, because V , is a simple FG-module. Thus eV, is a simple eFGe-module and so a; is an irreducible character of eFGe. Furthermore, deg(a;) = dimeV, = ni by Lemma 1.1 (ii). Let a! be the character of a simple eFGe-module. Then a is afforded by a left ideal (eFGe)u where u is a primitive idempotent of eFGe. By Lemma 1.1 (iii), u is primitive in FG . Hence FGu is a simple FG-module affording some irreducible character x. Applying the argument in the preceding paragraph to FGu, we see that the restriction of x to eFGe is the character of the simple eFGe-module eFGu = (eFGe)u; hence this restriction must be equal to a. Since x(e) = a(e) # 0, it follows from Lemma 1.1 (ii) that x = x ; for some i E (1,. . . ,r } , which shows that a! = a;. Assume that i , j E {l,.. . , T } are such that a!i = aj. By the preceding paragraph, there exists an irreducible character x of G afforded by a minimal left ideal F G u with u E eFGe such that the restriction of x to eFGe is equal to a;.Let V , be a simple FG-module affording x i . Then x;(u) = a ; ( u ) # 0, whence uV, # 0 and so FGu 2 K. Thus x = xi and, by the same argument,

Characters of Centralizer Rings

1026

x = xj which shows that i = j. The find assertion is a consequence of the fact that F is a splitting field for eFGe (Lemma 1 . 1 ) which implies that eFGe is commutative if and only if each n; = 1. (ii) Let p be a matrix representation of G affording x and let C be the conjugacy class of g . Then eze = ze (because z E Z ( F G ) ) and p ( z ) = XI for some X E (I:. In fact, by taking traces of both sides, we have = IClx(s)x(l>-'

Then

P ( 4 =p ( M e )=XP(4 and taking traces again, we obtain x ( e 4 = Ax@) = Iclx(s)x(e)x(l)-'

(1)

Now let E

= X(l)lGl-'

x(x-')s xEG

be the block idempotent of FG corresponding to x (see Lemma 19.2.7). Then Ee = eEe = x(l)IGl-' X(x-l)ese (2) xEG

Writing e = &Hehh

with eh E F , it follows from e2 = e and ( 2 ) that zEG h,k

Setting y = hxk, we have

2-l

= ky-lh and

Accordingly

Ee = x(1)lGl-l

c

= X(l)IGI-'

ekkz-lehh)ese h,kcH x(ez-le)exe

x(

xEG

zEG

Next we apply p and take traces. Because p ( ~ =) I, we have

x ( e > = x(1)lGl-l

x xEG

x(ef-le)x(eze>

(3)

1 Characters of eFGe

1027

Substituting this expression for x(e) in (l),the desired assertion follows.

Corollary 1.3. Let e # 0 be an idempotent of F H and let $ be the character of H afforded by F H e . If x is an irreducible character of G such that < $',x >= 1, then the restriction of x to eFGe is a homomorphism of eFGe into F. Conversely, every homomorphism of eFGe into F is the restriction to eFGe of a unique irreducible character x of the group G such that < t+hG,x>= 1. Proof. This is a direct consequence of Theorem 1.2 (i). W Corollary 1.4. Let e # 0 be an idempotent of F H , let $ be the character of H afforded by F H e and let

where each ni is a positive integer and the acters of G. Let

x; 's are distinct irreducible char-

be the block idempotent of F G corresponding to

x;.

Then

eel, ee2,. . . ,ee, are all block idempotents of eFGe.

Proof. Let V, be an FG-module affording x;. Then, by Theorem 1.2 (i), eV1,. . . ,eV, are all nonisomorphic simple eFGe-modules. Since ee; acts as the identity on e K and annihilates all e& with j # i, the result is established. B. Bases and block idempotents We shall present here a more detailed analysis of the structure of an F-algebra eFGe in case $ is a linear character, and hence

Characters of Centralizer Rings

1028

is the primitive idempotent of FH such that $ is afforded by F H e . For convenience, for each g E G , we put

ind(g) = ( H : ( g H n H ) ) where, as usual, gH = gHg-l. It is easily seen that ind(g) is the number of left (or right) cosets of H contained in H g H .

Theorem 1.5. (Curtis and Fossum (1968)). Further to the notation and assumptions above, let G = UiEIDi, where the Di are the distinct ( H ,H)-double cosets HgH in G , and let J I be the set of indices j such that $ = g $ on H n gH for some g E Dj (this condition depends only on the double cosets and not their representatives). For each j E J , pick g j E Dj and put aj = ind(gj)egje. Then (i) The elements {ajlj E J } form an F-basis for eFGe. Moreover, the basis elements { a j } are determined independently of the double coset representatives if 11, = 1 ~and , up to multiplication by a root of unity if $ # 1 ~ . (ii) The structure constants x ; j k E F defined by

U'a' ' 3 = CXijkak

(i,j E J )

kc J

are given by xijk

=

Ui(z)aj(x-'gk) ZED,ngkD;'

where a;(.) is the coefficient of x in a; E F G . Furthermore, each & j k is integral over 21; , (iii) The set { D j ( j E J } is invariant under the mapping D + D-' so that corresponding to { a j l j E J ) there is a second basis {iijlj E J } , where iLj = ind(gj)eg;'e. (iv) The block idempotents ee; of eFGe (see Corollary 1.4) are given b y ee; = x ; ( l ) ( G: H ) - l C [ i n d ( g j ) ] - l ~ ; ( i L j ) a j j €J

Proof. (i) and (iii) We first show that for z E G, exe # 0 if and only if 11, = "11, on H n " H . Observe that for h E H , he = eh = $(h)e. Now assume that h E H n " H , say h = "hl for some hl E H . Then ("$)(h)eze = $(hl)eze = exehl = e x h l e = ehxe = +(h)eze

1 Characters of eFGe

1029

and so ("$)(h) = $ ( h ) if e z e # 0. Conversely, assume that $ = "$ on H n " H . By the definition of e , we have eze = I H I - ~ $(y>-'$(z)-'yzt (4)

C

YJEH

Now y z z = z for y,z E H if and only if y E H fl " H . Because $ = "$ on H n " H , the equality yzz = z forces $(y) = + ( z ) - l . Hence, by (4) the coefficient of z in e z e is IHJ-' IH fl " H I ; in particular, e x e # 0 and aj # 0 for j E J. Moreover, given y,z E H , we have e y z z e = $ ( y ) $ ( z ) e s e . Hence the coefficient of yzz in e z e is $ ( y ) - ' q ! ~ ( z ) - ' I H l - ~( H n "HI and therefore aj =

c

p1-l

$(g)-l$(h)-lggjh

( j E J)

(5)

99,hEHg, H

where the sum is taken over distinct elements of Dj = H g j H . Note also that e z e # 0 if and only if e y e # 0 when z and y belong t o the same (H,H)-double coset, and e z e # 0 if and only if e z - l e # 0. Thus { D j J jE J } is well defined and invariant under D -, D-'. It is now obvious that { u j l j E J} is an F-basis for e F G e and that the same is true for { l i j l j E J}. It is also palin that changing the double coset representatives will multiply each basis element aj by some root of unity, which is equal to 1 if = 1 ~ . (ii) w e have ajaj = E k E J Xjjkak. Hence evaluating the coefficients of gk, k E J , at both sides and using ( 5 ) , we have Xijk

= IHI(w)(glC) =

C

IHI

ai(g)aj(g-lgk)

gEDtngkD,-'

This gives us a desired formula for the structure constants & j k . We now show that each Xijk is integral over 21; . By (i), it is easily verified that for all h E H , g E G,

ai(g)aj(g-'gk) = ai(gh)aj((gh)-lgk) Hence it follows from (4)that Xijk

= / H I 2IHI-'A

for some X E F which is integral over 23 . (iv) Owing to Corollary 1.4, e e l , e e 2 , . . . , e e , are all block idempotents of

1030

Characters of Centralizer Rings

eFGe. By (3), we have

Let g = hgjk E Dj,where h 7 k E H . Then g-' definitions of {aj} and {iij}, we have

= k-lg:3 'h-' and, by the

It therefore follows that

because [HgjHl = lDjl = ind(gj)lHI. Thus (iv) holds and the result is established.

F be a Corollary 1.6. (Curtis and Fossum (1968)). Let cp : e F G e homomorphism. Then cp is the restriction to eFGe of a unique irreducible character x of G such that < + G , ~>= 1. Moreover, in the notation of Theorem 1.5, --f

is a primitive idempotent of F G such that F G u affords x . Proof. The first statement follows from Corollary 1.3. Let

be the block idempotent of F G corresponding to x. Since < +',x >= 1, it follows from Theorem 18.4.11 that Ee is a primitive idempotent of F G such

1 Characters of eFGe

that FGEe affords

x.

1031

On the other hand, by Theorem 1.5 (iv),

Ee = x ( l ) ( G: H)-' E [ i n d ( g j ) ] - ' x ( & j ) a j je J = x ( l ) ( G: H ) - l E [ i n d ( g j ) ] - ' c p ( & j ) a j -

j6J 21,

as required. The next theorem may by interpreted as giving some orthogonality relations in the centralizer ring. Theorem 1.7. (Curtis and Fossum (1968)). Let II, be a linear character of H and let x,x' be irreducible characters of G both appearing in $G with positive multiplicity. Let e = 1HI-l ChEH $ ( h - ' ) h and let { a j l j E J } , { l i j l j E J } be two bases for eFGe given by Theorem 1.5. Then

x ( l ) ( G: H ) - l x [ i n d ( g j ) ] - ' x ( & j ) x ' ( a j )=

if

x#

X I

jEJ

+

Proof. Write $G = nlxl -t -.. nTXTwhere each n; is a positive integer and the xi's are distinct irreducible characters of G. Denote by ai the character of e F G e which is the restriction of xi to eFGe. Then, by Theorem 1.2 (i), a1,. . . , a , are all distinct irreducible characters of eFGe. Let ei = xi(l)lGl-' CBEG xi(g-l)g, 1 < - i 5 r. Then, by Corollary 1.4, e e l , ee2,. . . ,eer are all block idempotents of eFGe. Since ee; is the block idempotent of eFGe corresponding to a*,we have

where ayi(ee;) = a;(.). Hence, substituting the formula for ee; given by Theorem 1.5 (iv) into this expression, we obtain the desired assertion.

C. Applications to degrees of characters Let H be a subgroup of G and let $ be a linear character of H . Our aim is to compute the degrees of irreducible characters x of G such that $JG,X ># 0.

Characters of Centralizer Rings

1032

Theorem 1.8. (Curtis and Fossum (1968)). Let H be a subgroup of a finite group G, let F be a field of characteristic 0 which is a splitting field for H and G and let II, be a linear F-character of H . If x is an irreducible F-character of G such that < + G , ~># 0 , then in the notation of Theorem 1.5, we have : (i) ~ ( 1 =) (G : H ) < II,G,~ > (CjEJ[ind(gj)I-'~(Bj)x(aj))-l (iz) x(1) divides (G : H ) n , where n is the least common multiple of i n d ( g j ) as j ranges over J. In particular, x(1) divides (G : H ) if H 4 G . Proof. (i) Applying Theorem 1.7 and the fact that x ( e ) =< (Lemma 1.1 (ii)), we have

>

+ G , ~

proving (i). (ii) We first observe that if H 4 G, then each i n d ( g j ) = 1 and hence n = 1. To prove the remaining assertion, we may harmlessly assume that F is an algebraic number field which is a splitting field for G and all of its subgroups. Owing to Lemma 1.1 (iv), F is a splitting field for e F G e , where e = [HI-'ChEH +(h-l)h. Let R be the integral closure of Z in F . Then R is a Dedekind domain (Theorem 16.2.2) with quotient field F . Let P be a nonzero prime ideal of R and let Rp be the local ring of R at P. By Lemma 16.3.3, Rp is a discrete valuation ring and, by Proposition 16.2.19 (ii), R is the intersection of ad such Rp. We claim that

n(G : H)x(l)-' E Rp

for each

P

(6)

which will imply that

n(G : H ) x ( l ) - ' E R n Q = Z, as desired. From now on, we put S = R p for a fixed nonzero prime ideal P of R. Then S is a principal ideal domain (Theorem 16.3.2) with quotient field F . Now put r = Saj

C

j €J

where { a j } is the basis of eFGe given by Theorem 1.5. Then S C Z ( r ) and r is a ring because the structure constants { X i j k } lie in R 5 S by Theorem

1 Characters of eFGe

1033

1.5 (ii). Moreover, iij E I? for all j E J , because iij either coincides with one of the { a j } or differs from one by multiplication by a root of unity. Put E = eFGe. Then, by Theorem 1.2 (i), XE is afforded by a simple E-module M . Since F is a splitting field for E , M is absolutely simple. Now S is a principal ideal domain with quotient field F so there exists a r-module N C_ M such that F N = M and N is a free S-module with S-basis wl,. .. ,V k which is also an F-basis for M . Let p : E + M k ( F ) be the matrix representation of E afforded by M with respect t o ~ 1 . ., . ,Vk. Then p ( a ) E &(S)

for all

a E

r

Then y E I? since iij E I? and ~ ( i i j = ) t r ( i i j ) ( o n M )E S. By Theorem 1.5 (iv), y belongs t o the centre of I?, so

.

p(y) = s I

( I = identity matrix)

for some s E S, since F is a splitting field for E and y E I?. Taking traces of both sides, we obtain X(Y) = S X @ >

(7)

By Theorem 1.5 (iv), y = n(G : H ) x ( l > - ' & e

where

E

is the block idempotent of FG associated with

x. Then

and so, by (7),

n(G : H)x(

=s E S

This proves (6) and hence the result. We close by remarking that if H 4 G and H is abelian, then Theorem 1.8 (ii) reduces t o Ito's theorem (Theorem 17.10.4). However a direct proof of Ito's theorem due t o Knorr is much more elegant.

Characters of Centralizer Rings

1034

2

Characters of C F G ( H )

In what follows, H denotes a subgroup of a finite group G and F is any field of characteristic 0 which is a splitting field for H and G. As in the previuos section, the F-characters of G and H will be identified with the corresponding characters of F G and F H . We write Irr(G) and I r r ( H ) for the set of irreducible F-characters of G and H , respectively. Most of the results presented below are based on a work of Travis (1974). Let 11, E I r r ( H ) and let e be a primitive idempotent of F H such that 11, is afforded by F H e . By Theorem 1.2, the irreducible constituents of the induced character ll,G determine (by restriction) all distinct irreducible characters of eFGe, while the degrees of these characters are precisely the multiplicities of the irreducible constituents of ll,G. It is natural to enquire what happens if we replace the process of induction by the dual process of restriction. It turns out, as we shall see below, the role of eFGe will be played by the centralizer C F G ( H )of H in FG. We begin by determining an F-basis and the dimension of C F G ( H ) . Lemma 2.1. Let C1,. . . , C , be all H-conjugacy classes of G, let C$ E F G be the sum of all elements in C;, 1 5 i 5 T , and let

XH =

CX$$

(11, E I r r ( H ) , x E Irr(G))

1LE I r r ( H )

where the cX$ are nonnegative integers. Then (i) The elements C t , . .. ,C$ form an F-basis for C F G ( H ) . (ii) = &Elrr(G),$Elrr(H) c&

Proof. (i) Let x = CgEG xgg, xg E F, be an element of FG. Then, for any given h E H , h-lxh = C g E G x g ( h - ' g h ) . Hence x h = hx if and only if x h - l g h = xg for all g E G . This shows that x E C F G ( H )if and only if the coefficients xg are constant on the H-conjugacy classes of G , proving (i). (ii) The group H acts as a permutation group on G by conjugation and the corresponding permutation character X of H is given by X(h) = \ C ~ ( h ) l for h E H . Since the H-orbits of G are precisely the H-conjugacy classes and the number of orbits is < A, 1~ >, we have

= JHJ-' hGH X E lr r ( G )

x(h)x(h-') (by Theorem 19.2.3 (iii))

2 Characters of

CFG(H)

1035

as desired. Let A be a finite-dimensional algebra over a field of characteristic 0. A character x of A is said to be multiplicity-free if x is the sum of distinct irreducible characters of A. We remind the reader that, by Theorem 14.5.1, the irreducible characters of A are linearly independent. Thus if V is an A-module which affords x , then x is multiplicity-free if and only if the composition factors of V are mutually nonisomorphic.

Theorem 2.2. Let V, be a simple FG-module which affords x E I r r ( G ) , let V , be a simple FH-module which affords E I r r ( H ) and write ~

where the cX+ are nonnegative integers. Then (i) Each cX,V+ with cx+ # 0 is a CFG(H)-module which is a direct sum of $(1) copies of a simple CFG(H)-module, say W,,, and dimWx, = cx+. (ii) C F G ( H ) is semisimple and F is a splitting field for C F G ( H ) . (iii) The { W,,} are all nonisomorphic simple CFc(H)-modules. (iv) C F G ( H ) is commutative if and only if for all x E I r r ( G ) , x ~is multiplicity-free. (v) H is commutative i f and only if for each x E I r r ( G ) the restriction of x to CFG(H ) is multiplicity-free. Proof. (i) Assume that cx, # 0. Then cx,V, is a homogeneous component of ( V x ) ~Hence, . by Proposition 1.4.8, f(c,,V+) C cx,V, for all f E E n d F H ( V x ) . Now let x E C F G ( H ) . The map f, : Vx + Vx given by f,(v) = x v is an FH-homomorphism since

f,(hv) = x h v = h ( x v ) = hf,(v)

for all h E H,v E V,

Thus z(cx+V+) C cx+V+ which shows that cx$V$ is a CFG(H)-mOdUk. Moreover, x acts as the element A, of EndFH(c,+V,), where A, is the restriction of f, to cx,V,.

Characters of Centralizer Rings

1036

Conversely, we wish t o show that for any f E E n d ~ ~ ( c , , v + ) ,there exists 2 E CFG(H) such that zv = f ( v ) for all v E c x ~ V x .We may regard f as an element of E n d ~ ~ ( v , ) E n d ~ ( v , ) . Since F is a splitting field for G and V, is a simple FG-module, there exists y E F G such that yv = f ( w )

for all v E V,

Now put 2

=

p1-1

c

hyh-1

&H

Then

2

E CFG(H)and, for all v E V,, 22,

=

pq-1

C(hyh-1). = p 1 - 1 h€H

c

hf(h-'2,)

h W

= f(v), as required. By the foregoing, CFG(H)acts as the full ring E n d ~ ~ ( ~ , , j , v +Hence ). the CFG(H)-submodules of cx+v+ are precisely the EndFH(c,+v,)-submodules of cx+Vx. Since F is a splitting field for H , we have EndFH(cx4V1)

Mxt,b(F)

Thus cx,V+ is a direct sum of dim V, = 11,( 1) copies of an absolutely simple CFG(H)-module, say Wx+ and dimWx+= cx+. (ii) and (iii). Let ex,e+ be the block idempotents of FG and F H , respectively, corresponding to x E ITT(G), II, E I T T ( H ) .Then ex,e+ E C F G ( H ) . Assume that Wx+ 2 WX,,pfor some X I E I T T ( G )1,,' E I T T ( H ) .If x # X I , then ex, annihilates V,, hence annihilates Wxq,hence annihilates It',,+, E V,,, which is impossible since ex, acts as identity operator on V,,. Thus x = X I and, by a similar argument, 11, = 11,'. Hence the are mutually nonisomorphic. The rest of the proof will be based on the following equality

XllL

which is a consequence of Lemma 2.1 and (i). Put C = C F G ( H )and denote by 121,. . . ,n, the dimensions of all nonisomorphic simple C-modules. We first consider the case where F is algebraically closed. Then by (1)

dimC 2 d i m ( C / J ( C ) )=

n: 2 c ( d i m W x + ) 2= dimC i=l

2 Characters of C F G ( H )

1037

which shows that J ( C ) = 0 and that the { W x ~are } all nonisomorphic simple CFG(H)-modules. Turning t o the general case, let E be the algebraic closure of F . By Lemma 2.1 (i), @F C F G ( H ) = C E G ( H )

and hence J ( C F G ( H ) )E J ( C E G ( H ) )= 0. Since the modules { W X $ }are mutually nonisomorphic and absolutely simple, it follows from (1) that they represent all nonisomorphic simple CFG(H)-modules, proving (ii) and (iii). (iv) The condition that for all x E I r r ( G ) , X H is multiplicity-free is equivalent t o the requirement that each nonzero cx$ is equal to 1. The desired conclusion is therefore a consequence of (i), (ii) and (iii). (v) By (i), the restriction of x t o C F G ( H is ) multiplicity-free if and only if $(1) = 1 for each $ E I r r ( H ) with cx4 # O+ Observe also that, by Frobenius reciprocity, for any $ E I r r ( H ) there exists x E Irr(G) with cX$ # 0. Since H is commutative if and only if $(1) = 1 for each 1c, E I r r ( H ) , the desired assertion follows. H We next determine explicitly the irreducible characters of C F G ( H ) Given .

x E I r r ( G ) and $ E I r r ( H ) , define the function by ax&) = IHI-'

c

x(gh)$(h-'>

( 9 E G)

hEH

Theorem 2.3. Let Ax+ be the irreducible character of C F G ( H )afforded Then the {Ax$} are all by Wx$, where WX$is as in Theorem 2.2 (i). irreducible characters of C F G ( H )and for any z =

z b ) g 6 CFG(H)

E F)

gEG

we have Ax&)

=

c

z(s>ax$(s>

gEG

Proof. That the {Ax+} are all irreducible characters of C F G ( H )is a consequence of Theorem 2.2 (iii). Let e$ = lHI-'w

c

hEH

$W1)h

Characters of Centralizer Rings

1038

+

be the block idempotent of F H corresponding to E I r r ( H ) (see Lemma 19.2.7). Then eg acts as identity operator on V, and annihilates each VGI with +' # Hence

+.

(as CFG(H)-modules)

cxgVG = eGVx = + ( l ) W x g

The desired assertion is therefore a consequence of (2). W We close by determining all block idempotents of C F G ( H ) . In what follows, for each x E Irr(G) and each E I r r ( H ) , the elements e x and eg of F G are defined by

+

Recall that, by Lemma 19.2.7, { e x l x E I r r ( G ) } and { e ~ \ +E I r r ( H ) } are the sets of all block idempotents of F G and F H , respectively.

Theorem 2.4. For each x E I r r ( G ) and each < X H , + ># 0 define ex$ E F G b y

+ E I r r ( H ) such that

2 Characters of CFG(H)

1039

Then e x + = exe* and the { e x + } are all block idempotents of C F G ( H ) .

Proof. We first note that

Hence e x e+ is an idempotent of Z ( C F G ( H ) ) .Note also that ex+ ax+(1) = IHI-'

c

# 0 since

x@)+@-l)

h€H

= #O Since the { e x e + } are mutually orthogonal, it suffices, by Theorem 2.2 (ii), (iii), to show that ex+ = e x e + . The latter being a consequence of the equality

the result follows. We close by remarking that the maps ax+,< X H , + ># 0 with F = (c , are called spherical functions . Some further information on the behaviour of spherical functions can be found in works of Travis (1974) and Gallagher (1975).

This Page Intentionally Left Blank

Chapter 34

Characters and Relative Normal Complements Let H be a subgroup of a finite group G and let Ho d H . We wish to examine character-theoretic conditions which guarantee the existence (and uniqueness) of a normal complement of H over Ho. The main tool in our investigations will be a lifting operator introduced by Leonard (1983). The essential part of the main result (Theorem 3.3) is included in Dade’s argument (Dade (1975, Theorem 1)). Our first application is Dade’s theorem (Dade (1975)) which generalizes a classical result due to Brauer (1964b) and Suzuki (1963). We then establish a classical theorem due to Frobenius (1907) and provide a number of important generalized characters contained in a work of Robinson (1985). The find section is devoted entirely to a work of Ferguson (1984). One of the results presented asserts that if n is the set of all prime divisors of ( H : H o ) , every n-element of G is conjugate to an element of H and every irreducible (C-character of H having Ho in its kernel can be extended to a (I: -character of G, then there is a unique normal complement of H over Ho. As an easy application, we show that if n is a set of primes, then G has a normal n-complement if and only if (i) G has a Hall n-subgroup H ; (ii) every irreducible (I: -character of H extends to a (I:-character of G; and (iii) every r-element of G is conjugate to an element of H . Our final result gives a partial solution of the following problem posed by Ferguson (1984): For what groups Ho a H C G is the existence of extensions of all irreducible (I: -characters of H / H o to G both necessary and sufficient to guar-

1041

Characters and Relative Normal Complements

1042

antee a normal complement in G of H over Ho ? We show that if H / H o is solvable with ((G : H ) , ( H : H o ) ) = 1, then G has a normal complement of H over HO if and only if each irreducible (I: -character x of H with HO C K e r x extends to a (I:-character of G.

1

Relative normal complements

In what follows, all groups are assumed to be finite and all characters to be (I:-characters. Let H be a subgroup of a group G and let HO a H . Following Leonard (1983), we say that a subgroup Go of G is a normal complement of H over Ho if the following conditions hold : (i) Go a G. (ii) G = G o H . (iii) HO = Go t lH. Note that the above conditions imply that G/Go 2 H / H o . Of course, in the special case where Ho = 1, we come to the previously defined notion of a normal complement. Lemma 1.1. Let H be a subgroup of a group G and let H o d H . Suppose that there exists a normal complement Go of H over Ho. Then (i) Every character of H having Ho in its kernel can be extended to a character of G. (ii) If x , y E H and if x and y are G-conjugate, then X H O and yHo are H / Ho -conjugate.

Proof. (i) This is a direct consequence of the fact that G/Go E H / H o . (ii) Our hypothesis implies that xGo and yGo are G/Go-conjugate. Since the map

{

H/Ho hHo

--t

c--)

GIGO hGo

is an isomorphism, the desired assertion follows. H Note that, even in the case Ho = 1, neither (i) nor (ii) is sufficient to guarantee the existence of a normal complement of H over H o , e.g. take G to be the cyclic group of order 4 and H the subgroup of order 2. The question of what must be added to (ii) will be examined later.

2

A-

Sections

1043

The following result, due to Leonard (1983), provides circumstances under which a relative normal complement, if it exists, must necessarily be unique. Proposition 1.2. Let H be a subgroup of a group G and let Ho a H . Denote by n the set of all prime divisors of ( H : H o ) and assume that

( ( G: H ) , ( H : H o ) ) = 1 Then there exists at most one normal complement Go of H over Ho, namely Go = G1 - Ho, where G1 is the subgroup of G generated by all #-elements of the group G . Proof. Assume that Go is a normal complement of H over Ho. Then

and therefore the a’-part IGo(,t of \Go1 is the +-part of IGI. Thus every r’-element x of G belongs to Go, since otherwise I < x > Go1 is not a divisor of IGI. Let GI be the subgroup of G generated by all n’-elements of G. Then G1 a G and GI C Go. Moreover, IGllnt = IG~J,I, since if p E n‘ then G1 contains a Sylow p-subgroup of G. Finally, G/Go 2 H / H o and so (Go : Ho) is a n‘-number. Consequently, IHol, = IGoJn. Since GlHo Go, it therefore follows that [Go1 divides IG1HoI. Thus Go = G l H o and the result is established.

2

T-

Sections

Let n be a set of primes and let G be a finite group. As usual, the complementary set of primes will be denoted by n’. We say that an element x E G is a n-element if the cyclic subgroup < x > which it generates is a r-group (by definition, 1 is always a n-element). Lemma 2.1.

Every element x of G has a unique decomposition

x = x,x,t

= XAtX,

into a n-element x, and a n’-element x,r. Furthermore, both x, and X,I are powers of x.

Characters and Relative Normal Complements

1044

Proof. Write o(x) = mn where every prime divisor of m is in n and every prime divisor of n is in n'. Then ( m ,n ) = 1 and hence km s n = 1 for some k , s E 1z . Put x, = xSn and x,! = x k m . Then x = x,x+ = x,tx,, both x, and x,r are powers of x and

+

(x*)m = 1 = ( 2 , y

Now suppose that x = uz) = vu,where u is a n-element and v is a n'-element. Let X(x) denote the set of prime divisors of o(x). Since u E CG(Z)G CG(X,), we have X(x,u-') X(x,) U X(u) T . Similarly,

X(xc,,lv) & X(v) u X(5,)) Because xi/w = x,u-l, established.

we have z,u-l

n'

1 = 1 = x-! n v and uniqueness is

We refer to x, and X,I as the n-part and n'-part of x, respectively. Two elements x and y of G are said t o belong t o the same n-section of G if their n-parts x, and y , are conjugate in G. It is clear from the definition that : (i) G is a disjoint union of n-sections. (ii) Each n-section is an invariant subset of G and so is a union of some conjugacy classes of G. If S is a subset of G, we let SG~"denote the union of all n-sections of G that intersect S. Thus

xE

x,

N

y,

for some y E

s

where indicates conjugacy in G. Note that the n-section containing x E G is {x}GJ, that N

and that {x}Gi" is a union of conjugacy classes of G, exactly one of which, namely containing x,, consists of n-elements, Observe also that {l}Gq" is the set of all n'-elements of G. For any group G, T,(G) denotes the number of n'-elements of G, that is

Lemma 2.2.

For any n-element x of G,

3 A lifting operator

1045

Proof. If g E {x}Gy", then we have (G : CG(Z))possibilities for g,. If x' is one of them, we have g , = x' if and only if g = x'y where y is a n'-element of Cc(x') 2 CG(Z). Thus we have r,(CG(z)) possibilities for y , which implies the required assertion.

We next demonstrate importance of the sets SG!" in case S = H - Ho. Lemma 2.3. Let H be a subgroup of a group G, let Ho a H and let n be the set of all prime divisors of (H : Ho). Suppose that there exists a normal complement Go of H over Ho. Then (i) Go C G - ( H - Ho)'?". (ii) Go = G - ( H - HO)'~" (in particular, GO is unique), provided

Proof. (i) Assume by way of contradiction that there exists z E G~ n (H -

~

~

Then x, E Go and, since H/Ho is a n-group, y-'z,y y E G. Since y-lx,y E Go, we have y-'z,y

)

~

l

~

E H - HO for some

E Go n (H - Ho)

which contradicts the fact that Go n H = Ho. (ii) Since G/Go 2 H/Ho, we have [Go[ = (G : H)IHol. The desired conclusion is therefore a consequence of (i).

3

A lifting operator

All groups are assumed to be finite and all characters are (c -characters. Let H be a subgroup of G, let Ho a H and let n be the set of all prime divisors of ( H : Ho). Assume that there exists a normal complement Go of H over Ho. By Lemma 1.1 (ii), the following property holds : (A) If two n-elements z and y of H - Ho are G-conjugate, then zHo and yHo are H/Ho-conjugate.

Lemma 3.1. Let H be a subgroup of G, let HOa H and let n be the set of all prime divisors of ( H : Ho). Let c1 = {l},c2,. . . ,cn be all conjugacy classes of H/Ho and let C1 = Ho, C2,.. . ,Cn be the inverse images of

1046

Characters and Relative Normal Complements

. . ,c, respectively, in H . Then (i) H = U?==,Ci(disjoint union). (ii) Each set CF'T,1 5 i 5 n, is a union of Ir-sections of G (and hence of conjugacy classes of G). (iii) u$2c;7 = ( H - HO)G,?' Furthermore, if condition (A) holds, then (iv) The sets CF'T,.. . ,C:'" are mutually disjoint. (v) C; = cGtr n ( H - H ~ for ) all i E (2,. . . ,n ) .

c1, c 2 , .

Proof. (i) and (ii). The required assertions are direct consequences of the definitions of Ci and CFVT. (iii) We have H - HO = U;==,C;, from which the required assertion follows by the definition of SG*". (iv) Suppose that 5 E CF7"n Cy'" with i , j > 1. Note that if t E Ci, then t , E C; since t,r E Ho. Thus 5, is G-conjugate to some Ir-element y of Ci and also to some x-element z of Cj. Hence y and z are G-conjugate, so condition (A) implies that yH0 and t H o are H/Ho-conjugate. Hence i = j , as required. (v) It is clear from the definition of Cy'" that

cic c;"?" n(H -H

~ )

(2 5 i 5 n )

Conversely, assume that h E Cf'"n ( H - Ho). Then h, is G-conjugate t o some n-element hi of Ci. Hence, by (A) , h,Ho and hiHo are H / H o conjugate. Thus h, E C;. Since h,rHo is a n'-element of H / H o , we have h,, E Ho. Hence h = h,h+ E C, as desired. Keeping the notation of Lemma 3.1, assume that condition (A) holds and that the map

f:H-+(C is such that (a) f is constant on each subset Ci, 2 5 i 5 n. (b) f P o ) = 0 Following Leonard (1983), we define the lifting operator L( f ) of f to be the map

L(f) : G

-+

(c

3 A lifting operator

1047

given by

f(h) where

N

if g, h for some h E H - Ho otherwise N

indicates conjugacy in G.

Lemma 3.2. The map L( f) is well defined and satisfies the following properties : (i) L ( f ) constant on each Cy',, i > 1 and vanishes on

G-(H -HO)~', (ii) L ( f ) ( h )= f ( h ) for all h E H - Ho. Furthermore, i f H 0 4 G, then L ( f ) ( h )= f(h) for all h E H . Proof. Assume that h l , h2 E H - H o are such that g , hl and g, h2. Then hl and h2 are G-conjugate and so condition ( A ) implies that hlHo and h2Ho are H/Ho-conjugate. The latter means that hl, h2 E Cj for some j > 1, so by (a), L(f)is well defined. Assume that a,y E CyVff, i > 1. Then a , hl and y, h2 for some hi, hz E Ci. Hence N

N

N

N

proving that L ( f ) is constant on each Cy7., i > 1. Assume that g E G ( H - H o ) ~ Then, ~ ~ . by Lemma 3.1 (iii), g 6 Cy7fffor all i E (2,. . . , n } which means that gT is not conjugate to any element of H - Ho. Hence, by definitio, L ( f ) ( g ) = 0, proving (i). Assume that h E H - Ho. Then h,t E HO by the definition of I and hence, by (a), f(h) = f(h7rhd = f ( h A ) Since h, E H - Ho, it follows that

Finally, assume that HO I C 2~ 5~ ,i 5 n, and

IG - ( H - H O ) ~= ,(G ~ :~H)IHol

-

(ii) L ( x i - xi( 1) 1 ~t xi( ) 1) l ~2 ,5 i 5 n, is an irreducible character of G.

(iii) There exists a unique normal complement Go of H over Ho and Go = G - ( H - H O ) ~ ’ ~ Proof. Put CT = C:’.,

2 5 i 5 n, and C; = G - UY==,C:.Also let

Then, by hypothesis, each aj,1 5 i 5 n, is a generalized character of G. By Lemma 3.1 (iii), C; = G - ( H - Ho)GT” (1)

3 A lifting operator

1049

Note also that, by Lemma 3.2, each aj is constant on each C: and that, by the definition of x j , each xj is constant on each C;, 1 5 i , j 5 TI. Furthermore, by Lemmas 3.2 and 3.1 (v), (1 6 i , j I ).

aj (Cf ) = x j (Ci)

(2)

Here a j ( C t )means aj(z)for z E C3) and a similar remark applies t o xj(C;). With those preliminary observations, we now establish (i). To this end, for any i E { l , 2 , . . . , n } , put n

(hi E ci)

pi = Cxj(h;')aj j=1

If g E C;, then by (2), a j ( g ) = xj(hk). Hence, by orthogonality relations,

c n

Pi(d

=

Xj(hT1>Xj(W

j=l

= &kICH/Ho(hiHO)J

This shows that

Accordingly,

Taking into account that

we deduce that

In the definition of p;, the aj are generalized characters and the coefficients are algebraic integers. Hence < P i , l ~> is an algebraic integer. Thus (G : H ) ( H o l Ic;I divides IC;'l for each i E (1,. . . ,n}. In particular,

(G : H)IHol ICiI I ICfl

(1 6 i 6

4

(3)

Characters and Relative Normal Complements

1050

Adding these inequalities, we obtain

Therefore each inequality (3) must be equality. Since IC;l = lHol IC,~, (i) is established by applying (1). To prove (ii), we first note that aj is a generalized character of G such that aj(1) = xj(1) is a positive integer. Hence it suffices to verify that < a j , a j >= 1. Since

=

Xj,Xj > H I H ~ = 1,

property (ii) is established. To prove (iii), it suffices, by (1) and Lemma 2.3 (i), to show that Go = Ci is a normal complement of H over Ho. To this end, note that njn,lIierxj = Ho = C1

and therefore, by (2),

Go = C; = n~="=,i'eraj Hence Go a G and, by (i), lGol = (G : H)IHol. Now

and, by definition, Ci and Uy=2Ccfare disjoint. Hence

Go n H = C; n H

Ho

and therefore

It follows that G = GoH and Go n H = Ho, thus completing the proof.

