[Simon R. Blackburn, Peter M. Neumann, Geetha Venk-Enumeration of Finite Groups

7/21/2019 [Simon R. Blackburn, Peter M. Neumann, Geetha Venk-Enumeration of Finite Groups

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http://www.cambridge.org/9780521882170


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C A M B R I D G E T R A C T S I N M A T H E M A T I C SGeneral Editors

B . B O L L O B S , W . F U L T O N , A . K A T O K , F . K I R W A N , P . S A R N A K ,

B . S I M O N

173 Enumeration of Finite Groups


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S I M O N R B L A C K B U R NRoyal Holloway, University of London

P E T E R M N E U M A N NThe Queens College, Oxford

G E E T H A V E N K A T A R A M A NSt Stephens College, University of Delhi

Enumeration of Finite Groups


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CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, So Paulo

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

First published in print format

ISBN-13 978-0-521-88217-0

ISBN-13 978-0-511-35487-8

Simon R Blackburn, Peter M Neumann and Geetha Venkataraman 2007

2007

Information on this title: www.cambridge.org/9780521882170

This publication is in copyright. Subject to statutory exception and to the provision ofrelevant collective licensing agreements, no reproduction of any part may take placewithout the written permission of Cambridge University Press.

ISBN-10 0-511-35487-8

ISBN-10 0-521-88217-6

Cambridge University Press has no responsibility for the persistence or accuracy of urlsfor external or third-party internet websites referred to in this publication, and does notguarantee that any content on such websites is, or will remain, accurate or appropriate.

Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

hardback

eBook (EBL)

eBook (EBL)

hardback
http://www.cambridge.org/9780521882170http://www.cambridge.org/http://www.cambridge.org/9780521882170http://www.cambridge.org/


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ToKay and Keith Blackburn,

Sylvia Neumann,and Uttara Rangajaran


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Contents

Preface xi

1 Introduction 1

I E LE ME NT AR Y R ES U L TS 3

2 Some basic observations 5

II G R O U P S O F P R I M E P O W E R O R D E R 9

3 Preliminaries 11

3.1 Tensor products and exterior squares of abelian

groups 113.2 Commutators and nilpotent groups 123.3 The Frattini subgroup 173.4 Linear algebra 19

4 Enumeratingp-groups: a lower bound 23

4.1 Relatively free groups 234.2 Proof of the lower bound 26

5 Enumeratingp-groups: upper bounds 28

5.1 An elementary upper bound 285.2 An overview of the Sims approach 305.3 Linearising the problem 315.4 A small set of relations 355.5 Proof of the upper bound 40

vii


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viii Contents

III P Y B E R S T H E O R E M 45

6 Some more preliminaries 47

6.1 Hall subgroups and Sylow systems 476.2 The Fitting subgroup 506.3 Permutations and primitivity 52

7 Group extensions and cohomology 60

7.1 Group extensions 607.2 Cohomology 677.3 Restriction and transfer 737.4 The McIver and Neumann bound 75

8 Some representation theory 78

8.1 Semisimple algebras 788.2 Cliffords theorem 808.3 The SkolemNoether theorem 818.4 Every finite skew field is a field 85

9 Primitive soluble linear groups 88

9.1 Some basic structure theory 889.2 The subgroup B 90

10 The orders of groups 94

11 Conjugacy classes of maximal soluble subgroups ofsymmetric groups 98

12 Enumeration of finite groups with abelian Sylow

subgroups 102

12.1 Counting soluble A-groups: an overview 10312.2 Soluble A-subgroups of the general linear group and the

symmetric groups 103

12.3 Maximal solublep-A-subgroups 10812.4 Enumeration of soluble A-groups 109

13 Maximal soluble linear groups 113

13.1 The fieldKand a subfield ofK 11313.2 The quotientG/Cand the algebraC 11413.3 The quotientB/A 116


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Contents ix

13.4 The subgroupB 11913.5 Structure ofG determined byB 125

14 Conjugacy classes of maximal soluble subgroups of the

general linear groups 127

15 Pybers theorem: the soluble case 132

15.1 Extensions and soluble subgroups 13315.2 Pybers theorem 135

16 Pybers theorem: the general case 14016.1 Three theorems on group generation 14016.2 Universal central extensions and covering groups 14616.3 The generalised Fitting subgroup 15016.4 The general case of Pybers theorem 154

IV O T H E R T O P I C S 161

17 Enumeration within varieties of abelian groups 163

17.1 Varieties of abelian groups 16417.2 Enumerating partitions 16717.3 Further results on abelian groups 173

18 Enumeration within small varieties of A-groups 174

18.1 A minimal variety of A-groups 175

18.2 The join of minimal varieties 184

19 Enumeration within small varieties ofp-groups 187

19.1 Enumerating two small varieties 18919.2 The ratio of two enumeration functions 191

20 Miscellanea 195

20.1 Enumerating d-generator groups 195

20.2 Groups with few non-abelian composition factors 20620.3 Enumerating graded Lie rings 21120.4 Groups of nilpotency class 3 216

21 Survey of other results 222

21.1 Graham Higmans PORC conjecture 22221.2 Isoclinism classes ofp-groups 224


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x Contents

21.3 Groups of square-free order 22721.4 Groups of cube-free order 23321.5 Groups of arithmetically small orders 236

21.6 Surjectivity of the enumeration function 23821.7 Densities of certain sets of group orders 24621.8 Enumerating perfect groups 256

22 Some open problems 259

Appendix A: Maximising two functions 269

References 275

Index 280


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Preface

This book has grown out of a series of lectures given in the Advanced AlgebraClass at Oxford in Michaelmas Term 1991 and Hilary Term 1992, that is tosay from October 1991 to March 1992. The focus wasand isthe bigquestion

how many groups of order n are there?

Two of the lectures were given by Professor Graham Higman, FRS, two bySimon R. Blackburn and the rest by Peter M. Neumann. Notes were written upweek by week by Simon Blackburn and Geetha Venkataraman and those notesformed the original basis of this work. They have, however, been re-workedand updated to include recent developments.

The lectures were designed for graduate students in algebra and the book hasbeen drafted with a similar readership in mind. It presupposes undergraduateknowledge of group theoryup to and including Sylows theorems, a littleknowledge of how a group may be presented by generators and relations, avery little representation theory from the perspective of module theory anda very little cohomology theorybut most of the basics are expounded hereand the book should therefore be found to be more or less self-contained.Although it remains a work principally devoted to connected exposition ofan agreeable theory, it does also contain some material that has not hithertobeen published, particularly in Part IV.

We owe thanks to a number of friends and colleagues: to Graham Higmanfor his contribution to the lectures; to members of the original audience fortheir interest and their comments; to Laci Pyber for comments on an earlydraft; to Mike Newman for permission to include unpublished work of himselfand Craig Seeley; to Eira Scourfield for guidance on the literature of analytic

xi


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xii Preface

number theory; to Juliette White for comments on the earlier chapters ofthe book and for help with proofreading our first draft. We would also liketo acknowledge the support of the Mathematical Sciences Foundation, St.

Stephens College, Delhi, The Indian Institute of Science, Bangalore and ourrespective home institutions. Geetha Venkataraman would also like to ac-knowledge the encouragement and support extended by Uttara, Mahesh andShantha Rangarajan and her parents WgCdr P. S. Venkataraman and Visalak-shi Venkataraman. Professor Dinesh Singh has been a mentor providingmuch needed support, encouragement and intellectual fellowship. We recordour gratitude to an anonymous friendly referee for constructive suggestions

and for drawing our attention to some recent references that we had missed.We are also very grateful to the editorial staff of Cambridge University Pressfor their great courtesy, enthusiasm and helpfulness.