4 Theorems of Dade, Brauer and Suzuki

4

1051

Theorems of Dade, Brauer and Suzuki

We continue our assumption that all groups are finite and all characters are (c -characters. Here we shall illustrate the power of Theorem 3.3 by deriving some classical results due to Dade, Brauer and Suzuki.

Theorem 4.1. (Dade (1975)). Let H be a subgmup of a group G I let HO d H and let 'IF be the set of all prime divisors of ( H : H o ) . Assume that the following two conditions hold : (i) If two n-elements x and y of H - HO are G-conjugate, then X H Oand yHo are HIHo-conjugate. (ii) Every elementary n-subgroup E of G is G- conjugate to a subgroup of H . Then there exists a unique normal complement Go of H over Ho, namely Go = G - ( H - Ho)'~?'

Proof. If H = Ho, then there is nothing to prove. Let x # 1~ be an irreducible character of H whose kernel contains Ho. By Theorem 3.3, it suffices to verify that L ( x - x ( l ) l ~is) a generalized character of G. We will show that a = L(x - x ( l ) l ~-t)x ( l ) l G is a generalized character of G, which will complete the proof. Now by the definition of a , we have a(g) = a(g,)

for all g E G

(1)

and by equality (2) in the proof of Theorem 3.3,

a ( h ) = X(h)

for all h E H

(2)

Assume that E is an elementary subgroup of G. By Theorem 20.2.1, it suffices to show that C ~ Eis a generalized character of E . The group E , being nilpotent, is the direct product E = E , x E,I of its Hall x- and n'subgroups E, and E,I (respectively), both of which are also elementary. By ( l ) ,the restriction a~ is just the composition of the restriction LYE, with the projection of E onto E,. Hence it suffices to show that the latter restriction is a generalized character of En, i.e. we may assume that E = E, is an elementary n-subgroup of G. By (ii), E is G-conjugate to a subgroup of H . Since a is a class function of G, we may replace E by its conjugate and suppose that E C H . But then

1052

Characters and Relative Normal Complements

by (2) and x is a character of H by hypothesis. Hence character of E and the result follows. W C ~ E = XE

QE

is a

Let n be a set of primes and let G be a group. We say that G has a normal n-complement, if G has a normal n’-subgroup whose factor group is a n-group. As an application, we now prove the following classical result which provides necessary and sufficient conditions for a finite group to contain a normal n-complement.

Theorem 4.2. (Brauer (1964b), Suzuki (1963)). Let G be a group and let n be a set of primes. Then G contains a normal n-complement if and only if the following conditions are satisfied : (i) G contains a Hall n-subgroup H . (ii) If two elements of H are G-conjugate, then they are H-conjugate. (iii) Every elementary n-subgroup of G is conjugate to a subgroup of H . Proof. Assume that (i), (ii) and (iii) hold. Then H satisfies the hypothesis of Theorem 4.1 for Ho = 1. Thus G has a normal n-complement. Conversely, assume that G has a normal n-complement N . By the SchurZassenhaus theorem (Theorem 26.1.4), G = N H for some Hall n-subgroup H of G with N n H = 1. By Lemma 1.1 (with Ho = l ) , condition (ii) is satisfied. Finally, since G is n-separable, condition (iii) is also satisfied by virtue of Theorem 26.1.6. W The following easy consequence of Theorem 4.2 is due to Frobenius.

Corollary 4.3. Let p be a prime and let P be a Sylow p-subgroup of G. Then G has a normal p-complement if and only if whenever two elements in P are conjugate in G they are conjugate in P . Proof. This is an immediate consequence of Theorem 4.2 applied to n = { p } , since conditions (i) and (iii) of Theorem 4.2 are automatically satisfied by the Sylow theorems. W Corollary 4.4. (Sutuki (1963)). Let G be a group, let n be a set of primes and let H be a Hall n-subgroup of G . Then G has a normal ncomplement if and only if the following conditions are satisfied : (i) Every elementary n-subgroup of G is conjugate to a subgroup of H .

4 Theorems of Dade, Brauer and Suzuki

1053

(ii) There exists a left transversal T of H in G such that h-lTh = T for all h E H. Proof. Assume that Go is a normal n-complement in G. Then (ii) is satisfied for T = Go, while (i) follows from Theorem 4.2 (iii). Conversely, assume that (i) and (ii) are satisfied. Then conditions (i) and (iii) of Theorem 4.2 are satisfied. Let h l , h2 E H be such that g-lhlg = hz for some g E G. By Theorem 4.2, we are left to verify that hl and h2 are H-conjugate. To this end, write g = th with t E T , h E H . Then h-lt-lhlth = h2, so t-lhlt = hh2h-l = h3, say. Thus hlth,'

= th3h;' E T

n tH = { t }

and therefore h1t = th1. Hence hl = h3 as desired. H Whilst on the subject, we shall prove another classical result of Frobenius which will be required for future use. As a preliminary, we record the following observation.

Lemma 4.5. Let p be a prime and let a group G have a normal p complement. Then, for any two Sylow p-subgroups P and Q of G , Q = g-lpg

for some g E CG(Pn Q )

Proof. Let N be a normal p-complement in G. By Sylow theorem, we may write g-'Pg = Q for some g E G. Since G = P N , we may assume that g E N . If 2 E P n Q , then { x , g - ' x g } g-'Pg = Q . Hence x-lg-lzg E Q n N = 1. Thus g E C G ( Pn Q ) , as required. We are now ready to prove the following result due to Frobenius.

Theorem 4.6. Let p be a prime and let P be a Sylow p-subgroup of G. Suppose that for every subgroup Q of P ,

Then G has a normal p-complement.

Proof. For the sake of clarity, we divide the proof into a number of steps.

1054

Characters and Relative Normal Complements

Step 1. Reduction to the case where 1 is the only normal p-subgroup of G and every proper subgroup of G satisfies the conclusion of Lemma 4.5. We argue by induction on \GI. If IGI = 1, then there is nothing to prove. Observe that the hypothesis of the theorem is satisfied by every subgroup of G. Therefore, by induction, we may assume that G has no nonidentity pfactor group. If N # 1 is a normal p-subgroup of G, then GIN satisfies the assumption of the theorem. Therefore, by induction, GIN has a normal p-complement. But G has no nonidentity pfactor group, hence N = P and C c ( N ) = G. Thus, by Theorem 26.1.15, G has a normal pcomplement. Hence we may assume that 1 is the only normal psubgroup of G. Moreover, by induction and Lemma 4.5, every proper subgroup of G satisfies the conclusion of Lemma 4.5. Step 2. Assume that G satisfies the reduction condition described in Step 1. Here we prove that G satisfies the conclusion of Lemma 4.5. Let P and Q be Sylow psubgroups of G. We argue by induction on ( P : ( Q n P ) ) . If (P : (Q fl P ) ) = 1, then P = Q and the conclusion of Lemma 4.5 is satisfied with g = 1, Now put S = PI n P2 and G; = Pi fl N G ( S ) ,i = 1,2, where PI = P , Pz = Q . Let S; be a Sylow p-subgroup of N G ( S ) with G; S; and let Q ; Q i , i = 1,2. If S = 1, then there be a Sylow p-subgroup of G with 5'; is nothing to prove. Assume that 1 # S # PI. Then N G ( S ) # S. Since G; E P;n Q i , it follows by induction that there exists 9; E Cc(4 n Q i ) C CG(S)

such that g;'P;gj = Qi for i = 1,2. By hypothesis, there exists h E C G ( S ) such that h-'Slh = S2. Because S2 2 h-lQ1h n Q2, it follows by induction that there exists g E C ~ ( h - l Q l hn Q z ) & CG(S)such that Q2 = g-lh-'Qlhg. Setting z = glhgg,' and t = hgg,', we have

I - ~ P= ~ t-'Qlt Z = gzQ2gi = P2 and I E CG(S),as desired. Step 3. Conclusion of the proof. By Steps 1 and 2, we may assume that G satisfies the conclusion of Lemma 4.5.Let 2,y E P be such that y = g - ' z g for some g E G. By Corollary 4.3, it suffices to show that s,y are P-conjugate. Note that g - ' z g E P n g-lPg and that, by the conclusion of Lemma 4.5, there exists t E CG(Pn g-'Pg) such that g-'Pg = z - ' P z . Hence gz-' E N G ( P ) and (gz-')-'x(gz-') = g - ' z g . Write g t - ' = g l g 2 = g2g1,

5 Some generalized characters

1055

where g1 is a p-element and g2 is a p'-element. Because 9192 E N , ( P ) , this implies that g1 E P and, by assumption, g2 E C G ( P ) CG(Z).Thus Y = 9 - b = (9291)-149291) = 9+91,

and the result follows.

5

Some generalized characters

Let R be a subring of (c . Recall that an R-generalized character of a group G is defined t o be an R-linear combination of irreducible (C-characters of G and that the case R = 22 corresponds t o the notion of a generalized character. In this section, we exhibit some important R-generalized characters. As one of the applications, we demonstrate that if n is a positive integer and C a conjugacy class of G, then the number of elements x E G for which x n E C is a multiple of (!GI,ICln). All characters considered below are assumed t o be (I= -characters and all groups are finite. An invariant subset of a group G is defined t o be a union of some conjugacy classes of G (equivalently, a subset which is preserved under conjugations by all elements of G ) . Our chief goal is t o prove the following classical result, a generalization of which can be found in a work of P. Hall (1936).

Theorem 5.1. (Fmbenius (1907)). Let G be a group, let n be a positive integer and let R be a subring of (c such that R contains a primitive IGI-th root of unity. For any conjugacy class C of G, define the map fC,G

G

--f

by

Then fC,G is an R-generalized character of G.

Proof. For the sake of clarity, we divide the proof into a number of steps. Step 1. Reduction t o the case where G is elementary. Let H be a subgroup of G. Then C n H is an invariant subset of G, so

c nH

= U,",~C;

1056

Characters and Relative Normal Complements

where C1,. . . ,Ck are some conjugacy classes of H . It is clear that

for some mi E Z . Hence, for each h E H ,

which shows that

k i=l

Assume that the result is true for elementary groups. Then, by (l),the restriction of fc,cto each elementary subgroup of G is an R-generalized character. Hence, by Theorem 20.2.4, fC,Gis an R-generalized character, as required. Step 2. Reduction to the case where C # { 1) and G is a p-group for some prime p. Let C1 = {l},C2,. . . ,C, be all conjugacy classes of G. Then, for each i E {1,..., r } , there exists z; E Z such that (IGl,ICiln) = .zi(IGI,n) and, in particular, z1 = 1. Hence

which shows that it suffices to prove that each f c i , G with i # 1 is an Rgeneralized character. Thus we may assume that C # 1. Assume that G = G1 x G2 with (lG11,1G21) = 1. Then C = ClC2 where Ca is a conjugacy class of G;, i = 1,2. Since (IG11, IG2l) = 1, we have

and therefore

Because G is elementary, we may thus assume that G is a p-group. Step 3. Conclusion of the proof. Setting f = fC,G,we now verify that

< f,x > E R for all

x E ITT(G)

5 Some generalized characters

1057

which will complete the proof. Since G is nilpotent, we may, by Corollary 18.12.4, write x = AG where A is a linear character of a subgroup H of G. Consequently, by Frobenius reciprocity (Theorem 19.3.9) we deduce that < f , x >=< f H , X >. Thus we are left t o verify that

< fH,A > € R To this end, put [GI = p a , (HI = pb and n = p d n o with ( p , n o ) = 1. Then

< f H , A >= p W b

f(h)A(h-')

h€H

and therefore

Now write C n H as a disjoint union of some conjugacy classes of H :

c nH

= uf=,Ci

Then

Therefore, it suffices t o show that, for each conjugacy class S # ( 1 ) of H ,

Assume that h E H satisfies h" E S and let o(h) = p e . Since S # { I } , we have e > d . Let x E < h > be such that x n = h". Setting x = hp, we then have h n p = h", whence n p 5 n ( m o d p e ) , that is p = l ( m o d p e - d ) . Hence p does not divide p and so < x >=< h >. The conclusion is that the set of all elements h E H with h" E S is a disjoint union of subsets such that the elements h and x belong to some subset if and only if < h >=< x > and h" = 5". Moreover, the subset X ( h ) , containing h, is given by

X ( h ) = { hI + t P c - d 1 0 5 t 5 pd - 1) Now write S = {hrlwhl,.. . ,hL'wh,} with h" = v, the sets

X(h:'hhl),

with hl = 1. For any fixed h E H

. . . ,X(h,'hh,)

(3)

Characters and Relative Normal Complements

1058

are mutually disjoint. Further, if h and h are elements of H with h" = hn = v , then the sets X(hTl hhl ),.. . ,X(h;;l'hhm) either coincide with those in (3) or do not intersect them. Thus, the number given by (2), is a sum of the numbers of the form

Now X is a linear character of H ,so setting E

- ~=

X(hpe-d),(4)becomes

Since o(h) = p e , E is a pd-th root of unity and therefore

t=O

if if

~ = l ~ # 1

Hence the number given by (5) is either zero or

Let k = [ C ~ ( v and ) l k1 = [ C ~ ( v )Then l . kllk and IC( = IGl/k, m = I H l / k l . Substituting these values in (6), we obtain

But k is a power of p , so ( k , n ) = ( k , p d ) l p d . Hence the number in the right-hand side of (7) belongs t o R, thus completing the proof.

Corollary 5.2. Let G be a group, let n be a positive integer and let R be a subring of (E such that R contains a primitive IGI-th root of unity. For any invariant subset S of G , define the map

5 Some generalized characters

1059

Then OS,G is an R-generalized character of G.

Proof. First assume that 5' = C is a conjugacy class of G. There exists z E 52 such that (IGI,ICln) = z((G1,n). Hence, by Theorem 5.1, OS,G = ZfC,G is an R-generalized character. In the general case, write S = Uik,lC; as a disjoint union of some conjugacy classes of G. Then k i=l

and the result follows. Corollary 5.3. Further to the notation of Theorem 5.1, denote by x a generalized character of G. Then

where E is a primitive (GI-th root of unity in CC

.

Proof. Owing t o Theorem 5.1, < x , f c ,>E ~ zll [ E ] . Since

the result follows. Corollary 5.4. Let C be a conjugacy class of a group G and let n be a positive integer. Then the number of elements x E G for which x n E C is a multiple of (IGl,ICln). In particular, the number of elements x E G for which xn = 1 is a multiple of (IGl,n).

Proof. Apply Corollary 5.3 for

x=1

~ .

Characters and Relative Normal Complements

1060

Corollary 5.5. Let G be a group, let n be a positive integer and let be a primitive IGI-th root of unity in CC . Then (i) The map f : G -, Z given by

E

is a genemlized character. (ii) For any g E G and any generalized character x of G,

( n ,IcG(g>l)-'

X(z) E

[El

xEG,x" =g

Proof. (i) Let us put R = 21; [ E ] . By definition, f = f { l } , G . Hence, by Theorem 5.1, f is an R-generalized character of G. We claim that for any irreducible character x of G , < f , x >E Q ; if sustained, it will follow that

< f,x>EQ n R = Z as required. To substantiate the claim, we first put X

If u E Gal(Q ( E ) / Q ), then

C ( E )= EM

=< f , x > so that

with ( p , IGl) = 1 and

u(x(s-')) = x(x-") Accordingly, xEG

Since f ( z ) = f(z"), it follows that .(A) = X and hence X E Q . (ii) Let C be the conjugacy class of G containing g. Then [GI = ICI lCG(g)l and ([GI,IC(n)-' = ~ C ~ - ' ( ~ C G ( g ) ~ , On n ) -the ' . other hand,

c

xEG,xnEC

X W = ICI

c

x(4

xEG,x"=g

The desired conclusion is therefore a consequence of Corollary 5.3. H

Corollary 5.6. Let n be a positive integer and let S be an invariant subset of G. If x is a generalized character of G , then

5 Some generalized characters

1061

Proof. Property (ii) follows from (i) by taking Corollary 5.2, < X , ~ S , G>E 23 [ E ] . Since

x

= 1 ~ Owing . to

the result follows. H

We next record the following lemma which will enable us to take full advantage of Corollary 5.6. Lemma 5.7. Let S be a nonempty subset of G with sp E S f o r all s E S and all integers p coprime to /GI.Then (i) For any generalized character x of G,

(ii) x( 1)1CsES Ix(s)[~,

provided

x

is irreducible and S is invariant.

x

Proof. (i) Let E E (c be a primitive [GI-th root of 1 and let :G C be defined by x ( g ) = x ( g ) for all g E G. Then X is a generalized character of G and hence so is xx. We now put 01 = CsEs lx(s)I2. Then --$

On the other hand, if u E G a l ( Q ( E ) / Q ), say u(&)= E ” , ( p , [GI) = 1, then

a ( a )= &(s”)x(s”)

=Q

SES

which shows that Q E Z [ [ E=iz ]~Q

1062

Characters and Relative Normal Complements

as required. (ii) Write S = U,",,Ci as a disjoint union of conjugacy classes of G and let gi E C;. Then, by Theorem 21.1.1 (iv),

is an algebraic integer. Since, by (i), Q E

Z the desired assertion follows.

H

It is now an easy matter to prove the following application of Corollary 5.6 which is contained in a work of Feit (1967). Theorem 5.8. Let \GI = mn with ( m ,n ) = 1 and let x be an irreducible character of G. (i) If rnlx(1) and g E G is such that gn # 1, then x(g) = 0 . (ii) If x(g) = 0 for all 1 g E G with gm = 1, then mlx(1).

+

Proof. (i) Let S = {x E GIx" = 1) and let

= CsEsIx(s)12. Then, by Lemma 5.7, Q E Z and x(l)la. Since Q E Z , it follows from Corollary 5.6 (applied for XX instead of x and for S = {l}), that nla. Since mlx(l), we see that mnla. Hence Q

which shows that CgEG-S lx(g)I2 = 0. Thus x(g) = 0 for all g E G - S, proving (i). (ii) Assume that x(g) = 0 for all 1 # g E G with gm = 1. We may assume that m # 1. Let p be a prime dividing m and let P be a Sylow p-subgroup of G. Then x(g) = 0 for all 1 # g E P . Hence, by Lemma 26.3.3 [PI divides x( 1). Thus mix( 1) and the result follows. H

Corollary 5.9. Let IG( = mn with ( m , n ) = 1 and let x be a n irreducible character of G. Assume that x(g) = 0 for all 1 # g E G with gm = 1. Then x(g) = 0 for all g E G with gn # 1. Proof. By Theorem 5.8 (ii), mlx(1). Now apply Theorem 5.8 (i). H

6 Generalized characters and *-sections

6

1063

Generalized characters and 7r-sections

Our aim here is to exhibit a number of generalized characters of a finite group G which will be encountered in our future work. In what follows, all groups are assumed to be finite and their characters to be C -characters.

Proposition 6.1. Let G be a group, let T be a set ofprimes and let R be a subring of (I: . Then (i) Iff : G C is a class function which is constant on Ir-sections of G and if the restriction of f to every elementary Ir-subgroup of G is an R-generalized character, then f is an R-generalized character. (ii) For every R-generalized character x of G , the class function xa : G + (I: defined by --$

is also an R-generalized character of G.

Proof. Let E be an elementary subgroup of G and let x be an irreducible character of E. Owing to Theorem 20.2.4, it suffices to show that < fE,x > E R. Now E = El x E2 where El is a Ir-group and E2 is a TIgroup. Since E = El x E2,x = x1 x x2 for some irreducible character xi of E;, i = 1,2. By hypothesis,

Hence

ti

EEi

as required. (ii) The function f = xn : G -, xn is an R-generalized character.

(I:

satisfies the hypothesis of (i). Hence

Characters and Relative Normal Complements

1064

+

Let n be a set of primes and let g 1 be an element of a group G. If g is a n'-element, then g is said to be n-regular; otherwise, g is called 7r-singular (by definition, 1 is a n-regular element). Corollary 6.2. Let n be a set of primes and let G be a group. Then the map f : G + (E given by

(GI,

if

if

g g

is n-regular is n-singular

is a generalized character of G . Proof. Let H be a n-subgroup of G. Then f(h) = 0 for all 1 # h E H and f(1) = IGIT. Hence fH = (JGIT/IHl)p,where p is a regular character of H . Since /HI divides IGla, the desired assertion follows by applying Proposition 6.1 (i) for R = Z ,

As before, we write {z}~*" for the n-section of G containing z. Proposition 6.3. (Robinson (1985)). Let n be a set of primes, let G be a group and let x be a n-element of G . Then the map f : G + (I: given by IcG(z)l~ if 9 E {x}GtT f ( s )= 0 otherwise

{

is a Z [&I-linearcombination of characters of G, where I < z > 1-th root of unity.

E

is a primitive

Proof. Let a be the class function of CG(Z)which takes constant value ICG(Z)I~ on 7r-regular elements of CG(Z)and takes the value zero elsewhere. Owing to Corollary 6.2, Q = '&n;X; where each n; E Iz and x l , .. . ,xr are all irreducible characters of CG(X). Now consider the class function p : CG(Z)--f CC given by

Since z E Z(Cc(z)),we have xj(z) = & x i ( l ) , xi(s-l) = Gflxi(1) for some S; E < E >. Hence /3 is a 24 [&]-linearcombination of characters of CG(Z). Note also that, since x E Z(CG(Z)), g E {z}CG(G),?T if and only if gn = z.

6 Generalized characters and *-sections

1065

Assume that y E CG(Z). Since y = x(x-ly) and x E Z ( C c ( x ) ) ,we have xi(y) = 6;x;(x-'y). Hence

c T

P(y) =

= a(s-'y)

nis;'6iX;(x-'y)

i=l

and so, by the definition of a , P assumes the value I C G ( X on ) ~ the ~ ?r-section of 5 in CG(Z),and zero elsewhere. We now claim that the induced class function PG of G assumes the value ICG(Z)I~ on { Z } ~ J , and zero elsewhere; if sustained, it will follow that f = PG has the required form. By definition,

PG(d = I C G ( W

c

P(Y-lgY)

YEG

where, by convention, P(y-'gy) = 0 if y-'gy 4 CG(S). If P(y-lgy) # 0 for some y E G , then y-lgy E CG(Z)and (y-lgy)* = y -1 gny = x. In particular, ga = yxy-' and P(y-'gy) = P(x) = ICc(x)ln. This shows that PG vanishes off {z}~*" and that, for g E { x } G t " ,

P G ( 9 ) = ICG(W I{Y

E G1yxy-l = sn}l I C G ( 4 l n

= ICc(x)ln,

as desired.

Proposition 6.4. (Robinson (1985)). Let ?r be a set ofprimes, let G be a group and let x be a ?r-elementof G. Let a be the class function of G if gn is conjugate to a generator of < x >, and defined by a(g) = ICG(X)I~ a(g) = 0 otherwise. Then a is a generalized character of G. Proof. For each p coprime to the order of x, let f,, : G .--, (c be defined as in Proposition 6.3 with respect to x p . Then each f,, is a Z [ E ] linear combination of characters of G. Let {xplp E I} be all nonconjugate f,,, it follows that a is a Z [&]-linear generators of < x >. Since a = &I combination of characters of G. Hence we are left to verify that (Y is a Q generalized character of G. Let S be a primitive n-th root of unity in (c , where n is the exponent of G. By the definition, (Y assumes values in Q . Given a E Gal((c / Q ), write a(6) = P(")for some uniquely determined integer with 1 5 m ( a ) 5 n. By

Characters and Relative Normal Complements

1066

Lemma 31.4.10, it suffices to verify that a ( g ) = a(gm(u)) for all g E G. Since the latter is a consequence of the definition of a,the result follows.

Proposition 6.5. (Robinson (1985)). Let n be a set of primes, let G be a group and let S be the ring of algebraic integers in a . Put R = Z or R = S and assume that f : G 3 (I: is a class function which vanishes on n-singular elements of G and satisfies the following properties : (i) \ c G ( g ) l i ' f ( g ) E s for all E G. (ii) The restriction off to every elementary r'-subgroup of G is a n Rgeneralized character. Then f is an R-generalized character of G. Proof. Keeping the notation of the proof of Proposition 6.1, it suffices to show that < f E , x > € R. Since f vanishes on n-singular elements of G, we have

= I~11-'xr(

Hence it suffices to show that

< fE2,XZ >= T I E 1

for some r E R

By (ii), < f ~ ~ , x>E2 R. Also, by (i), \E11-'f(q) E S for all El C C G ( Q ) for each 2 2 E Ez. Hence

22

E Ez as

Since R = Z or R = S, it follows that IE1J divides < fE2,x2 > in R. Thus, by Lemma 23.4.3, \Ell divides < f E 2 , x 2 > in R and the result is established. The following result, due t o Leonard (1987), is originally stated in a weaker form.

Proposition 6.6. Let T be a set of primes, let H be a n-subgroup of G and let .1c, : H + C be a class function of H . Assume that if two elements

6 Generalized characters and *-sections

1067

of H are G-conjugate, then they are H-conjugate. Let the map $* : G -+ be given by

-

$*(d=

{

(c

-

if g, h for some h E H otherwise

$(h) 0

indicates conjugacy with respect to G. Then $* is a well-defined where class function of G such that : (i) $*(h) = $ ( h ) for all h E H (ii) If $ is a Q -generalized character of H , then $* is a Q -generalized character of G .

-

-

Proof. If g, hl and g , h2 for some h l , h2 E H , then hl and h2 are G-conjugate and hence H-conjugate. Thus $ ( h l ) = $(h2) which shows that $* is well defined. Since $* is constant on n-sections of G and each n-section is a union of conjugacy classes, we see that $ is a class function of G. If h E H then h, = h since H is a n-group. Hence $*(h) = $ ( h ) , proving (i). To prove (ii), assume that $ is a Q -generalized character of H . Let E be a primitive n-th root of unity in (c, where n is the exponent of G. Fix (T E G d ( C / Q ) and write a(&)= E"(") for some uniquely determined integer 1 5 m ( a ) 5 n. By Lemma 31.4.10, it suffices t o verify that

a($*(g))= +*(g"("))

for all g E G

(1)

To this end, we first note that since $* is constant on n-section of G and since (g"("))?r = g$"),

we have

$*(g"(")) = $*(g,"'"') In case neither g , nor g$")

(2)

is conjugate to an element of H , we have

$*(g"(")) = 0 = $*(g) = .($*(g))

Next assume that g E G is such that g , is conjugate t o an element h of H . Since $ is a Q -generalized character of H , Lemma 31.4.10 tells us that cT($(h))= $(h"(")). Hence, by (2) and (i),

o($*(g))= a ( $ ( h ) )= $(hm("))= $*(g"(")) which proves (1) in this case. Finally, assume that g E G is such that gY(") is conjugate to an element h of H . Since < g, >=< g$"' >, g, is conjugate to a power of h. Hence the desired assertion follows by the preceding paragraph.

1068

7

Characters and Relative Normal Complements

Complements and character extensions

Again, all groups are assumed to be finite and all characters are (I:-characters. Let H be a subgroup of a group G, let Ho 4 H and suppose that there exists a normal complement Go of H over Ho. Then, by Lemma 1.1, every character of H having Ho in its kernel can be extended to a character of G. Our aim here is to present a number of results which may be viewed as partial converses to this observation. Lemma 7.1. Let H be a subgroup of G, let Ho a H and let every irreducible character of H having Ho in its kernel be extended to a character of G. If two elements x and y of H are G-conjugate, then XHOand yHo are H / Ho -conjugate.

Proof. If SHOand yHo are not H/Ho-conjugate, then there exists an irreducible character x of H/Ho such that x(xH0) # x(yH0). Let X be the character of H corresponding to x. Then X has Ho in its kernel and X(x) = x(xH0) # x(yH0) = X(y). By hypothesis, A extends to a character p of G. Since p ( x ) # p(y), we see that x and y are not G-conjugate, as required. I Lemma 7.2. Let H be a subgroup of G, let Ho a H and let N be a normal subgroup of G such that H n N = Ho. If Go/N is a normal complement of H N I N in GIN, then the group Go is a normal complement of H over Ho. Proof. Since GIN = Go/N HN/N and N C Go, we clearly have G = GoH. Note also that Ho & N c Go and so HO c Go n If. Finally, assume that g E Go n H. Then gN E Go/N n H N / N = N and so g E N n GOn H = Ho n Go c Ho, as required. I Theorem 7.3. (Ferguson (1984)). Let H be a subgroup of a group G, let Ho 4 H and let R be the set of all prime divisors of (H : Ho). Assume that the following two conditions hold : ( i ) Every irreducible character of H having Ho in its kernel can be extended to a character of G .

7 Complements and character extensions

1069

(ii) Every n-element of G is conjugate to an element of H. Then there is a unique normal complement Go of H over Ho, namely Go = G - ( H

-H O ) ~ ’ ~

Proof. We may, of course, assume that H # Ho. By Lemma 7.1, if two n-elements x and y of H - Ho are G-conjugate, then Z H Oand yH0 are H/Ho-conjugate. Let x # 1~ be an irreducible character of H having Ho in its kernel. By (i), there is an irreducible character X of G extending x. By Proposition 6.1 (ii), the class function A, : G -, CC defined by X T ( g ) = X ( g a ) , g E G, is a generalized character of G. We claim that L ( x - x( 1) ’ 1 H ) = AT - x( 1) ’ 1G

If sustained, it will follow that L ( x - x( 1) a 1 ~ is) a generalized character of G. Hence, by Theorem 3.3, Go is a required unique normal complement of H over Ho. Fix g E G. By (ii), ga is conjugate to an element h of H. Setting f = x - ~ ( 1 ) l . ~ it ,follows from the definition of L ( f ) that

Since ~ ( h = ) x ( 1 ) for h E Ho, we see that L ( f ) ( g )= x ( h ) other hand, (X7r

- ~ ( 1 ) .On

the

- x(1) - l c ) ( g ) = X(g7r) - x(1) = w-4- x ( 1 ) = X(h)- x(1),

thus completing the proof.

Corollary 7.4. (Ferguson (1984)). Let G be a group and let n be a set of primes. Then G has a normal n-complement if and only if the following conditions hold : (i) G has a Hall n-subgroup H. (ii) Every irreducible character of H extends to a character of G. (iii) Every A-element of G is conjugate to an element of H. Proof. Assume that G has a normal n-complement. Then, by Theorem 4.2, (i) and (iii) hold, while by Lemma 1.1 (i) condition (ii) also holds.

Characters and Relative Normal Complements

1070

Conversely, assume that (i), (ii) and (iii) hold. Then, by Theorem 7.3 (with Ho = l),there is a normal complement Go of H . Since GonH = 1, condition (iii) ensures that Go is a d-group. Hence Go is a normal Ir-complement. W

Theorem 7.5. (Ferguson (1984)). Let H be a subgroup of a group G , let Ho d H and let K be the set of all prime divisors of ( H : Ho). Assume that the following three conditions hold : (i) Every irreducible character of H having Ho in its kernel can be extended to a character of G. (ii) ((G : H ) , ( H : H o ) ) = 1 (iii) [ ( H - H o ) ~=~(G~ :(H)IH - Hol Then Go = G - ( H - H o ) ~ Jis a unique normal complement of H over Ho. Proof. By Lemma 2.3 (i), it suffices t o show that G - ( H - H O ) ~ ~ * is a normal complement of H over Ho. Let XI,.. . , x m be all irreducible characters of H whose kernels contain Ho. For each i E (1,. . . ,m}, let A; be a character of G extending x; and let

N = flK"=,KerA; For any subset S of G, let S be the image of S in G I N . If x E H - Ho, there exists i E (1,. . . ,m} such that x ; ( x ) # xi( 1). It follows that N fl H = Ho so that 2 H / H o and every irreducible character of H extends to a character of G. Let GOconsist of all elements of G whose images in G are 7r'-elements. It will be shown that GO is a normal complement of H over Ho and that Go = G - ( H - H o ) ~which + ~ will complete the proof. Because N 2 HO and E H / H o , we have

n

In particular, (G : I?) divides (G : H ) so that ((G : H),IHI) = 1. Put n = [GIT,so that n = lGITt/lNITt. By definition, G o is the set of all d-elements in G. Hence

G o = (x E G o p = 1) Applying Corollary 5.6 (ii) (with S = (1) and G = G), we see that lGol = t n for some positive integer t.

7 Complements and character extensions

1071

It will next be shown that

and

Go G - ( H - HO)Gyl To prove (l),note that Go is a union of cosets of N whence

(2)

Because ((G : H ) , ( H : H o ) ) = 1 and H / H o is a r-group, we have

proving (1). To prove (2), assume by way of contradiction that x E Go n ( H - H o ) ~ ? " . Then < x n N where n = IG~,J.Now (I < 2, > I,n) = 1 implies < x, x" >C N . Hence x, is conjugate t o an element of H - Ho and x, E N . Because N a G, there exists g E G with g-*x,g E N f l ( H - H o ) , which is impossible since N n H = Ho. Since, by hypothesis,

>s
z

I(H - HO)G1aI= (G : H ) I H - Ho[ it follows from (1) and (2) that t = 1 and

G = Go U ( H - HO)GiT

(disjoint union)

We are therefore left t o verify that the group Go is a normal complement of H over Ho. Let us show next that every r-element of G is conjugate t o an element in B. If g is not a d-element in G, then g $ GO.Hence g = xN for some

1072

Characters and Relative Normal Complements

Therefore g = x N = a: and a: E (H - {l})GJ, which shows that every Telement of G is conjugate to an element in 17. By the foregoing, we may apply Theorem 7.3 to deduce that G has a unique normal complement X of 17 over I. In fact, since II7oI = 1, it follows from Theorem 7.3 that X = G - ( R - {l})d~n = G o . Hence, by Lemma 7.2, X = Go is a normal complement of H over Ho. This completes the proof of the theorem. H

Corollary 7.6. (Ferguson (1984)). Let G be a group and let T be a set of primes. Then G has a normal n-complement if and only if the following conditions hold :

(i) G has a Hall n-subgroup H . (ii) Every irreducible character of H extends to a character of G. (iii) [ ( H - {l})G*"I = ( G : H ) ( H - {1}1 Proof. Assume that G has a normal T-complement Go. Then (i) and (ii) hold by virtue of Corollary 7.4. Since G is a T-separable group, it follows from Theorem 26.1.6 that every n-subgroup of G is conjugate to a subgroup of H . Since Go consists of all T'-elements of G, if g E G - Go, then gn # 1. It follows that gn is conjugate to an element of H - (1). Hence G - Go = ( H - {l})G~", proving (iii). Conversely, if G satisfies (i), (ii) and (iii), then G has a normal T-complement by virtue of Theorem 7.5. H

Our final result gives a partial solution for the following problem posed by Ferguson (1984).