Turning lecture notes into a book involves much hard work. Inevitably thatwork has fallen unequally on the three authors. The senior author, too happyto have relied on the excellent principlejuniores ad labores(which he admitsto having embraced less enthusiastically when he was younger), is glad tohave the opportunity to acknowledge that all the hard work has been done by

his two colleagues, whom he thanks very warmly.

SRB,MN, GV: 25.xi.2006


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1

Introduction

The focus of this book is the question how many groups of ordern are there?This is to be interpreted in the natural way: we define fnto be the numberof groups of order n up to isomorphism and ask for information about thefunctionf.

The values offnfor small values ofnare:

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 fn 1 1 1 2 1 2 1 5 2 2 1 5 1 2 1 14

For 1 n 16 the groups of order n were classified well over a hundredyears ago, and the value offnclearly follows from this classification. Theeasiest case is when nis a primeLagranges Theorem shows that a groupof ordernmust be cyclic, and sofn = 1. Whennis in the range of the tableabove, only n = 16 requires a lengthy argument to establish a classification.Note that f15 = 1 even though 15 is not prime.

Asnincreases, the problem of classifying groups of order nbecomes hard.The groups of order 210 have only recently been classified, by Besche, Eickand OBrien [6]. An appendix to their paper lists fnwhen 1 n 2000;in particular when n = 210 they count 49487365422 groups! However, thegroups of order 211 have not been classified and it is not known how manygroups of order 211 there are. (We will show in Chapter4that f211 > 244.)So if we are to say anything about fnwhen nis large, we must resort togiving estimates for fnrather than calculatingfnexactly.

Graham Higman [45] showed in 1960 that

fpm p 227 m

3Om2

Charles Sims [86] proved in 1965 that

fpm p 227 m

3+Om8/3

1


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2 Introduction

and, as the culmination of a long line of development, Laszlo Pyber [82]proved in 1991 (published in 1993) that

fn n

2

27

n2

+On5/3

wherenis the highest power to which any prime dividesn. Amplificationof these results, and their proofs, forms the main part of the work: the resultsof Higman and Sims are expounded in Part II (incorporating a modification ofSims argument due to Mike Newman and Craig Seeley [77], which improvesthe error term significantly) and Pybers theorem is the subject of Part III.The proofs use a large amount of very attractive theory that is just beyond

the scope of an undergraduate course in algebra. All that theory is expoundedhere, so that our treatment of the theorems of Higman and Sims and of Pyberstheorem in the soluble case is self-contained. Our treatment of the general caseof Pybers theorem in Chapter16is not self-contained, however, because itrelies ultimately upon the Classification of the Finite Simple Groups (CFSG).

The asymptotics of the function ftell us much, but far from everything,about the groups of order n. To get a clearer picture we consider relatedmatters. For example, context is given by the questions how many semigroupsand how many latin squares of order nare there? These questions are treatedbriefly in Chapter2. Detail is given by such questions as: how many abeliangroups of ordernare there? how many of the groups of ordernhave abelianSylow subgroups? how many of the groups of ordernsatisfy a given identicalrelation? how many are soluble? how many are nilpotent? Questions of thistype are treated in PartIV.

Standing conventions:

most groups considered are finiteif at any point finiteness is not men-tioned but seems desirable, the reader is invited to assume it;

fhas already been introduced as the group enumeration function; for a class Xof groups (or of other structures) fXndenotes the number

of members ofXof ordern, up to isomorphism;

logarithms are to the base 2;

maps are on the left; palways denotes a prime number; ifn = p11 p22 pkk , wherep1 p2 pkare distinct prime numbers, then

n = 1 + 2 + + k and n = maxi 1 i k.

Other notation and conventions are introduced where they are needed.


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I

Elementary results


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2

Some basic observations

This chapter is devoted to elementary estimates for fn, the number of groupsof order n(up to isomorphism). We begin by looking at some enumerationfunctions for weaker objects than groups.

Since a binary system is determined by its multiplication table, we findthat

fn fbinary systemsn nn2

At most n!of these multiplication tables are isomorphic to any fixed binarysystem, since an isomorphism is one of only n!permutations. Hence

nn2n

nn

2

n! f

binary systemsn nn

2

If we consider binary systems with a unit element, we have

nn23n+O1

fbinary systems with 1n nn12 = nn22n+1

Recall that a semigroup is a set with an associative multiplication definedon it. For all > 0,

n1n2 fsemigroupsn n

n2

if n n0. To see this, consider the binary systems on 0 1 n 1described by tables of the following form:

5


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6 Some basic observations

0 1 m 2 m 1 m m +1 n 2 n 10 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 0

m 1 0 0 0 0 0 0 0 0m 0 0 0 0

m +1 0 0 0 0

n 1 0 0 0 0 Here the starred entries are arbitrary subject to being at most m 1. Theassociative law holds for this table, since

aiajak= 0 = aiajak

Hence

fsemigroupsn mnm2

(Notice here that we should divide by n!, but this again does not make asignificant difference.) Settingmto be approximatelyn1

12 we have

fsemigroupsn n1 12 nn

1 12 2

For sufficiently largen,

n112 nn

1 12 2 n1n

2

Thus we get the requisite lower bound.If we add the condition that all our semigroups contain a unit element, we

have similar results to the above.Daniel Kleitman, Bruce Rothschild and Joel Spencer enumerate semigroups

more precisely in [55]. They show that most semigroups can be split into twosubsetsA and B having the following property: there exists an element 0 B

such that ifx y Athenxy Bbut ifx Bor y Bthenxy = 0. They thenuse this fact to prove

fsemigroupsn =

nt=1

gt

1+ O1 where

gt =

n

t

t1+nt

2


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Some basic observations 7

The function gt maximises at t0, where t0 n/2loge n. Thus we mayimprove the lower bound we gave above to

fsemigroupsn nn21

loglog n/ log n

O1/ log n

where (for this inequality only) log should denote the natural logarithmalthough, as the astute reader will realise, in fact the base of the logarithmsdoes not matter here.

A multiplication table with inverses is a latin square (i.e., in each row andcolumn of the table, an element appears only once). We have

n12 n

2On flatin squaresn n

n2

The lower bound was proved by Marshall Hall [40]. Using less elementarymethods, the lower bound may be improved: Henryk Minc showed in [69]that

n!2n/nn2 flatin squaresnHis proof uses the EgorycevFalikman theorem [26,32,70] establishing thevan der Waerden conjecture on permanents. Note that there is a constant c

such thatn! > c n/en

, and so flatin squares n > c2nn2

11/ log n. Much remainsto be discovered about this enumeration function. In 2005, Brendan McKayand Ian Wanless [68] state At the time of writing, not even the asymptoticvalue offlatin squaresnis known.

Returning to the group enumeration function, we see that even very ele-mentary methods are enough to show that there are seriously fewer groupsthan semigroups or latin squares:

Observation 2.1

fn nn log n

Proof: For a group G, define

dG = mink g1 gk Gsuch that G = g1 gkWe first show that ifG = nthendG log n. Let

1 = G0< G1< G2< < Gr= Gbe a maximal chain of subgroups. Let giGi \ Gi1 for 1 i r. Theng1 gi =Gi, as one easily sees by induction. In particular, G can begenerated byrelements. Now by Lagranges Theorem

G =r

i=1Gi Gi1 2r (2.1)


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8 Some basic observations

Hencer log n. Then by Cayleys theorem G Symnand sofn number of subgroups of order nin Symn

number oflog n-generator subgroups of Symn number oflog n-element subsets of Symn n!log n

nn log n

and the result follows.

Recall the notation nand nthat we introduced at the end of Chap-ter1. A factorisation ofnhas at most nnon-trivial factors. Equation (2.1)shows that r n, and therefore the bound for dGcan be sharpened tosay thatdG n. We remark that in fact dG n+1, as we will seein Corollary16.7, but ignoring this for the moment and feeding the simplebound fordGinto the above argument we get that

fn nnn

That is about as far as one can go with elementary methods. Nevertheless,it already shows that the associative law and the existence of inverses areseparately very much weaker than is their combination.