Problem. For what groups Ho 4 H G is the existence of extensions of all irreducible characters of H / H o to G both necessary and sufficient to guarantee a normal complement in G of H over Ho ? Theorem 7.7. (Ferguson (1984)). Let H be a subgroup of a group G, let Ho a H with H / H o solvable and let ((G : H ) , ( H : H o ) ) = 1. Then G has a normal complement of H over Ho if and only if each irreducible character x of H with HO K e r x extends to a character of G.

Proof. The necessity follows from Lemma 1.1 (i). To prove sufficiency, let XI,. . . ,x m be all irreducible characters of H whose kernels contain Ho. For each i E (1,. . . ,m } ,let A i be a character of G extending xi and let N =

7 Complements and character extensions

1073

ng,KerXi. For any subset S of G, let S be the image of S in G I N . Then, by the proof of Theorem 7.5, H n N = Ho, H / H o S H,every irreducible character of l? extends to a character of G and ((G : R),I@I)= 1. Since fl is solvable, it follows from Theorem 26.2.2 that G contains a normal complement of H . Thus, by Lemma 7.2, G has a normal complement of H over Ho.

This Page Intentionally Left Blank

Chapter 35

Isornetries and Generalized Characters Let H be a subgroup of a finite group G, let T be a set of primes and let D be a union of n-sections of H . Let V H , ~ ( Dbe) the vector space of all (C-valued functions on H which are constant on A-sections of H and which vanish on H - D. Denote by VG,=the vector space of all (C -valued functions on G which are constant on n-sections of G. When does there exist a linear into ) VG,*preserving generalized characters ? The main isometry of V H , ~ ( D result of this chapter, namely Theorem 3.4, provides a detailed study of this question. All correspondences presented are in terms of n-induction introduced by Reynolds (1967). The first section provides a detailed background of properties of n-induction together with the motivation for the necessity of their use. In particular, we derive an explicit formula for n-induced functions and show that when G is a n-group, then n-induction becomes an ordinary induction. We finally remark that the main theorems presented in this chapter are generalizations, due to Reynolds (1967), of Dade's earlier results contained in Dade (1964). Some related results are also contained in Niccolai (1974) and Leonard and McKelvey (1967).

1

dnduction

In what follows, all groups are assumed to be finite and all characters are (C -characters. As usual, we denote by n a set of prime numbers and by A' the complementary set of primes. Let H be a subgroup of G and let the function a : H + a! be constant 1075

Isometries and Generalized Characters

1076

on n-sections of H . Then the induced class function'a : G -, (c need not be constant on n-sections of G. This motivates an effort to introduce a new function :G 02 aGln

--f

which we shall call n-induced from a, and which will be constant on nsections of G. Of course, if G is a n-group, then n-sections of G are precisely the conjugacy classes of G. For this reason, it is desirable to define aGgnin such a manner that 'a = whenever G is a n-group. It is the purpose of this section to introduce n-induction and examine some relevant properties. We warn the reader, that although induction always maps generalized characters to generalized characters, the same need not be true of r-induction. Given a group G, we write n(G) for the set of all primes dividing the order of G. If x E G, then { x } ~denotes the conjugacy class of x in G. All information and notation pertaining to n-sections of G is contained in Sec. 2 of Chapter 34. Recall, in particular, that every element x of G has a unique factorization x = X n X a I = x,rxn, where x, is a n-element and x f f 1 is a n'-element of G. The group G is a disjoint union of subsets called n-sections by defining that elements x and y of G are in the same n-section of G if and only if x, and y, are conjugate in G. If S is a subset of G, we let SG*" denote the union of all n-sections of G that intersect S. In particular, is the n-section of G containing x E G. For any subgroup K of G, we write r,(K) for the number of n'-elements of K , that is r , ( K ) = [{l}Kq Recall that, by Lemma 34.2.2, for every n-element x of G,

Let X be any subset of G. We denote by V c I K ( Xthe ) (c -space of all functions f : G + C such that : (i) f is constant on each n-section of G. (ii) f vanishes on G - X.

1 *-Induct ion

1077

Similarly, V G ( X )is defined to be the C-space of all functions f : G t CC such that : (iii) f is constant on each conjugacy class of G. (iv) f vanishes on G - X . We also put VG,?,(G) = V G , ~Thus . V G ,consists ~ of all functions f : G + C which are constant on r-sections of G. Note that VG(G)= C f ( G ) ,where as usual Cf (G) denotes the CC -space of all class functions of G. It is clear that

and

c v G ( X ) & Cf(G)

vG,r(x)

Recall that Cf ( G ) has the usual positive definite hermitian form :

< f1,fi >= IW

c fi(.)fio

( f d 2

E Cf(G))

xEG

which is inherited by the above subspaces of Cf ( G ) . In particular, all these subspaces have orthonormal bases. Let H be a subgroup of G. Given f E C f ( G ) , we write f H for the restriction o f f to H . This gives us a C-linear map

{

VG,n

-+

vH,n

f * f H

The following lemma will enable us to define a dual of this restriction map.

Lemma 1.1. Let P I , . . . ,pn be an orthonormal basis of V G , ~let, H be a subgroup of G and, for any given a E V H , ~let,

Then aGJ is a unique element of V G ,such ~ that

Moreover, the map

VG,a

vH,r

a

H

aGJ

Isometries and Generalized Characters

1078

is

([: -linear.

Proof. Since each y E V G ,is~ uniquely determined by all < r,P; >, 1 5 i 5 n, there exists at most one such By definition of aG*=, the map a H is (c -linear. Finally, let /3 E V G , ~Then . ,f3 = A;@; for some A; E (c and so aGIX.

xy=l

aGjR

n

n

< Q , ( P ; ) H > p;,cAip; >

< aGYn,p> = < i=l

i=l

n

i=l n

as desired. H Following Reynolds (1967), we shall refer to the map

{

VH,n a

+

H

VG,a aGJ

as n-induction from H to G. If G is a n-group, then the n-sections of G are precisely the conjugacy classes of G, and hence n-induction becomes the usual induction map a H aG by virtue of Frobenius reciprocity (Proposition 19.3.12 (iv)). Lemma 1.2. (Transitivity of n-induction). Let H G. Then, for any a E V H , ~ ,

S be subgroups of

G,= - aG,= 0 -

This proves the required assertion, by applying Lemma 1.1. H

1 *-Induction

1079

Our next aim is to obtain an explicit formula for n-induction. Let H be a subgroup of G and let { h i } and {gj} be sets of representatives of the distinct conjugacy classes of H and G, respectively, which consist of n-elements. Recall that each n-section of G is a union of conjugacy classes of G, exactly one of which consists of n-elements. Hence {hi} and {gj} are sets of representatives of the n-sections of H and G, respectively. Let

a;:H-+(C

and

pj:G-+(C

be the characteristic functions of {h;}Hi" and {gj}Gj" respectively, (thus a;(h) = 1 if h E a;(h) = 0 otherwise and, similarly, pj(g) = 1 if g E {gj}GJ and pj(g) = 0 otherwise). Then { c Y ~and } { p j } form C -bases of V H ,and ~ VG,~ respectively. , The following preliminary result computes n-induction using these bases.

Lemma 1.3. Further to the notation above, for each i choose a unique j(i) such that {h;}GJ= {gj(j)}Gy". Then, for all i, we have :

Proof. We first note that (ii) follows from (i) and (1). To prove (i), let y; denote the right-hand side of (i). By Lemma 1.1, it suffices to show that

< yi,Pj >=< a i , ( P j ) H > for all i and j. Write { h i } = {hl,hz,. . . ,h,} and {gj} = {g1,g2,. . . ,gt}. Then

< ri,Pj >

=

[GI-'

C ri(g>pjo g€G

t

=

P1-l

c

l{gk}G~*17z(gk)m

k=l

= IGI-' I{gj}G" Iyi(gj>

Isometries and Generalized Characters

1080

On the other hand,

k=1

Hence < a ; , ( p j ) ~ > also equals to lH[-ll{h;}H*Tl if j = j ( i ) and t o zero otherwise, as required.

beI , the set of all ?r-elements of a subgroup H Proposition 1.4. Let € of G and let a E VH,,. Then, for all x E G , (i) aGJyx) = IHI-1r7r(cG(z7r))-1 C y E G r,(cH(y-lx,Y))~(Y-l~,Y) where, by convention, a(y-lz,y) = 0 if y-'x,y $H. (ii) We also have : ( ~ ~ ' ~ = ( z )(G : H)I{z,}G'al-l

T,(c~(h))a(h) h€Hnn{xn}G9n

= (G: H ) ~ { x , } ~ ' " I - ~

C

a(h)

h€Hft{xn}Gpn

where, by convention, both right-hand sides are zero if H ,

n {x,}~,~ = 8.

Proof. (i) Since, by Lemma 1.1, the map VH,, + V G , ~a, I+ aG is suffices to take a = a;,where a;is as in Lemma 1.3. By Lemma

(c -linear, it

= { g j ( i ~ } ' * then ~ , aFyn(z)= 0. On the other hand, Hence, if x $! if z $ { h ; } G J ,then y-'z,y # { h ; } H Jfor all y E G. Hence the right-hand Then the side of (i) is also zero. Thus we may assume that x E summation on the right-hand side of (i) has nonzero terms for just those y Now there are precisely for which y-lz,y E ICG(hi)I I{hi}HI = ( ( ~ G ( h i:)~ H ( h i )IHI )

such terms, each equal to r,(CH(hi)). it follows, by comparison with (2), that (i) is valid.

1 *-Induction

1081

(ii) Put X = H , n { z , } ~ J and observe that y-'z,y E H if and only if y-ls,y E X. Hence if X = 0, then by (i) and the convention on the right-hand side, both sides are zero. On the other hand, if h E X then as y runs over G, y-'z,y = h for exactly /CG(Z,)( values of y. Applying (i) and (l),we deduce that

.>

aGq

=

~,(cH(Y-ls,Y))~(Y-lz,Y

YEG

1H 1 -

=

c

JH I - l c r ( CG( %>)-' T r ( c G ( %))-'

ICC ( 5 , ) 1

Tr ( C H (h))a( h)

C ~,(C~(h))a(h)

= (G : H)({z,}G7"I-1

h€X

proving the first equality. By the foregoing, we are left to verify that

To this end, write H n { z , } ~ ? "= {hl}Ht"u

. . . u {hn}Hyn

(disjoint union)

for some r-elements h l , . . . ,h, in H . Then

X = C h l u ... u c h , where

Chi

(disjoint union)

is the conjugacy class of hi in H . Hence

as required.

Isometries and Generalized Characters

1082

The next result provides a simple formula involving restriction, a-induction and induction.

Proposition 1.5. Let K a E VH,,. Then, for all x E G,

CH

and K C L be subgroups of G and let

where

Yg = {y E G(y-'x,y E K and y-lzy E L) (by convention, the right-hand side is zero if Y, is empty). Proof. Given g E G, put x g = g - ' x g and let O , = ah'. Then, by the formula for induction and Proposition 1.4 (i), [pL*"]G(x> =

p1-l C[PL*"](x") zEG

= ILI-'

c

IKI-lr,(CL(z:>)-l

zEG

=

IL I-'IK I-

c

rA(CK(x:t))P(x:t)

t€G

C

r, (CL (x i )1-l r , (CK

( x i t))a(x :t )

(zJIE.4

where A = { ( z , t ) l z E G , t E L , x* E L , z:t E K}. Because xz E L if and only if xZt E L, and since r,(CL(zf,)) = rr(CL(xgt)),we can convert to a sum over y = zt E Yz with each term repeated ILI times. This gives the desired formula.

2

Normal d-subgroups

This section contains some preliminary results of a group-theoretic nature. All conventions and notation introduced in Sec. 1 remain in force. In particular, G denotes a finite group and a a set of primes.

Lemma 2.1. Let N be a normal a'-subgroup of G. If x and y are a-elements of G such that Nx = Ny, then y = t-lxz for some z E N.

Proof. Owing to the Schur-Zassenhaus theorem, applied to the group N < x >= N < y >, < y >= g-' < x > g for some g E N < x >. Writing

2 Normal r'-subgroups

1083

g = xi, for some z E N and some i 2 1, we have < y >= Hence y = z-*xnz for some n 2 1. Then

2-l

< z > z.

N X = N y = Nx" and so zn = z,as required. Let N be a normal a'-subgroup of G and let z,y E G .

Lemma 2.2. Then

(i) The elements 2 and y are in the same a-section of G if and only if N x and N y are in the same n-section of G I N . (ii) Each a-section of G is a union of cosets of N in G and the natural homomorphism G 4 G I N induces a bijective correspondence between the Ir-sections o f G and those of G I N . GIN a (iii)1 - J =1*J Proof. (i) If z and y are in the same ?r-section of G, then obviously N z and N y are in the same n-section of G I N . Conversely, assume that { N x } ~ / ~=* {"N Since {Nx)'IN*" = { ( N x ) , } G / N * n ={NZ,}~/~?"

we then have N y , = ( N z ) - ' ( N z , ) N z = Nz-lz,z

for some z E G. Hence, by Lemma 2.1, yn = t-lz-lz.r,zt for some t E N . Thus x and y are in the same a-section of G , as required. (ii) Let g E G and let S be the union of all cosets N z of N with x E { g } G - T .We claim that {g}G*"= S. Indeed, it is clear that {g}G*" C S. On the other hand, if x E {g}GI" then N x C { g } G J by virtue of (i). Hence S E { g } G J as claimed. Let {gl)G*",. .,{g,}G*a be all distinct r-sections of G. Then, by (i), { N g l } G / N J , .. . ,{Ngn}G/NIRare distinct n-sections of G I N . On the other hand, if g E G, then g E { g i } G J for some i E { 1,...,n ) and hence N g E { Ngj}G/Ni",as required. (iii) By (ii), we may write

.

{ x } ~ ! "= Ngl U . . . U Ngt

(disjoint union)

for some 9 1 , . . . , g t E G. Then, by (i),

{ N x } ~ I ~ *=" { N g l , . . . ,Ngt }

Isometries and Generalized Characters

1084

and therefore

thus completing the proof.

Lemma 2.3. Let H be a subgroup of G such that G = N H for some normal n'-subgroup N of G . Then (i) There is a bijection of the n-sections of H onto those of G : for each h E H , { h } H J corresponds to {h}'*" = N { h } H J (ii) For all h E H , I{hlH1"I- I{hIG9"I PI IGI Proof. (i) By Lemma 2.2 (ii), the maps

{g}G1"H {gN}GIN*", {h}*I"

H

{ h ( H n N ) }H/(HnN),x

are bijections between n-sections of G and G I N and H and H / ( H n N ) , respectively. Since the map H / ( H n N ) + G / N , h ( H n N ) H hN is a group isomorphism, it follows that {h}H*TH {h}'*" is a bijection of n-sections of H onto those of G. By Lemma 2.2 (i), N{h}Hp" & {h}Gl". Conversely, assume that g E {h}'*" and write g = nhl for some n E N , hl E H . Then gN = h l N and hN are in the same n-section of G I N . Hence h l ( H n N ) and h ( H n N ) are in the same n-section of H ( H n N ) . Therefore, by Lemma 2.2 (i), hl E { h } H J . Thus g E N{h}H*",proving that {h}Gi" = N{h}HI". (ii) Since G I N H / ( H n N ) and { h } G J = N { h } H * Tit, suffices to show that

By Lemma 2.2 (ii), we may write {h}Hl" = ( N n H ) h l U . . . U ( N n H ) h , (disjoint union). Then

N{h}H1"= N h l

U,

.. U N h ,

(disjoint union)

which obviously implies (1).

Lemma 2.4. Let H be a subgroup of G such that G = N H for some normal n'-subgroup N of G. Then, for any n-element h of H ,

3 Isometries and generalized characters

1085

Proof. It is clear that CN(h)CH(h)C C G ( ~ Conversely, ). assume that g E Cc(h). Since G = N H , we have g = x y with x E N , y E H . Also

g = h-lgh = (h'lxh)(h--lyh) and h-lxh E N , h-'yh E H , whence

x-'(h-'xh) = y(h-'yh)-* E H n N Then in H / ( H

n N ) , y(H n N ) commutes with h(H n N ) , so that (y-lhy)(~ n N ) = h(H nN )

Applying Lemma 2.1 for H n N and H , we see that

y - l h y = z-'hz Thus z-ly E H

for some z E H n N

n C G ( ~=)C H ( ~ Since ) . g = ( z z ) ( ~ - ~and y) zz

E N n C G ( ~=)C N ( ~ ) ,

the result follows.

Lemma 2.5. Let H be a subgroup of G and let D be a union of nsections of H . Then the following conditions are equivalent : (i) Any two r-elements of D that are conjugate in G are conjugate in H . (ii) Any two elements of D which are in the same n-section of G are in the same n-section of H . Proof. (i) + (ii) : Assume that x , y E D are in the same n-section of G. Then yn = g-'z,g for some g E G. Since xT,y, are n-elements of D,it follows from (i) that yn = h-'s,h for some h E H . Hence x and y are in the same n-section of H . (ii) =+ (i) : Let z , y be two n-elements of D with y = g-'xg for some g E G. Then x , y are in the same n-section of G. Hence, by hypothesis, x and y are in the same n-section of H . Therefore y = h-'xh for some h E H , as desired.

3

Isometries and generalized characters

In what follows, A is a set of primes, G is a finite group and all characters are assumed to be (I: -characters. Let U and V be (I: -spaces on which some

1086

Isometries and Generalized Characters

hermitian forms are defined, By a linear isometry of U into V , we mean any CC -linear map f:U+V such that

Let Cf ( G ) be the (I: -space of all class functions G + (I: . All hermitain forms on the subspaces of Cf(G) considered below are inherited from the positive definite hermitian form on C f ( G ) given by

Of course, the same remark applies to C f ( H ) , where H is a subgroup of G. Let H be a subgroup of G. Recall that VG,=denotes the (I:-space of all functions f : G -, (I: which are constant on n-sections of G. The space V H , ~ is defined in a similar manner with H playing the role of G. The following preliminary result provides circumstances under which T induction is a linear isometry.

Lemma 3.1. Let H be a subgroup of G such that G = N H for some normal d-subgroup N of G . Then Ir-induction a H aGvn is a linear isometry of VH,* onto VG,=which maps all generalized characters of H in VH,= onto all generalized characters of G in V G , ~Furthermore, . (a) aGy"(xh) = a ( h ) for all a E VH,",h E H , x E N

where, for any su'bgmup Ir' of G , r,(K) denotes the number of d-elements in #.

Proof. Define the inflation map inf : C f (GIN) --t C f (G) by

inf(a)(g= ) a ( g N ) for all a E C f ( G / N ) , gE G By Lemma 2.2 (ii), the inflation map induces a bijection of VG/N,,onto V G , ~ . Let P I , . . .pn be all irreducible characters of GIN and let x; = i n f ( p ; ) , 1 5 i 5 n. Then X I , . . . ,xn are some distinct irreducible characters of G.

3 Isometries and generalized characters

1087

n

< a,p > =

&bj

i=l

=

n

n

i=l

i= 1

< inf(cy),inf(p) >

Thus the inflation map induces a linear isometry of VG,,, onto V G , ~Note . also that, since i n f ( a )= C7=laj//,i, a E V G ~ ,is, a generalized character if and only if i n f ( a )E VG,"is a generalized character. Owing to Lemma 2.3 (ii), for all h E H

which proves (ii) by taking h = 1. Let {hl}*J,. . . ,{h,}HJ be all distinct n-sections of H where each hi is a n-element. Then, by Lemma 2.3 (i), {hl}Gt", . . . ,{hr}'9" are all distinct CI: be the characteristic n-sections of G. Let ai : H + (c and p; : G functions of and {hj}Gt", respectively. Then, by Lemma 1.3(i) and (l),we have aiG," = pj (1 5 2 5 T ) --f

Hence, for any a E VH,",say a = Cr=lXja;, Xj E r

(c

, we have

c

Since Q i ( h )= &(h) for all h E H and, for x E N , xh and h belong to the same n-section of G, we have

proving (i). Finally, let us identify GIN with H / ( H n N ) via hN H h ( H n N ) , h E H , Then, by (i), the same element of VG,,," which maps to a E VH," under inflation from H / ( H n N ) to H maps to actn under inflation from GIN to G. Hence, applying the first paragraph, the result follows.

Isometries and Generalized Characters

1088

Let X be any subset of G. Recall that V G , ~ ( Xdenotes ) the (I:-space of all functions f : G --t (I: such that f is constant on each n-section of G and f vanishes on G - X. Recall also that XG7" denotes the union of all n- sections of G that intersect X. In particular {x}'J is the n-section of G containing x E G. We are now ready to prove our main results.

Theorem 3.2. (Reynolds (1967)). Let H be a subgroup of a group G and let D be a union of n-sections of H which satisfies the following two properties : (Q) Any two n-elements of D that are conjugate in G are already conjugate in H . (b) For each n-element d E D ,CG(d) has a normal n'-subgroup N ( d ) such that C G ( d ) = N(d)CH(d) Then (i) The map a H aGJ is a linear isometry of VH,"(D)onto V G , ~ ( D ~ * * ) . (ii) For all a E V H , ~ ( Dand ) all x E G, QG,"(X)

=

a(d)

if if

z E {d}G*",d E D x E G - DG9"

Proof. Let {hl}Hfn,, . . ,{hs}H*nbe all distinct n-sections of H whose representatives hi are n-elements of D. By (a) and Lemma 2.5, are distinct n-sections of G. Hence D and DGJ can be written as the following disjoint unions :

D = U:=l{h;}H*" and DG*" = U:=l{hi}G~" Let a; : H + C and pi : G + (I: be the characteristic functions of and {hi}'*", respectively. Then, by (2), 0 1 , .

PI,. . . ,pT is a basis of V G , ~ ( D ~ J ) .

,

(2)

. ,aTis a basis of V H , = ( Dand )

By (b), we may apply Lemma 3.1 (ii) t o find that

for all n-elements d E D. Hence, applying Lemma 1.3 (ii) (with j(i) = i) and equality (3) (with d = hi), we deduce that "i G," = p;

for all i E { I , ...,r }

(4)

3 Isometries and generalized characters

1089

Now fix a E VH,,(D) and z E G. Then cr = C;'=lcr(hj)a; and so, by (4),aG,n = CT=la(hi)P;. If z E G - DG>*then each pi(.) = 0 and so aGir(z)= 0. On the other hand, if z E DG*,, then z E for some d E D . By (2), we have {d}G*" = {h;}Gi" for some i, 1 5 i 5 s. Then d , and h; are G-conjugate, hence H-conjugate, by virtue of (a). Therefore a ( d )= a ( & ) = a ( h ; ) and so a G J ( z )= cr(h;)= a(d),proving (ii). Since a1, . . . ,aT and PI,. . . ,PT are bases of VH,,(D) and V G , , ( D ~ ~ ~ ) , respectively, it follows from (4) that a H aGrTis a linear isomorphism of VH,,(D) onto Vc,,(DGln). To prove that the map a H aGJ is an isometry, we use the fact that a1, . . . ,a,. and PI,. . . ,P,. are orthogonal bases of VH,,(D) and VG,,(D~*,).By this fact and (4),it suffices to show that < a;,ai >=< P;,,B; >, i.e. that (HI-1 I{ h1). H l=l = lGl-ll{hzlG~~l (5) {d}G3"

Now, by Lemma 1.3 (i), (ii) we have

which equals 1, by applying (3) with d = hi. This proves (5) and hence the result.

Theorem 3.3. (Reynolds (1967)). Let H be a subgroup of G, let D be a union of n-sections of H and let D , be the set of all n-elements of D . Assume that for each d E D,, Cc(d) has a normal Hall Ir'-subgroup N ( d ) such that C G ( d ) = N ( d ) C H ( d ) . For each nonempty subset S of D,, put

where S runs through d l nonempty subsets of D,. (ii) If a is a generalized character of H , then aCyTis a generalized character of G.

Proof. If S = 0, then put

K ( S )= H

and

L(S)= G

Isometries and Generalized Characters

1090

Then (i) is equivalent to

where S runs through alI subsets of D, including 0. For the sake of clarity, we divide the rest of the proof into a number of steps.

Step 1. Let S be a nonempty subset of D, and let I ( S ) = n,EsN(s). Here we show that K ( S ) normalizes the r'-group I ( S ) . This will imply that L ( S ) is a group and that the groups L ( S ) and K ( S ) satisfy the hypothesis of Lemmas 3.1 and 2.4,with L ( S ) , K ( S ) and I ( S ) playing the roles of G, H and N , respectively. Let h E K ( S ) and let s E S. Then N(h-lsh)CH(h-'sh) = CG(h-'sh) = h-lCG(S)h = (h-1 N (3)h)CH ( h-1 sh) Since, for each d E D,, N ( d ) is the set of all #-elements of C G ( ~=) N ( d ) c ~ ( d )we , deduce that N(h-lsh) = h-'N(s)h. Thus K ( S ) normalizes the d-group I ( S ) .

Step 2. We now prove that, for any subset S of D and any d E D,, the following conditions are equivalent : (i) S D, and d E K ( S ) (ii) S U { d } C D, and d E II(S U { d } ) First assume that S # 0. Then, since d E N H ( S )if and only if d E N H ( S U { d } ) conditions (i) and (ii) are equivalent. If S = 0, we must show that d E H if and only if d E li({d}) = C H ( ~which ) , is obviously true. Step 3. It will next be shown that (i) implies both of the following conditions : (iii) C ~ ( s ) ( dand > C~,-(sqd))(d) contain the same n'-elements. C L ( q ( d ) and CL(Su{d))(d) contain the same ?r'-elements. (iv) Indeed, assume that (i) holds. By the definitions,

N H ( S )n G ( d ) = N H ( Su (4)n C H ( ~ ) whence

Ch-(S)(d>= C K ( S U { d } ) ( d )

3 Isometries and generalized characters

1091

which implies (iii). If S = 8, then CL(s)(d) = Cc(d). On the other hand, L ( { d } ) = N ( d ) c ~ ( d=) Cc(d) and so

This proves (iv) in case S = 8. Now assume that S we may apply Lemma 2.4 to deduce that

# 8.

Then, by Step 1,

Since, for any s E S, N ( s ) is the set of all +-elements of CG(S), I ( S ) is the set of all n'-elements of Cc(S). Then

I ( s )n C L ( S ) ( d )= I ( s )n CG(d) is the set of all d-elements of CG(S)n Cc(d) = CG(SU { d } ) , namely I ( S U { d } ) . Hence, by (i),

which clearly implies (iv). Step 4. Here we establish (6). Given z,y E G, let 29 = y-lzy. Using Proposition 1.5 and reversing the order of summation, we can express the value on the left side of (6) at an arbitrary z E G as

the inner sum being over the set S(z,y) = { S

DTlz; E I i ( S ) , X Y E L ( S ) )

It will now be shown that for fixed z and y, the inner sum vanishes which will complete the proof. First assume that zY, E G - H . Since K ( S ) C H , we then have S(z, y) = 8. Hence we may assume that z: E H . If zY, E H - D ,then each term of the inner sum has the vanishing factor a ( z g ) ,because a E V H , ~ ( D We ) . are therefore left to consider the case where z: E D. In this case, we pair the elements of S(z,y) in the following manner.

Isometries and Generalized Characters

1092

Assume that S E S(x,y) and d = xi @ S. Because, by Step 2, (i) and (ii) are equivalent, S U { d } D, and d E K ( S U { d } ) . Because

xy = dx;, = x $ d we have x z l E CL(s)(d).Hence, applying Step 3, it follows from (iv) that

We conclude therefore that if S E S ( x , y ) and xY, @ S, then S U {z:} E S ( x , y ) . A similar argument proves the converse and thus the elements of S ( x ,y ) fall into pairs {S,S U {z:}}. Applying (iii) and (iv), it follows that the summands corresponding to the elements of such a pair cancel. This demonstrates that in all cases the inner sum vanishes, thus completing the proof of (6). Step 5. Here we prove (ii). Assume that a E VH,*(D)is a generalized character of H . Then, for each nonempty subset S of D,, ( a ~ ( ~ ) )is~a ( ~ ) ~ " generalized character of L ( S ) by virtue of Lemma 3.1. Hence

is a generalized character of G. Now let S be a nonempty subset of D, and let h E H . Then h-'Sh D, and S and h-lSh contribute equal terms to the right side of (i). Hence, by (i) 1

where we sum over one representative S of each equivalence class of nonempty H-conjugate members of D,. Thus aGynis a generalized character of G.

A careful examination of the proof of (6), reveals that (6) is valid under more general circumstances. Namely, we can replace the set of all subsets of D, by any collection X of subsets of D for which the following properties hold : (a) For each S E X, there are defined groups K ( S ) and L ( S ) such that

K ( S ) C H , K ( S ) C L ( S ) C G. (b) Conditions (i) and (ii) in Step 2 (with S C D, replaced by S E X ) are equivalent. (c) Condition (i) implies both of the conditions (iii) and (iv) in Step 3. This abstract formulation is the content of Theorem 4 in a work of Reynolds

3 Isometries and generalized characters

1093

(1967). The following theorem is the main result of this section.

Theorem 3.4. (Reynolds (1967)). Let H be a subgroup of G , let n be a set of primes and let D be a union of n-sections of H . Assume that the following two conditions hold : (a) Any two n-elements of D that are conjugate in G are already conjugate in H . (b) For each n-element d E D , C G ( d ) has a normal Hall n'-subgroup N ( d ) such that C G ( d ) = N(d)CH(d). Then (i) The map a H aGvnis a linear isometry of VH,=(D)onto V G , ~ ( D ~ I ~ ) . (ii) For all a E V H , ~ ( Dand ) all x E G , aGyx)=

a(d)

if if

x E { d } G * T , dE D x E G-D ~ J

(iii) If a E VH,=(D)is a generalized character of H , then aGia is a generalized character of G. (iv) If a E VH,=(D)is an irreducible character of H , then actn is an irreducible character of G . Proof. Properties (i), (ii) and (iii) follow from Theorems 3.2 and 3.3 (ii). To prove (iv), assume that a E VH,=(D)is an irreducible character of H . Then, by (i),

< a G J , a G ' R >=< a,a >= 1 and, by (iii), actn is a generalized character of G. Since, by Porposition 1.4 (i), aG9"(1) > 0, we deduce that aG?=is an irreducible character of G. H We close by summarizing the situation in which G = N H for some normal n'-subgroup N of G.

Corollary 3.5. Assume that G = N H for some normal n'-subgroup N of G and some subgroup H of G. Then ~ V G ,which ~ maps (i) The map a H aGgnis a linear isometry of V H ,onto all generalized characters of H in V H ,onto ~ all generalized characters of G in V G , ~ . (ii) If a E V H ,is~ an irreducible character of H , then is an irreducible character of G. aGln

Isometries and Generalized Characters

1094

(iii) For all a E V H ,and ~ all x E G,

&yx)

=

a(h)

if if

x E {h}Gp",h E H x E G - HGjr

Proof. (i) This is Lemma 3.1. (ii) Apply (i) and the argument in the proof of Theorem 3.4 (iv). (iii) By Theorem 3.2, it suffices to show that D = H satisfies properties (a) and (b) of that theorem. That D = H satisfies (a) is a consequence of Lemma 2.3 (i). By Lemma 2.4,for any x-element h of H ,

This proves property (b) by setting N ( h ) = C N ( ~ for ) each x-element h of H. H

Chapter 36

Exceptional Characters Exceptional characters have been a powerful tool in its many applications. For example, the theory of coherent sets of characters played a crucial role in the proof of solvability of groups of odd order. This chapter provides a detailed account of the general theory of exceptional characters. An important part of the theory pertaining to Frobenius groups will be presented in the next chapter. Exceptional characters were first used by Brauer in his original version of the work published in the joint paper of Brauer, Suzuki and Wall (1958). A more elementary approach was given by Suzuki (1955). For a self-contained elementary treatment of some aspects of the theory of exceptional characters we refer to a work of Brauer and Leonard (1962). It should be pointed out that some important generalizations of the Brauer-Suzuki theory were given by Feit (1960, 1962).

1

Trivial intersection sets

In what follows, all groups are assumed t o be finite and all characters are (E -characters. A nonempty subset S of a group G is called a trivial intersection set in G if S fl z-'Sz ( 1) for all z E G - N G ( S ) Thus, if 1 @ S, then S is a trivial intersection set in G if and only if

1095

Exceptional Characters

1096

Example 1.1. Let H # 1 be a subgroup of G such that Hng-lHg=l

forall g E G - H

Then H is a trivial intersection set in G with N G ( H ) = H .

Proof. It clearly suffices to show that N G ( H ) = H . If g E N G ( H )- H , then by hypothesis H fl g-'Hg = H = 1, a contradiction. H

Example 1.2. Let g be a nonidentity element of a group G and let S be defined by S = {x E G(g E < x >}

Then S is a trivial intersection set in G with 1 $! S , N G ( S ) = N G ( < g >) and S-l = S .

Proof. It is clear that S is a nonempty subset of G such that 1 and S-' = S. Let x E N G ( < g >) and s E S. Then < g s > and

>c
= x-l < g > x 2 x-' < s > x =< x-lsx > Hence x-'sx E S and so N G ( < g >) N G ( S ) . It therefore suffices to verify that Sn x-'Sx = 0 for all 2 E G - N G ( < g >)

If x E G - N G ( < g >) and y E S n x - l S x , then < y > contains both < g > and < x-'gx >= x-' < g > x #< g >. This is impossible, since the cyclic group < y > contains only one subgroup of order I < g > I = 1x-l < g > 21. Thus S n x-'Sx = 0 and the result follows. H

Example 1.3. Let A # 1 be an abelian subgroup of G such that for all 1 # x E A , C G ( X= ) A . Then A is a trivial intersection set in G . Proof. Assume that 1 # x E A n y-'Ay for some y E G. Then x = y-'zy for some 1 # z E A. By assumption, we have

and therefore

1 Trivial intersection sets

1097

This proves that y E N G ( A ) . Hence

A ny-lAy = 1

for all y E G - N G ( A )

as required. H We next provide some elementary properties of trivial intersection sets.

Lemma 1.4. Let S be a trivial intersection set in G. Then (i) NG(S). (ii) CG(S)G N G ( S )for all 1 # s E S . (iii) Two elements of S are G-conjugate if and only if they a r t N G ( S ) conjugate.

sc

Proof. (i) Assume that x E S - N G ( S ) . Then

z = X - ~ X Z E s n X-'SX

c { 1)

whence x = 1, a contradiction. Thus S & N G ( S ) . (ii) Assume that x E CG(S)for some 1 # s E S. Then 1# s=5

- l E ~ S ~

n~

S

Z

and hence x E N c ( S ) . (iii) Assume that ~ 1 , sE ~S are G-conjugate, say s2 = g-lslg. It suffices t o show that s1, s;! are NG(S)-conjugate. We may assume that s;! # 1. Since s2 E

s n9 - 9 s

it follows that g E N G ( S ) ,as desired. H The following result shows that the converse of Lemma 1.1 also holds provided S # { 1).

Lemma 1.5. Let S # (1) be a nonempty subset of G and let H be a subgroup of G such that the following three properties hold : (i) S is an invariant subset of H . (ii) CG(S)& H for all 1 # s E S . (iii) Two elements of S are G-conjugate if and only if they are H conjugate. Then S is a trivial intersection set in G and H = N c ( S ) .

Exceptional Characters

1098

Proof. Assume that x E G is such that 1 # y E S f l z - l S x . Then y E S and y = x-lsx for some 1 # s E S. By (iii), y = h-lsh for some h E H . Then xh-' E CG(S).Since s # 1, it follows from (ii) that CG(S) C H . Hence x = (xh-l )h E H

which shows that S n x-lSx (1) for all x E G - H . By the foregoing, we are left to verify that H = N c ( S ) . By (i), we have H G N G ( S ) . Conversely, assume that x E N G ( S ) . If x # H , then by the above s = Sn x - l s ~G (1) contrary to the assumption that S is nonempty and S # (1). Thus x E H and the result follows. Given a subset S of G , let C f ( G ,S ) be the space of all functions of G which vanish on G - S. Also put

(L: -valued

class

SG = UgEGq-1Sq In the theorem below, the assumption that 1 $i S forces a(1) = 0 and so a cannot be a character.