The aim of the remainder of the book is to prove the better bounds on fngiven in the Introduction, using more sophisticated methods.


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II

Groups of prime power order


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3

Preliminaries

This chapter contains a brief account of some of the results we will need inthe next two chapters. More specifically, we review some basic commutatoridentities and results on nilpotent groups, discuss the Frattini subgroup of agroup and prove some simple enumeration results concerning vector spaces,general linear groups and symplectic groups. We emphasise that all groupsare finite in this sectionsome of the results (and definitions) differ in the

infinite case. We assume that the reader has already met a few commutatoridentities and the idea of a nilpotent group, and so we have included sketchproofs rather than full detail for some of the results. For more detail, seeGorenstein [36, Sections 2.2 and 2.3].

3.1 Tensor products and exterior squares of abelian groups

As preparation for some of our treatment of commutators we recall (withoutproofs) the definition of tensor product and exterior square of abelian groups.IfA,Bare abelian groups (which we write additively here) thetensor productA B is defined to be the abelian group which is generated by all symbolsa bfora Aand b Bsubject to the relations

a1 + a2 b a1 b a2 b= 0 a b1 + b2 a b1 a b2= 0

which make the operation bilinear. We identify a b with its imagemodulo the relations and then the map A BA B (where here A Bsimply denotes thesetof pairs),ab a b, is bilinear. IfAis generatedbya1 arand Bis generated by b1 bs then

ai bj 1 i r 1 j s

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12 Preliminaries

will be a generating set for A B; moreover, the order ofa b divides thegreatest common divisor of the orders ofaandb.

The exterior square A2 (sometimes written2A) of A is defined to bethe abelian group generated by all symbols a bfora b Awith the samebilinearity relations as the tensor product and, in addition, the relations

a a = 0 for all a A

which make an alternating function of its arguments. Again, we identifya bwith its image modulo the relations and then the two-variable functionab

a

b is an alternating bilinear map A

A

A2. Note that the

equation b a = a bfollows easily from the defining relations for theexterior square, and that ifAis generated by a1 ar then

ai aj 1 i < j r

is a generating set for A2.One of the main properties of the tensor product is that it is universal for

bilinear maps. That is, ifCis an abelian group andf A B Cis a bilinearmapthenthereisauniquehomomorphismf AB Csuchthat fab =fab for all aA, bB. Similarly, the exterior square is universal foralternating bilinear maps in the sense that iff AA Cis bilinear and suchthatfaa = 0 for allathen there is a unique homomorphismf A2 Csuch thatfab = fabfor alla b A. Another fundamental property isfunctoriality: the tensor product and exterior square are functorial in the sensethat ifA1 , A2 , B1 , B2 are abelian groups and f A1 A2, g B1 B2 arehomomorphisms then there are homomorphisms f

g A

1 B

1A

2 B

2and f2 A 21 A 22 such thatf ga b = fa gbfor all a A1,b B1and f2a b = fa fbfor alla b A1.

3.2 Commutators and nilpotent groups

Let G be a group and let x y

G. The commutator xy of x and y is

defined byxy = x1y1xy. Forx y z G, we definexyz = xyz.Throughout this section, we will write xy to meany1xy.

Lemma 3.1 LetGbe a group.

(1) For all x y G,xy = yx1 (3.1)


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3.2 Commutators and nilpotent groups 13

(2) For all x y z G,xyz = xzyyz = xzxzyyz (3.2)xyz = xzxyz = xzxyxyz (3.3)

(3) For all x y z G,xy1 zyyz1 xzzx1 yx = 1 (3.4)

The proof of this lemma is easy: just use the definition of a commutator toexpress each side of the above equalities as a product of xyz and their

inverses.

Corollary 3.2 LetGbe a group.

(1) For all x y z G,

x1 y = xy1x1 = xyx11xy1 (3.5)xy1 =

xy1

y1 = xyy11xy1 (3.6)

(2) For all x y z G,xyz = zx1 y11xy y1 z1 x1zy (3.7)

Proof: The corollary follows from Lemma3.1by making the appropriatesubstitutions. To derive (3.5), replaceyby x1 andzby yin (3.2). For (3.6),replacezby y1 in (3.3). To derive (3.7), replaceyby y1 in (3.4).

Lemma 3.3 LetG be a group. Letx y G. Suppose that yx commuteswith bothxandy. Then for all positive integers n

yxn = yn x = yxn (3.8)xyn = xnynyx 12 nn1 (3.9)

Proof: The equality (3.8) follows by induction onn, using (3.2) and (3.3) in

the inductive step. To establish (3.9), use the fact that yi

x = xyi

yi

x.We will now consider a collection of results related to nilpotency of groups.

Let H and Kbe subgroups of a group G. Then HK is defined to be thesubgroup generated by all elements of the form hk where hH andk K. Note that HK = KH, by Equation (3.1). The subgroupHKLis defined by HKL=H K L. The following lemma, known as theThree Subgroup Lemma, is often useful.


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14 Preliminaries

Lemma 3.4 LetK, LandMbe subgroups of a group G. Then KLM MKLLMKwheneverMKLandL M Kare normal subgroupsofG.

Proof: Suppose that MKL and LMK are normal subgroups of G.The subgroup KLM is generated by elements of the form gh whereg KLand h M. We may express gas a product of commutators of theformg hwhereg Kandh L, and then use Equations (3.2) and (3.3)to express ghas a product of conjugates of elements of the form xyzwhere xK, yL and zM. But (3.7) expresses xyz as a product

of a conjugate of an element of MKL and a conjugate of an elementof LMK. Since MKL and LMK are normal, we find that eachgenerator ofKLMlies in M K LL M K, so the lemma follows.

Thelower central seriesG1 G2 G3 of a groupGis defined byG1= GandGi+1 = Gi Gfor every positive integeri. From now on, we will alwaysuseGito denote theith term of the lower central series ofG. It is not difficultto see, using the definition of the lower central series, that the subgroups Gi

are characteristic subgroups ofG. ClearlyGi/Gi+1is central inG/Gi+1. Forall normal subgroupsNofG, we have that G/Ni= GiN/N. Moreover, ifHis a subgroup ofGthenHi is a subgroup ofGi for all positive integers i.

Proposition 3.5 LetGbe a group. LetA = G/G2 = G/GandAi = Gi/Gi+1.Then A2 is a homomorphic image ofA

2 andAi+1 is a homomorphic imageofAi Afor all i 1.

Proof: It follows immediately from Lemma3.1that the map A A A2,aG bG abG3is well-defined and bilinear. It is also alternating sinceaa = 1 for alla G. Therefore there is a homomorphism A2 A2suchthat aG bGabG3 for all a bG. This is surjective since G2 isgenerated by the commutators ab for a bG, and therefore A2 is ahomomorphic image ofA2. The proof that Ai+1is a homomorphic image ofAi Afor all i 1 is similar and we omit it.

Proposition 3.6 LetGbe a group. For all positive integers i andj, we havethatGi Gj Gi+j.

Proof: We use induction on j. The case when j= 1 follows by definition ofthe lower central series. Assume that j > 1 and that Gi Gj1is a subgroupofGi+j1for any groupGand anyi 0. We prove thatGi Gjis a subgroupofGi+jas follows.


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3.2 Commutators and nilpotent groups 15

By replacing Gby the quotient G/Gi+jif necessary, we may assume thatGi+j= 1. Our inductive hypothesis implies that

Gi Gj1 G Gi+j1 G = Gi+j= 1 andGGi Gj1 = GiGGj1 = Gi+1 Gj1 Gi+j= 1

In particular, Gi Gj1 G and GGi Gj1 are normal. But now Lemma3.4implies that

Gi Gj = Gj Gi = Gj1 G Gi

Gi Gj1GGGi Gj1 = 1 = Gi+j Hence the proposition follows by induction on j.

Proposition 3.7 Let ibe a positive integer. LetGbe a group generated bya setS. LetTbe a subset ofGiwhose image inGi/Gi+1generatesGi/Gi+1.ThenGi+1/Gi+2 is generated by the set

tsGi+2 t T s S

Proof: WriteS for the image of S in G/G and T for the image of T inGi/Gi+1. ThenSgenerates G/G and, by assumption,T generates Gi/Gi+1.It follows that the set t s tT sS generates the tensor productGi/Gi+1 G/G and the result now follows immediately fromProposition3.5.

The final result of this section will form one of the key steps in Simsupper bound for the number of isomorphism classes ofp-groups of a givenorder. Recall that a group GisnilpotentifGr= 1for some integer r. Ifris the smallest such integer, we say that Gisnilpotent of class r1. A finitegroup is nilpotent if and only if it is the direct product of its Sylow subgroups(see [36, Theorem 2.3.5]). In particular, all p-groups are nilpotent.

Proposition 3.8 LetGbe a nilpotent group, and letHbe a subgroup ofG.IfH2G3= G2, thenHi= Gi for all i 2.

Proof: We prove first that HrGr+1 = Grfor allr 2. We prove this equalityusing induction onr: it is true when r= 2, by assumption. Assume thatr > 2and that HsGs+1= Gs whenever 2 s < r. Clearly HrGr+1 Gr, and so wemust prove that Gr HrGr+1. We need two preliminary results. Our first


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16 Preliminaries

result asserts that whenever K, Land Mare subgroups ofGsuch that Lisnormal in Gand such that KML Gr+1then

KLM KMLMGr+1 (3.10)(Note that the right-hand side of (3.10) is indeed a subgroup, for LM Las Lis normal, and so KMLM KML Gr+1.) To prove thisresult, observe that the subgroup KLM is generated by elements of the formxyz where x K, y Land z M. But (3.2) expresses this commutatoras the product of three commutators, the first in KM, the second in Gr+1and the third in LM and so the result follows. We will use this result

three times: with K= Hr1, L = Gr and M= G; with K= H2, L = G3 andM=Hr2 and with K=Hr1, L=Gr and M=H. In all three cases, thecondition KML Gr+1 follows by the fact that Hi Gi for all positiveintegers iand by Proposition3.6.

The second result we require asserts that

Hr1 G GHr2HHGHr2Gr+1 (3.11)

(The right-hand side of (3.11) is a subgroup, since the subgroups G Hr

2 H

andH G Hr2lie inGrand so commute moduloGr+1.) To see why (3.11)holds, first note that Hr1is generated by elements of the form h

hwhereh Handh Hr2. We may use (3.2) and (3.5) to express a typical elementof Hr1 G as a product of various elements ofGr+1 and elements of theformhhgwhereh H, h Hr2 and g G. Here we use the fact that

Hr1 G H r1 Gr1GGr1 G2r1 Gr+1

by Proposition3.6. So our second result holds if we can show that elementsof the form hhgall lie in GHr2HHGHr2Gr+1. But Equation(3.7) shows thathhgis equal to the product of a conjugate of an elementof GHr2 H and a conjugate of an element of HGHr2. Since bothGHr2 HandH G Hr2are contained inGr, and sinceGr G = Gr+1,these conjugates differ from the original elements by members ofGr+1. Hencethe result follows.

Using these two results, we may establish our inductive step:

Gr= Gr1 G= Hr1Gr G(by the inductive hypothesis)= Hr1GGr+1(by (3.10)) GHr2HHGHr2Gr+1 (by (3.11))

Gr1HG2 Hr2Gr+1(using Hi Giand Proposition3.6)


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3.3 The Frattini subgroup 17

= Hr1GrHH2G3 Hr2Gr+1 (by the inductive hypothesis)= Hr1HGrHH2 Hr2G3 Hr2Gr+1 (by (3.10))

= HrGr+1(by Proposition3.6)So, by induction on r, we find that Gr= HrGr+1 forr 2 as required.

Next we prove that Gi= HiGi+k for k 1 and i 2. This has just beenproved when k = 1. Moreover, ifk > 1 and Gi= HiGi+k1then

Gi= HiGi+k1= HiHi+k1Gi+k= HiGi+kthe second equality using the case r= i+ k 1 of the fact that Gr= HrGr+1forr 2. So the claim follows by induction on k.WeareassumingthatG isnilpotent,andsoGi+k = 1 forallsufficientlylargek. Hence the previous paragraph implies that Gi= Hifor i 2, as required.

3.3 The Frattini subgroup

Let G be a group. The Frattini subgroup G of G is the characteristicsubgroup of G defined as the intersection of all the maximal subgroups ofG. This section investigates the Frattini subgroup of a group, with an eye tousing the results in Section4.2.

Lemma 3.9 LetGbe a finite group, and letXbe a subset ofG. ThenGandXtogether generateGif and only ifXgenerates G.

We may express this lemma more informally by saying that Gis the setof non-generators ofG.

Proof: Clearly G= X implies that G= X G. For the converse,suppose thatG = X. ThenXis contained in some maximal subgroup MofG. SinceGis the intersection of all maximal subgroups ofG, we havethat G M. But nowX G M < G, and soG = X G.

When G is a p-group, the maximal subgroups of G have an especiallysimple form. This is a consequence of the next lemma.

Lemma 3.10 LetGbe a finite group, and letHbe a subgroup ofG. SupposethatHis a p-group, and thatpdivides the index ofHinG. ThenHis strictlycontained in its normaliserNGH. Moreover, there exists a subgroup KofGsuch thatHhas index pinK.


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18 Preliminaries

Proof: There is a natural action of the subgroupHon the set of cosets ofH in G. A coset is fixed by Hunder this action if and only if it is containedinNGH. So the index ofHinNGHis equal to the number of trivial orbits

ofHacting on . Now, pdivides. SinceH is a p-group, pdivides thelength of any non-trivial orbit in . Hence p divides the number of trivialorbits. Sopdivides the index ofHinNGHand thusHis strictly containedinNGH. Letx NGHbe such thatxHhas orderpin NGH/H. The lastassertion of the lemma now follows by taking Kto be the group generatedby Hand x.

Corollary 3.11 LetGbe a finite p-group. Then every maximal subgroup MofGis normal and has index pin G.

Lemma 3.12 Let G be a finite p-group. Then G/G is an elementaryabelian p-group of orderpd, where d is the minimal number of generatorsofG. Moreover G = GpG, where Gp is the subgroup generated by thesetxp x G.

Proof: By Corollary3.11, every maximal subgroup M ofGis normal andhas index p in G. It follows that every maximal subgroup contains GpG,and so GpG G and G/G is an elementary abelian group. SinceG is generated by d elements, so is G/G and soG/G pd. IfG/G < pd thenG/Gmay be generated byd 1 elements, and soGmay be generated byGtogether with the inverse image of these elementsunder the natural map from G to G/G. By Lemma 3.9, this impliesthat G may be generated by d

1 elements. This contradiction shows that

G/G = pd.We have seen that G GpG. SinceG/GpGis elementary abelian,

the intersection of all maximal subgroups ofG/GpGis trivial. The inverseimages in G of these maximal subgroups under the natural homomorphismfromGto G/GpGare maximal in Gand their intersection is GpG. ThusG GpG and so G = GpG, as required.

Lemma3.12implies that the Frattini subgroup of ap-group is generated bythe set of values of the words xp andx yinG. (ThusGis an example ofa verbal subgroup; see [75].) This implies, in particular, that for any normalsubgroupNof ap-groupG, we have G/N = G/N.

We will use the next lemma in the proof of Pybers theorem.