Theorem 1.6. Let S be a trivial intersection set in G with 1 # S and let H = N G ( S ) . Then (i) The map cr H crG is a linear isometry of C f ( H , S ) onto Cf(G,SG). (ii) For all cr E Cf ( H ,S ) and all x E G, (Y

G

(x) =

~ ( s )

if x is G-conjugate to s E S if x E G - S G

Proof. Let n be the set of all prime divisors of IGI. Since G is a T group, n-induction coincides with ordinary induction. Hence, by Theorem 35.3.4, it suffices to verify that conditions (a) and (b) of that theorem are fulfilled. Condition (a) says that any two elements of S that are conjugate in G are conjugate in H . Therefore, (a) follows from Lemma 1.4 (iii). Condition (b) says that for any s E S, CG(S)has a normal Hall n'-subgroup N ( s ) such that CG(S)= N ( s ) C H ( S ) Hence . it suffices to verify that C G ( S = ) CH(S) for all s E S. But, since by hypothesis 1 $i S, it follows from Lemma 1.4 (ii) that CG(S)C_ H for all s E S, as desired.

1 Trivial intersection sets

1099

Although the next result is an easy consequence of Theorem 1.6, we shall provide a direct proof due to Brauer and Suzuki (see Suzuki (1955)).

Theorem 1.7. Let S be a trivial intersection set in G , let H = N G ( S ) and let a,p be class functions of H vanishing on H - S. Then (i) a G ( x ) = a(x)for all x E - (1). (ii) Ifa(1) = 0, then < aG,PG>=< a,P >. (iii) Ifa(1) = 0 and SG n H G S , then a = (aG),.

s

Proof. (i) Let a' : G + CC be given by a'(x) = 0 for x E G - H and a'(h) = a ( h ) for all h E H . Then, by the definition of aG,

If y-'xy E H - S, then a'(y-'xy) = a(y-'sy) = 0. Similarly, by the defini= 0 if y-'xy E G- H . Finally, assume that y-'xy E S. tion of a', a'(y-'xy) Then y E H , since otherwise x E S f l ySy-' C {l},a contradiction. Hence aG(x) = p 1 - l

c

d(y-by) = a(z),

y€H

proving (i). P Since P van(ii) By Frobenius reciprocity, < aG,PG>=< ( a G ) ~ , >. ishes on H - S and, by (i) and the assumption that a(1) = 0, ( a G )-~a vanishes on S, we conclude that

Therefore < aG,PG>=< a,P >, as desired. (iii) Since a(1) = 0, we have aG(1) = (G : H)a(1) = 0. Hence, by (i), it suffices to show that for any 1 # h E H - S, aG(h)= 0. Since a vanishes on H - S, the latter will follow provided we show that g-lhg 4 S for all g E G. But if g - l h g E S for some g E G, then

h E gsg-' n H

sGn H

c S,

a contradiction. This proves (iii) and the result follows. H

We now illustrate the use of Theorem 1.7 by proving the following classical result due to Frobenius (1901).

Exceptional Characters

1100

Theorem 1.8. Let H be a subgroup of G such that

H n g - l H g = 1 for all g E G - h and let

N = (G - UgEGg-lHg) U {I} Then N is a unique normal complement of H in G . Proof. If H = 1, then there is nothing t o prove. Hence we may assume that H # 1, in which case H is a trivial intersection set in G with H = N G ( H ) (see Example 1.1). Let X I , . . . ,xr be all nonprincipal irreducible characters of H and let ai = xi(1)lH - xi, 1 I i 5 r. Since ai(1) = @(I) = 0, it follows from Theorem 1.7 that

and ( a G ) H = ai

(1 I i

Ir)

(2)

By F'robenius reciprocity, we have

< a;G ,1G >=< a;,1H >= Xi(1)

(1 6 i 6

T)

(3)

Because @(1) = 0, it follows from (1) and (3) that = xi( 1)lG - Xi

(1 5 i 5 r )

for some irreducible character X i of G. Note that, by (2), we have ( X ~ ) H = xi, l l i l r . Now put Ii = n;==,KerXi.If g E G and g is not conjugate t o an element of H , then a y ( g ) = 0 and so X;(g) = xi(1) = Xi(1) for each i E { l , .. . , r } , which means that g E K. This shows that N K . We now claim that Ir' is a normal complement of H in G; if sustained, it will follow that K N and hence Ir' = N . Since any other normal complement M of H in G is contained in N , the result will follow. Let a = 1Gtx:=1 Xi(l)&. Since ( A ; ) H = xi, O H is the regular character ) [ H I . If g $! K , then g is of H . Hence for all g E K , a ( g ) = ~ ( 1 = conjugate to an element of H and, since CYH is the regular character of H , a ( g ) = 0. Hence IHI-'(G : K ) a is the regular character of G / I i which forces (G : K )= [HI. Since Ii n H = n;==,Kerxi= 1, the result follows.

2 Coherent sets of characters

1101

Let H be a subgroup of G with 1 c H C G and with

Hng-lHg=i

forall g E G - H

Then H is called a Frobenius complement in G. A group which contains a Frobenius complement is called a Frobenius group. By Theorem 1.8, H has a unique normal complement in G and we refer to it as the Frobenius kernel of G . A detailed study of Frobenius groups will be postponed until the next chapter. We close by providing a further example of a trivial intersection set.

Example 1.9. Let H be a subgroup of G which satisfies the following properties : (i) C G ( ~E)H for all 1 # h E H (ii) Z ( H ) # 1 Then H is a trivial intersection set in G. Proof. Let z,y E G be such that y E H n z - ' H z and y # 1. It suffices to show that z E N G ( H ) . By (i), for any 1 # z E z - l H x , &(z) & z - l H z . Hence Z ( H )u Z ( X - ' H ~ E ) cG(Y) G z " ~ z n H By (ii), we may choose 1 # TL E Z ( z - l H z ) . Then u E H and CG(U) = z - ' H z C H . Thus z-'Hz = H , as required. I

2

Coherent sets of characters

In what follows, G denotes a finite group. All notation and conventions introduced in Sec. 1 remain in force. In addition, I T T ( G )denotes the set of all irreducible characters of G. Given a subset X of I w ( G ) , we write Z ( X ) for the Z-linear combinations of elements of X . The 23 -submodule 24 (X)"of Z ( X ) is defined by

Iz ( X ) O = {a E Z ( X ) ( Q ( l=) 0) A. Preliminary results

Our first task is to exhibit some distinguished elements in 23 ( X ) " such that each Z -linear map from Z ( X ) " into Z ( I r r ( G ) )is uniquely determined by the images of these elements.

Exceptional Characters

1102

Lemma 2.1. Let X = {xl,.. , ,xn}, n 2 2, be a set of irreducible characters of a subgroup H of G and, for each i E ( 2 , . . . ,n } , let

Then any Z -linear map

f : Z( X ) " + Z(Irr(G)) is uniquely determined by f ( a ; ) ,2 6 i 5 n.

Proof. It is clear that a2,. , . ,a, are Z -linearly independent elements of Z ( X ) " . Hence Z ( X ) " is afree 24 -module of rank r , where n-1 5 T 5 n. If r = n, then there exists n Z -linearly independent elements XI,. . . ,A, in Z ( X ) " . Let Q ( X ) be the Q -space of all Q -linear combinations of elements of X . Then XI,. . . ,A, is a Q -basis for Q ( X ) consisting of elements which vanish on 1, which is impossible. Thus r = n - 1 and we may choose a basis P 2 , ...,Pn of Z ( X ) " . Then there exists an integer t > 0 such that t P 2 , . . . ,tP, are Z -linear combinations of 0 2 , . . . ,a,. Hence, if h : Z (X)'

-+

Z( I r r ( G ) )

is another Z-linear map with h ( a ; )= f ( a i ) , then h(tPi) = f ( t P ; ) and so h(P;)= 2 5 i 5 n. Thus h = f, as required. H

f(P;),

Assume that H is a subgroup of G and that X is a set of some irreducible characters of H . Suppose further that there exists a linear isometry

* : Z ( X ) Z (IrT(G)),a H a* --f

i.e.

(1)

* is a Z -linear map for which < a*,@*>=< a,P >

Then, for any >: E X ,

for all a,P E Z ( X )

< x*,x*>=< x,x >= 1 and so

for a uniquely determined E(X) E (1, -1). The characters E(X)X* are called the exceptional characters associated with X and *, while E ( X ) X * is called the exceptional character corresponding to x.

2 Coherent sets of characters

Lemma 2.2. (1) is such that

Assume that

1103

1x1 2 2 and

a*(l)= 0

that the linear isometry

*

in

if a(1) = 0

Then the following properties hold : (i) There exists E E (1, -1) such that EX* E I r r ( G ) for all x E X . (ii) The map X + I r r ( G ) , x H E X * is injective and the image of it is { A E Irr(G)I

and

< a*,X ># 0

for some

a E Z (X)")

(iii) The set { E X * ~ X E X } of exceptional characters associated with X *, is uniquely determined by X and the restriction of * to Z ( X ) " .

Proof. (i) We must show that &(A) = ~ ( p for ) all X,p E X. To this end, put a = A( 1)p - p ( 1)X. Then a( 1) = 0 and so, by hypothesis, 0 = a*(l)= X(l)p*(1)- p(l)X*(l) This shows that X*(l) and p'(1) have the same sign. Thus &(A) = ~ ( p )as, desired. (ii) To show that x I+ EX* is injective, it suffices t o verify that * has trivial kernel. But if a* = 0, then 0 =< a*,a* >=< a , a > and hence a = 0. It remains t o find the image of x H EX*. Put X = EX* for some x E X and choose x1 E X with x # xl. Then p = EX; E I r r ( G ) with p # A. Also a = EX(1)Xl - EXl(1)X E Z (X)" and a* = x(1)p - xl(1)X. Hence < a*,X ># 0. Conversely, assume that A E I T T ( G is ) such that < a*,X ># 0 for some a E Z ( X ) " . Write a = CxEX n , and ~ a* = CxEX nxx*, nx E Z . Then

0

#< x*,X >= E < E X * , X >

for some x E X. Hence X = EX*, proving (ii). (iii) This is a direct consequence of (ii). The construction of exceptional characters is closely related to the notion of coherence due t o Feit (1967, p. 158). Assume that X is a set of some irreducible characters of a subgroup H of G, let 2 2 and let

1x1

f : Z (X)"+ Z ( I r r ( G ) ) "

Exceptional Characters

1104

be a Z -linear map. We say that X is f-coherent if f is a linear isometry which extends to a linear isometry

* : Z( X ) + Z( I r r ( G ) ) In case f is the induction map, we simply say that X is coherent. Note that if X is coherent, the map * need not be the induction map. If X is f-coherent, then the exceptional characters associated with X and * are also called the exceptional characters associated with X and f (or, simply, with X if f is the induction map). This is justified by the fact that, by Lemma 2.2, the exceptional characters E ( X ) X * are uniquely determined by X and f and do not depend on the choice of *. Under certain circumstances, the choice of * is also uniquely determined (see Lemma 2.6). Note also that, by Lemma 2.2, all ~ ( x )x ,E X , are equal and the exceptional characters E ( X ) X * are in bijective correspondence with X . The following is a typical example of coherence. Example 2.3. Let H be a Frobenius complement in G. Then (i) The set I r r ( H ) is cohewnt. (ii) The exceptional chamcters p1, . . . ,pn associated with Irr( H ) are certain extensions to G of irreducible characters of H . (iii) rl;=lKerpj is the unique normal complement of H in G .

Proof. Let

XI,,

. . ,xr be

all nonprincipd irreducible characters of H

and let ai

= X i ( l ) l H - xj

(1 5 i

5 T)

By Theorem 1.7 and Example 1.1, the induction map

Z (Irr(H))' + Z (Irr(G))' is a linear isometry. As we have seen in the proof of Theorem 1.8, = Xj(1)lG

- A;

(1 5 i 5 r )

for some X i E I r r ( G ) with ( X ~ ) H = xi, 1 5 i 5

T.

Hence the Z -linear map

* : Z ( I r r ( H ) )-, Z ( I r r ( G ) ) determined by 1~ H l ~xi ,H A j is a linear isometry. Since a: = X'(1)lG - Xi = aiG

2 Coherent sets of characters

1105

it follows from Lemma 1.1 that I r r ( H ) is coherent. Moreover, by the definition of *, p1 = lc, p2 = XI,. . . ,p r + l = A, are exceptional characters associated with I r r ( H ) . Since, by Theorem 1.8, the unique normal complement of H in G is n;==,KerX;= ni=+,1Kerp; the result follows. The next result provides necessary and sufficient conditions for f -coherence and identifies the corresponding exceptional characters. Lemma 2.4. Let X = ( X I , . . . , x ~ n} ~2 2, be a set of irreducible characters of a subgroup H of G and let

f : 25 (X)"-+ 25 (Irr(G))" be a Z -linear map. (i) If X is f -coherent and A; is the exceptional character corresponding to xil then there exists E E { 1, -1) such that

for all i E (1,. . . , n } . (ii) If (2) holds for some E E (1, -1) and some distinct irreducible characters XI,. . . ,A, of G, then X is f -coherent and X i is the exceptional character of G corresponding to xi, 1 i n.

<
= 2. By the form of a j k , the only possibility is that X j , XI, are constituents of a j k of multiplicities E , - E , respectively, whence where

A3

2 5 j , k 5 12 Hence, applying (6) and (7) together with ajj = 0, the result follows. ajk

= E(Xj - Xk)

(7)

Lemma 2.6. Let X = {XI,... , x n } , n 2 2, be a set of irreducible characters of a subgroup H of G and let

f : Z (X)'

-+

Z (Irr(G))O

be a linear isometry. Assume that all characters of X have the same degree. Then X is f-coherent and, for n > 2 , the extension * of f to 2Z ( X ) is uniquely determined. Moreover, for n = 2, any choice of E E ( 1 , - 1 ) is associated with an appropriate extension

Proof. P u t

a;j

*.

= f ( x ; - x j ) , 1 2 i, j 5 n . Then

Exceptional Characters

1108

Furthermore, ajj(1) = 0 by the definition of aij. Hence, by Lemma 2.5, there exists E E (1, -1) and distinct irreducible characters XI,. . . ,A, of G such that a;j=~(A;-Aj) forall i , j ~ { l..., , n} (8) Moreover, by Lemma 2.5, if n > 2, then E is uniquely determined and for n = 2, E is arbitrary. By (8) and Lemma 2.4 (ii), X is f-coherent, A; is the exceptional character corresponding to x; and the extension * is given by xf = &Xi, 1 5 i 5 n. Assume that * and ** are two extensions of f to the linear isometry Iz ( X ) + Z ( I r r ( G ) ) Then there exist 6, 6' E {l,-1} such that A; = 6x7 = S'xf*, 1 5 i 5 n. Since the degrees of all x; are the same, it follows from the proof of Lemma 2.4 (i) that 6 and 6' satisfy (8). Hence, if n > 2, then 6 = 6' and so xr = xf'. This proves that both extensions are the same, as required. Corollary 2.7. Let H be a subgroup of G , let X = {XI,.. . ,x,}, n 2 2, be a set of irreducible characters of H and let

f : Z (X)"+ Z (Irr(G))" be a linear isometry. Assume that X is f -coherent and x;(1) = z;x1( 1) for some integers z;, 1 5 i 5 n . Then the extension * off to Iz ( X ) is uniquely determined unless n = 2 and z1 = z2.

Proof. If each z; = 1, then the result follows by Lemma 2.6. Hence we may assume that z, > z1 = 1 for some s > 1. Since xi(1) = zixl(l), we have z;xl - x; E Z ( X ) " . Since GX; - x;" = f (zix1- xi)

we see that each zix; - xf is uniquely determined by f . Now I < x;,z8 x ; - x: > I = z, > 1 On the other hand, for any j

#

1,

I < xj*,z,x; - x: > I = 0 Hence

or

1

xi is uniquely determined by f . Since, for any i E { 1, . . . ,n }

x;"= zix; - (GX;

- Xl),

2 Coherent sets of characters

1109

the result follows. H We now apply Lemma 2.6 to prove the following classical result due to Brauer and Suzuki (see Suzuki (1955, 1986)).

Theorem 2.8. Let S be a trivial intersection set in G , let H = NG(S) and let X = {XI,. . . ,xn}, n 2 2 be a set of irreducible characters of H which have the same degree. Assume further that x ; ( x ) = xj(X)

for all x E H - S

Then the set X is coherent and the exceptional characters X I , . . . ,A, where A; corresponds to x;, satisfy the following properties : (i) There exists E E (1, -1) and, 6 = 0 or a character of G, such that - ~ j ) (1 5 i , j 5 n ) (a) (xi - ~ j = E) ( A ~~ (b) = E A -t ~6 (1 5 i 5 n ) (ii) The characters X I , . . . ,An are of the same degree. (iii) A;(g) = Aj(g) f o r all g E G - SG,i , j E (1,. . . ,n } . (iv) If A E Irr(G) and A $! { A l , . . . , A n } , then < AH,x; >=< 6,A > for a l l i E {l,..., n}. (v) For each i E { l , . . . , n } ,

XC

Proof. Let a = ~ 1 x t 1 . - .t znxn, z; E 21; , be such that a(1) = 0. Then CZ1z; = 0 since x;(l) = xl(l), 1 5 i 5 n. Hence n-I

and so, by hypothesis, a vanishes on H - S. Therefore, by Theorem 1.7, induction is a linear isometry of Z ( X ) " into 21; (Irr(G))".Hence, by Lemma 2.6, X is coherent. Therefore, by Lemma 2.4 (i), there exists E E {l,-1} satisfying (a). Taking the value of both sides of (a) at 1, we see that A;(l) = Aj(l), 1 5 i,j 5 n. By (a), we have xc - E A = ~ x? - E A ~ . Setting this common value equal to 6, we see that 6 is independent of i and satisfies (b). If E = -1, then 6 = t A; is a character. If E = 1, then x? A 1 = xy t A:! and so A1 is a constituent of xp. Hence 6 = xp - A 1 is a character or zero. Thus in all cases 6 = 0 or 6 is a character of G, proving

+

XC

Exceptional Characters

1110

(i) and (ii). Since xi-xj vanishes on H - S , we see that (xi-xj)’ vanishes on G-S’. Hence, applying (a), (iii) follows. If X E I r r ( G ) and X $! {XI,. . .,A,}, then by (b) and Frobenius reciprocity,

>=< x;G , A >=< 6,X >,

< proving (iv). Finally, for i

#j, < (Xi)H,xj >=< X F , X i >=< 6,X; >

and

< (&)H,Xi >=< X y , X i >= &+ < 6,Xi > where we again applied (b) and Frobenius reciprocity. This proves (v) and hence the result. W Corollary 2.0. Further to the notation and assumptions in Theorem 2.8, assume that for each i E {l,.. . , n } , x;(x) E Z for all x E H - S . Then the exceptional character X i corresponding to x; satisfies Xi(g)

EZ

for all g E G - SG

(1 5 i 5 n)

Proof. Let a E (I: be a primitive e-th root of 1, where e is the exponent of G. For each X E I r r ( G ) and u E Gal(Q ( a ) / Q), let “A be the Galois conjugate of A, i.e. ( “ X ) ( g ) = u(X(g))for all g E G . It suffices to show that ( “ X j ) ( g ) = X ; ( g ) for all g E G - SG,1 5 i

5n

To this end, we first observe that (“xi)(l) = xi(1) and, for all x E H (“Xi)(X>

(9) - S,

= 4Xi(X)) = X i ( 4

since xi(x) E Z . Hence, by enlarging the set X , if necessary, we may assume that X contains all Galois conjugates of xi, 1 5 i 5 n. Let pi be the exceptional character corresponding to “xi. Since induction commutes with taking Galois conjugates, it follows from Theorem 2.8 (i)(a) that &(Pi

- P j ) = (“Xi - “ X j y = “[(xi - x ~ > ~ = &(“Xi - “ X j )

I

2 Coherent sets of characters

1111

Hence p ; =' A; and so X j = XI, for some k E { 1 , . . . ,n}. This proves (9) by applying Theorem 2.8 (iii), as desired.

B. The main theorem and its applications Our principal goal is to find some sufficient conditions for coherence of a given set of irreducible characters of a subgroup H of a finite group G. In particular, we are interested to investigate when the union of coherent subsets is coherent. The following generalization of Lemma 2.6 is the main result of this chapter.

Theorem 2.10. (Feit (1960), Feit and Thompson (1963)). Let H be a subgroup of G, let X be a set of irreducible characters of H and let

f : Z ( X ) " .+ Z (ITr(G))O be a linear isometry. Assume that X satisfies the following conditions : (i) X = U;"=,Xi, IXjl = n;, X; = (xjSIs = 1,. . .,n;}, where for each i E (1,. . .,k}, either X ; is fj-coherent or all the characters in X i have the same degree, f; = flZ ( X ; ) " . (ii) There ezist integers z ; ~ 1, 5 i 5 k, 1 5 s 5 n;, such that

for all i E (1,. . .,k} and s E ( 1 , . . .,n;}. (iii) n1 2 2 and, for any m E ( 2 , . . . ,k}, m-1 ni

i=l s=1

Then X is f-coherent and the corresponding extension * o f f to Z ( X ) is uniquely determined unless k = 1, nl = 2 and zll = z12.

Proof. We argue by induction on k. For convenience, we divide the proof into a number of steps. Step 1. Here we establish the case k = 1. Since n1 2 2, we have First assume that k = 1 and x l l ( 1 ) = x 1 2 ( 1 ) = . - . = x l n l ( l ) . Then, by Lemma 2.6, X is f-coherent and, for n1 > 2, the extension * of

1x1 2 2.

Exceptional Characters

1112

f t o 2% (X) is uniquely determined. If 72.1 = 2, then ~ 1 = 1 212 = 1, which proves the given case. If k = 1 and X is f-coherent, then the required assertion follows from Corollary 2.7. This proves the case k = 1. Step 2. Assume that k > 1. It is clear that Y = $=;'Xi satisfies the assumptions of the theorem. Hence, by induction, Y is g-coherent, where g = jlZ ( Y ) " ,and the corresponding extension h : Z ( Y ) -, Z ( I w ( G ) ) of g is uniquely determined, unless k - 1 = 1, nl = 2 and 211 = 212, i.e. k = n1 = 2 and z l l = 212 = 1. By (iii), with m = 2, we have

s=l

and so the above case cannot occur. Thus the extension 23 ( Y ) 5 is uniquely determined.

Z( I T T ( G ) )

Step 3. For convenience, let us put Pis

= Zisxll - Xis,

ais

= f(Pis)

(1

i i I k, 1 I s I ni)

(1)

Given xis E Y , let A;, be the exceptional character corresponding to x i s . Then, by Lemma 2.4 (i) and the assumption that xi,(l) = z ; s x l l ( l ) , there exists E E { l , - 1 } such that 0j.q

ForeachsE(1,

...,nk},defineAk, aks

Then

Aks

( I 5 i 5 k - 1, 1 5 s 5 n;)

= ~ ( ~ i s X 1-1 Xi,)

(2)

by

= E(ZksA11 - A k s )

(3)

is a generalized character of G with Aks(1) = Z k , x 1 1 ( 1 )

> 0.

Step 4. We now show that X is f-coherent provided

< x l l t a k s >= Ezks (1 5 s 5 Indeed assume that (4)holds. Since k > 1, we have

nk)

(4)

2 Coherent sets of characters

1113

Hence each Xks is an irreducible character of G. We now show that the irreducible characters Ail, 1 5 i 5 k, 1 5 t 5 are distinct. Indeed, by (2) and the definition of Xks, we have

Hence, if

X;t

722,

= Xjs, then

1 ( = 0. Hence zit = zjs and ail = Q j s . Taking values at 1, we get ( ~ 2 t - ~ j ~ ) A 1 1) But then /?it = /?js, i.e. z i t x l l - Xit = zj$xll - xjs. Hence x i t = X j s and (i, t ) = ( j ,s), proving that the Xit are distinct. The equalities (2) and (3) can be written as one equality :

f(ZisX11

Since xis(l) = f-coherent.

(1 5 i

- X i s ) = E(ZisX11 - X i s )

2i8x11(1),

L 171 2 s 5 71;)

(6)

it follows from (6) and Lemma 2.4 (ii) that X is

Step 5. Here we prove the case nk = 1. First of all, if * is any extension of f making X f-coherent, then

=

ZklEXll

-xi1

Hence, by (3), xil = z k l & X 1 1 - Q k l = determined. By Step 4, we are left to verify that

&Xkl,

proving that

*

is uniquely

Exceptional Characters

1114

-

-&ZZjs

(9)

Now, by (I), ~ k =i f ( ~ k i ~ 1-1 ~ k i ) . If a k i E Z (h(Y))"= h(Z ( Y ) ) " , then akl = f ( a ) for some a E Z (Y)". But f is injective, so z k l x 1 1 X k l E Z (YO), a contradiction. Thus a k l # Z (h(Y))".Hence there exists an irreducible character x of G such that x # h ( Y ) and < a k l , ~># 0. Counting contributions to < a k l , a k l > coming from (8), (9) and applying ( 5 ) (with s = l),we deduce that k-1 n; i=l s=l

This is equivalent to

i = l s=l

If z

# 0, then applying condition

(iii) with m = k, it follows from (10) that

cc

k-1 2Z2Zk1

< z2(

n,

Zi",)

I 2ZZk1

i=l s=l

which forces z2 < z. This is impossible, since s E

Z , and therefore

z = 0,

as required.

Step 6 . From now on, we assume that n k > 1. By the case 6 = 1, X k is $-coherent (where $ = fllz ( X k ) " ) and the corresponding extension :Z( X k )+ Z(Irr(G))

2 Coherent sets of characters

1115

of ?/I is uniquely determined unless n k = 2 and Z k l = 2 k 2 . Hence, by Step 2, if X is f-coherent, then the corresponding extension * of f to Z ( X ) is uniquely determined, except possibly in the case where 7Lk = 2 and Z k l = 2 4 2 . Let p k s be the exceptional character corresponding to X k s , 1 5 s 5 n k . Then there exists & k E (1, -1) such that Y(xks)

= Ekpks

(1 5

2

nk)

(11)

It will next be shown that

=

-&&k(z12zkt

which is impossible since

t z k s < A 1 2 , P k t >>,

< A 1 2 , p k t >> 0. Thus (12) is established.

Step 7. Completion of the proof. So far we have not used the assumption that z;l(z;, for all i E ( 1 , . . . ,k}, s E ( 1 , . . . ,n;}. We now apply this assumption for i = k. Then Z k s / Z k l is an integer for each s E (1,. . . ,ni} and, by ( l ) , we have

where each summand in (14) belongs to 23 ( X ) " . Hence

Exceptional Characters

1116

It follows that for all s E ( 1 , .

. .,n k } ,

proving (4). Hence, by Step 4, X is f-coherent. We are left to verify uniqueness of the extension * of f. By the first paragraph of Step 6, we need only to consider the case where n k = 2 and Zkl = Zk2. Let * be any extension of f . Then, by the first paragraph of Step 5, we have )(El = & X k l . On the other hand, z k l X k 1 - z k 2 x k 2 E Hence f ( Z k l X k 1 - zk2Xk2) = ZklEXkl - z k 2 x i 2

(x)".

which shows that xz2 is uniquely determined. Thus both xil and xi2 are uniquely determined. Therefore, by Step 2, * is uniquely determined by f and the result follows.

Corollary 2.11. Let H be a subgroup of G , let X be a set of irreducible characters of H and let

f : Z (X)"+ Z ( I r r ( G ) ) " be a linear isometry. Assume that X satisfies the following conditions : (i) X = X 1 U {x}, where X 1 is fl-coherent with f 1 = flZ (X,)", and 1 x 1 1 2 2. (ii) There exists 11 E X 1 such that $( 1)Ip(1) for all p E X and such that

Then X is f-coherent and the corresponding extension uniquely determined.

*

o f f to

Z ( X ) is

2 Coherent sets of characters

1117

( ~ 1 1 , .. . ,xlnl}and X2 = (x21}, where x11 = $, = x. By hypothesis, x;,(l) = zisxll(l), i E {1,2}, s E (1, ...,n;}, n2 = 1. We wish t o apply Theorem 2.10 with k = 2 and n k = 1. The proof of this case given by Step 5 did not use the assumption that zil(z;,. Hence it suffices t o verify that

Proof. Write X1 =

x21

which holds by virtue of (ii). I Corollary 2.12. Let H be a subgroup of G , let X be a set of irreducible characters of H and let

f : Z ( X ) " + ZZ (ITr(G))O be a linear isometry. Assume that X satisfies the following conditions : (i) X = Uf=lX;, lXil = ni, X i = (x;,ls = 1,.. . ,ni}, where the characters in each Xi have the same degree. (ii) There exist integers z1,. . . ,zk such that xi,( 1) = z;x11( 1)) 1 5 i 5 k, 15 s 5 n;. (iii) n1 2 2 and, for any m E ( 2 , . . . ,k}, m- 1 i=1

Then X is f-coherent and the corresponding extension uniquely determined unless k = 1 and n1 = 2.

*

o f f to

Z ( X ) is

Proof. Keeping the notation of Theorem 2.10, we have z;, = z;, 1 5 i 5 k, 1 5 s 5 n;; in particular, z;lJz;, for all 1 5 i 5 k, 1 5 s 5 n;. Furthermore, for any m E (2,. . . ,k}, m-1

n,

m-1

n,

m-1

i=l

Hence the desired conclusion follows by virtue of Theorem 2.10. W

Exceptional Characters

1118

Corollary 2.13. Let S be a trivial intersection set in G , let H = N G ( S ) and let X be a set of irreducible characters of H which all vanish on H - S. Assume that X satisfies the following conditions : (i) X = U;"=,Xi, lXil = ni, X i = {xi81s = 1,..., n;}, where the characters in each X i have the same degree. (ii) There exist integers 21,. . ,zk such that xi8(l) = ZiX11(1), 1 I i 5 k, 1 5 s 5 n;. (iii) n1 1 2 and, for any m E ( 2 , . . . ,k},

.

Then X is coherent and the corresponding extension of induciton to Z ( X ) is uniquely determined unless k = 1 and 72.1 = 2.

Proof. The induction map 1z (X)O + zc1 (Irr(G))O

is a linear isometry by virtue of Theorem 1.7. The desired conclusion is therefore a consequence of Corollary 2.12. W

Theorem 2.14. Further to the notation and assumptions of Corollary 2.19, let Xit be the exceptional character of G corresponding to Xit, 1 5 i 5 k, 1 5 t 5 n;, let P I , . ..,pm be all irreducible characters of G distinct from the Xit and let

cc k

x=

n,

ZiXit

i=l t=l

Then there exist integers d , d l , ... ,d , and generalized characters

2 Coherent sets of characters

1119

Proof. By Corollary 2.13, X is coherent. Hence, by Lemma 2.4 (i), there exists E E (1, -1} such that

Applying (15) and Frobenius reciprocity, we have

and we denote this integer by dj (note that dj does not depend on the choice of i and t , but only on j ) . Then < ( p j ) H , X i t >= d j z ; . Setting aj = ( p j ) -~ djx, it follows that aj is a generalized character satisfying (ii) and orthogonal to each Xit. Again, applying (15) and Frobenius reciprocity, we have

for some generalized character particular, (i) holds. Applying (16), we have

oit

of H which is orthogonal to each

xj,; in

, from Theorem 1.7 (i) Since, by (15), Xjt - ziX11 = E(X;t - z i ~ l l ) it~ follows that ( X i t - z i X 1 1 ) ( 2 ) = E(Xit - z;x11)(x) for all x E S - (1) (18)

Exceptional Characters

1120

Furthermore Xit(1)

- ~ i X 1 1 ( 1 )=

(G : H ) & ( ~ i t (l ) G X I I ( ~ ) )= 0

Applying (16) - (19), we see that the generalized character Pit

Pit

(19)

defined by

= (sit - allZi)X t ait - Gall

vanishes on S . Since, by hypothesis, xjs vanishes on H - S , we must have < & , x j S >= 0, whence a;t = alla;. It follows from (16) that ( W H

= Exit t UllZiX t ait

and so, setting a l l = d , (iii) is established. Since Pit vanishes on S and ait = ullz;, we have (ait)S

proving (iv). Finally, assume that SG n H we have (Xit)H

=(~ia11)~,

S. Then, by (15) and Theorem 1.7 (iii),

- (ZiX11)H =

= Hence, applying (20) and (iii), we have

3

HXit

Exit ait

- .ZiX1l)GIH

- WXll

(20)

= zia11, proving (v).

Exceptional subsets

All groups considered below are assumed t o be finite and all characters are assumed to be C-characters. Elements of order 2 in any given group are called involutions. We begin by introducing the following terminology. A nonempty subset S of a group G is said to be exceptional in G if it satisfies the following two properties : (i) S n z:-'Sz = 0 for all z E G - N G ( S ) (ii) For any two involutions z, y E G , if z y E S then yz E S. Note that (i) is equivalent to the requirement that S is a trivial intersection set such that 1 # S if N G ( S )# G. Example 3.1. Let S satisfy property (i). Then S is exceptional in G under either of the following hypotheses :

3 Exceptional subsets

(a)

s-'

=

1121

s.

(b) S = H - H1 for some subgroups H , H I of G.

Proof. If (b) holds, then obviously S-l = S. On the other hand, if z , y are involutions in G, then yz = (zy)-'. Hence (a) implies that S is exceptional in G. H Our aim is to prove a useful formula, due to Suzuki (1959), which involves exceptional subsets. A number of applications will be provided in our subsequent investigations. One application, presented in this section, is Theorem 3.6 whose proof relies on a useful lemma due t o Belonogov (see Belonogov and Fomin (1976, Lemma 3.3)). The main theorem, namely Theorem 3.3, is actually a generalized version of Suzuki's result which can be found in Belonogov and Fomin (1976, Theorem 17). In what follows, given a nonempty subset S of G, we denote by S+ the element of (I= G given by

S+=C. XES

If S = 0, then we put Sf = 0. If x is a character of G, then we view character of (I: G. With this convention, we have

x(s+)=

Cx(s)

if

x as a

S+ 0

XES

and

x(S+) = 0 if S = 0.

Lemma 3.2. Let S be an invariant subset of G, let g E G and let X I , . . ,x+ be all irreducible characters of G . Denote by ns(g) the number of pairs (z,y) E S x S such that z y = g. Then

.

Proof. Let C1,. . .,C, be all conjugacy classes of G, let g; E C; and h; = IC;l. Then T

k= 1

where aijk is the number of pairs if z k = g , then

( 2 , ~E)

Ci X Cj with gk = zy. Therefore,

Exceptional Characters

1122

By Burnside's formula (Theorem 21,1.2), we have

Hence, for g = g k , we have

as desired.

We say that an element x E G inverts y E G if z - l y x = y-'. In what follows, an invariant subset is assumed to be nonempty. Theorem 3.3. (Suzuki's formula, Suzuki (1959)). Let S be an exceptional subset of a group G , let H = N G ( S ) and let I be an invariant subset of involutions in G. Let X I , . . . ,xr be all irreducible characters of G, let P I , . , p s be all irzdwible characters of H and let a : H ---i (c be Q class function of H which vanishes on H - S. Denote by Jo the set of all elements in I n H which invert an element of S (note that if JO # 0, then JO is an invariant subset of H). Then, for any invariant subset J of H with JO J C I n H , we have :

..

Moreover, if Jo = 8, then the left side of (1) is zero. Proof. For any subset X of G and any g E G, let nx(g) denote the number of pairs ( a , b ) E X x X with ab = g . Then, by the definition of the

3 Exceptional subsets

1123

exceptional subset, n r ( s ) = nr(s-') for all s E S. Hence, by Lemma 3.2,

Similarly, if J is any invariant subset of involutions in H , then

We now claim that nI(s) = n~,,(s)

for all

sE

S

(4)

Indeed, if n r ( s ) = 0, then obviously ~ J , ( S )= 0 since JO g I . Assume that nr(s) # 0. Then there exists ( a , b ) E I x I such that a6 = s. Since S is an exceptional subset in G, ba = $1 E S . Hence s1 = a-lsa E a-'Sa n S and so a E I n H . Similarly, b E I n H. Moreover,

a-lsa = ba = 8-l

and

b-lsb = ba = s-'

which shows that a and b invert s E S, i.e. that a , b E Jo. This proves (4). Now assume that J is an invariant subset of H with Jo & J C I n H . Then, by (4), n ~ ( s=) n ~ ( sfor ) all s E S. Hence applying (2) and (3), we see that the class function II,of H given by

vanishes on S . Since, by hypothesis, o vanishes on H - S , we have 0. Hence

< II,,a >=

Takinginto account that, by Frobenius reciprocity, < (Xj)H, o >=< xi,& >, the required formula follows. Finally, assume that Jo = 8. Then, by (2) and (4),the left side of ( 5 ) is zero. Hence, by Frobenius reciprocity, the left side of (1) is also zero.

As an application, we prove the following lemma due to Belonogov (see Belonogov and Fomin (1976, Lemma 33)). In the statement below, the subgroup generated by an empty set is understood to be equal to 1.

Exceptional Characters

1124

Lemma 3.4. Let H = N c ( S ) , where S is an exceptional subset of a group G, let X be the set of all involutions in G and let X o be the set of all involutions in H which invert some element of S . Assume that there exist two nonprincipal irreducible characters p 2 and p3 of H such that for P1 = 1Hj p1 t p2 - p3 vanishes on H - S

Then

N= is a normal subgroup of G with N

# G.

Proof. If X - Xz = 0, then N = 1 and, since G # 1, the lemma is proved. Hence we may assume that X - Xf # 0. If 1 E S, then H = G and X = Xo. Hence X - Xf = 8, a contradiction. Thus 1 # S . Since a = p1 t p 2 - p 3 vanishes on H - S and 1 # S , we have a(1) = aG(1) = 0

(6)

< a G , 8 >=< a , a >= 3

(7)

Hence, by Theorem 1.7 (ii),

We claim that ac

= x1 t x 2 - x 3

where x 1 = l ~and , x 2 and x 3 are distinct nonprincipal irreducible characters of G . Indeed, by Frobenius reciprocity < aG,lc>=< a , 1 >= ~ 1. Hence, by ( 6 ) and (7), aG has precisely three irreducible constituents, one of which is lc and the other two appear with opposite signs, proving (8). Now choose z E X - Xf. Denote by C the G-conjugacy class of z and put 1 = C. Then G,H,S,I and a satisfy the hypotheses of Theorem 3.3 and Jo = 0. Hence, if X I , ~ 2 , ...,xr are all irreducible characters of G, then

It follows that

3 Exceptional subsets

1125

and therefore, by (8),

If 5 = g-'sg for some s E S , g E G, then s-lss = s-l and 2 E Xg,which is impossible. Thus 5 $! SG and so aG(.) = 0. Applying (6) and (8), we then have

1

+ x2(1) - x3(1) = 0

1t x 2 ( 4 - x 3 ( 4 = 0 It follows from (9), (10) and (11) that

(10) (11)

whence (x2(4

- x2(1))2

X 2 ( W

Thus

~ 4 2= ) xz(1)

+

= E Ir'erx2. Hence

x2W)

which means that

2

X -Xf C K e r x 2 # G Finally, since X - X$ is an invariant subset of G, it follows that N is a normal subgroup of G with N # G.

Corollary 3.5. Let G be a simple group, let S be an exceptional subset in G and let H = N G ( S ) . Assume that there exist two nonprincipal irreducible characters X and p of H such that 1~

+A -p

vanishes on H - S

Then each involution in G is conjugate to an involution of H which inverts an element of S . Proof. Since G is simple, the normal subgroup N of G given by Lemma 3.4 must be identity. The latter is possible only in the case X - X: = 8, which is equivalent to the assertion of the corollary. H As a consequence of Corollary 3.5,we prove the following result contained in Belonogov and Fomin (1976). The special case of this result in which M = 1 is due t o Harada (1967).

Exceptional Characters

1126

Theorem 3.6. Let G be a simple group and let Ii be a subgroup of odd order such that the following two properties hold ; (i) 'h n g-lKg = 1 for alE g E G - N G ( K ) . (ii) N ~ ( 1 i = ) K < x > x M , whew K Q li < x >, Iin < x >= 1, ( [ I < \ IMI) , = 1 and z is an involution of G which inverts each element of li. Then each involution in G is conjugate to an involution in xM.In particular, if ]MI is odd, then all involutions of G are conjugate to x.

Proof. Put S = K x M - M and H = N G ( K ) . We first claim that S is an exceptional subset of G with N G ( S )= H . Indeed, H

NG(s)

Nc(
) = NG(K x M )

H

and

(ICxM)ng-'(KxM)ggM

forall g c G - H

This shows that NG(S)= H and that S n g-lSg = 0 for all g E G - H . Hence, by Example 3.1 (b), S is an exceptional subset in G. Our next aim is to find two nonprincipal irreducible characters A and p of H which satisfy the hypothesis of Corollary 3.5. To this end, put L = K < x > and consider the irreducible characters of L . Since L' = I 2, of L have degree 2. Indeed, since the number s of conjugacy classes of L is equal to (1/2)(IKI - 1) t 2, then the equality

i=3

forces

$j(

1) = 2 for all i

> 2. By the second orthogonality relation, we have 8

ICL/K(xlOI = ICL(Z>I-

C i=3

I$i(x)12

(12)

Since \Cr,(z)l = 2 = ~ C L / K ( S # )it( ,follows from (12) that &(z) = 0 for all i 1 3. On the other hand, kz is an involution for all k E K which is L-conjugate to 2. Hence

$i(kx) = 0

for d l k E Ii,i 2 3

(13)

Now H = L x M and so $ 1 , . . ., can be viewed as irreducible characters of H with $j(lm) = $j(l) for all 1 E L, rn E M . Setting X = $2 and p = $3,

4 Special classes

1127

it follows that X and p are distinct irreducible characters of H and we claim that 1,y + A - p vanishes on H - S (14) Indeed, H - S = 1 r ' z M ~ M If. h E M , then X ( h ) - p ( h ) = $2(1)-$3(1) = -1 and (14) holds. If h E K z M , say h = kxm, k E #, rn E M , then by (13)

A(h) - ~ ( h=)+ 2 ( k 5 ) - $ s ( k x ) = -1 and (14) holds, as claimed. By (14) and Corollary 3.5, any given involution t in G is conjugate to an involution, say h, of H which inverts an element of S. Since no element of M inverts an element of S, we see that h is conjugate to some element of zM.Thus t is conjugate to an involution in z M and the result follows.

4

Special classes

All characters considered below are assumed to be C -characters. Let H be a subgroup of a finite group G. Our aim is to describe a situation where part of the character table of G can be recovered from a corresponding part of the character table of H . The method which will be presented here is due to Suzuki (1959). An invariant subset S of H is said to be special (with respect to G) if it satisfies the following conditions : (a) CG(S)& H for all s E S. (b) Any two elements of S which are conjugate in G are already conjugate in H . (c) If s E S and < s >=< h > for some h E H , then h E S. The conjugacy classes of H which constitute such S are called special classes of H . Lemma 4.1. If S is a special subset of H , then S is a trivial intersection set in G which satisfies the following properties : (a') H = NG(s) (b') If s E S and k E Z is coprime to [ H I , then sk E S. Conversely, if 1 # S and S is a trivial intersection set in G satisfying (a') and (b'), then S is a special subset of H .

.

Proof. First assume that 1 E S. Then H = G = N G ( S ) and S is a trivial intersection set in G. Furthermore, property (b') follows from prop-

1128

Exceptional Characters

erty (c), Next assume that 1 !ij S. Then, by Lemma 1.5, S is a trivial intersection set in G and H = N G ( S ) . By Lemma 1.4 (i), S E H and hence property (b’) follows from property (c). Conversely, assume that S is a trivial intersection set in G with 1 @ S and let S satisfy (a’) and (b’). Then, by Lemma 1.4, S C N G ( S ) = H and properties (a) and (b) hold. Finally, assume that s E S, h E H and < s >=< h >. Then h = st for some integer t coprime to the order n of s. Hence h = Pnrn for any integer m. By choosing m to b e the product of a l l primes dividing [ H I , but not dividing t , we see that k = t t nm is coprime to [HI and h = sk. Hence, by (b’), h E S and the result follows. I We next provide a simple example of a special subset.

Example 4.2. Let A # 1 be an abelian subgroup of G such that for all 1 # z E A , CG(Z)= A . Then A - (1) is a special subset of H = N G ( A ) . Proof. By Example 1.3, A is a trivial intersection set in G . Hence S = A-(1) is also a trivial intersection set in G. Since N G ( S )= N G ( A ) = H and property (b’) in Lemma 4.1 is obviously satisfied, the desired assertion follows by virtue of Lemma 4.1. W Lemma 4.3. Let S be a special subset of H and let a,P be class functions of H vanishing on H - S . Then (i) a G ( s ) = a ( s )for all s E S . (ii) < aG,PG>=< a,/? > Proof. By Lemma 4.1, S is a trivial intersection set with H = NG(S). If 1 E S, then H = G and there is nothing to prove. Hence we may assume that 1 $ S, in which case the result follows by Theorem 1.7 (i), (ii). W Given a subset S of H , we write C f ( H , S) for the space of all (I: -valued class functions of H which vanish on H - S. As usual, C f ( H ) denotes the space of all C! -valued class functions of H . We denote by VH the Q -space of all Q -linear combinations of irreducible characters of H and by V,y(S) the subspace of VH consisting of elements in VHwhich vanish on H - S. An invariant subset S of H is called closed if, for any s E S, h E H , < s >=< h > implies h E S. It is clear that if S is closed, then so is H - S.

4 Special classes

1129

.

Lemma 4.4. Let S be a closed subset of H , let X I , . . ,xT be all irreducible characters of H and let E be a primitive (HI-th root of 1 in a!. Assume that a = C;'=la;X;, a; 6 Q ( E ) vanishes on H - S and, for any u E Gal(Q ( E ) / $ ) define o(a) by a(&)= Ci==, a(a;)x;. Then (i) .(a) vanishes on H - S.

(ii)

C a E G a l ( Q (.)/Q )

44 E VH(S)

Proof. (i) Write U ( E ) = E~ for some k coprime to IHI. Then a ( x ; ( h ) )= Xj(hk) for all h E H , i E (1,. . . , r } (see Lemma 17.5.1 (i)). Since S is a closed subset of H, then so is H - S . Hence, for any 2 E H - S , there exists h E H - S such that z = hk. It follows that

proving (i). (ii) The element in (ii) obviously belongs to on H - S , the required assertion follows.

VH. Since, by (i), it vanishes

Lemma 4.5. Let S be a closed subset of H consisting of n conjugacy classes of H . Then (i) VH = vH(s)@ VH(H - s) (ii) dimQ VH(S)= n and each Q -basis of VH(S)is a (I= -basis o f C f ( H , S ) .

(iii) C f ( H , S ) has a 6!-basis which consists of n generalized characters of H . Proof. (i) Choose representatives hl, . . . ,h, of the conjugacy classes of H such that h l , . ..,h, E S. For each i E (1,. . .,r}, define A; E C f ( H )by X j ( h j ) = 6;j, 1 5 j 5 r . Then XI,. . . ,Ar form a basis of C f ( H ) . Given Q E V H ,we may write a = C:=l a(hi)Ai. Hence Q = p + y, where p = CZ1cr(hi)X; vanishes on H - S and y = C:==n+l a(h;)Ai vanishes on S. Let XI,.. .,xT be all irreducible characters of H and let E be a primitive IH(-th root of 1 in a! . Since XI,. . .,xT is a Q -basis for VH,we may write B = a ; ~with ; a; E Q . Since X I , . . . ,xTis a C! -basis of C f ( H ) , we may also write p = E (c . Since each pix;,y = Cr=lyix; with

Exceptional Characters

1130

(~(hE ; )Q ( E ) , we have

and similarly 7i E Q ( E ) , 1 5 i 5. T . Given CT E r = Gal(Q ( c ) / Q ), we have .(a) = a(P) t 4 7 ) . Hence

and, by Lemma 4.4,the terms on the right belong to VH(S)and VH(H- S), respectively. Since, by the definition of V H ( S ) ,V H ( S )n VH(H- S ) = 0, property (i) is established. ,A, is a (1: -basis of Cf ( H , S) and Xn+l,. . ,A, (ii) We first observe that XI,. is a Q) -basis of C f ( H ,H - S). Note also that

..

.

Now VH has a Q -basis, namely XI,.. . ,xr which is a CC -basis of C f ( H ) . This means that C f ( H ) = CC ~ J VH Q and hence any Q -basis of VH is a C-basis of C f ( H ) . If X and Y are Q-bases of V H ( S )and VH(H - S), respectively, then, by (i), X n Y = 0 and X U Y is a Q -basis of V H . Hence X U Y is a CC -basis of C f ( H ) . Therefore, by (l),

1x1

Since T = IX UYl = 1x1 t IYI, it follows from (2) that = n , proving (ii). (iii) Let P I , . . , o n be a Q -basis of V H ( S) .Then, by (2), P I , . . . ,Pn is a (1: -basis of Cf ( H , S). We may choose an integer o such that z & , , ..,zPn are generalized characters of H . Since &,. . . ,Z P n is a (I= -basis of C f ( H ,S), the desired assertion follows.

.

Theorem 4.6. Let H be a subgroup of Q finite group G , let S be a special subset of H consisting of n conjvgacy classes of H with representatives s1,. ,S n and let X I , . . . ,x, be all irreducible characters of H . Then (i) There exists an n x matria: A = (ajj) with each a;j E Z such that 1 5 i 5. n , form a (I=-basisof the generalized characters cwi = &a,jxj, C f (H, S).

..

4 Special classes

c

1131

(ii) For any given i E { l , . .. , T } , there exists exactly one n x n matrix = (cjk), cjk E a= such h i !

(iii) If Q? = & b ; j X j , where XI,. . . ,A t are all irreducible characters of G , then B = (bij) is an n x t matrix with each bij E Z and such that BB' = AA' where B' and A' are transposed of B and A, respectively. Moreover, if A is fixed, then there am only finitely many B with BB' = AA' up to deleting or inserting .zero columns. (iv) X;(sj) = C;E=l cjkbki, 1 5 i 5 t, 1 5 j 5 n

Proof. (i) This is a special case of Lemma 4.5 (iii). (ii) P u t P j = C & = l x m ( s j ) ~ m1, 5 j 5 n. Then, by the orthogonality C j p k for uniquely deterrelation, p j E C f ( H ,s). Hence, by (i), pj = mined cjk € (I: , 1 5 j 5 n, 1 5 k 5 n. It follows from (i) that n

T

which obviously implies (ii). (iii) Because a; is a generalized character, so is a? and therefore each bij E Z , 1 5 i 5 n, 1 5 j 5 t . If 5 E G - SG,then obviously aQ(5) = 0. Hence, by Lemma 4.3,

< af,aF >=< a;,aj >

(3)

and aiG ( 2 ) =

ai(z)

Equality (3) yields t

r

k= 1

k= 1

if if

xES xEG-SG

(4)

Exceptional Characters

1132

which means that BB’ = AA‘. If matrix A is fixed and i E (1,. . . ,n},then by ( 5 ) with j = i, C;=,b:k is an integer depending on A only. Hence there are only finitely many nonzero values for b:l, . . . ,b;t and therefore for b i l , . . . , b i t , proving (iii). (iv) Applying (4),we have t

T

Also, for any s E S, CG(S)= C H ( S )Hence, . by the orthogonality relation (Theorem 19.2.3 (iii)) t

r

Now write

n

k(Sj)

=

C

cjk61i-i

+

Eij

k=l

Then we are left to verify that each ~ i isj zero. Substituting (8) in (6) and applying ( 5 ) , we have

Substituting (8) in the left hand side of (7), we obtain t

t

n

i = l p=l

i=l

For the “cross terms”, we have t

n

i=l

q=1

by virtue of (9). Similarly, we have t

n

i=l p=l

n q=l

4 Special classes

1133

For the “leading terms”, we have by ( 5 ) ,

i=l

which is the right hand side of (7). Thus

Finally, if

sk

is H-conjugate t o

sj’, then

Therefore r

It follows that

T

Exceptional Characters

1134

and hence &;k = E;j. Therefore, by (lo), t

i=l

for all j E (1,. . . ,n}, and hence ~ i=j 0 for all i and j. This completes the proof of the theorem. I

Chapter 37

F'robenius Groups Frobenius groups arise naturally in many contexts. This chapter is concerned with a detailed investigation of properties of Frobenius groups and their characters. Historically, these groups were first studied by Frobenius in terms of permutation groups. Namely, Frobenius groups constitute the class of transitive permutation groups in which only the identity fixes more than one point, but the subgroup fixing a point is nontrivial. Many problems arising in group theory stem from this class of groups. A fundamental result of Frobenius, whose proof relies on the character theory, asserts that if G is a Frobenius group and H a subgroup fixing a point, then the subset of G consisting of the identity together with those elements which fix no points forms a normal subgroup N of G of order (G : H).A profound discovery, due to Thompson, asserts that N is a nilpotent group. We shall derive this fact as a consequence of some properties of fixed-point-free automorphisms of groups. The chapter contains a number of other important group- theoretic facts together with a detailed analysis of coherence of characters arising from the notion of Frobenius groups.

1

Preliminary results

All groups considered below are assumed to be finite. We begin by recalling the following definition. Let H be a subgroup of a group G with 1 c H c G and with Hng-'Hg=l forall g E G - H Then H is called a Frobenius complement in G. A group which contains a Frobenius complement is called a Frobenius group . By Theorem 36.1.8, 1135

F’robenius Groups

1136

H has a unique normal complement in G and we refer to it as the F’robenius kernel of G. It will be shown later (see Theorem 4.2 (ii)) that Frobenius kernels are unique (i.e. they do not depend upon the choice of a Frobenius complement) and that Frobenius complements are unique up to conjugation. In this section, we shall record a number of elementary properties of Frobenius groups. Lemma 1.1. Let N and H fl N = 1. Then

a G and

let H be a subgroup o f G with G = N H

N G ( H )n N = C N ( H ) N , G ( H )= H C N ( H ) Proof. It is clear that N G ( H ) n N 4 N G ( H ) . Since H follows that [ H ,N G ( H )fl N ] C H f l N = 1

nN

= 1, it

Hence N G ( H ) n N C_ C N ( H ) . Since the opposite containment is obvious, this proves the first formula. Applying Lemma 27.5.1 (with X = H , Y = N , H = N G ( H ) ) , we have

N G ( H ) = G f l N G ( H ) = H ( N fl N G ( H ) ) Hence the first formula implies the second. W Theorem 1.2. Let G be a Frobenius group with Frobenius complement

H and Frobenius kernel N . Then (i) C&) c N for all 1 # 2 E N . (ii) /HI divides IN1 - 1 and, in particular, N is a normal Hall subgroup of G . (iii) If IH I is even, then N is abelian and H has a unique involution. Conversely, if N is a normal subgroup of G with 1 c N c G and N satisfies (i), then G is a Frobenius group and N is a Fmbenius kernel of G. Proof. For any y E G , y-lHy is a Frobenius subgroup of G. Hence, if 1 # z E y - l H y , then CC(Z)

c y-lHy

Therefore, no element of N - { 1) commutes with any element of y-’Hy- { 1). Thus, by Theorem 36.1.8, (i) holds. The group H acts on N - (1) by conjugation. Since N fi H = 1, it follows from (i) that 1 E H is the only element of H which fixes any given

1 Preliminary results

1137

z E N - (1). Hence the size of each H-orbit in N - (1) is equal to [ H I , proving (ii). Let h be an involution in H . Suppose that n,l(h-'nlh) = ny1(h-'n2h)

for some

~ ~ 1 , 7 1E2 N

Then n2nT1 = h - l ( n 2 n l ' ) h and so, by (i), n1 = 712. Hence there are IN1 distinct elements of the form n-'(h-'nh), n E N . Consequently, if g E N , then g = n-'(h-'nh) for some n E N . Thus /-&-Ig/.&

Therefore, if

g1,g2

= h-ln-'h-lnhh = h-ln-1h-ln

= g-l

E N , then

9192 = h-'(!&'g,')h

= g2g1

proving that N is abelian. If hl is another involution in H , then hh,' E C G ( N )C N . Hence h = h l , proving (iii). Conversely, assume that N is a normal subgroup of G with 1 c N C G and that N satisfies (i). If 5 # N , then Clv(x) = 1. Choose an element x # N of prime order p . Then the subgroup L =< N , x > is a semidirect product of N and < z >. Since C N ( ~=)1, < z > coincides with its normalizer in L by virtue of Lemma 1.1. Hence < z > is a Sylow psubgroup of L. Thus p does not divide the order of N . Hence N is a Hall subgroup of G and so, by Theorem 26.1.4, there exists a subgroup H of G with G = N H and N n H = 1. We will show that H is a Frobenius complement in G, which will complete the proof. Since N # 1 and N # G, we have H # 1, H # G. Assume that z E G is such that H n x - l H z # 1. Then z - l y x E H for some 1 # y E H . We must show that x E H . Since G = H N , we can write z = hn with h E H, n E N . Set u = h-lyh and D = [u,n].Then u E H - {l}, x - * y z = n-lun and D = u - ' ( z - l y x ) E H . On the other hand, n E N implies D E N . Hence D = 1. This shows that u E Cc (n) with 1 # u E H . Hence n = 1 and z = h E H , thus completing the proof. W

Corollary 1.3. Let K # 1 be a subgroup ofG such that Ii # N G ( K ) and &(z) C K for all 1 # z E Ir'. Then N G ( K ) is a Frobenius group with the Frobenius kernel I n# = 1 and 0 is of order p , we see that j = 1 and cr = a1,as claimed, By the foregoing, for any given i E { 1,. . . , p - l}, {(oo)"laE

Therefore,

n cuEK

(..)"a)

I. If r = p , then p would have a fixed point in R, since R < p > would be nilpotent. Thus T # p. We now claim that G = A R for some prime T # q. Indeed, assume that for all primes T # q, G # AR. Since R A is padmissible, the minimality of G forces R A to be nilpotent and so R C C c ( A ) since (lAl,lRl) = 1. Since this is true for all T # q , the group G / C c ( A ) ,and hence the corresponding semidirect product A ( G / C G ( A ) ) would , be a q-group and thus nilpotent. But then A n Z(G) # 1, contrary to Step 1. This substantiates our claim. Let u = pIA and let A' be the group of automorphisms of A that arise from conjugacy by elements of R . Then S = Ii < u > is a subgroup of Aut(A), K d S and < u > nK = 1. We claim that A , S and K satisfy the hypotheses of Lemma 3.2; if sustained, it will follow that Ii = 1. But then 1 # A C Z ( G ) , a contradiction. Let g E R and let i, E K be defined by i g ( a ) = g-'ag, a E A . It will be shown that i,a is conjugate to u in Aut(A); this will imply that iga is a fixed-point-free automorphism of order p . Hence, since (IAI,IKl) = 1, the proof will be complete. Since p-' is fixed-point-free on R , by Lemma 3.1 (ii) there exists z E R such that cp-'(g) = z-'cp-'(z). Hence g = cp(z-')x. The automorphism iZuiG1sends a E A to 6 = z - ' ~ ( z a z - ~ ) zNow .

o ( z a z - l ) = 'p(zaz-') = cp(z)cp(a)cp(z)-' so that

b = z-'&)$9(u)$7(x)-lx = g-"P(a)g = g - ' a ( a ) g Hence iZui;' = iga,as required. Step 3. Here we complete the proof by treating the case where G is not solvable. Choose an odd prime divisor q of (GI. By Lemma 3.1 (v), there is a Sylow q-subgroup Q of G such that v(Q) = Q. Hence ( p ( J ( Q ) ) = J(Q), which implies that N c ( J ( Q ) )is padmissible. If J ( Q ) d G , then as in Step 1, we see that G / J ( Q ) satisfies the hypotheses on G. This would imply that G / J ( Q )is nilpotent and hence G solvable, a contradiction. Thus

Frobenius Groups

1150

N c ( J ( Q ) ) is a proper subgroup of G. Therefore, being cp-admissible, it must be nilpotent. By the same argument, CG(Z(Q))is nilpotent. Hence, by Theorem 2.3, G is q-nilpotent which implies that G/O,t(G) is a q-group. Since O,,(G) is a proper subgroup of G which is cp-admissible, it must be nilpotent. Thus G is solvable, a desired contradiction. Our next aim is to examine the structure of fixed-point-free automorphism groups. First, we introduce an important type of a finite 2-group. By the generalized quaternion group Q p , n 2 3, we understand the group of order 2" with a presentation of the form

< z,y1x2"-' = l , y 2 = x

2"-2

,y-'xy

= 2-l

>

The group Qs of order 8 is known as the quaternion group. The dihedral group D2n of order 271, n 2 2, is defined by

Writing z = xy, we see that

is another presentation of Dzn. Lemma 3.4. Let G be a group with and any x,y E G,

(ZY)"

m m

=2 Y

G' C Z ( G ) . Then, for any rn 2 1

[Y,4

Proof. The case m = 1 being obvious, we argue by induction on m. Applying the induction hypothesis and the fact that [y,x] E Z ( G ) ,we have

Again, since G' E Z ( G ) , we easily see that [y",x] = [ y , ~ ] " , Thus y m s = xyn[y,x]m. Since

(

m;1)

=

( ;)

+m

3 Fixed-point-free automorphisms

1151

it follows that (xy)m+l

= x m f l Ym+l [ Y , X l

as required. W

Lemma 3.5. Let p be a prime and let G be a group of order p n , n 2 1. Then G has a cyclic subgroup of index p if and only if G is of one of the following types : (i) G is cyclic. (ii) G is the dimct product of a cyclic group of order p*-' and one of order p . n-1 (iii) G =< X , Y ~ X P = yp - 1, x-lys = yl+p"-' >, n 2 3. (iv) G is the dihedral group D p , n 2 3. (v) G is the generalized quaternion group Q p , n 2 3. (vi) G is the semidihedral group given by

Proof. It is clear that all groups G described below have a cyclic subgroup of index p . Conversely, assume N =< y > is a cyclic subgroup of G of index p . Then N d G and we write GIN =< z N >. Then G =< x,y > with o ( y ) = pn-' and xp E N . I f G is abelian and xp = z p for some z E N , then ( x z - ' ) ~= 1 and G =< z z - l > x N ; otherwise zp = y' with ( i , p ) = 1 and G =< x >. Hence, if G is abelian, then G is of type (i) or (ii). From now on, we assume that G is nonabelian so that n 2 3. Since G is nonabelian, the element x induces an automorphism of N of

order p. Therefore s - ' y x = yp where p p = l(rnodp*-') and 1 < p < pn-'. Since, by Fermat's theorem 3 l(rnodp), we see that p 5 l(rnodp). We now show that if p is odd, then G is of type (iii). Indeed, write p = 1 t kp' where ( p , k ) = 1 and 0 < i < n - 1. Since pp = ( 1 t kp"P = 1

we see that pP

+

+ kpit1 + ( 1 / 2 ) ( p- 1)k2p2i+' +

* * *

+

1 kpi+1(modp'+2). However, pP -= l(rnodpn-'), so that kpi+l + tpi+2 = spn-l for some t , s E Z3

E

-+

Because i 1 5 n - 1 and ( p , l c ) = 1, it follows that i 1 = n - 1 and i = n - 2. Hence p = 1 Now choose T such that kr E l ( r n o d p ) so that (ltlCPn-2)' = 1tpn-2 x-TyZ' = y Y

+

Fkobenius Groups

1152

Then we may replace x by 'x and assume that p = 1 t p n A 2 . Since x p commutes with x, we see that o(xP) divides pn-2 and x p E < y p >, say XP = u p where u E N . Also G' Z(G) since [ y , ~ = ] yp"-'. Hence, by Lemma 3.4, (zu-')" = xpu-p = 1. Replacing z by m-l, we can assume that xp = 1 and so G is of type (iii). From now on, we consider the case p = 2. Then p is odd, say p = 2k t 1. Since p2 l(mod 2n-1), we see that k(k 1) = O(rn0d 2n-3) and k 3 0 or -lmod ( Y - ~ ) . Hence we have two possibilities for p : p = 2 n - 2 ~ 1, where s is odd, and p = 2 n - 2 ~- 1. In the first case, replacing z by a suitable power, we may assume that p = 2n-2 -t 1. In the second case, either s is even and p = 2n-1 - 1 or s is odd and we may take p = 2n-2 - 1. We are therefore left t o examine three cases. Assume that p = 2'+l - 1. Then x-'yz = y-l. Because x commutes 2"-2 with z2 E N , we see that o ( x 2 ) = 1 or 2. Hence x2 = 1 or x2 = y and G 2 D 2 n or Q p , respectively. Next assume that p = 2n-2 -l-1. Because N #< x 2 >, we have x 2 = y2k for some k 2 1. Setting z = yk(2"-3-1), we have

-+

+

If n 2 4, this power of y equals 1 and G is of the type (iii). If n = 3, then x-lyz = y-' and x 2 = 1 or y2, so that G E Ds or Q 8 . Finally, assume that p = 2n-2 - 1. If x2 = y2k, then

whence 2k

O ( r n 0 d 2 ~ - ~and ) x2 = 1 or y2n-2. If x2 # 1, then

) -Y

-1 2

(ZY

273-2

Y

-2

Y

-(2n--2-2)

=1

and G is of type (vi), thus completing the proof. prime and let G be Q group o j order pn, n 2 1. Then G has exactly one subgroup of order p if and only i j G is cyclic or Q generalized quaternion group.

Lemma 3.6. Let p be

Q

Proof. It is clear that cyclic p-groups and generalized quaternion groups have exactly one subgroup of the required order. Conversely, assume that G has exactly one subgroup of order p . If G is abelian, then G is obviously cyclic. We may thus assume that G is nonabelian. Suppose that p is odd

3 Fixed-point-free automorphisms

1153

and let H be a subgroup of index p . By induction, H is cyclic. However, examining the list of groups given by Lemma 3.5, we see that none of them qualify. Thus we conclude that p = 2. Let A be a maximal normal abelian subgroup of G. Then A = C G ( A ) (since G is nilpotent) and A is cyclic, say generated by y . Let z A be an involution of G/A. Then < A , z > is not abelian and it has a cyclic subgroup of index 2. Hence, by Lemma 3.5, < A , z > is a generalized quaternion group. Therefore z-'yz = y-' which shows that G / A has exactly one involution. Now G / A is isomorphic to a subgroup of A % t ( A ) . Hence G / A is abelian and therefore cyclic. But -1 is not a square modulo 2", JAl = 2m, unless m = 1 which is impossible since it would force A Z(G). Thus G/A has order 2 and G is a generalized quaternion group.

We are now in a position to provide important information concerning the structure of a fixed-point-free automorphism group.

Theorem 3.7, Let H be a fixed-point-free automorphism group of a finite group G and let p , q be (not necessarily distinct) primes. Then (i) Every subgroup of H of order p q is cyclic. (ii) The Sylow p-subgroups of H are cyclic if p is odd. (iii) The Sylow 2-subgmups of H are cyclic or generalized quaternion.

Proof. (i) Assume that S is a noncyclic subgroup of H of order p q . Then S = KP where llil = q, /PI = p , li n P = 1 and a t least one of P,K is normal in S, say li a S. By Theorem 3.3, G is nilpotent. Let R be a nontrivial Sylow r-subgroup of G . Then a ( R ) = R for all Q E S. If T = q, then the semidirect product R K would be a q-group and hence R n Z(R1i) # 1, which would prevent H from being fixed-point-free. Thus T

# Q.

Now write P =< Q > and choose a E I. Since GIG' is isomorphic to a Hall r-subgroup, say H , of G, we see that H is abeIian and hence cyclic. Write H =< y > for some y E G. Then G =< z , y > with zm = 1, y n = 1 and ( m , n ) = 1. Since < z > a G , we have y-'zy = x f l for some p E {0,1,. . . , m - l}, with ( p , m ) = 1. Then z = y-"xy" = zpn and so pn = l(modm). Note also that G' =< [2,y] >=< z >. Since [z,y ] = s-'y-'zy = P-', we must have ( p - 1 , m ) = 1. If m is even, then p is odd and p - 1 is even, contrary to ( p - 1,m) = 1. Thus m is odd. Conversely, assume that G has the given presentation and that P is a Sylow psubgroup of G. Then < z > 4 G and G is of order mn with (m,n) = 1. Hence P c< z > or Pn < z >= 1. In either case P i s cyclic, as required. H Corollary 4.7. Let G be a Frobenius group with Fmbenius complement of odd order. Then G is a split extension of a nilpotent group by a split

metacyclic group. Proof. Apply Theorem 4.2 and Lemma 4.6. H Lemma 4.8. Let Fit(G) be the Fitting subgroup of a solvable group G . Then CG(Fit(G))= Z ( F i t ( G ) )

1158

F'robenius Groups

Proof. It suffices to show that C c ( F i t ( G ) ) Fit(G). The case [GI = 1 being obvious, we argue by induction on [GI. So assume that [GI > 1 and put F = Fit(G), 2 = Z ( F ) and N = Cc(2). Since G i s solvable, F # 1. Now N 4 G and N 2 F . Hence Fit(N) 2 F and F i t ( N ) d G. Therefore F = F i t ( N ) . Because C G ( F ) c CG(Z) = N the result will follow by induction if N # G. Hence we may assume that N = G in which case 2 = Z(G). Let K / Z = Fit(G/Z). Then K is a nilpotent normal subgroup of G and therefore li' c F . Observe also that F / Z is a nilpotent normal subgroup of G / Z . Hence F c li and so F = I ( , proving that F / Z = F i t ( G / Z ) . Since

and by induction C G ~ ( F / ZC) F / Z , the result follows. We are now ready to investigate the structure of solvable Frobenius complements. The following result is due to Zassenhaus (see Passman (1968, p. 196)).

Theorem 4.0. Let H be a solvable Frobenius complement. Then H has a normal subgmup Ho such that : (i) H / H o is isomorphic to a subgroup of S,. (ii) HO =< z, ylzm = 1,yn = 1,y-lzy = z@> where ( p , m ) = ( p - 1 , m ) = (m,n) = 1, p n 5 l(modm), m is odd and 0 5 p < m. Moreover, pn't i I( mod m ) where t denotes the product of the distinct prime divisors of n.