Lemma 3.13 LetGbe a finitep-group. LetBbe the group of automorphismsofGthat induce the identity automorphism onG/G. ThenBis ap-group.


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3.4 Linear algebra 19

Proof: Suppose that B is not a p-group. So there exists xB such that xhas order not a power ofp. By replacingxby a suitable power ofx, we mayassume that xhas order q, whereqis a prime distinct from p.

Let gG. Since x induces the identity automorphism on G/G, wehave that x permutes the elements in the coset gG. Each cycle ofx haslength either 1 orq. SincegGhas order a power ofp(and since a powerof p is not a multiple ofq), not all these cycles can have length q. Hencex fixes an element in every coset gG of G in G. In particular theelements inGthat are fixed by xgenerateGmoduloG, and so generateGby Lemma3.9. Sincexfixes a generating set for G, we find that xis the

identity automorphism. This contradicts the fact that x has order q, and sothe lemma follows.

To finish this section, we prove (as an aside) a result due to Wielandt thatgives a characterisation of nilpotency in terms of the Frattini subgroup.

Proposition 3.14 LetGbe a finite group. Then Gis nilpotent if and only ifG G.

Proof: Suppose that G is nilpotent. Any proper subgroup Hof a nilpotentgroupGis strictly contained in its normaliser. (To see this, letibe the largestinteger such that Giis not contained inH. ThenGi H Gi+1 H, and soGi NGH.) In particular, any maximal subgroup MofGis normal. SinceG/M is non-trivial and nilpotent, G/M is a proper subgroup of G/M.Hence, sinceMis maximal, G/Mis trivial and therefore G/Mis abelian.Thus G Mfor any maximal subgroup MofGand so G is contained in

the intersectionGof all maximal subgroups ofG.To prove the converse, suppose thatG G(and so in particular every

maximal subgroup ofG is normal). Let Pbe a Sylow subgroup ofG, andassume, seeking a contradiction, thatNGPis a proper subgroup ofG. ThenNGP Mfor some maximal subgroup MofG. Now P M, and so P isa Sylow subgroup of the normal subgroup M. By the Frattini argument (seeGorenstein [36, page 12, Theorem 3.7]), NGPM= G. But NGP Mandso we have our required contradiction. Thus every Sylow subgroup is normalin Gand so Gis nilpotent.

3.4 Linear algebra

This section contains some enumeration results from linear algebra, rehearsessome of the standard material on alternating forms and draws out some of the


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20 Preliminaries

immediate consequences for elementary abelian p-groups. Throughout thissection, qwill be a power of a prime number, and Fqwill be the finite fieldwith qelements.

Proposition 3.15 LetVbe a vector space of dimension d over the finite fieldFq. There are

qd 1qd q qd qk1choices for a sequence v1 v2 vk of linearly independent vectors in V. Inparticular,

GLdq

=qd

1qd

q

qd

qd1 qd

2

Proof: Assume that we have already chosen linearly independent vectorsv1 v2 vk1 V. The result follows if we can show that there are qd qk1choices for vk. But vk can be any vector of V that is not in the k 1-dimensional subspaceUspanned byv1 v2 vk1. There areq

d vectors inV and qk1 vectors in U, and so there are qd qk1 possibilities for vk andthe first statement of the proposition follows.

The second statement of the proposition follows from the first statementtogether with the observation that a ddmatrix with entries in Fqis invertibleif and only if its rows are linearly independent vectors in Fq

d. The finalinequality follows from the fact thatqd qi < qd, but this bound can be seendirectly by observing that a ddmatrix is specified by choosing its d2 entries.Proposition 3.16 Let V be a vector space over Fq of dimension d. For0 k d, letndk be the number of subspaces ofVof dimension k. Then

ndk= qd 1qd q qd qk1qk 1qk q qk qk1 (3.12)

Moreover,

qkdk ndk qkdk+1 (3.13)

Proof: A subspace of V is determined by choosing a basis for it. ByProposition 3.15, a basis v1 v2 vk for a k-dimensional subspace may

be chosen in qd

1qd

q qd

qk

1

ways. Each subspace U hasqk 1qk q qk qk1bases (again by Proposition3.15), and so thetotal number ndk of subspaces satisfies (3.12), as required. The bounds forndk given in (3.13) are obtained by noting that for 0 i < k d,

qdk qd qiqk qi q

dk+1

and so the proposition is proved.


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3.4 Linear algebra 21

Since an elementary abelian p-group may be thought of as a vector spaceover Fp, we have the following corollary to Proposition3.16.

Corollary 3.17 Letpbe a prime number, and letPbe an elementary abeliangroup of orderpd. ThenPhas exactlyndk subgroups of orderp

k, where

ndk=pd 1pd p pd pk1pk 1pk p pk pk1

We now turn to some standard results concerning alternating forms. Let Vand Wbe vector spaces over a field F. Just as for abelian groups, a bilinear

map VV Wis said to bealternatingifvv = 0 for allv V. Notethat uv= vu for any vectors u vV when is an alternatingmap: to see this, expand u + v u + v. IfWhas dimension 1 over F, wesay that is analternating formon V.

Let be an alternating form on V. For a subspace U of V, we defineUby

U= v V u v = 0 for all u UThe radical R of an alternating form on V is the subspace defined byR = V, so

R = u V u v = 0 for all v VIf the radical R of an alternating form is trivial, we say that is non-degenerate.

We say that a basis u1 u2 ur v1 v2 vr of V is symplectic (with

respect to the alternating form ) if for all i j 1 2 r ui uj = 0vi vj = 0 and

ui vj =

1 wheni = j0 otherwise.

Proposition 3.18 LetVbe a finite-dimensional vector space over a fieldF,and let V V Fbe a non-degenerate alternating form on V. Thenthere exists a basis u1 u2 ur v1 v2 vr ofV that is symplectic withrespect to. In particular, the dimension ofVmust be even.

This is a standard result in linear algebra. Here is a sketch of the proof. Asymplectic basis forVmay be built up as follows. Firstly choose any non-zerovector u1 V. Since is non-degenerate, the linear map V Fdefined by


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22 Preliminaries

v = u1 vis not zero, and so we may choosev1 Vsuch thatv1 =u1 v1 = 1. It is now not difficult to show that vectorsu1andv1are linearlyindependent, and so they span a subspace Uof dimension 2. The subspaceUis a complement to U inV, and the restriction of to Uis again a non-degenerate alternating form. A symplectic basis u2 u3 ur v2 v3 vrfor U is chosen, and the resulting basis u1 u2 ur v1 v2 vr is asymplectic basis for V. It is not difficult to show that every symplectic basismay be constructed in this way.

Let Vbe a vector space of dimension 2rover a finite field Fq, and let be a non-degenerate alternating form on V. We write Sp2rqfor the group

of all linear transformations ofVthat preserve the alternating form, soSp2rq = g GLV uv = gugvfor all u v V

Proposition 3.19 LetVbe a vector space of dimension 2rover a finite fieldFq, and letbe a non-degenerate alternating form on V. Then the numberof bases forVthat are symplectic with respect to is

q2r

1q2r1q2r2

1q2r3

q2

1q

=qr

2q2r

1

q2r2 1 q2 1Moreover, this is the order ofSp2rq.