Proof. By Theorem 4.2, the Sylow 2-subgroups of H are cyclic or generalized quaternion. For the sake of clarity, we divide the proof into a number of steps. Step 1. Here we treat the case where the Sylow 2-subgroups of H are cyclic. By Theorem 4.2, d l Sylow subgroups of H are cyclic. Hence, by Lemma 4.6, H = Ho satisfies (ii) except possibly the last assertion concerning p. Let p and q be primes with pln, qlm. Then < xmlq,ynlp > is a group of order p q and therefore, by Theorem 4.2, must be cyclic. Hence y n l p centralizes all elements of < z > of prime order and so, by Lemma 4.4,

5 Characters of Fbobenius groups

1159

< x >. Because this is true for all pln, it follows that ynlt centralizes < x > and so p n / l = l(modrn). yn/p centralizes

Step 2. From now on, we assume that the Sylow 2-subgroups of H are generalized quaternion. Put F = F i t ( H ) and let Fp be the Sylow p-subgroup of F . If p > 2, then by Theorem 4.2, Fp is cyclic and for p = 2, Fp is cyclic or generalized quaternion. Denote by S a Sylow 2-subgroup of H so that S 2 Fz. In this step, we establish the case where Fz is cyclic. Since F2 is cyclic and F is nilpotent with all other Sylow subgroups cyclic, we see that F is cyclic. Moreover, by Lemma 4.8, H / F acts faithfully on F and so H / F is abelian. Hence S' F and, since S/S' is elementary abelian of order 4, it follows that H / F has a subgroup H o / F of odd order with H / H o of order 2 or H / H o elementary abelian of order 4. Because all Sylow subgroups of Bo are cyclic, the result follows here. Step 3. Here we prove the case where Fz is the quaternion group of order 8. To this end, put Ho = C 4 F 2 ) . It is an easy exercise to show that Aut(F2) E 5'4. Hence H / H o is an isomorphic image of a subgroup of Sq. Since we obviously have

it follows that all Sylow subgroups of HO are cyclic, as required. Step 4. Here we complete the proof by treating the case where F2 is a generalized quaternion group of order at least 16. It is clear that F2 has a characteristic cyclic subgroup Ii of index 2. Then K Q H and if Ho = CH(IL'), then Ho Q H . Since A u t ( K ) is a 2-group, so is H / H o . Since IIil 2 8, we obviously have (S : Cs(1i))= 2. Therefore ( H : Ho) = 2 and, because Cs(1i') is cyclic, all Sylow subgroups of Ho are cyclic. This completes the proof of the theorem.

5

Characters of Frobenius groups

In what follows, all groups are assumed to be finite and all characters to be (I: -characters. If G is a Frobenius group with Frobenius kernel N and Frobenius complement H , then each irreducible character of H extends uniquely to an irreducible character of G containing N in its kernel. Conversely, any irreducible character of G containing N in its kernel determines a unique irreducible character of H . For this reason, we shall identify these characters. The following theorem provides a complete description of irreducible

Frobenius Groups

1160

characters of G in terms of those of N and H .

Theorem 5.1. Let G be a Frobenius group with Frobenius kernel N and Frobenius complement H. Then (i) If x is a nonprincipal irreducible character of N , then xG is an irreducible character of G . (ii) If X is an irreducible character of G with N Ir'er A, then X = ,yG for some nonprincipal irreducible character x of N. Furthermore, for any 1 # h E H , X(h) = 0 and X(1) = IHIX(1). (iii) If X I , . . . ,x,, are all representatives of the G-conjugacy classes of nonprincipal irreducible characters of N and XI,. . . ,Am are all irreducible characters of H , then (a) xp, . .. ,x$ are all distinct irreducible chamcters of G induced from irreducible characters of N . (b) . . . ,x:, XI,. . . ,Am are all distinct irreducible chamcters of G . (c) n is the number of nonidentity conjugacy classes of G contained in N . In particular, if N is abelian, then n = (IN1 - l)/lH/.

x?,

Proof. (i) By Theorem 1.2 (i), we have CG(Z)& N for all 1 # z E N . Hence the desired conclusion follows by virtue of a much more general result, namely Theorem 18.7.6 (i). (ii) By Clifford's theorem AN has an irreducible constituent x # 1 ~By. Frobenius reciprocity X is a constituent of xG. Since, by (i), xG is irreducible, we deduce that X = xG. If 1 # h E H, then h # N and so, by the definition of induced characters we have X(h) = x G ( h )= 0. On the other hand, as required. (iii) If Q and p are irreducible characters of N, then by Theorem 18.6.12 (ii) and Corollary 17.1.8 (ii),'a = PG if and only if Q and p are G-conjugate. Hence, by (i), xf ,. . . ,x: are all distinct irreducible characters of G induced from irreducible characters of N. By Theorem 18.7.6 (ii), n is equal to the number of nonidentity conjugacy classes of G contained in N. Since H acts regularly on N - {l},it follows that if N is abelian, then n = (IN1 - l)/lHl. The remaining assertion being a consequence of (ii), the result follows.

Our next aim is to provide various characterizations of Frobenius groups in terms of their irreducible characters. The following preliminary observation will clear our path.

5 Characters of F’robenius groups

1161

Lemma 5.2. Let G be a group and let tl IG]with t > 1, If K = n K e r x, where x ranges all irreducible characters of G with t not dividing ~ ( l )then ,

(G : K )2 ( t 2 ,[GI)

. . ,xT be all irreducible characters of G where t does not divide xi( 1) and let x r + l , .. . ,xn be the remaining irreducible characters Proof. Let

XI,.

of G. Put

n

f

a=

~ ; ( 1 )and ~ b= i=r+l

Then a E O ( m o d t 2 )and so a

C~i(l)~ i=l

3

O(rnod(t2,IGl). Therefore

b = [GI - u

O(mod(t2,]GI))

(1)

Since 1~ E {XI,. . . ,x7},it follows that b # 0 and so, by (l),b 2 ( t 2 ,IGl). Since K = n ; = = , K e r x i we , also have (G : K ) 2 b. So the lemma is true.

Theorem 5.3. (Berkovich (1990)). Let G be a group, let t > 1 be a proper divisor of IGI and let K = n K e r x , where x ranges all irreducible characters of G with t not dividing ~ ( 1 ) .Then the following conditions are equivalent : (i) G is a Frobenius group with Frobenius complement of order t . (ii) (G : K ) = t and G is a Frobenius group with Frobenius kernel K. (iii) (G : K ) = t . Proof. (i)

* (ii) : Applying Theorem 5.1 (iii), we see that K = nzllierX; = N

proving (ii). (ii) j (iii) : This is obvious. (iii) =+ (i) : Since (G : K ) = t , it follows from Lemma 5.2 that ( t 2 ,]GI) 5 t . Hence (t 2 ,IG]) = t , ( t ,IGl/t) = 1 and Ii is a normal Hall subgroup of G. Let p be a nonprincipal irreducible character of K and let x be an irreducible constituent of pG. Then p is an irreducible constituent of x~ and so, by Clifford’s theorem, p ( 1 ) divides ~ ( 1 )Since . p # lh: and p is a constituent of x ~ we, have K I i e r x . Hence, by the definition of K , tlx(1). But then p(1)t divides ~ ( l )since , ( p ( l ) , t ) = 1. Thus pG = x is irreducible. This shows that for any nonprincipal irreducible character p of K , pG is an irreducible character of G.

F'robenius Groups

1162

Let 1 # x E I 0. Therefore a = t and so zi = xi(1) for all i E ( 1 , . . . ,T}. Thus

IC Q G. Since this is true for and, setting IC = n:='=,I= X z ( 1 ) < X H , X ~> which shows that

whence

X2(1)((X;)H - X l ) = Xl(l)((X;)H - x2>, proving (ii). (iii) Let p E

X and let ,b' = x(l)p - p ( 1 ) x . Then P E Z ( X ) O and

< ( X * ) H , P >=< X * , p G >=< x * , p * >=< x , p > so that

< (x*)H - x,P >= 0. Hence

Denoting this common value by r x , we have

as desired. H Lemma 6.4. Let x be a character of G which is constant on G - ( 1 ) . Then x = a l G t bpG where a , b E Z and p~ is the regular character of

6 Coherence

1167

G. I n particular, if G constituent of x.

#

K e r x , then each nonprincipal character of G is a

Proof. P u t a = x ( g ) , g E G - { 1). Then x - alc vanishes on G - { 1). Hence x - a l c = b p c for some b E C . Thus x = U l G t bpG for some a , b E (c . Since < x, 1~ >= a t b and x( 1) = a -k blGI, we see that a E Q . But x ( g ) is integral over Z , so a E Z and hence b =< x , 1 > ~ -a E Z . So the lemma is true. The following result is contained in Isaacs (1976) and Gorenstein (1968).

Theorem 6.5. Let a subgmup S of a group G be tamely embedded in G , let H = NG(S)and let the set X = {x E I r r ( H ) [ S K e r x ) be w h e m n t . Let Y C I r r ( G ) be the corresponding set of exceptional characters of G, let * : Z ( X ) -, Z ( I r r ( G ) ) be a linear isometry which extends the induction map Z(X)"-, Z ( I r r ( G ) ) " and let E E ( 1 , -1) be such that Y = {&X*IX E X } Then

(i) If X

s - (1).

E I r r ( G ) and X !$

Y , then X has a constant integral value on

(ii) Y = {A E Irr(G)IA is not constant on S - (1)). (iii) There exists an integer m E Z such that

X*W

= x ( 4 -t mx(l)

for all s E

s - {l},X E x

(iv) I f g E G is not conjugate to a n element o f S - {l}, then X(g)/X(l) is independent of the choice of X E Y . (v) If x E X , then x*(l)/x(l)is independent of the choice of x . Proof. (i) Let a = CXEXx(1)x. Then a = p~ - P H I S where p~ and P H I S are the regular characters of H and H I S , respectively. In particular, a has a constant integral value on S - { 1). Assume that X E I r r ( G ) - Y . Then, by Lemma 6.3 (i), there exists r E Q such that < AH - r a , x >= 0 for all x E X . Hence S is in the kernel of every irreducible constituent of AH - ra. Therefore AH - r a has a constant rational value on S. Since AH = ra t (AH - r a ) , we deduce that X has a constant rational value on S - (1). But, for any g E G , X ( g ) is integral

Frobenius Groups

1168

over Z . Hence A has a constant integral value on S - { 1). (ii) Assume that A E Y , say A = EX* with x E X . Let r = rx be as in Lemma 6.3 (iii) and let p = &AH - x - ra. Then E X H = x r a -I-/3 and < p , p >= 0 for all p E X . However, x E X cannot be constant on S - {l}, otherwise, by Lemma 6.4, each nonprincipal irreducible character of S is a constituent of xs. This would force = 1 which is not the case. Thus A is not constant on S - (1). The converse being a consequence of (i), (ii) is established . (iii) If X = EX' E Y , then as we have seen in (ii), Xs - E X S has a constant rational (and hence integral) value on S - ( 1 ) . Hence x* - x has a constant integral value on S - (1). To evaluate the constant, choose x o E X with xo(1) = ( H : S), that is xo = p H for some nonprincipal linear character p of S. Then (x;f)s- ( ~ 0 ) has s a constant integral value, say m, on S - (1). If x E X is arbitrary, then Lemma 6.3 (ii) yields that m'/x(l) = m / x o ( l ) where m' is the constant value taken on by (x*)s- xs on S - (1). Since XO( 1) = ( H : S), we have m' = mx(1)/(H : S), proving (iii). (iv) Given x, cp E X , we have cp(l)x*- x(l)cp* = (cp(1)x - X(l)y)G, which vanishes on elements g E G not conjugate to elements of S - (1). Hence cp(l)x*(s> = x(l)cp*(g) = 0

+

1x1

and (iv) follows. (v) This is a direct consequence of Lemma 6.3 (ii). W Our next aim is to generalize Corollary 6.2 by dropping the assumption that S is abelian. The main results presented in the rest of this section are due to Sibley and Feit (see Isaacs (1976)). Our exposition is based on the presentation due to Isaacs (1976).

Lemma 8.8. Let H be a subgroup of G , let X C I r r ( H ) and let f : Z(X)"+ Z ( I r r ( G ) ) " be a linear isometry. Assume that A , B C X with A n B = 8 and both A and B am f -coherent. Let a and /3 be isornetries on %(A) and Z ( B ) ,respectively, whem a and p am QS implied by f-coherence.

Then (i) < a(x),P(A) >= 0 f o r all x E A , A E B . (ii) Assume that xo E A and A0 E B are such that

Then A U B is f -coherent.

6 Coherence

1169

x,

Proof. Givenp17p2 E Put A(Pl,P2) = p1(1)p2-p2(1)p1 E q X ) " . Choose 6 ,=~f l such that 6a(x) and E , B ( X ) E I r r ( G ) for all x E A , X E B. Assume by way of contradiction that < a(x),p(A)># 0 for some x E A , X E B . Then

ba(x) = y = &/?(A)

for some y E I r r ( G )

Choosex1 E A-{x} and A1 E B-{A}, and put II,= 6cr(x1) and cp = &/?(XI). Then II,# y # cp and we have

0 = < A(X,Xl),A(X,Al) > = < f (A(x, X d ) , f ( A ( k X l ) ) > = E 6 < x ( W - Xl(1)7,X(l)V - X l ( 1 ) Y > = E6(Xl(Vl(l) X(1)W) < $79 >>

+

contrary t o the fact that < $ , y >> 0. This proves (i). To prove (ii), assume that f(A(x0,Xo)) = x o ( l ) / ? ( X ~ ) Xo(l)a(xo). Define * on A U B by x* = a(x) for x E A and A* = /?(A) for X E B and extend * by linearity to Q ( A U B ) . Then, by (i) and the fact that a and p are isometries, we see that * is a linear isometry mapping Z ( A U B ) into Z ( I r r ( G ) ) . We are therefore left to verify that * agrees with the linear extension of f t o Q ( A U B)". But * agrees with f on Q ( A ) " , Q ( B ) " and A(x0,Xo). Because these span Q ( A U B)" over Q , the result follows. Lemma 6.7. Let p be an odd prime, let a Sylow p-subgroup P of G be tamely embedded in G and let P be nonabelian. Let H = N G ( P ) and let 1 # Z E Z ( P ) with 2 a H . Then the set

X = {x E Irr(H)IZ

Kerx}

is coherent.

Proof. P u t n = ( H : P).Since P is nonabelian, it follows from Theorem 1.2 (iii) that n is odd. Since p is also odd, we deduce that IHI is odd. Note also that, since Z # 1, X # 0. Let $ E X have minimum possible degree and let x 1

=

{x E XlX(1) = W))

Because IHI is odd, 4 # $, and so IX,( 2 2. By Lemma 36.2.6, coherent. Now define sets Xi for i > 1 by

xi = xi-1 u {Xi}

X1

is

F’robenius Groups

1170

where the

x; are chosen from X - XI XI

so that x i ( 1 ) 5 x ; + l ( l ) . Thus

c x2 &

Xt =

x

We use induction on i to show that X;is coherent. First we note that the degrees of the x E X are of the form np’. Thus + ( l ) l x ( l )for all x E X. Therefore, by Corollary 36.2.11, in order to prove that Xi,i 2 2, is coherent, it suffices to show that

To this end, observe that

PI =

c

x(1)2 = ( H : 2 )t

xElrr(H)

cX(q2

XEX

and therefore ( P : 2 ) divides C x E X x( 1 ) 2 . Consequently, n 2 ( P : 2) divides

~(1)’ XEX

x E X, then x

= p H for some p E I r r ( P ) and, by Theorem 21.2.3 (iii), ~ ( 1 5) ( ~P : 2). Therefore ~ ( 1 =) n2p(1)2 ~ divides n 2 ( P : 2 ) and so

If

~ ; ( l divides ) ~

~ ( 1 ) ~ XEX

Since, for all j 2 i, ~ i ( 1divides ) ~ ~ j ( 1we ) ~deduce that ~ ; ( 1 ) divides ~

~ ( 1 ) ~ XEXi-1

In particular, because 2$(1)

< p + ( 1 ) 5 x ; ( l ) , we finally conclude that

2$(1)Xi(l)

< Xi(U2 L

c X(V

XEXI-1

thus completing the proof. I

Theorem 6.8. Let p be an odd prime, let a Sylow p-subgroup P of G be tamely embedded in G and let H = N G ( P ) , X = { x E I r r ( H ) ( P K e r x } . Then one of the following occurs :

6 Coherence

1171

(i) 1 x1 = 1, P is elementary abelian and [PI = ( H : P )t 1. (ii) X is coherent. Proof. If P is abelian, then the result follows by virtue of Corollary 6.2. Suppose that P is nonabelian and put n = ( H : P). By Theorem 1.2 (iii), IH( is odd. Put Z = P' n Z ( P ) (hence 2 # 1) and let

Then, by Lemma 6.7, A is coherent. Furthermore, by definition, every x E B satisfies x(1) = n. Because x # R for x E B,the set B is coherent by Lemma 36.2.6. Since 2 P', we also have A n B = 8. For the sake of clarity, we divide the rest of the proof into a number of steps. Step 1. Let * : Z(B)+ Z ( I r r ( G ) )be a linear isometry which extends induction on Z(B)O, and let T be a linear isometry on Z ( A ) which extends induction on Z(A)O. Because A n B = 8, it follows from Lemma 6.6 that < x*,X' >= 0 for all x E B , X E A , where AT = .(A), x* = *(x). P u t p = ( l / n ) CXEA X(1)X. Because nlX(1) for all X E A , p is a character of H. If x E B,then x* # fX' for any X E A and by Lemma 6.3 (i),

where < B,,X >= 0 for all X E A . Now the degrees of characters in B are all equal. Hence, by Lemma 6.3 (ii), (X*)H- x = 8, - x T X P

+

is independent of of x E B and

x E B . Therefore 8, - x = A and rX= T are independent (X*)H=

Step 2.

x +A +TP

(1)

We now show that

x*(I) - x*(z) = TIPI for all

t

E 2 - {I}

(2)

For any x E B, Z P' C Iierx and 2 is contained in the kernel of every irreducible constituent of A. Therefore, for any z E 2 , x(z) = x(1) and A(z) = A(1). Hence, by ( l ) , we have

Frobenius Groups

1172

Note also that p~ = P H / Z t np, where p~ and Thus,for any 1 # t E 2, we have

pH/Z

are regular characters.

and therefore

x*(l) - x * ( z ) = r l H l / n = TIP), proving (2). Step 3. Here we prove that 1" E Z . First, we compare x*(l) and x * ( z ) in a different way. Let CO = {l}, C1,. . . be the conjugacy classes of G numbered so that Ci n Z # 0 if and only if i 5 t and C; n P # 0 if and only if i 5 s. Because P is a trivial intersection set with H = N G ( P ) , C; n P is a conjugacy class of H for i 5 s and C; n P C_ 2 d H for i 5 t . Let C: E C G be the sum of all elements in Ci and write

c+c+ 3 =

(aijk E

aijkC,+

)

k

Given i , j 5 t , II, E I w ( G ) , write o(Ck) = ? , b ( Z ) I C k l / $ ( l ) for z E Ck. If R denotes the ring of algebraic integers, then W ( c k ) E R. Now write $(1) = mpa with ( p , m )= 1, and put q = (Pl/pa. We have w(c:)W(cf) =

a;jkw(Ck+) k

Let us show next that w(C,') E qR if k > s. Indeed, we know that CknP = 8 whence IPI divides Ickl and so m(w(C,')/q) E R . Because we also have q(w(Ck+)/q)E R and ( m , q )= 1, it follows that w ( C t ) / q E R. Hence 9

C

w(C~)W(C G ~ ) a;jkw(Ct)(modqR) k=O

Suppose that t

< k 5 s and 5 E ck n P.

Put C = C p ( z ) and let

R = {(u,v)Iu E c;,vE c j , u v = z} Then = U ; j k and C acts on via (u,v)'= ( u c , v c ) ,uc = C - ~ Z L C . If 1 # c E C and ( u , ~ ) ' = ( u , v ) , then u , v E CG(C)C P and therefore u E ci n P & Z and similarly v E 2. But uv = z E c k and Ck r l Z = 8, a contradiction. Thus all orbits of C on R have size ICl and hence ICI divides aijk.

6 Coherence

1173

Because CG(Z)E P , we have CG(Z) = C and ( P : C) divides ICkl. Hence m(u;jkw(Ct)/q)E R. Because q(u;jkw(Ck+)/q) E R , we have aijkw(C;) E qR and

=

W(C;')W(Cj')

c t

uijkw(Ck+)(modyR)

k=O

Assume that X E I r r ( G ) is such that X is constant on 2 - (1). Then all w(C,') are equal to w , say, for 1 5 k 5 t and w(C0) = 1. We claim that w

TOprove (3), write

Uij

=

= IGl/lPI modqR

c;=,

a i j k , SO

w2 = uijo

(3)

that

+ uijw mod qR

Because [HI is odd, the nonidentity elements of 2 are not conjugate in H (hence not in G either) t o their inverses. Hence a110 = 0. Suppose that C2 is the conjugacy class of inverses of C1 so that a120 = lCll = lGl/lPl. Then ~

1 G o2 1 ~

lGl/lPl+ ~ l z o ( m o d q R )

(4)

Taking X = l ~we, have w = lGl/lPl and q = [PIso that

which proves (3). We apply (3) to the character X = E X * E Irr(G) with E E {l,-1} and x E B . Observe that by (2), X is constant on 2 - ( 1 ) and so (3) applies. Given z E 2 - {l},we have

x*(z)(G: P)/x*(l)= o = (G : P)(modqR) Because [PI divides q x * ( l ) ,it follows that

x * ( z ) ( G: P ) By (2), we have

X*(l)(G: P ) (mod IPIR)

Frobenius Groups

1174

and thus (G : P ) T E R. Because lPlr = x*(l) - x * ( z ) E R and because ( [ P I(G , : P)) = 1, we deduce that r E R n Q = 23 , as required. Step 4. Here we show that the set B U A is coherent. Let x , x I E B and 8 E A. Observe that x(1) = TZ divides e(1) and we put c = e(l)/n. Setting Q = c x - 0 E Z(X>', we have

< ( P ~ , x>=< ; ( P , ( x ; ) H >=< V,XI t A t ~p > by virtue of (1). Because < x,p >= 0 =< B,xl + A >, we have < vG,x;>= c < x,xl t A > - T < e,p > But < 0 , p >= c by definition of p and since T E Z 3 , we have c] < pG,x;>. It follows also that for x # XI,

< (PG,X* >= ct < ( P G , x ; > This implies that

where b E Z and Note also that

< y,X* >= 0 for all X E B . < qG,@ >=< (P,q >= 1 t c2

which shows that we have to examine two cases : (i) b = 0 and < y , y >= 1. (ii) b = -1, (BI = 2 and < y,y >= 1. In case (ii), we may replace * by ** where x;' = -x; and x;* = -x; where B = {x1,x2}. With this change, we are in case (i). Hence we may assume that (CX - e)G = CX* - y. Given 81 E A - {O}, we have

e1(q

= = =

< v,e(i)e1- el(i)e >

< tpG,e(i)e; - el(i)eT> < y,el(i)eT - e(i)e; >

which shows that, again we have two possibilities : (i) y = 8' (ii) y = -6; and e(1) = & ( l )

6 Coherence

1175

If 7 # P, then 0; = -7 for all O1 E A - (6) and so IAl = 2 and A = {O,O1} with e(1) = &(l).In this situation, we can redefine T and so we may assume that (i) holds. Hence (cx - e) G = -eT and therefore, by Lemma 6.6 (ii), B U A is coherent. Step 5. Completion of the proof. We first note that

C e(q2=

- ( H : z)

&A

and so is divisible by ( P : 2 ) and therefore by n 2 ( P : 2 ) . Next choose any $ E X - ( A U B ) . Then, as in Lemma 6.7, we have $ = PH for some p E I r r ( P ) and p(1)25 ( P : 2). Therefore

+(112 = n2p(1125 n 2 ( P : Z ) 5

C~ ( 1 ) ~ PEA

Let

xEB

p(1) 2

so that p . Then

x( 1) = n and $(1) 2 px( 1) > 2x( 1) since 1c, # B 2X(l)W)

< $(V I

and so

c m2

pEAUB

Because A u B is coherent and x(l)l$(l) for all $ E X ,repeated application of Corollary 36.2.11 shows that X is coherent. This completes the proof of the theorem. I

2 G, = 1 is called a A chain of subgroups G = Go 2 G1 2 Gz 2 normal series of G provided each G; 4 Gi-1, 1 5 i 5 n. A normal series G;, 0 5 i 5 n, of a group G is called a chief series provided each Gi is a proper subgroup of Gi-1, 1 5 i 5 n, chosen maximal subject to being normal in G. The factor groups Gi-l/Gi, 1 5 i 5 n, are called the chief factors of the series. It is clear that in a solvable group the factors of every chief series are elementary abelian of prime power order. Theorem 6.9. Let a subgroup S of a group G be tamely embedded in G , let H = N G ( S ) and let X = {x E I r r ( H ) I S Iierx}. Then one of the

following occurs : (i) 1x1 = 1, S is an elementary abelian p-group for some prime p and ( H : S) = IS1 - 1. (ii) S is a nonabelian Sylow 2-subgmup of G with (S : S’) < 4 ( H : S ) 2 . (iii) X is coherent.

Frobenius Groups

1176

Proof. If S is abelian, then the result follows by virtue of Corollary 6.2. From now on, we assume that S is nonabelian and put n = ( H : S). By Theorem 1.2 (iii), n is odd. Assume that X has precisely one character x with S' E. K e r x . Then all nonprincipal linear characters of S form a regular HIS-orbit. Hence (S : S') - 1 = n and S/S' is an elementary abelian pgroup. Since S is nilpotent, S is a pgroup and, since n -t 1 is even, p = 2 which shows that (ii) holds. By the foregoing, we may assume that I{x E XIS' 2 IierX}I 2 2. Hence, by Lemma 36.2.6, this set is coherent. Let N Q H with N S' minimal such that Y = { x E XIN C ILerx} is coherent. Assume that X is not coherent so that N # 1. Let M a H be such that N / M is a chief factor of H . Because

is not coherent, it follows from repeated application of Corollary 36.2.11 that there exists I) E Irr(H) such that

2n11,(1) 5

c

x(1)2

= ( H : N ) - ( H : S)

XEY

= n ( ( S : N ) - 1) and M K e r 11. Now define 2 by Z / M = Z ( S / M ) . Then ZnN 2 M and so N 2 2. Observe also that 11, = aH for some cy E I T T ( S )and ~ ( 15)(S ~:2) so that $ ~ ( 15) ~ n2(S : 2). Hence

4n4(S : 2 ) 2 4 ~ ~ ~ 1 1 ,2( 1n2((S ) ~ : N) - 1)2 Now put k = (S : N ) and t = (2 : N ) . Then t ( S : 2 ) = (S : N ) = k and

4n2k = 4n2t( s : 2 ) 2 t ( ( S : N ) - 1)2 = t ( k - 1)2 If 4n2 5 i ( k - 2), this would yield t k ( k - 2) 2 t ( k - 1)2, which is not the case. Hence t ( k - 2) < 4n2. We now claim that if 2 = N , then (ii) holds. Indeed, if 2 = N , then Z ( S / M ) = N / M is a p-group for some prime p and so S / M is a p-group since S is nilpotent. Because M S', it follows that S is a p-group. By Theorem 6.8, we have p = 2. Observe also that k - 2 < 4n2 so that ( S : S')5 ( S : N ) = k

5 4n2 t 1

6 Coherence

1177

But, since (5’ : S’) is a power of 2 and n > 1 is odd, we have (S : 5’‘) < 4n2 and (ii) holds, as claimed. Finally, assume that 2 2 N . Then n l ( ( 2 : N ) - 1) and t 2 n 1. Observe also that S / M is nonabelian and hence 2 c S and nl((S : 2 ) - 1). Therefore k 2 ( n 1)t 2 ( n 112 2 n2 2

+

+

+

+

+

and so 4n2 > t ( k - 2) 2 ( n l)n2 2 4n2 , since n 2 3. This gives a desired contradiction and so the proof is complete. H

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Chapter 38

Applications of Characters This chapter is devoted exclusively to applications of characters. Our applications are confined to two areas : group theory and integral group rings. Among other results, we prove the famous solvability criterion of Burnside and some classical theorems of Wielandt pertaining to existence and uniqueness of relative normal complements. We then establish a theorem due to Brauer and Suzuki which shows that if G is a group with a generalized ) z is the inquaternion Sylow 2-subgroup S , then G = 0 2 t ( G ) C ~ ( zwhere volution of s. In particular, any such group G cannot be simple. The proof of the most difficult case where IS1 = 8 is based on a work of Glauberman (1974). As a final group-theoretic application, we provide a characterization of PSL2(q), q odd, due to Brauer, Suzuki and Wall (1958). The second part of the chapter is devoted exclusively to the study of U ( Z G ) ,where G is a finite abelian group. Special attention is drawn to the problem of effective construction of units of !&'which is of fundamental importance in algebraic topology (see Milnor (1966, Theorem 12.8 and Corollary 12.9)) and unitary K-theory (see Novikov (1970, p.502))

1

Burnside's paqb theorem

In this section, we prove the famous solvability criterion of Burnside. All groups below are assumed to be finite and all characters are C-characters. The essence of the argument is contained in the following result.

Theorem 1.1. (Burnside (1904)). Let p be a prime and let a group G have a conjugacy class C with ICl = p" f o r some n 2 1. Then G is not 1179

Applications of Characters

1180

simple.

Proof. Assume by way of contradiction that G is simple. Fix g E C and let x be a nonprincipal irreducible character of G. Since G is simple, K e r x = 1. Since ICI > 1, G is nonabelian and so Z ( x ) = Z(G) = 1. Let I r r ( G ) be the set of all irreducible characters of G, let

x = {x f W G ) I P 1 x(%,

y = {A E WG)lPIX(1)}, and let p be the regular character of G. If x E X ,then x ( g ) = 0 by virtue of Theorem 21.2.3 (v). Hence

Denoting by R the ring of algebraic integers of C , it follows that -1 = p a , where

This contradiction completes the proof of the theorem.

Theorem 1.2. (Burnside (1904)). Let G be a group of order paqb where p and q Q W primes. Then G is solvable.

Proof. We argue by induction on (GI. We may assume that G is nonabelian and that G has a nonidentity Sylow q-subgroup Q. Then there exists 1 # g E Z ( Q ) and if C is the conjugacy class of G containing g , then (C(= (G : C&)) divides pa. If (C( = 1, then 1 # g E Z(G). Hence Z(G) is a nontrivial normal subgroup of G. If ( C (# 1 , then by Theorem 1.1 G has a nontrivial normal subgroup. Thus, in all cases, G has a nontrivial normal subgroup N . Since both N and GIN satisfy the hypothesis of the theorem, it follows by induction that both N and GIN are solvable. Thus G is solvable and the result follows. W

2

Wielandt's theorems

This section provides some classical results of Wielandt (1958). Our first aim is to prove a generalized version of the Frobenius theorem which guarantees the existence of the F'robenius kernel in Frobenius groups (see Theorem 36.1.8). All groups considered below are finite and all characters are (c -characters.

2 Wielandt's theorems

1181

Lemma 2.1. Let H o d H be subgroups of a group G and let S be a subset of G . Assume that every irreducible character of H with Ho C K e r x can be extended to a character X of G with S K e r A. Then there exists N d G such that S C N and H n N = H o .

x

Proof. As usual, we identify the characters of H / H o with characters of H having HO in their kernels. Let p be the regular character of H / H o . Then, by hypothesis, p can be extended to a character p of G with S C I i e r p . Setting N = Ii'erp, it follows that N a G, S C N and H n N = Ho, as required. I Let H be a subgroup of a group G and let H o d H . Recall that a subgroup Go of G is said to be a normal complement of H over Ho if Go d G, G = GoH and Ho = Go n H .

Theorem 2.2. G such that

(Wielandt (1958)). Let Ho -t 2 C xi(za> zEB

zEBi

and therefore

GIBI = xi@) t 2

c

x(za>

zEBi

Since IBI and E ; are odd and Xi(za) E from (8),we have 2

G

IBI =

Z ,we see that x;(a)is odd. Similarly,

CX;(Z~) ZEB

3 Generalized quaternion Sylow subgroups

1193

which shows that < x?,XG> is odd. Assume that there exists i E {1,2,3} with x ; ( a ) # ~ i .Then, since x i ( a ) is an odd integer, Ix;(a)l > 1 or x;(a) = - E ; . In any case, we have x?(a) > E;x;(u). Since the character x;(i = 1,2,3) assumes only integral values, we have x:((y) 2 E i x j ( y ) for all y E G. Thus, applying (8),we obtain

=

C xi(za>)

Ei(IBI-*

zEB

< Xi,XG >

=

E;

=

EiE;

= 1,

i.e. < x?,XG >> 1. Since < xf,XG > is odd, we have < x?,XG > 2 3. Because < x i , lc >= 0, we obtain CyEGxi(y) = 0 and therefore

Bearing in mind that if if and that x?(1)

Thus

> -&;x;(l), we deduce that

ED^ yEG-DG

Applications of Characters

1194

a contradiction. Thus xi(a) = E; for all i E {1,2,3). Note further that for any i E {1,2,3},

YES

= n; t x; t 6xi(a) = n; +xi t 6 ~ ;

+

and therefore nj zj 4- 6 ~ =i O(mod8). Since = n; - y; and 6 ~ ; -2ej(mod8), we have 2( n; - E ; )

- y;

O( mod 8)

As we noted above, 41~;. Therefore nj - E ; is even and n; is odd. Moreover, if y; is divisible by 8, then n; - E* E O(mod4), i = 1,2,3. Let 2k be the highest power of 2 which divides each of the numbers y1, y2, y3, k 2 2. To reach the final contradiction, we put m; = ~ ; / 2 i~=, 1,2,3. Then, by (7), we have

o = ~ 1 n 2 n 32mt~~2n1n3mit ~3n1n2m:

(10)

Therefore, since E ; , ni(i = 1,2,3) are odd, we may harmlessly assume that ml is even (and hence 81yl) and m2,m3 are odd. Invoking (lo), we deduce the following congruences modulo 4 :

+

Thus ~ 2 n 2 &3n2= O(rnod4), But, by (3),

3 Generalized quaternion Sylow subgroups

1195

+

and so 41(nl ~ 1 ) .It follows from the fact that 81yl that 4)(nl - ~ 1 ) .But then 4 divides (n1 -t ~ 1 -) (n1 - ~ 1 =) 2 ~ 1 a, contradiction. Step 3. We now complete the proof by treating the easier case where IS[ 2 16. In this case

S =< a,bla2n = 1, b2 =

b-'ab = a-1

>

(n 1 3 )

Setting Q =< a2"-* >, we see that Q is a subgroup of S of order 4. It is clear that CG(Q)= B < a > where B is a normal subgroup of CG(Q)of odd order. Setting N = N G ( Q ) ,it follows from

that N = BS with B a N . It is clear that ( B x where x is the involution of 5'. Now put

D =B

< 2 >)aN and B < a > Q N ,

-Bx < x >

Then D is an invariant subset of N. Let g E G and D n g-'Dg # 8. Then g-' < dl > g =< d2 > for some dl,d2 E D. Because the Sylow 2-subgroups of < dl > and < d2 > contain Q ,we have g E N . This shows that D is a trivial intersection set in G and N = N c ( D ) . Moreover, by Example 36.3.1, D is an exceptional subset in G . Now any element of N can be written in the form ta%' with t E B and b , s integers. Let A1

:N

+ (I: *

and

X2

: N + GL2((I:)

be given by

Xl(takba) = (-1)'

where E is a primitive 2n-1-th root of unity. It is clear that A 1 and X 2 are (I: -representations of N . Of course, X1 has degree 1 and so A1 is irreducible. We claim that is also irreducible. Indeed, if not, then X 2 is a sum of one-dimensional representations and so & ( N ) is abelian. But, since 6 # E - ~ and K e r A2 = B < z >, Az(a)Az(b) # A2(b)X2(a),a contradiction. Thus A 2 is an irreducible representation.

Applications of Characters

1196

Consider the generalized character a = 1~ -t x 1 - x 2 of N , where x; is the character of Xi, i = 1,2. By the definitions of D and N, each element from N - D is of the form t , t x or ta'b, where t E B. This easily implies that cr vanishes on N - D. Note further that, in view of the structure of S , no involution inverts an element of D. Hence, by Lemma 36.3.4, all involutions in G generate a proper normal subgroup of G. Thus G is not simple and the result is established.