Proof: We prove the result by induction on the dimension 2r ofV. Whenr=0 the result is trivial. Assume that r is positive and, as an inductivehypothesis, assume the result to be true for vector spaces of smaller dimension

than V. Consider the construction of a symplectic basis given above. Thereare clearly q2r 1 choices for the non-zero vector u1. The kernel of themap V Fhas codimension 1 and therefore contains q2r1 vectors. Thepossible choices for the vector v1 make up a particular coset of the kernel,and so there are q2r1 choices for v1. So the number of symplectic basesfor V is q2r 1q2r1 times the number of symplectic bases for U. Ourinductive hypothesis now implies that the number of symplectic bases ofV agrees with the formula above. The statement follows by induction on

r. The final statement of the proposition now follows fairly easily. For letu1 u2 ur v1 v2 vrbe a fixed symplectic basis ofV. Then it is easyto see that an (invertible) linear transformationglies in Sp2rqif and onlyif the image of u1 u2 ur v1 v2 vr under g is a symplectic basis.So there is a bijection between the elements of Sp2rq and the set ofsymplectic bases ofV, given by associating the elementg Gwith the imageofu1 u2 ur v1 v2 vrunder g.


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4

Enumeratingp-groups: a lower bound

Let p be a fixed prime number. For any positive integer m, fpm is thenumber of (isomorphism classes of) groups of order pm. This chapter willshow that

fpm p 227 m

2m6

This bound, and the proof we shall give, are due to Graham Higman [45]. Thechapter is divided into two sections. The first section investigates a specificclass of groups that we will need in our enumeration. In fact, this class isthe set of relatively free groups in a certain variety ofp-groups. LetGbe agroup from this class. We will show that the isomorphism classes of certainquotients ofGare in one-to-one correspondence with the orbits of Aut Gonthe set Xof subgroups of the Frattini subgroup ofG. In the second section,

we give an upper bound on the length of an orbit of Aut Gacting onX. Thisenables us to show that a large number of isomorphism classes of groupsoccur as quotients of the groupG, and this will provide the lower bound weneed.

4.1 Relatively free groups

Let rbe a positive integer. Let Frbe a free group of rank r, generated byx1 x2 xr. Let Grbe the quotient ofFrby the subgroup Ngenerated byall words of the formxp

2,xyp andxyz. Note that all words of the form

xp y lie in N, by Lemma3.3. The group Gr is the relatively free groupin the variety of p-groups of -class 2; see Hanna Neumann [75] for anintroduction to varieties of groups in general. A p-group G is said to have-class2 if there exists a central elementary abelian subgroup HofGsuch

23


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24 Enumerating p-groups: a lower bound

thatG/His elementary abelian. We identify the elements xiwith their imagesin Gr. Using this identification,x1 x2 xrgenerateGr.

Lemma 4.1 LetHbe a group of-class2, and lety1 y2 yr H. Thereis a homomorphism Gr Hsuch thatxi = yi fori 1 2 r .

Proof: Since Fr is free, there is a unique homomorphism Fr H suchthat xi=yi for all i1 2 r . By our restriction on H, we haveN ker and so induces a homomorphism Gr Hsuch that xi = yifori 1 2 r , as required.

Lemma 4.2 The group Gr is a p-group. The Frattini subgroup Gr ofGr is central of order p

12 rr+1 and index pr. Moreover, any automorphism

Aut Gr that induces the identity mapping on Gr/Gr fixes Grpointwise.

Proof: Since any commutator or pth power in Gris central of order p, wehave that GprG

r is a central elementary abelian p-group. Since Gr/G

prG

r is

also an elementary abelian p-group,Gr is ap-group.By Lemma3.12,Gr = GprGr, and soGris central. It is not difficult

to show that Gris generated by the elements xpi for i 1 2 r and

xj xiwhere 1 i < j r. These elements form a minimal generating set ofGr; we can see this as follows. Suppose there existai 0 1 p1fori = 1 2 r and bij 0 1 p1for 1 i < j rsuch that

r

i=1 xpi ai 1i


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4.1 Relatively free groups 25

of the group H of 3 3 upper unitriangular matrices over Fp and where maps all other generators to 1. Hence the generating set ofGris minimal,and

Gr

=pr+

12 rr1.

NowGr/Gris elementary abelian, and has the images ofx1 x2 xras a minimal generating set. Therefore Grhas indexp

r.Let Aut Gr be an automorphism that induces the identity mapping

on G/Gr. So there exist h1 h2 hrGr such that xi=xihi fori = 1 2 r . SinceGris central and of exponent p,

xpi = xip = xihip = xpihpi= xpi

Also, sinceGris central,

xj xi = xjxi = xjhj xihi = xj xiHencefixesGrpointwise, as required.

Lemma 4.3 LetN1 N2 Gr. Then Gr/N1 Gr/N2 if and only if thereexists AutGrsuch thatN1 = N2.

Proof: Note that quotients by N1 and N2 make sense, since Gr is con-tained in the centre ofGr.

An element Aut GrmappingN1to N2induces an isomorphismfromGr/N1 toGr/N2. We need to show the converse.

Let Gr/N1 Gr/N2be an isomorphism. Let y1 y2 yr Grbe suchthat xiN1 = yiN2. SinceGrhas -class 2, Lemma4.1implies that thereexists a homomorphism Gr

Grsuch thatxi

=yi. Now, since

is anisomorphism,y1 y2 yrand N2 together generateGr. SinceN2 Gr,the elementsy1 y2 yrandGrtogether generateGrand so Lemma3.9tells us that y1 y2 yr generate Gr. But the elements yi are contained inthe image of, and so is surjective. Since Gris finite, this implies thatAut Gr. Finally, we need to show that N1=N2. The definition of shows that xN2=xN1 when x is one of the generators xi, andso xN2= xN1 for all x Gr. In other words, the following diagramcommutes, where the vertical maps are natural:

Gr Gr

Gr/N1

Gr/N2It is easy to check from this diagram that N1=N2, and so the lemmafollows.


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26 Enumerating p-groups: a lower bound

4.2 Proof of the lower bound

This section contains a proof of Graham Higmans lower bound for the

number of isomorphism classes ofp-groups of a given order.

Proposition 4.4 Letrbe a positive integer, and letsbe an integer such that1 s 12 rr+1. Then there are at leastp

12 rsr+1r2s2 isomorphism classes

of groups of orderpr+s.

Proof: Let Grbe the group defined in Section4.1. Let Xbe the set of sub-groupsN Grof indexp

s inGr. Each subgroupN

Xgives rise toa groupGr/Nof orderp

r+s. Moreover, by Lemma4.3the set of isomorphismclasses of groups that arise in this way is in one-to-one correspondence withthe set of orbits of Aut Gracting in the natural way on X.

Let be the natural homomorphism from Aut Grto Aut Gr/Gr. ByLemma4.2, any automorphism ker fixes Grpointwise and so actstrivially on X. Thus ker is contained in the stabiliser of every elementof X, and so the length of any orbit of Aut Gr acting on X is at most

Aut Gr/ker AutGr/Gr.We may regard Gr/Gras a vector space over Fp, and group automor-

phisms of Gr/Gr correspond exactly with invertible linear transforma-tions in this setting. HenceAutGr/Gr = GLrp pr2 by Propo-sition3.15. So every orbit on Xhas length at most pr

2. By Corollary3.17,

X p 12 rr+1ss. Hence there are at least p 12 rr+1ss/pr2 orbits of Aut GronX. Thus there are at leastp

12 rsr+1r2s2 isomorphism classes of groups of

orderpr+s, as required.

Theorem 4.5 The numberfpmofp-groups of orderpm is at least

p 227 m

2m6

Proof: The theorem is trivially true in the case when m 6. When m >6,define the integersby

s = 13 m ifm 0 mod 313 m +2 ifm 1 mod 313 m +1 ifm 2 mod 3

and definer= m s. Then, by Proposition4.4,fpm p

12 rsr+1r2s2 p

227 m

2m6

as required.


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4.2 Proof of the lower bound 27

The proof of Theorem4.5gives an indication as to why the constant 2/27appears in the group enumeration function: Proposition 4.4yields approx-imately p

12 a

2bm3 groups having a Frattini subgroup of index pam and orderpbm, and 2/27 is the maximum (at a = 2/3) of the function 12 a2bsubject toa + b = 1.