4

The Brauer-Suzuki-Wall theorem

We begin by introducing the following terminology. Given a prime power q, let GL,(q) be the group of all 2 x 2 nonsingular matrices with entries in the finite field of q elements. This group with certain closely related groups plays an important role in group theory. The group GL2(q) is known as the general linear group (in dimension 2); its subgroup SL2(q) consisting of matrices of determinant 1 is c d e d the special linear g r o u p . The centre of GL2(q) consists of scalar matrices and the corresponding factor group PGL2(q) is called the projective linear g r o u p . Finally, the image PSL,(q) of SL2(q) in PGL2(q) is called the projective special linear group Our aim is to provide a characterization of PSLz(q) for q odd. This characterization constitutes an important part of the classical paper of Brauer, Suzuki and Wall (1958). By combining this characterization with some other results pertaining to the case where q is a power of 2, Brauer, Suzuki and Wall proved the following assertion. Let G be a group of even order having no subgroups of index 2. Assume that if A , B are cyclic subgroups of the same even order and A n B # 1, then A = B. Then either G E P S L 2 ( q ) for some q 1 3 or the Sylow 2-subgroup of G is normal and elementary abelian. A simplified version of the characterization of PSL2(q) for q odd was given by Bender (1974). Our presentation is based on Bender's approach.

.

T h e o r e m 4.1. (Brauer, Suzuki and Wall (1958)). Let t be an involution of a finite group G , let H = &(t) and let A be an abelian subgroup of H containing t and satisfying the following properties : (i) H = A < s > with an involution s # A . (ii) CA(S)=< t > and s-'as = a-* for aEI a E A . (iii) A n g-'Ag = 1 for all g E G - H .

4 The Brauer-Suzuki-Wall theorem

1197

(iv) All involutions in G are conjugate. Then G S P S L 2 ( q ) for some odd prime power q.

Proof. We wish to show that G has a subgroup Q of odd order q such that the following properties hold : ( 4 IGI = d!l4- l ) ( q (b) CG(X) = Q for all x E Q - (1). (c) NG(Q)= QB with an abelian subgroup B of order (l/2)(q - 1). (d) If B # 1, then for any subgroup B1 # 1 of B , Nc(B1) = N c ( B ) = B < u > with an involution u $ B inverting B . At the end of the proof, by appealing to a result of Zassenhaus (1936), we will show that this will complete the proof. In what follows, Steps 1-10 treat exclusively the most difficult case where IAl > 4. Step 1. Since A is abelian, it has exactly IAI irreducible characters which are all linear. Furthermore, each linear character of A takes values 1 and -1 on the involution t . Since the group A / < t > has precisely IAl/2 linear characters, we see that A has ]A1/2 linear characters x with X(t) = 1 and the remaining IA)/2linear characters of A satisfy ~ ( t=) -1. Given a character x of A , define xs by ~ " ( a=) x ( s - ' a s ) for all a E A . Then, if x E I r r ( A ) , then x" = x if and only if x ( a 2 )= 1 for all a E A since by hypothesis s-las = a-l for all a E A. Setting A0 =< a E Ala2 = 1 >, we see that x" = x if and only if A0 K e r x . By (i) and (ii), t is a unique involution in A . Hence the Sylow 2-subgroup of A is cyclic and so ( A : Ao) = 2. In particular, A has precisely two irreducible characters containing A0 in their kernel. Therefore A has precisely two irreducible characters x with x = x". Because, by hypothesis, IA1/2 2 3 we deduce that there exist linear characters X and p of A with X # A", p # p", X ( t ) = 1 and p ( t ) = -1. Step 2. Here we examine the induced characters A H and p H . Firts of all, by definition, we have

w.

X(h)+ X(h-')

if if

hE A hEH-A

p(h)t p(h-')

if if

hEA hEH-A

In particular, XH(t )= 2, p H ( t )= -2 and so AH # p H . Furthermore, since X # As and p # p", the induced characters AH and p H are irreducible.

Applications of Characters

1198

Step 3. Put D = A - { l}, a1 = ( 1 ~ -) AH~ and a 2 = AH - p H . Then it is easy to see that D is an exceptional subset of G, H = N G ( D ) and a l ,a2 vanish on H - D . Our aim is to show that

where xi = E B ~i, = 1,2,3, E E (1, -1) and 81,82,83 are distinct nonprincipal irreducible characters of G . First of all, we note that

(lA>H(x) =

[

2 0

if if

xEA X E H - A

and therefore

~ >= , 1. Hence ( 1 ~ =) 1~ ~ and < ( l ~ )1~ character of H and

+ 7,where 7 is an irreeducible

It now follows that q ( 1 ) = 0, q ( t ) = 0, 4 1 ) = 0, a2(t) = 4, y ( t ) = 1 7 ( s )= 7 ( s t ) = -1, P ( t ) = 2, P ( s ) = A H ( s t ) = 0 pH(t) = -2, pH(s) = pH(st) = 0 ctl

= lHt Y - x ~ ,a2 = X~ - p H

where l ~7, ,AH and p H are distinct irreducible characters of H . We also have

< a 2 , a 2 >= 2, < a1,CYz >= -1

4 The Brauer-Suzuki-Wall theorem

1199

Applying Frobenius reciprocity together with Theorem 36.1.7, we have a1G (1) = a f ( 1 ) = 0,

G "1 ( t ) =

a&) = 0

G

( t ) = a2(t)= 4, < l& >= 1, < lc,af >= 0 < ay,a:: >= 3, < "2G ,a2G >-- 2, < aF,af >= -1, "2

which proves (1). Step 4. Here we demonstrate that

X3(q2

-Xl(q2

=0

(3)

To this end, we first apply Suzuki's formula (Theorem 36.3.3) with a = al, to obtain

Note that O;(t)E = x;(t), i = 1,2; p ( I + ) = B P ( t ) , since I = tG and 111 = (G : H ) ( p is any character of G). For any character T of H , we have T ( ( I ~H ) + ) =

~ ( tt) +s) IAl

t TI4~ ( ~ t )

since I r l H consists of three H-conjugacy classes of involutions in H , with representatives t , s and st. The sizes of these classes are 1, respectively. Hence, substituting the values of characters, we obtain

y,y,

Applications of Characters

1200

proving (2). Similarly, applying Suzuki's formula for Q = a2 and taking into account that x3(1) = xl(1) (since af(1) = 0), we obtain (3). Step 5. Our aim is to show that there exists 6 E {-1, l} such that

(G : H ) = (21-41

+ S)(lAl t 6)

To this end, we first note that from a f ( t ) = xs(t) have xs(t) = 2 and xl(t) = -2. Because

- xl(t) = 4 and

(4) (3), we

= 1 t x d t ) - X2(t) = 0,

we have xz(t) = -1. Hence (2) can be rewritten in the form 21A12Xl(l)X2(1)= (G : H)m

(5)

where

m = X1(1)Xz(1) t 4X2(1) - Xl(1)

(6) Because ay(1) = 1 t xl(1) - xz(1) = 0, we see that xl(1) and xz(1) are coprime. Thus (m,xz(l)) = 1 and so, by ( 5 ) , xz(1) divides (G : H). Consequently, ~ 4 1 is) odd and so 81rn by ( 5 ) . Since 81m and x2(1) is odd, we see that x1(1)(~2(1)- 1) is not divisible by 8. Hence xl(1) is not divisible by 4. The equality 1 t xl(1) - xz(1) = 0 implies that xl(1) is even. Therefore, applying (6) and the fact that (xl(l),Xz(l)) = 1, we see that (xl(l),rn) = 2. By (iv), H contains a Sylow 2-subgroup of G, so (G : H ) is odd. Furthermore, by (iii) and (iv), H is a Hall subgroup of G. Since A H , we therefore deduce that (21AI2, (G : H)) = 1. Applying ( 5 ) , we see that

(G : H ) = (1/2)Xl(l)X2(1)

(7)

and, since xl(1) = x ~ ( 1-) 1, 4[AI2 = m = xl(l)(x2(1) - 1) t 4 x ~ ( 1 ) = (Xz(1) - 1)' 4Xz(1)

+

= (XZ(1) t

Therefore xz(1) t 1 = -21A16, xl(1) = ~ 4 1 -) 1 = -21A16 - 2 for some 6 E {-1,1), which implies (4)by virtue of (7). Step 6. Setting q = 2lAl 6, it follows from ]HI = 21AI and (4) that

+

IGl = 4(4 t l)(q - 1)/2

4 The Brauer-Suzuki-Wall theorem

1201

The rest of the proof is based on group-theoretic arguments. Because H A tG, we have H G = AG and AG contains the centralizer of any of its nonidentity element. Let p ( z ) be defined by p ( 5 ) = I{(u,?I)Iuv= 5 , o ( u ) = .(?I)

= 2}1

Then p ( z ) = 1{u E Glu #

5, O(U)

= 2,

U-'ZU

= x-'}/

and p(1) = ItGI = (G : H ) . If a E A - {l},then {U

E

G~u-'uu= a-', u #

U,

O(U) =

2) = A3

and therefore p(u) = IAl. Now choose y E G - H G with maximal p ( y ) . Then

( { ( a , b ) ( ~E bG - H G , O ( U ) = ~ ( b=) 2}( JG- H G J (G : H ) 2 - I{(a,b)lubE A', O ( U ) = ~ ( b =) 2}1 IGI - lAGl (G : H ) 2 ((G : H ) lAl(lA1- 1)(G : H)) [GI - 1 - ( [ A ]- 1)(G : H ) (G : H ) - 1 - lAl(lAl - 1) > IHI - (IAI - 1) - ( 2 1 4 t S)(lAl t 6) - 1 - lAl(lAl - 1) 2 1 4 - (I4 - 1) -

+

Since p(y) > 0, then by (iv) we may assume that t-lyt = y-'. Step 7 . P u t K = Cc(y), M = N ~ ( l i )k, = IKI,n = ( M : K ) and Mo the set of all involutions in M . It is clear that t E Mo. Because K fl H G = 1, we have (IKI,[ H I )= 1. In particular, k is odd and, for u E Mo, CK(U) = 1. Hence any element of K is a commutator of the form X - ~ U ~with U 2 E K and so u inverts K. It follows that K is an abelian group and for any a E K - (1) and any involution u of Mo, I< = c G ( a ) , U 'a?/,= (8)

Applications of Characters

1202

It follows from (8) that Ii < t > d M , MO = K t , M = KCn,l(t) and C,v(t) contains precisely one involution t . Since t E M n A and As consists of involutions, we have C M ( ~ =)M n H = M n ( A U As) = M n A . Therefore M is the semidirect product

Because ( ( I { ( ,IAI) = 1, it follows from (8) and (9) that

Step 8 . The set of all involutions inverting y coincides with K t . Therefore, by Step 6,

and

Now consider the set

L = Uv&t4oMCG(V) We wish to find the number of cosets of M and the number of involutions in L. If w E Mo, then M = KCM(V),K 4 M , K n C M ( V = ) 1, and IC,v(v)l = n. Because v E t M and each subgroup of A is normal in Cc(t), we have Cn,l(w) d CG(D).Hence, applying (lo), we have cM(v)

= M n g - ' ~ g for a~ g E CG(V) - CM(V)

Moreover, for u,w E

Mo,u # w, MCG(Un ) MCc(v)= M ,

since M

# Mz = M g ( z E c G ( U ) , g E C G ( V )implies )

and u = v since Cn,l(u)contains only one involution. It follows from the above that for all v E Mo,

4 The Brauer-Suzuki-Wall theorem

1203

and L consists of

cosets with respect to M (since lM0l = \.lit1 = k). The set C G ( ~-)C M ( ~has ) precisely IAl involutions (these are the elements of As). Hence L - M contains precisely klAl involutions and L contains precisely kIAI k involutions. Step 9. Here we show the existence of Q satisfying (a) - (d) under the assumption that L = G. So assume that L = G. Then

+

(G : H ) = k(IAI t 1) (13) Invoking (4) and (13), we see that [ A ] 1 divides (21AI t & ) ( [ A [ 6). Hence, taking into account that the even integer 1AI is at least 6, we obtain :

+

6 = 1, k = 2lAl+ 1 = q

+

(14)

Because [MI = kn, IHI = 21AI it follows from (12), (13) and (14) that n = IAl. Now put Q = li and B = A. Then it is easy to verify that Q satisfies (a) - (d). Step 10. Here we show the existence of Q satisfying (a) - (d) under the assumption that L # G. By (lo), for any z E G - M ,the set {Mzglg E K } contains precisely k cosets modulo M . Moreover, if g E K ,u E Mo,we have MCG(u)g= MCG(g-'ug) L. Therefore G - L contains at least Ic cosets modulo M . Accordingly,

( G : M ) L ( w - k + l ) + k = - 2klAl + 1 n n Now any coset M z E G- L contains at most one involution. Indeed, if v1,02 are distinct involutions in Ms,then 1 # I = 0102 E M and vi'tv1 = z-'. From (10) and (9) it follows that C M ( Z n) MO # 8. Hence either o ( z ) is even and q centralizes an involution of < z >or v1 inverts CG(Z)and therefore centralizes C M ( Zn) Mo.It follows easily from (10) and (9) that M z = Mvl C L , a contradiction. Thus the number of involutions in G - L is at most IG - LI/IMI. Therefore (G : H ) - k(lAl+ 1) I (G M ) - ILI/IMI

= ( G : M ) - ( Un - k + l )

Applications of Characters

1204

which implies

(-

(G: H ) - (G : M) 5 k(lAl+ 1) - 2k'A' - k + 1) n We now claim that

k = /A\ - 1, = 2 , 6 = -1

(17)

Indeed, if IM 5 IHI, then

This implies (17) by applying (11) and the fact that k is odd and 1AI is even. Assume that [MI> [ H I , i.e. l M l / l H l - 1 > 0. Then, by (15) and (16), we have n

(fi

- 1)


has [CG(Z)~ involutions inverting x. Thus

and so as I x < s > is a Sylow 2-subgroup of G. Setting N = N G ( H ) , we see that N = H < b > where o(b) = 3. We may assume that G # N (otherwise, for q = 3 conditions (a) - (d) are fulfilled). Let ~1,212E Hs be involutions. Then 01 centralizes ~1212and so 01 E H . Hence, if Ha: # H then Ha: contains at most one involution. Note also that H b U Hb2 has no involutions and that G contains precisely (1/4)IGI involutions. This easily implies that each Hs in G - N contains exactly one involution. Let u be an arbitrary involution. If H b u has no involutions, then Hbu N and u E N. On the other hand, if v E Hbu is an involution, then u inverts the element wu of order 3 and H b C N . Thus, for any involution u E G - N , 2 = u-'Nu n N has order 3 and is inverted by u. Now { u } = H u n N G ( Z ) , for otherwise N G ( Z )n H # 1. Hence

C G ( Z )= 2, N G ( Z ) = Z < u

>, INc(2)l= 6

Each subgroup of order 3 of N is inverted by precisely 3 involutions. Note also that N contains 4 subgroups of order 3. Therefore G - N contains precisely 12 involutions, while G contains 15 involutions. Thus 1GI = 15 !HI = 60. It is now clear that for q = 5 conditions (a) - (d) are fulfilled. Step 12. Here we treat the case IAl = 4. We apply the same argument as in the previous step. Put N = NG(< t > x < s >). Then one easily verifies that N 2 Sq. Assume that NG(< b >) C N , where b is an element of order 3 in N . Then counting the number of involutions in cosets of N , we see that [GI = 8 3 . 7 and that for q = 7 conditions (a) - (d) are fulfilled.

-

5 Applications to U ( Z G )

1207

Assume that N G ( < b >) $ N .lThen N G ( < b >) = C ~ ( b ) uwhere , u is an involution which inverts Cc(b). Put q = ICc(b)l,q 2 9. One easily verifies that N G ( < b >)x, x !$ N G ( < b >), contains at most two involutions. Denote by ,s the number of cosets NG( < b >)z containing precisely rn involutions. Since, by the structure of H , any involution of N G ( < b >) commutes with four involutions outside of N c ( < b >) and since the number of involutions in N G ( < b >) is q, we have s2 = 2q. Hence

ltGl = (1/8)(GI = q

+ 292 + si = 5q + si

(G : N G ( < b >)) = IG1/2q = 1 t ~2 t si t SO = 1 t 2q t s1 t so It follows that 0 5 s1 5 ( l / q - 4)q(19 - 2q). Thus q = 9, s1 = 0 and . the conditions (a) - (d) are fulfilled for Q = Cc(b). IGl = 8 9 ~ 5 Now Step 13. Completion of the proof. Here we use the following grouptheoretic fact due to Zassenhaus (1936). Let G be a 2-transitive permutation group of degree q t 1 which has no nonidentity permutation fixing 3 points. If the stabilizer B of two points is abelian of order (1/2)(q- 1)and is inverted by some involution in G, then G E PSL2(q) for some odd prime power q. Let Q and B satisfy (a) - (d). For each x E G , put Q" = z-lQz. Consider the permutation representation n of G given by

Then one easily verifies that n(G) Z G and that either q = 3, G Z PSL2(3) or n(G) satisfies the conditions of Zassenhaus mentioned above. This completes the proof of the theorem. W

5

Applications to U ( Z G )

Let G be a finite abelian group. A number of results of algebraic topology (see Milnor (1966, Theorem 12.8 and Corollary 12.9)) and unitary I 1, let and let g,,, =< E~ >.

ambe the set of all primitive m-th mots of 1 over Q

Applications of Characters

1210

If

a,, E E

am,are

integers such that (i) a, = a c - ] , (ii) x u c = 0 , and (iii)

for all linear CC -characters

x of g,,,,

n e(x(W

=1

,€Om

then all a, = 0. Proof. See Franz (1935). As a preliminary to our next result, it will be convenient to record the following notation : an is the set of all primitive n-th roots of 1 over Q . !I!a is the group of all n-th roots of 1 over Q

.

Q n = Q (En). Lemma 5.7. Let n = ps, where p is a prime, and let N = NQ ,/Q *. (i) Assume p [ s, 1 # E E !I!#,and 6 E !DP. Then N ( ~ - E )= ( l - ~ ) ~ - ' N(1- ~ 6 ) = (1 - E P ) ( ~ - c)-'

(ii) Assume PIS,

and let 1 #

N(1-

E)

=

{

E

E

Qn.

Then

(1-E)P

if

EE!Ps

1 - 9

if

E$!%

Proof. In both cases, the first formula is simply a question of degree :

For the proof of the second formula, we require the following identity : 1

-en

=

n

(1 -OX)

8

aroot of I

X€Qn

To prove (l),we set X = 8-1 in nxEQn(X - A) = X n- 1 to infer that

r n -1

=

n n

(e-I

-A)

A€'€",

=

&Qn

=

e-n

d-l(l-eX)

n (i-ex)

(1)

5 Applications to U ( Z G )

1211

Now multiply by en and we obtain ( 1 ) . Returning to the second formula of (i), we note that

( & 6 ) P = &P E Q s It follows that the conjugates of €6 are all &A, X E

N(1 -&a)

=

n

QP.

Therefore

(1 - E X )

A€@,

n

=

(1 - E X )

[A€..

1

(l-&)-l

= (1 - EP)( 1 - &)'I

using (1). In case (ii)) E P E Q and \vp C Q s. Hence, if E 4 Q s, the conjugates of E are all E X , X E \vp. Applying ( l ) ,we therefore deduce that

n

N(1-€) =

(1-EX)

A€O,

= 1-&P as required.

Lemma 5.8. Let R C S be commutatiwe rings such that S is integral over R . Then T E R is a unit of R if and only if T is a unit of S . Proof. If T E R is a unit of R, then obviously versely, assume that T € R is a unit of s. Say T-"

+ T ~ T - ( ~ - ' +)

6 .

+

T,

=0

T

(Ti

is a unit of S. ConE R)

exhibits the fact that r - l is integral over R. Multiplying this equation by P , we obtain T(TI T 2 f t * * * t TnTn-l) = -1,

+

thus completing the proof.

B. Torsion units. Throughout, G denotes a finite group and R a commutative ring. A unit u of RG is said to be trivial if u = rg for some T E U ( R ) ,g E G. Our aim

Applications of Characters

1212

here is to exhibit a class of rings R for which all central units of finite order in RG are trivial.

Lemma 5.0. Let E ~ , E z , . . , ~ tbe n-th roots of unity over Q and let a l , . . . ,at be positive rational numbers such that a1 - at = 1 . If

+ - +

a = al&l t ... t at&t is an algebraic integer, then either cr = 0 or E I = ~2 = - - - = € 1 .

Et.

Proof. Observe first that la1 5 1 and la1 = 1 if and only if ~1 = - = It therefore suffices to show that if a # 0, then la1 = 1. Assume by

way of contradiction that a # 0 and that la1 < 1. Let E be a primitive n-th root of unity over Q Then, for any u E G d ( Q ( E ) / Q ), In(cr)I < 1. Hence INQ (a)[< 1, contrary to the fact that the norm of cr is a nonzero integer. o the lemma is true.

(e)q

.

Lemma 5.10. Let (Y be an algebraic number and let n be a natural number such that ncr is an algebraic integer. If (a1 = a,a2,. , a t } is the set of all Q -conjugates of a , then either a is an algebraic integer or in the ring Z [ a l ,a2,.. . ,at]at least one prime divisor of n is a unit.

..

Proof. Let E;, 1 5 i 5 t , be the elementary symmetric function of t variables and degree i. If

f ( X )= X t t

t ' ' t at *

(a; E Q )

is the irreducible monic polynomial satisfied by a , then

Assume that a is not an algebraic integer. Then there exists i E { 1 , . . . ,t } such that E;(cr1,cr2, ...,at) = a / b for some integers a and b with 6 > 1. Hence, denoting by p a prime divisor of 6 , we see that a l p E Z [ q ,0 2 , . , , a t ] .Furthermore, p divides n because

.

E ; ( n a l ,na2,.

. . ,nat) = na(a/b) E z

Finally, since ( a , p ) = 1, there exist c,d E 23 such that

uc

+ dp = 1

5 Applications to U ( Z G )

Therefore

1213

1 1 a - = -(ac+ d p ) = c . - t d E P P P

Z[Cul,rrZ .,at] ,..

as desired. W The following result for the case where R is the ring of algebraic integers was established by G. Higman (1940).

Theorem 5.11. (Saksonov (1971)). Let R be an integral domain of characteristic 0 and let G be Q finite group. Assume that no prime dividing the order of G is a unit of R. If u = CgEG u9g is Q unit of finite order in RG with u1 # 0 , then u = u1 - 1. In particular, all central units of finite order in RG are trivial.

Proof. Let z = CgeG zgg E RG be a central unit of finite order. Then zt # 0 for some t E G and therefore u = zt-' is a unit of finite order with u1 = zt # 0. Hence it suffices to show that if urn = 1 and u1 # 0, then 21 = 211 * 1. To prove this assertion, let F be the quotient field of R and let p be the regular matrix representation of RG. Denote by x the character of p. Since urn = 1, p ( u ) is similar to diag(E1,. . . , E ~ ) ,n = IG(,where each E ; is an m-th root of 1 over F . Hence

x ( u ) = u1 - n= ~1

+. - . + E ~

By looking at the ~ ( u " )where , ( p , r n ) = 1, it is also clear that the set {Pl = U l r P Z , .

..

, P S I

of all Q -conjugates of u1 belongs to R. Thus Z[Pl,P2,...,PSl

GR

Because nu1 is an algebraic integer, Lemma 5.10 ensures that u1 is also an algebraic integer. But then, by Lemma 5.9, ~1 = - = E" which implies that p(u) = diag(E, . . . , E ) = P(E l), where E = E I . Thus u = E - 1 as desired.

-

-

C. The isomorphism class of U ( Z G ) .

For any commutative ring R and any integer n 2 1, we write R" for the direct product of n copies of R. As before, E~ denotes a primitive n-th root of 1 over Q .

Applications of Characters

1214

Lemma 5.12. Let G be a finite abelian group of exponent n and, for any dln, let ad be the number of cyclic subgroups of order d in G. Then the integral closure of ZG'in Q G can be identified with

n

z[&dIud

dln

Proof. We keep the notation of the proof of Theorem 19.4.2. For any given subring R of F , the isomorphism of Theorem 19.4.2 (ii) induces an isomorphism

RG[zx,

7

3

* *

7

zx3]

+

n

R[&dIad

dln

of R-algebras. Thus, if we identify RG with its image, then ndlnR[~dIad can be regarded as an integral ring extension of RG. In particular, by integrally closed taking F = Q , R = Z3 and applying the fact that Z [ Eis ~ (Proposition 5.1 (ii)), we infer that the integral closure of Iu: in Q G can Z [ ~ d ] ~ d . be identified with

ndln

Lemma 5.13. Let S be a subring of a commutative ring R. Assume that the additive group R+ of R is finitely generated and that the torsionfree ranks of S+ and R+ are equal. Then the torsion-free ranks of U ( R )and U ( S ) are also equal. Proof. The hypothesis on ranks of S+ and R+ ensures that S+ has finite index, say m, in R+ and so mR E S . Since R+ is finitely generated, we also have (R+ : m R + ) < 00. Therefore it suffices to show that for any Z7Y

E

W),

- y E mR

implies x-'y E U(S) Because x - y E mR implies both 1 - x-'y and y-'x - 1 E m R C S , the result follows. W 2

The following result is essentially due to G. Higman (1940) (see also R.G. Ayoub and C. Ayoub (1969)).

Theorem 5.14. Let G be a finite abelian group. Then

U ( Z G )= f G x F where F is a free abelian group of rank 1

-(\GI - 21 t m 2

+ 1)

5 Applications to U ( Z G )

1215

Here m (respectively, I) is the number of cyclic subgroups of G of order 2 (respectively, the number of cyclic subgroups of G).

Proof. Let a d be the number of cyclic subgroups of order d in G and let n be the exponent of G. By Lemma 5.12, we have

n

ZG

Z[&dlad

= R, ,say

dln

and by Corollary 19.4.3,

Since, by Proposition 5.1 (ii), ( Z [ E: ~ Z]) = (Q( ~ d :) Q ), it follows that and R+ have the same torsion-free rank, equal to [GI. the groups (ZG)+ Because R+ is free, we may apply Lemma 5.13 to conclude that U ( m ) and U ( R )have the same torsion-free rank. Furthermore, by (1) and Proposition 5.2, U(2zG)is finitely generated. Hence, by Theorem 5.11,

U ( Z ) = fG x F for some free abelian group F . By the foregoing, we are left to verify that 1 r a n k ( F ) = -(]GI - 21 2

+ m + 1)

By Corollary 5.3, if d > 2 then the torsion-free rank of u ( z [ & d ] ) is equal to (1/2)(p(d) - 1. By the preceding paragraph, ranL(F) = rank(U(R)). Thus

1

= -(IGl 2 as desired.

- 21 + TTZ + l ) ,

Applications of Characters

1216

D. Effective construction of units of ZG. Throughout this subsection, we employ the following notation : n = the exponent of a finite abelian group G. ad = the number of cyclic subgroups of G of order d. r = (1/2)(IGI - 21 t m t l), where m and 1 are given by Theorem 5.14. E~ = a primitive n-th root of 1 over Q . All characters of G are assumed to be linear (c -characters. Thus any character x of G is a group homomorphism

Recall that the augmentation map is a homomorphism

given by a u g ( C w) = g€G

c

zg

(% f

z1

g€G

We denote by V(=) the subgroup of U ( = ) consisting of units u with aug(u) = 1. For convenience, we shall identify any character x of G with its extension to the ring homomorphism

In particular, the principal character will be identified with the augmentation map. As we mentioned earlier, a number of results of algebraic topology suggest the importance of finding out precisely what units exist within the ring ZG. Recall that, by Theorem 5.14, there exists a system of r units

uI,u2, ...,u, in Z G such that every unit of Z G is represented uniquely in the form

% ' and refer to We call (u1, . .. ,u r } a f u n d a m e n t a l system of units in 2 each u;as a f u n d a m e n t a l u n i t , The problem we have alluded to earlier, may be stated as follows :

5 Applications to U ( Z G )

1217

Problem. Given a finite abelian group G, find a specific fundamental system of units in 1zG. This is an exceptionally difficult problem. For example, if G is cyclic of prime order p , then as we shall see below the given problem is equivalent to one of finding a specific fundamental system of units of the subgroup of U ( Z [ E ~consisting ]) of all units which are congruent to 1 modulo E~ - 1. The latter is, needless to say, a formidable task. In general, the main difficulty seems to be the fact that effective calculations in U ( = ) are intimately ). into account that, in general, no connected with those in U ( Z [ E ~ ]Taking specific fundamental system of units of Z[E,]is known, any optimism in solving the corresponding problem for U ( Z G )must be guarded.

.

Lemma 5.15. Let X I , ~ 2 , . . ,xs be a full set of representatives of equivalence classes of Q -conjugate characters of G. (i) If x,y E ZGG, then x = y if and only if xi(.)

= xi(y) for all i E {I,. . .,s}

(ii) A n element x E ZG' is a unit if and only if for all i E (1,. . . ,s}, x ; ( x ) is a unit of

Z[E,]

Proof. (i) Direct consequence of Theorem 19.4.2. (ii) Owing to Lemmas 5.12 and 5.8, x is a unit of Z G if and only if for all i E { 1,. . .,s } , x ; ( x ) is a unit in Imx;. Since Z[E,]is integral over Imxi, it follows from Lemma 5.8 that xi(%) is a unit in I m x ; if and only if xi(%) is a unit of Z[E,]. So the lemma is true. 1 Turning to the case where G is cyclic of prime order p > 2, we now prove the following result.

Theorem 5.16. Let G be a cyclic group of prime order p > 2 and let

be the homomorphism which sends g to E ~ Then . (i) K e r X = ~ ( t 1 gt t gp-'). (ii) A n element u E ZG is a unit if and only if aug(u) = f l

and

x(u)E U ( Z [ E ~ ] )

Applications of Characters

1218

(iii) The induced homomorphism x* : U ( Z G ) U(ZCJ[E~]) is injective and x*(V (Z G ) ) is the subgroup of V (Z [ E ~ consisting ]) of all units which are congruent to 1 modulo ( E ~ 1). (iv) The indices of x*(V(ZG)) and x * ( U ( Z G ) )in U ( Z [ E ~ are ]) p -1 and (1/2)(p - l), respectively. --f

Proof. (i) This statement follows from the equality

and the fact that { l , c p , .. . , E ; - ~ } is a %basis for Z [ E ~ ] . (ii) We first note that { l ~x} ,is a full system of nonconjugate characters. Hence the desired conclusion is a consequence of Lemma 5.15 (ii). (iii) If u E U ( Z G )is such that ~ ( u=) 1, then x(u2)

= l&)

= aug(u2) = 1

so u2 = 1 by Lemma 5.15 (i). Because, by Theorem 5.14, f G is the torsion subgroup of U(ZE),the assumption that p is odd forces TA = fl, whence u = 1. Thus x* is injective. Let F be the field of p elements. We next claim that

Indeed, since I)-

v-2

2

the ring Z [ E ~ ]-/ 1) (E coincides ~ with its prime subring. On the other hand, p E (cp - 1) by Lemma 30.1.13,proving (1). By (ii), we have

Because P-1 i=O

i=O

5 Applications to U ( Z G )

1219

we see that each element of x*(V(ZG)) is congruent to 1 modulo ( E ~ 1). Conversely, let u = Cyzi A;&;(& E Z ) be a unit of Z [ Esuch ~ ] that u z l(mod(Ep- 1))

Because p E (2),

( E ~

l), we have ( c p - 1) n

(2)

Z = pZ . On the other hand, by

P-2

C Xi

1(mod (

E ~

1))

i=O

Thus A0

+- A1 + ... + Ap-2

=1

+ kp

for some k E 23

Setting 5

= (A,

- k) * 1 + (A, - k)g +

* * *

t (Ap-z - k)gp-2 - kgp-l

it follows that ~ ( 5=) u and aug(s) = 1. Hence u E x*(V(ZG)), proving (iii). (iv) Since U ( E ) = {*1} x V ( W ) ,it suffices to prove that the index of x*(V(ZG))in U(Z[E,]) is equal t o p - 1. By (iii), x*(V(W)) is the kernel of the natural homomorphism V(Z[EPl)

+

~(W,I/(EP

- 1))

Hence, by (l),it suffices to show that this homomorphism is surjective. Now a typical element of V ( Z [ E , ] / ( E ~1 ) ) is a i- ( E -~ 1) where a is a positive integer such that 1 5 a < p . On the other hand, the element &; - 1 -= 1t E p + Ep - 1

and is mapped to a t (

E ~

* * *

+ &;-I

E U(Z[Ep])

l),as desired.

Corollary 5.17. Let G be a cyclic group of prime order p > 3 , let r = (1/2)(p - 3 ) and let ~ 1 , ...,u, be units of 1zG. Then (211,. ..,u,} is a fundamental system if and only if the image of < f G , 111,. . . ,u, > in U ( Z [ c p ] is ) a subgroup of index ( 1 / 2 ) ( p- 1). Proof. By Theorem 5.14, U ( Z ) = f G x F where F is a free abelian group of rank T . Hence { u ~ ,...,u,} is a fundamental system of units if and only if

U ( ZG)=< f G , ~

. . ,U, >

1 , .

Applications of Characters

1220

Thus the desired conclusion follows by virtue of Theorem 5.16 (iii), (iv).

We close by providing three examples which illustrate the interplay between the unit group of Z G and that of Z [ E ~ ] . Example 5.18. (Kaplansky, unpublished). Let G be a cyclic group of order 5 with generator g . Then

Proof. We first observe that the torsion-free rank of U ( Z [ E ~is] )1 and that ~5 1 is a fundamental unit (see Proposition 5.5). On the other hand, the element

+

21

= (g t 1)2 - (1 t 9 t g 2 t g 3 t g 4 ) = -g(g2 t g3 - 1)

is mapped to ( E S + 1)2 and has augmentation -1. Hence u is a unit (Theorem 5.16 (ii)) such that the image of < f G , u > is a subgroup of U ( Z [ E ~ ] of ) index 5 2 . Thus, by Theorem 5.16 (iv), this index must be equal to 2. Now apply Corollary 5.17. The next example deals with a more involved situation.

Example 5.19. Let G be a cyclic group of order 7 with generator g . Then

Proof. By Proposition 5.5, { 1+&7, 1+&7+&;} is afundamental system of units of U ( Z [ E ~ ] ) Hence .

is also a fundamental system of units of U ( Z [ E ~ ] ) Observe . that the element u1

+

+

= (9 + - (1 t 9 t !I2 g3 t g4 g5 t g6) = g ( 2 t 29 - 9 3 - 94 - g5)

5 Applications to U ( Z G )

1221

is mapped to ( ~ t7 1)3and has augmentation -1. Hence, by Theorem 5.16 (ii), u1 is a unit. Similarly, the element

is mapped to

= (1 t E7)-l( 1 t E7

+

2 1 E7)-

and has augmentation -1. Thus, by Theorem 5.16 (ii), 212 is also a unit. Let V be the image of < fG,211, u2 > in U = U(Z[&7]). Because

it follows that ( U : V) 5 3. Therefore, by Theorem 5.16 (iv), ( U : V) = 3. Now apply Corollary 5.17.

Example 5.20. Let G be a cyclic group of order 8 with generator g. Then U ( Z G ) = ~ G

-+

is 1 and Proof. We first observe that the torsion-free rank of U(Z[&8]) that 1 &g t E: is a fundamental unit (see Proposition 5.5). Let

+

be the homomorphism which sends g to &8. Taking into account that { ~ , E ~ , E ; , E ; } is a Z b a s i s of Z [ & g ] , it is clear that

t~

~t g 7 ( )1~ ~E i

z31

(3)

We claim that the image of U(2G)is a proper subgroup of U ( Z [ E ] ) .To prove the claim, it suffices to verify that no unit in 2%'is mapped to 1 E8

+

+

&*:

Applications of Characters

1222

Assume by way of contradiction that x(z) = 1 + &8 + &; for some z E U(ZG). Then, by (3), there exist A; E Z such that 2

= 1t g t g2 t Xo(1 t g4) t h ( g t g5) t A2(g2 g6) t X3(g3 9')

+

+

Observe that {XI = ~ G , x J , x ~ , x } where , x2(g) = -1 and x3(g) = i, is a full set of nonconjugate characters of G. Thus x ; ( s )is a unit for i = 1,2,3. The latter is equivalent to the following system of equations :

It is routine to verify that this system does not admit integral solutions, which gives a desired contradiction. This substantiates OUT claim. Setting 21

= (1tgtg2)-(g+g5)-2(g2tg6)-(g3tg7) = -g(g6 t 295 t g4 - g2 - - 1)

it follows that

Therefore, by Lemma 5.15 (ii), '(L is a unit. Furthermore, x(< f G , u >) is a subgroup of U(Z[&g]) of index 2. Because the image of U ( = ) is a proper subgroup of U ( Z [ E ~ we ] ) , deduce that

We are therefore left to verify that the kernel of the induced homomorphism U ( Z G )-+ U(Z[&g]) is contained in fG. To this end, we employ Theorem 19.4.2 (i) to deduce that the map

ZG

+

1z x

z x z[i]x z q E 8 1

is an injective homomorphism. Since the unit group of the first three factors is finite, it follows that any 9 E U(zU=) for which x ( v ) = 1 is of finite order. This implies v E fG and so the proof is complete.

5 Applications to U ( Z G )

1223

E. Cyclic groups. Throughout, we employ the following notation : Qn is the set of all primitive n-th roots of 1 over Q . Q n is the group of n-th roots of 1 over Q . Q n ( X )= n,,,,(X - E ) , the n-th cyclotomic polynomial e(E)=

{

1;E

if if

E #

~

=

l

where E is a root of 1. By a character of Q n , we mean a homomorphism x : Q n --t In this subsection, we examine the following problem.

Qn.

Problem. Given a finite abelian group G, find a system of independent units of Z G of infinite order which generates a subgroup of finite index.

Our goal is to present a theorem, due to Bass (1966), which solves this problem for cyclic groups. The proof is based on the application of the following important result. Theorem 5.21. (Bass (1966)). Let m 2 1 be an integer. Assume that a,(& E \urn) an? integers such that

a, = a,-1

and such that, for all chamcters x of

(1)

Qm,

n 4x(aUc

=1

eE@m

Then a, = 0 for all

E

#

1.

Proof. We wish to show that conditions (1) and (2) imply that a, = 0 if E @ !Dm and that & ' a, a, = 0. Once this is accomplished, the result will follow by virtue of Proposition 5.6. A remark that will be used implicitly is the following. If x is the identity map, then (2) becomes

Suppose that x has trivial kernel. Then (3) implies (2) since an automorphism of the field Q = Q ( E ~ ) Em , E am.

x extends to

Applications of Characters

1224

The theorem being vacuous for m = 1, we argue by induction on m. For the sake of clarity, we divide the rest of the proof into three steps. Step 1. Our aim here is to prove

by showing that

and that

=0

(6)

To establish ( 5 ) , write m = ks and choose a character QErl, and such that ImX = !Pa. Setting

x such that K e r x =

Ifplm, pprime, then

ag S€@,

be =

C

ac

(0 E qS)

x(c)=e

it follows that ( 5 ) is equivalent to be = 0, for all 0 # 1. But be, 6 E Q 8 , satisfy the hypothesis of the theorem. Thus, by induction, be = 0 for all e + 1. To prove (6), write rn = p"q with (p,q) = 1 and put N = NQ / Q. It is a m consequence of Lemma 30.1.3 that if

Applying N to (3), we may thus neglect all but prime power order Segregating the latter according to the relevant prime, we deduce that

E'S.

Therefore Cy=lri CcEQp, a, = 0. If i > 1, aPpi is a union of nontrivial cosets a, = 0. Thus (6) is of Q p , so it follows from ( 5 ) (with k = p ) that CcEQp,

5 Applications to U ( Z G )

verified. Step 2.

1225

Assume that m = pt, ( p , t ) = 1, p prime. Here we prove that a,

= 0 for all 1 #

E

E

\Et

(7)

To this end, put N = NQ m / t .~ Applying N to (3) and using Lemma 5.7 (i) and Lemma 30.1.13, we obtain 1 =

n

N(e(~))"c

r€Qm

n E

E

E

#

e(E)(P-l)ac

n

[~(EP)~(E)-']"s~

6€@p

\Et

1

-

Pd

where

and

It follows from (6) that d = 0. By ( 5 ) (with k = p), we also have

Thus 1=

e(E)'C,

E & # E

c,P

= pa,p - a,

Qt

I

We next claim that to prove (7), it suffices to verify the hypothesis of the theorem for t and ce. For then we can appiy induction and conclude cb = 0 for E 1, i.e. a, = p e p for 1 # E E Qi. Selecting T such that pr = l(modt)

+

Applications of Characters

1226

and setting q = p', it follows that E B = E , whence a, = qa,q = qa,. This proves that a, = 0 and substantiates our claim. That t and c, satisfy (1) is obvious. To prove (2), suppose x is a character of 9t. If K e r x = 1, we can apply the remark of the first paragraph of the proof. So assume that K e r x = Q k , k > 1. Extend x to a character of !ljm = XPt x !Pp by the identity map on ! l j p . By (2), we then have 1 =

n

e(x(W

EEQm

n

[e(X(E)Yc

= { EE@t-*k

n

]

n

6E@p

1

e(X(E)6)Qc61

l-I e ( V 6

[Ec'@k6e@p

Since k > 1, we have &Qk a,6 = 0(6 E QP) by ( 5 ) ; hence the second factor is 1. We now apply N to the first factor and calculate as above :

(since e(1) = 1). This establishes (2) for the C,(E E Q,)and so (7) is proved. Step 3. Let m = p t , plt, p prime. Here we complete the proof by showing that a, = 0 for all 1 # E E Qt (8) Again, we apply N = to (3), this time applying Lemma 5.7 (ii) :

N~

m/Q

t

Collecting exponents in the second factor leads to sums over nontrivial cosets of Q p , and these sums are zero by ( 5 ) . Hence EEQt

5 Applications to

U(ZG)

1227

and these data clearly satisfy (1). For ( 2 ) , suppose x is a character of Q t and write rn = pns, ( p ,s) = 1. Extending x to a character of Q m , it follows from (2) that

If K e r x n Qpn # 1, then we can collect exponents modulo Q p to eliminate the second factor. Otherwise, we apply N to the last equation and calculate as above, applying Lemma 5.7 (ii). Hence, by induction, we see that pa, = 0 for all 1 # E E qt, proving (8). The following observation now completes the proof. For any E E Q m am,we can choose a prime p , m = p t , so that E E Q t . Then a, = 0, b y virtue of (7) and (8). Thus the theorem reduces to Proposition 5.6, by applying (4) and (2). Theorem 5.22. (Bass(1966)). Let G be a cyclic group of order n > 2 , let m be a multiple of y ( n ) and, for each dJn,d > 2, let Qd be a fixed element of order d in G. Put

Sd = {g:ll < p and, for any s = g$ E s d , put

< d/2

and

( p , d ) = 1)

Then Udln,d>2{"slS E S d ) is a system of independent units of ZG of infinite order which genemtes a subgroup of finite index in U ( Z G ) .

Proof. By assumption, cp(n)lm,and so cp(d)lmfor all dln. Since ( p , d ) = 1, we also have E l ( m o d d ) . Thus d l l - p m and therefore us E ZG. Let x be a linear character of G . If x(gd) = 1, then

If X(gd) = E

#

1, then Ed- 1 - - -0 E-1

Applications of Characters

1228

Thus X('1LS)

= (1 t E

t

=

* * *

(-)1l--&&P

E U(Z[E])

(9)

Hence, by Lemma 5.15 (ii), a, E U ( Z G ) . Since G is cyclic, G has exactly one subgroup of order dln. Hence, by the proof of Theorem 5.14, the torsion-free rank of U ( Z G )is

Since [Sdl = (1/2)(p(d) - 1, we are therefore left to verify that the units us are independent. To prove independence, suppose we have a relation

We must show that all as = 0. Our aim is to apply Theorem 5.21, so we must construct the necessary data. Denote by @d(G)the set of all elements of order d in G . If g E G, then g E @d(G)for a unique d)n. We now define the integer 6, as follows : if d 5 2 set 6, = 0; otherwise exactly one of the following four cases occurs : (a) g E sd,(b) 9-' E s d , (c) = gd, (d) 9-l = gd- Define

bg = and

[

mag

(-m>&s,,

as

if if

g Esd g = 9d

-

bg-1 = bg Now choose an isomorphism xo : G iJn and put b, = b, if E = xo(g). We claim that n and the be(& E S,) satisfy conditions (1) and (2) of Theorem 5.21; if sustained, it will follow that b, = 0 for all E # 1. In particular, for any s E s d , bX(+) = mas = 0 and thus a, = 0, as required. To substantiate our claim, we first observe that condition (1) follows from the construction. Hence, we are left to verify that for any character x of $,

5 Applications to U ( Z C )

1229

Let u: be the image of us under the automorphism of Z G which sends each element of G to its inverse. Then, by (lo), we have

and the product of this with (10) implies

Applying the character 0 = x o xo of G to the above, we find that

and this, by (9))is

the last equality being true by the definition of the 6,. This proves (11) and hence the result.

This Page Intentionally Left Blank

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van der Linden, F. (1082 ) Class number computations of real abelian number fields, Math. Comp. 39, 693-707. Villamayor, 0.E. (1959 ) On the semisimplicity of group algebras 11, Proc. Amer. Math. SOC.10, 27-31. Walter, J.H. (1966 ) Character theory of finite groups with trivial intersection sets, Nagoya Math. J. 27, 515-524. Ward, H.N. (1968 ) The analysis of representations induced from a normal subgroup, Michigan Math. J. 15, 417-428. Warfield, R.B. Jr. (1969 ) A Krull-Schmidt theorem for infinite sums of modules, Proc. Amer. Math. SOC.22, 460-465. Washington, L.C. (1982 ) Introduction to Cyclotomic Fields, Springer-Verlag, New YorkHeidelberg- Berlin.

Weil, A. (1938 ) L’Int6gration dans les Groupes Topologiques et ses Applications, Paris, Hermann and Cie.

Wielandt, H. (1958 ) Uber die Existenz von Normalteilern in endlichen Gruppe, Math. Nachrichten 18, 274-280. Willems, W. (1976 ) Bemerkungen zur modularen Darstellungstheorie 111, Induzierte und eingeschrankte Moduln, Arch. Math. (Basel) 26, 497-503.

1254

Witt, E. (1952 )

Die algebraische Struktur des Gruppenringes einer endlichen Gruppe iiber einem Zahlenkorper, J. fur Math. 190, 231-245.

Yoshida, T. (1078 ) Character-theoretic transfer, J. Algebra 52, 1-38. Zassenhaus, H. (1936 ) Kennzeichnung endlicher h e a r e r gruppen als Permutationsgruppen, Hamb. Abh. 11, 17-40. (1949 ) The Theory of Groups, Chelsea Publisihng Company, New York. (1950 ) An equation for the degrees of the absolutely irreducible representations of a group of finite order, Canad. J. Math. 2, 166-167. Zmud, E.M. (1058 ) On isomorphic linear representations of finite groups, Mat. Sb. N.S. 38 (80), 417-430. (1977 ) On zeros of group characters, Uspehi Mat. Nauk 32, No 6, 223224. (1081 ) On solvable finite groups that have an irreducible complex character with conjugate zeros, Vestn. Kharkov Univ. 221, 70-80. (1990 ) On Gallagher’s theorems concerning zeros of characters, Publ. Math. Debrecen 37, No 3-4, 345-353.

Notation Number Systems the natural numbers the rational integers the padic integers the rational numbers the real numbers the complex numbers the integers mod m

Set theory

C

E

1x1

x-Y

proper inclusion inclusion the cardinality of the set X the complement of Y in X

Number theory a divides b a does not divide b the greatest common divisor of a and b the p p a r t of n the p'-part of n

1255

1256

Group theory the subgroup generated by X the cyclic group of order n direct product of GIand G2 N is a normal subgroup of G the centralizer of X in G the normalizer of X in G the order of g E G

= z-ly-'xy the commutator subgroup of G the center of G the group of all nonsingular linear transformations of V the image of x under the action of g E G the stabilizer of x E X the automorphism group of X the maximal normal p-subgroup of G the maximal normal p'-subgroup of G the maximal normal x-subgroup of G = Oa,(G/~a~..,ai-1 (G)) the p-component of A = {U E Alan = 1) the group of all n x n unitary matrices unitary group pcommutator subgroup of G transfer homomorphism transfer homomorphism sign of permutation o n-part of 5 union of all r-sections of G that intersect S lifting operator of f group of all nonsingular n x n matrices over the field of p elements projective special linear group of degree 2 over the field of q elements

Notation

1257

Rings and modules R-homomorphisms of V into W the F-dimension of V projective cover of V the opposite ring of R the center of R tensor product the characteristic of R direct sum direct product the Jacobson radical of R the left socle of R the algebra of R-endomorphisms of V the principal ideal generated by T n x n matrices with entries in R

=L@)Fv =L@FA the Loewy length of V the annihilator of V conjugate of V the socle of V the radical of V the commutator subspace of A the block containing e. projective limit ring the kernel of f the image of f the lattice of submodules of V the left regular R-module the right regular R-module composition length of V the unit group of R matrix unit direct product of n copies of V the i-th socle of V the i-th socle of R tensor product of maps

1258

= H o ~ R ( VR, ) injective hull of V dual homomorphism = {$ E V*l+(U) = 0) transposed homomorphism induced module coinduced module the sum of all ideals contained in X chain complex cycles boundaries homology module of C = Kerd, = Imdn+l

= Zn(C)/Bn(C) the connecting homomorphism f is homotopic to g projective dimension of V injective dimension of V n-th left derived functor of F n-th right derived functor of F = R"HomA (v,-) = R"HomA(-,W)

= Ln(V @ A -) = Ln(- @ A W ) Heller operators iterated Heller operators projective homomorphisms of U into V group algebra of G over R support of z supporting subgroup of 2 augmentation map augmentation of x augmentation ideal of RG the sum of all elements of X in RG = ( 9 E G ) g - 1 E 1) projection map left annihilator of X

1259

Notation

right annihilator of X trace map quotient ring associated with S contracted ideal local ring of R at P commutator submodule graded units of A degree of a unit u centralizer of X in A crossed product of A1 over G induced module general semilinear group of V intertwining number for U and V Brauer group of F index of A the kernel of B T ( F ) B T ( E ) Schur index of I' = {u E E [ A , A ]for some n 2 1) class group of R the group of all R-automorphisms of V the group of all n x n nonsingular matrices over the ring of all n x n matrices over R Reynolds ideal of FG inner tensor product outer tensor product kernel of V dual of V contragredient of V outer tensor product of Vl and V, conjugate of V relative trace map tensor induced module --f

Cohomology theory

= EztE&, V ) = H " ( W ,V ) derivations of G into V

1260

inner derivations of G into V invariant elements of V standard n-cochains standard n-cocycles standard n-coboundaries coboundaxy restriction map inflation map conjugation map corestriction map transgression map

Field theory the multiplicative group of F field extension degree of F over K Galois group of F over K the smallest subfield containing S and K the smallest subfield containing K and 0 1 , . . ,an the trace map the norm of p over F

.

Character theory conjugate of x Galois conjugate of x induced character

X * ( d = x(s-') (xl x XS)(gl,gS)

= Xl(gl)x2(92)

inertia group of x induced class function restriction of x t o H class functions from G to R ring of generalized R-characters restriction of a to H

1261

Notation

= Cf(G,C ) =Ch(G,(c ) the ring of all R-linear combinations of F-characters of G principal character = {v E Vlgv = u } the set of irreducible characters of G determinant of x = ( 9 E GI Ix(s)l = X W = ( 9 E *(s) = x(1)) = x(d for all 9 E G Frobenius-Schur indicator of x determinantd order of x character- theoretic transfer tensor induced character tensor induced class function the ring of generalized permutation characters (c -space of all functions f : G ---* CC such that f is constant on each n-section and f vanishes on G - X n-induced function =linear combinations of characters in X = { a E Z ( X ) l a ( l )= 0)

m

This Page Intentionally Left Blank

Index abelian group 399 m-divisible 399 of finite exponent 399 p-component 399 torsion 399 uniquely divisible by m 399 acyclic complex 272 algebra 8 central simple 495 definable over a subfield 474 Frobenius 185,556 G-graded 427 homomorphism of 8 irreducible character of 549 matrix representation of 549 of finite representation type 200 quasi-Frobenius 563 representation of 108 separable 483 strongly G-graded 428 symmetric 185, 364, 556 algebraic integer 796, 1207 algebraic number field 1207 Artin 791 associated elements 593 Atiyah 768 augmentation 355 augmentation ideal 355 augmentation map 355 Auslander 327 automorphism system 438 1263

Ayoub C. 1214 Ayoub R.G. 1214 Azumaya 58, 59, 70, 86, 95, 97, 253 Azumaya’s decomposition theorem 70 balanced map 114 Banaschewski 766, 768 basic idempotent 256 basic set of idempotents 256 basis 5 Bass 163, 225, 1222, 1226 Beck 67 Belonogov 1121, 1123, 1125 Bender 1196 Berger 977 Berkovich 819,920,921,1161,1162 Berman 641, 691, 711, 733, 759, 770, 779, 781, 785, 786 bifunctor 287 contravarian t 287 covariant 287 bilinear form 178 automorphism of 181 matrix of 182 nonsingular 179 nonsingular on the left 179 nonsingular on the right 179 rank of 184 skew symmetric 876

1264

symmetric 181 bimodule 5 Blau 654,655,658, 949,951 Blichfeldt 881, 897, 901, 903 block decomposition 77 Boolean ring 1014 boundary 263 boundary operator 262 Brauer 374,500,636,637,639,645, 659, 665, 731, 732, 779, 787, 795, 804, 824, 845, 847,901,1005,1041,1051, 1052,1095,1099,1106,1109, 1179, 1187, 1196 Brauer’s characterization of characters 787 Brauer classes 503 Brauer group 503 Brauer homomorphism 374 Brauer’s permutation lemma 639 Burnside 732, 795, 798, 801, 804, 825, 837, ‘839, 892, 897, 900,1179,1180 Burnside’s formula 798 cancellation property 69 canonical homomorphism 122,146, 174, 366 cardinality 3 Cartan 307 Cartan invariants 103 Cartan matrix 103 category 285 Cauchy-Schwartz inequality 762 center of group algebra 356 central character 490, 796 centralizer 435 central product 811 central simple algebra 495

INDEX

Brauer classes of 503 Brauer group of 503 degree of 504 division component of 503 exponent of 516 index of 504 primary 517 similar 503 split 504 chain complex 262 chain homomorphism 264 character 111 degree of 111 character table 824 characteristic 4 characteristically simple subgroup 816 characteristic subgroup 816 character-theoretic transfer 966 chief factor 1175 chief series 1175 Chillag 949, 951 Chinese remainder theorem 73 class function 756 induced 756 n-induced 1078 Cliff 351, 371, 372 Clifford 211, 726 Clifford’s theorem 211, 725, 739 closed invariant subset 1128 coboundary 263, 395 operator 263 cocycle 263, 395 cochain 263, 394 standard 394 cochain complex 263 cochain homomorphism 264 cogenerator 559 Cohen 615, 618

INDEX

coherent set of characters 1104 coherent sequence 616 cohomologous cocycles 395 cohomology class 395 cohomology group 380 cohomology module 263, 379 coimage 5 coinduction 192 commutative diagram 4 commutator submodule 369 commuting subalgebras 121 compatible map 401 completion 43 composite map 285 composite of field extensions 547 equivalence 547 conjugate of a module 704 conjugation by g 402 conjugation map 402 connecting homomorphism 265 contracted ideal 367 contracting chain complex 272 contracting homotopy 272 control of strong fusion 413 coordinate functions 542 corestriction 403 criteria for algebra to be artinian 16 algebra to be crossed product 438 algebra to be Frobenius 186, 572 algebra to be noetherian 16 algebra to be quasi-Frobenius 563 algebra to be semilocal 93 algebra to be separabIe 483, 486

1265

algebra to be skew group ring 466 algebra to be symmetric 187, 557 characters to be linearly independent 542 equivalence of crossed products 51 1 every module to be semisimple 89 existence of composition series 20 field to be a splitting field 472 finiteness of cohomology groups 386 group algebra to be local 363 group algebra to be self-injective 365 group algebra to be semisimple 361 group to be p-nilpotent 1142 ideal to be direct summand 75 ideal to be nilpotent 478 ideal to be quasi-regular 36 ideal to be superfluous 36 idempotent to be primitive 17 induced module to be semisimple 678 isomorphism of simple modules 542 linear independence of characters 553 map to be inner derivation 384 module to be absolutely simple 472 module to be flat 122, 130 module t o be injective 158 module to be principal indecomposable 18

INDEX

1266

module to be progenerator 148 module to be projective 142 module to be semisimple 34 module to be simple 33 modules to be in the same block 490 module to have injective hull 165 representations to be of degree 1633 ring to be artinian 16 ring to be discrete valuation ring 611 ring to be idempotent-lifting 47 ring to be local 60 ring to be matrix ring 80 ring to be noetherian 16 ring to be primary 238 ring to be primitive artinian 91 ring to be regular 50 ring to be regular noetherian 89 ring to be semilocal 92 ring to be semiperfect 233,249 ring to be semiprimitive artinian 84, 89, 91 ring to be semisimple 24 ring to be simple artinian 91 submodule to be fully invariant 25 uniqueness of rank 21 crossed homomorphism 383 principal 383 crossed product 437 crossed system 437 Curtis 1023,1025,1028, 1030,1031, 1032

cycle 263 Dade 65, 433, 727, 733, 772, 774, 775, 776, 777, 1041, 1048, 1051, 1075 Dedekind 603 Dedekind domain 599 Dedekind law 922 deleted chain complex 275 deleted cochain complex 276 derivation 383 inner 383 determinantal order 913 Deuring-Noether theorem 493 diagonal subgroup 648 dimension shifting 301, 315, 336, 382 direct product of rings 4 Dirichlet 731 Dirichlet’s unit theorem 1207 discrete valuation 611 discrete valuation ring 611 division component 503 domain 162, 285 Dress 977 dual homomorphism 171 Eckmann 163 Eckmann-Shapiro lemma 389 Eilenberg 113, 185, 307, 320, 327, 330, 331, 332, 364, 397, 557 equivalent crossed products 462 equivalent representations 108 Ernest 936 essential extension 164 maximal 165 proper 164 essential submodule 30

INDEX

Evens 977 exact sequence 6 exact triangle 267 exceptional set 1120 expanded ideal 367 exponent of Brauer class 516 extension of groups 421 equivalence classes 423 equivalent 423 section of 423 split 421 extension of modules 322 factor through 341 F-conjugacy 639 Feit 327, 659, 732, 799, 822, 846, 1062,1095,1103,1111,1168 Ferguson 837, 840, 841, 881, 894, 1041,1068,1069,1070,1072 fibre product 156 fibre sum 155 finite injective resolution 278 length 278 finite projective resolution 278 length 278 Fitting 253 Fitting’s lemma 71 Fitting subgroup 1154 fixed point 1146 fixed-point-free automorphism 1146 fixed-point space 452 Fomin 1121, 1123, 1125 Fong 796, 818 Fossum 1023,1025,1028,1030,1031, 1032 fractional ideal 600 invertible 601 principal 601 Franz 1209

1267

Franz’s independence lemma 1209 Frattini’s argument 882 free resolution 275 Frobenius 691,731,732,1052,1053, 1055, 1099 Frobenius complement 1101 Frobenius kernel 1101 Frobenius reciprocity 691, 751, 760, 874 Frobenius-Schur indicator 874 fully invariant submodule 25 functor 287 balanced 310 coerasable 292 cont ravariant 287 contravariant cohomology 294 contravariant homology 294 covariant 287 covariant cohomology 292 covariant homology 294 erasable 292 left derived 303 right derived 306 universal 295 functorial isomorphism 288 functorial morphism 287, 294 fundamental system of units 1208 Gallagher 817, 820, 837, 839, 842, 844, 847, 855, 860, 861, 862, 863, 907, 910, 911, 913, 931, 932, 933, 935, 936, 937, 938, 953, 956, 957, 958, 960, 961, 964, 966, 1039 Gaschiitz 806 Gaschutz’s theorem 806 Gauss 731 GCD-domain 594

1268

generalized Brauer permutation lemma 642 generalized orthogonality relation 748 generalized permu tat ion character 997 generalized Schanuel’s lemma 277 generator 148 G-field 456 G-invariant ideal 403 Glauberman 914, 1179, 1188 Gluck 840, 978, 997, 999, 1000 Gorenstein 822, 1167 graded homomorphism 262, 431 degree 262 graded ideal 430 graded isomorphism 262, 431 graded subalgebra 430 graded submodule 262 graded unit 429 degree 429 greatest common divisor 594 Green 787 Grothendieck 261, 291, 297, 307 group 3 action 420 dihedral 1150 elementary 786 extra-special 811 F-elementary 780 F-normalizer 780 Frobenius 1101 general linear 1195 generalized quaternion 1150 general semilinear 451 metacyclic 1156 p-nilpotent 1140 p-normal 780 p-solvable 885

INDEX

Ir-element 1043 n-group 815 n-section 1044 n-separable 815 socle of 806 special linear 1195 split metacyclic 1156 unitary 868 X-regular conjugacy class of 933 X-regular element of 932 group algebra 351 group character 626 absolutely irreducible 631 afforded by V 627 degree of 628 exceptional 1102 faithful 800 Galois conjugate 640 G-conjugate 709, 738 generalized 756 G-invariant 738 induced 734 inertia group of 738 irreducible 627 kernel of 800 linear 786 linear independence of 626 multiplicity-free 1035 permutation 813 principal 790 rational valued 790 regular 747 R-generalized 788 tensor induced 985 group representation 623 absolutely irreducible 631 afforded by V 624 completely reducible 624 degree of 624

INDEX

faithful 623 equivalent 624 Galois conjugate 640 indecomposable 624 induced 708, 734 induction 672 irreducible 624 kernel of 679 matrix 624 regular 624 restriction 671 unitary 870 Griin 888 Griin’s first theorem 888 Gustafson 717 Hall P. 821, 1055 Hall ?r-subgroup 815, 882 Harada 1125 Hawkes 921 Heller 327 Heller operator 328 Hekster 808, 809 hermitian form 761 invariant 870 positive definite 762 Higman D. 719 Higman G. 824, 1212, 1214 Hochschild 191, 379, 418 Hochschild-Serre exact sequence 418 Holt 412 homogeneous component 25 homology module 263 homotopy 270 Humphreys 921 Huppert 729 idempotent 4 centrally primitive 4

1269

orthogonal 4 primitive 4 identity morphism 286 Ikeda 555 indecomposable ideal 75 index of division ring 508 induction 192 induction determinant property 988 inductive set 4 inductive source 774 inertia group 532, 704 inflation map 402 injective dimension 278 injective hull 165 injectively equivalent modules 280 injective resolution 275 inner tensor product 647 integral domain 162 integral element 56 intertwining number 479 invariant subset 1055 irreducible element 594 Isaacs 820, 837,840,841, 904,907, 913, 914, 927, 928, 940, 944, 956, 977, 978, 987, 988, 997, 999, 1000, 1002, 1167, 1168 isometric bijection 772 isometry 772 isomorphism 286 Ito 623, 669, 881, 897, 1033 Janusz 698, 1207, 1208 Jordan-Holder theorem 20 Kaplansky 600, 1219 Karpilovsky 453,455,459,519, 711, 727, 908 kernel of action 450

1270

Knorr 191,623,669, 681, 717, 722, 977, 978, 990, 996, 1005, 1006,1009,1020,1021,1033 Knorr’s generalized character 1008 Knorr’s theorem 722 Klyachko 695 Krull 603 Krull-Schmidt theorem 72 Lang 1207 lattice 7 homomorphism 7 isomorphism 7 left principal ideal domain 162 Leonard 1041, 1043, 1046, 1048, 1066, 1095 lift of idempotents 42 lifting operator 1046 limit 616 linear isometry 1086 linked idempotents 78 localization 369 local ring at P 369 Loewy factors 106 Lowey length 105 Loewy series 105 long exact sequence 268 Mackey 704, 705, 706, 907, 910, 977 Mackey decomposition 704, 739 Mackey intertwining number theorem 706 Mackey tensor product theorem 705 MacLane 397 Matlis 163 Matusita 603 matrix units 80 full system 80

INDEX

system 80 maximal submodule 5 McKelvey 1075 M-group 729 Michler 796, 819 Milnor 1179, 1207 minimal injective resolution 334 minimal normal subgroup 806 minimal projective resolution 330 modular law 7 module 5 absolutely simple 472 annihilator of 32 artinian 13 coinduced 192 composition length of 20 composition series of 19 cont ragredient 685 cyclic 161 direct product of 5 direct sum of 5 divisible 162 faithful 32 finitely cogenerated 13 finitely generated 13 finitely presented 139 flat 126 free 5 Galois conjugate 530 graded 261 homogeneous 25 H-projective 441 imprimitive 702 indecomposable 5 induced 192 injective 158 noetherian 13 of finite length 20 principal indecomposable 18

INDEX

projective 142 projective-free 223 radical of 26 realizable over a subfield 476 semisimple 22 series of 19 socle of 26 stably free 279 strongly indecomposable 65 monomial space 452 Morita 633 Morita-equivalent rings 255 morphism 285 multiplicative subset 366 mutual exchange property 67 Nagao 1005 Nakayama 113, 185,320, 364, 555, 557, 577, 692, 711, 746 Nakayama automorphism 577 Nakayama’s lemma 35, 57 Nakayama reciprocities 692 natural exact sequence 6 natural homomorphism 43, 117 natural isomorphism 288 natural map 356 natural transformation 288 Nesbitt 665, 847 Niccolai 1075 Nicholson 46, 49, 51 nil ideal 4 nilpotent element 4 nilpotent ideal 4 n-multilinear map 978 Noether 544, 603 Noether’s theorem 544 normal closure 972 normal p-complement 820

1271

normal n-complement 1052 normal section 375 normal series 1175 normal subalgebra 208 normalized basis 208 normalized cochain 397 Novikov 1179, 1207 object 285 opposite category 286 orthogonality relations 744 &ma 374, 745, 746 outer tensor product 648 P-adic valuation 612 p-adic field 614 padic integers 614 padic valuation 612 Passman 374, 810, 820, 1158 Perlis 733, 770 permutation isomorphism 828 permutation module 637 Pierce decomposition 12 p-modular system 662 sufficiently large 662 point 257 conjugate 257 positive complex 273, 274 p-regular conjugacy class 635 p-regular element 635 prime subring 4 principal ideal domain 162 product 286 progenerator 148 projective cover 225, 226 uniqueness 227 projective dimension 278 projective homomorphism 34 1 projective representation 908

1272

projective resolution 275 projectively equivalent modules 279 projection map 357 proper submodule 5 Puig 225, 257 pullback 156 diagram 156 pure submodule 651 pushout 155 diagram 155 quasi-regular element 35 left 35 right 35 quasi-regular subset 35 left 36 right 36 quotient field 366 quotient ring 366 ramification index 726 range 285 rank 21 real conjugacy class 641 refinement 19 regular orbit 452 element 452 relative normal complement 1042 relative trace map 713 relatively prime ideals 73 RG-lattice 663 full 663 representation 108 afforded by V 108 completely reducible 109 degree of 110 equivalent 108 indecomposable 109 irreducible 109 matrix 109

INDEX

regular 109 residue field 611 restriction 192 restriction-inflation sequence 406 restriction map 401 Reynolds 1075, 1078, 1088, 1089, 1093 ring 4 basic 256 block of 75 block idempotent of 75 complete 43, 52 G-graded 428 idempotent-lifting 46 indecomposable 5, 75 integral 56 integrally closed 592 integrally closure of 592 integral over subring 591 Jacobson radical of 31 local 60 opposite 4 primary 238 prime 32 primitive 32 regular 50 self-injective 239 semilocal 92 semiperfect 233 semiprimitive 34 semiregular 51 semisimple 24 strongly G-graded 428 Robinson 693, 695, 696, 697, 827, 855, 856, 857, 858, 859, 864,865,1041,1064,1065, 1066 Roquette 920

INDEX

Sah 889,892,895,896, 926, 936 Saksonov 764, 796, 822, 824, 1212 Sandling 375 sattelite 296 left 296 right 296 Schanuel’s lemma 277 Schopf 163 Schreir’s refinement theorem 19 Schur 62, 550, 874, 953, 956 Schur index 550 Schur’s lemma 62 Schur-Zassenhaus theorem 883 section 375 Seitz 808 semilinear monomial representation 452 semilinear transformation 451 nonsingular 451 Serre 379, 418 short exact sequence 6 Sibley 1168 sign 955 similar central simple algebras 503 simple constituent 479 multiplicity 479 skew group ring 437 Skolem- Noether theorem 499 socle length 107 socle series 107 Solomon 780, 781, 826 special class 1127 special invariant subset 1127 split injection 6 split surjection 6 splitting field 472 splitting homomorphism 6, 437 stable element 408 stable submodule 411

1273

stably equivalent modules 279 standard resolution 392, 394 subdirect product 34 subquotient 242 superfluous ideal 35 superfluous submodule 29 supporting subgroup 352 supporting subring 352 Suzuki 1041,1051,1052,1095,1099, 1106,1109,1121,1122,1127, 1179, 1187, 1196 Suzuki’s formula 1122 Swan 413 Swan’s theorem 413 Sylow tower 920 system of imprimitivity 702 syzygy 278 Tachikawa 558 tamely embedded subgroup 1164 tensor induced character 985 tensor induced class function 990 tensor induced module 980, 981 tensor product 113, 978 of algebras 122 of maps 116 of modules 113 Thhvenaz 519, 524, 526, 536, 588 Thompson 732,823,830,831,832, 846, 907, 908, 910, 916, 918, 919, 920, 1111, 1142, 1148 torsion-free rank 1207 trace map 364 transitivity of induction 673 transitivity of n-induction 1078 transitivity of tensor induction 982 transgression map 417 transposed homomorphism 176

INDEX

1274

Travis 1034, 1039 trivial intersection set 1095 trivial unit 1211 Tsushima 645 Turull 779, 792, 793 twisted group algebra 437 twisted group ring 437 unique decomposition property 72 unique factorization domain 595 unit 35 left 35 right 35 unitary basis 868 unitary-equivalent representations 871 unitary group 868 unitary matrix 868 valuation ideal 611 valuation ring 611 van der Linden 1208 van der W a d 808,809 Villamayor 678 Walker 733, 770 Wall 1095, 1179,1196 Warfield 67 Washington 1208 Weber 731 Wedderburn-Artin’s theorem 90 Wedderburn’s theorem 478 Weil 795, 799 WieIandt 845,953, 1179,1180,1181, 1183, 1185 Willems 678 Witt 547, 641, 711, 779, 781, 786 Witt-Berman’s induction theorem 781 Witt-Berman’s theorem 641

Witt-Fein’s theorem 547 Yoshida 953, 966, 968, 969, 971, 974 Zassenhaus 960, 1158, 1196, 1206 zero divisor 162 Zmud 837,853,857