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5

Enumeratingp-groups: upper bounds

Our aim in this chapter is to prove a good upper bound on the number fpmof isomorphism classes of groups of order pm. We will establish that

fpm p 227 m

3+Om5/2

The leading term is the best possible, by the results of Chapter4. This bound,

but with the larger error term ofOm8/3, is due to Charles Sims [86]. Themodification of Sims argument that leads to the improved error term is dueto Mike Newman and Craig Seeley [77].

Before proving the upper bound above, we give an argument that producesa weaker upper bound, but which is much more elementary than the Simsapproach.

5.1 An elementary upper bound

The upper bound for fpm we give in this section is due to GrahamHigman [45]. A better upper bound, using more complex techniques, wasalso given by Higman in the same paper, but he remarks of the boundwe will give It seems to me surprising that an upper bound obtainedso simply should be as near the truth as [Theorem 4.5] shows that this

must be.The strategy of the proof is to show that any group of order pm has a

presentation of a restricted type, and then to give an upper bound for thenumber of such presentations. This is also the basic strategy used to prove theSims bound, but Sims uses a class of smaller presentations, thus improvingthe upper bound but at the expense of having to use more sophisticatedarguments.

28


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5.1 An elementary upper bound 29

Theorem 5.1 Letp be a prime number and let m be an integer such thatm 1. Then

fpm p1

6 m

3

m

Proof: Let G be a group of order pm, and let G=G0 > G1 > >Gm 1 > Gm=1 be a chief series for G. Thus Gi is a normalsubgroup ofGof index pi in G.

For 1 i m, let gi Gi 1 \ Gi. Then every element g Gmay bewritten uniquely in the form

g= g11 g22 gmm (5.1)where 1 2 m0 1 p 1. Moreover, for i 1 we have thatg Giif and only if1= 2= = i= 0.

We have that gpi Gi since Gi 1/Gi has order p. Hence for 1 i mthere exist ii+1 ii+2 im 0 1 p 1such that

gpi= gii+1i+1 g

ii+2i+2 g

imm (5.2)

Now Gj 1/Gj is central in G/Gj since it has order p and is anormal subgroup of thep-groupG/Gj. Hence the commutator gj giliesin Gj for all integers i and jsuch that 1 i < j m. Therefore we maywrite

gj gi = gijj+1j+1 g

ijj+2j+2 g

ijmm (5.3)

for some elementsijk 0 1 p 1.It is not difficult to verify that the generators g1 g2 gm, the relations

(5.2) for 1 im and the relations (5.3) for 1 i < jm form a presentationforG. (This is a so-called power-commutator presentation for G.) To verifythe relations do form a presentation, it is only necessary to verify that theproduct of two words in normal form (5.1) can be brought into normalform using the relations given. This can be done by one of the collectionprocesses used in computational p-group theory. See Sims [87] for a detaileddiscussion of such processes. One method goes roughly as follows. Giventhe product of two words in normal form, all occurrences of the generatorg1 can be moved to the far left, using the relations (5.3) with i=1. Theresulting power ofg1 can then be made to lie in the correct range by using(5.2) with i= 1. This process results in a word of the form g11 g, whereg is a word in g2 g3 gm. The process is then repeated with g and thegenerator g2. After m applications of this process, the result is a word innormal form.


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30 Enumerating p-groups: upper bounds

As the relations (5.2) and (5.3) form a presentation forG, the isomorphismclass ofG is determined by the values of the elements ikwhere 1 i < kmandijkwhere 1 i < j < k m. There are at mostp choices for the values

of each of these 16 m3 m elements, and so the number of isomorphismclasses of groups of order pm is at most p

16 m

3m, as required.

5.2 An overview of the Sims approach

We need to improve the leading term of the upper bound given in Section5.1.

This is achieved by reducing the number of relations in our presentation fora groupGof order pm, by choosing the set of generators for the group morecarefully. In the bound of Section5.1, the power relations are specified by theelements ikwhere 1 i < km. Since there are at most p

12 mm1 choices for

the elementsik, the power relations do not affect the leading term of the enu-meration function, and so we must concentrate on choosing the presentationof a groupG in such a way that the number of commutator relations we needis reduced as much as possible. To this end, the role of the chief series ofGinthe previous section will be taken by the lower central seriesG1 G2 ofG.This will allow us to make more detailed use of the commutator structure ofthe group.

We define the Sims rank sG of a p-group G to be the smallest non-negative integer s such that there exists H G with s generators such thatG2= H2G3. (Recall that, by Proposition3.8, this condition onHimplies thatHi= Gi for i 2.) We will show that when the Sims rank of a p-group islarge, there are comparatively few possibilities for G/G3. When the Simsrank is small, there are comparatively few possibilities for Gonce G/G3 isfixed. This trade-off will be used to prove the upper bound on the number ofgroups of orderpm that we are seeking.

The proof is divided into three parts, each contained in one of the next threesections. Firstly, we will concentrate on bounding the number of possibilitiesforG/G3 when the Sims rank sGis fixed. It is the number of possibilitiesfor the commutator structure of G/G3 that is important, and we considera linearised version of the problem of counting the possible commutatorstructures that can arise. Secondly, we use our analysis of this linearisedproblem to show that we may always choose a presentation for G to havea special form, that has fewer commutator relations when compared withHigmans approach above. Finally, we give an upper bound for the numberof groups of orderpm by bounding the number of choices for the presentationsof this special form.


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5.3 Linearising the problem 31

5.3 Linearising the problem

Let V and Wbe vector spaces over some field. Recall from Chapter3that

a map VV W is alternating if it is bilinear and if xx=0 forall xV. Let U1 and U2 be subspaces of V. We define U1 U2 to be thesubspace ofWspanned by all vectors of the form xy where x U1 andy U2.

A p-group G naturally gives rise to an alternating map as follows. There is anatural function GG G2/G3G2, wherexy GGis mappedtoxyG3G2. The facts thatG2 G1 = G3and thatgp h = ghp modG

3for allg h

Gimply that multiplying x and y by elements ofGdoes

not changexy, and soinduces a map G/G G/G G2/G3G2. Now, bothG/Gand G2/G3G2are elementary abelianp-groups, and so we may think of them as vector spacesVandWover Fp. Itis not difficult to verify that V V W is bilinear and that xx = 0for allx V. Moreover, the Sims rank ofGissif and only ifsis the minimaldimension of a subspace UofVsuch that UU = W. (This follows since asubgroup HofGhas the property that H2G3

=G2 if and only if the image

UofHunder the natural homomorphism from Gto Vhas the property thatUU=W.) The aim of this section is to give a bound on the number ofpossibilities for the map and some structural information about this map,if we know the Sims rank ofG.

Proposition 5.2 Let V V Wbe an alternating map, and supposethatVV = W. Supposedim W >0. LetUbe a subspace ofVof minimal

dimension subject to the property that UU=W. Then U has a basisx1 x2 xssuch that if we define Vi = x1 x2 xithen for1 i s 1xi xi+1 Vi Vi

Proof: Define s=dim U. We show by induction that there exists a basisx1 x2 xs ofUsatisfying the proposition, the inductive hypothesis beingthat there exists a (linearly independent) set x1 x2 xk Usuch that

xi xi+1 Vi Vifor all i 1 2 k1 (5.4)The hypothesis is clearly true when k = 2, for since dim W > 0 and UU = Wwe have that dim U 2 and there exist x1 x2 Usuch that x1 x2 = 0.

To prove the inductive step, suppose that 2 < k s and there exists alinearly independent set y1 y2 yk1 Usuch that

yi yi+1 y1 y2 yi y1 y2 yifor all i 1 2 k 2


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For any i 1 2 k 1, define Ui= y1 y2 yi. We consider fourcases:

There exists xU\ Uk1 such that yk1 xUk1 Uk1. In this case,settingxi= yifor 1 i k1 and setting xk= x, we see that (5.4) holds.

yk1 x Uk1 Uk1for all x U\ Uk1, but there exists x U\ Uk1such that yj x Uk1 Uk1 for some j 1 2 k 2. Set xi= yifor all i 1 2 k 2, set xk1= yk1 + yjand set xk= x. Note thatVi=Ui=x1 x2 xi whenever i k 1. Hence Vi Vi=Ui Uiwhenever i k 1 and so (5.4) clearly holds when i < k 2. But (5.4)also holds when i

=k

2:

xk2 xk1 = yk2 yk1 + yk2 yj Uk2 Uk2sinceyk2 yk1 Uk2 Uk2and yk2 yj Uk2 Uk2. Furthermore,(5.4) holds wheni = k 1:

xk1 xk = yk1 x + yj x Uk1 Uk1since yk1 x Uk1 Uk1and yj x Uk1 Uk1. Hence (5.4) holdsin this case. Suppose that the above two possibilities do not occur. So we have thatUk1 U = Uk1 Uk1and there exist x y U\ Uk1 such that xy Uk1 Uk1. Suppose further that

yk2 yk1 + yk2 x Uk2 Uk2This implies that yk2 x Uk2 Uk2. Set xi= yi for all i 1 2 k

2, setxk

1

=x and xk

=y. SoVi

=Uifor 1 i k

2. Now, (5.4)

clearly holds wheni < k2, and also in the casei = k2 sinceyk2 x Uk2 Uk2. To see that the case i = k 1 also holds, we observe that

Vk1 Vk1 Uk2 Uk2 + Uk2 xk1 Uk1 Uk1Sincexy Uk1 Uk1,(5.4) holds in the case when i = k 1.

Finally, we suppose that Uk1 U=Uk1 Uk1 and there exist x yU\ Uk1such that xy Uk1 Uk1and such that

yk2 yk1 + yk2 x Uk2 Uk2Set xi= yi for all i 1 2 k 2, set xk1= yk1 + xand set xk= y.Again, it is clear that (5.4) holds when i < k 2. Since Vk2= Uk2and xk2 xk1 = yk2 yk1 + yk2 x Uk2 Uk2,(5.4) holds wheni = k 2. Finally,

Vk1 Vk1 Uk2 Uk2 + Uk2 yk1 + x Uk1 Uk1


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5.3 Linearising the problem 33

and so the fact that xk1 xk=yk1 y + xy implies that (5.4) holdswhen i = k 1.

The proposition now follows by induction on k.

Corollary 5.3 LetV andWbe vector spaces over some field, and let VV Wbe an alternating map such thatV V = W. Then there exists asubspaceUofVsuch thatUU = Wanddim U dim W+1.

Proof: The corollary is trivial when dim W= 0, so assume that dim W >0.Let Ube a subspace ofVof minimal dimension subject to the property thatUU = W. Proposition5.2shows thatUhas a basis x1 x2 xs such thatthe s 1 vectors x1 x2x2 x3 xs1 xsform a linearly independentset in W. Hence dim U 1 dim Wand so dim U dim W+1.

Corollary 5.4 LetV andWbe vector spaces over some field, and let VV Wbe an alternating map such thatV V = W. Letsbe the smallestinteger such that there exists an s-dimensional subspace U of V such that

UU = W. Then for any subspace KofV,dimKK dim K+dim W+1 s

Proof: The map induces an alternating map V VW/K K.SinceVV= W/KK, Corollary5.3implies that there is a subspace LofV such that dim L dimW/KK + 1 and LL= W/KK. But now,K

+L K

+L

=W. Thus

s dimK+ L dim K+dim L dim K+dim WdimKK+1and the corollary is proved.

We have established some information about the structure of alternatingmaps arising from p-groups with Sims rank s as defined on p. 30. Thefollowing proposition uses this information to provide two upper bounds onthe number of such maps. Each bound is good in certain situations, but bothbounds must be used together to provide the approximations we require. Theidea of using bounds on the number of alternating maps in this way is thecontribution of Newman and Seeley.

Proposition 5.5 LetVandWbe vector spaces overFpof dimensionsr1andr2 respectively, and lets be a positive integer. LetNpr1 r2 s Rbe suchthatpNpr1r2s is the number of alternating maps V V Wfor which


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there exists ans-dimensional subspaceUofVwith the property thatU U =W, but no s 1-dimensional subspace U0 exists such that U0 U0=W.Then

Npr1 r2 s 12 r

21 r2 s 1 + Or1 + r28/3and (5.5)

Npr1 r2 s 12 r

21 r2 s 1 + 12 r1r2s 1 + Or1 + r25/2 (5.6)

Proof: We begin by proving the bound (5.5). Define f= r2/31 and defineg= r1/f. Let x1 x2 xr1 be a basis of V. Note that the map isdetermined by xk x where 1 k < r1. Since there are at most p

r2

choices for each vector xk x, we have (trivially) that the number of mapsof the form we are looking for is at most p

r12 r2 . When s


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5.4 A small set of relations 35

We now establish the bound (5.6). We will use the same techniques asin the proof of the bound (5.5), except our choice of f will differ andour counting argument changes slightly. When s 2r1/21

+2, the trivial upper

bound ofpr12 r2 implies the bound (5.6). So we may assume thats > 2r1/21 +2.Let f= r1/21 and define g= r1/f. We choose a basis x1 x2 xr1 anddefine subspacesViand Wijjust as before. Note that Wijis well-defined, sincer2 s +1+2f < r2.

By Proposition 3.16, the number of d-dimensional subspaces of W isbounded above by pr2d+1d. Hence the number of possibilities for each sub-space Wij is at most p

r2r2s+1+2f+1r2s+1+2f =ps2fr2s+1+2f. Just asbefore, there areg2 subspaces Wij, and once these subspaces have beenchosen there are at most pr2s+1+2f choices for each of the

r12

images

xk x with 1 k < r1. So pNpr1r2s is bounded above by p to the

power g

2

s 2fr2 s + 1+ 2f +

r12

r2 s + 1+2f

But f= r1/21 + O1and g= r1/21 + O1, and so the bound (5.6) follows.

5.4 A small set of relations

In this section we show that we may always find a presentation for a p-group G of a restricted type. Our eventual enumeration in the next sectionwill depend on the fact that there are comparatively few possibilities forpresentations of this sort.

We begin by choosing our generating set for Gand finding a collection ofrelations this set satisfies. We then prove that we have a presentation for G.

Let G be a group of order pm. Let G = G1 G2 Gc Gc+1 = 1 bethe lower central series ofG. Define integersr1 r2 r c by setting ri to bethe rank ofGi/Gi+1. DefineV= G/Gand W= G2/G3G2. The char-acterisation of the Frattini subgroup of a p-group given in Lemma3.12impliesthat dim V= r1and dim W= r2. We choose a basisxiforVand a basisyjforWas follows. As before, the process of forming commutators inGinducesan alternating bilinear map V V W. Letsbe the Sims rank ofG. LetUbe a subspace ofVof minimal dimension such that U U = W; thenUhasdimension s. By Corollary5.3we have that s r2 + 1. By Proposition5.2,there exists a basis x1 x2 xs ofUsuch that for all i 1 2 s 1,xi xi+1 Vi Vi, where we define Vi= x1 x2 xi. Define

Wi= Vi Viand di= dim Wi


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Note that 0=d1 < d2 da1, the commutator g2k g1alies in Lbythe definition ofLso we may assume that 1 k da1. But in this case (5.7)implies that g2k= g1fk g1iwhere 1 fk < i a 1, and so

g2k g1a = g1fk g1i g1a= g1i g1a g1fk1g1a g1fk g1i1 modG4

Now, modulo G3 the commutators g1i g1a and g1a g1fk are equal to aproduct of the elementsg2j. Sincefk < aandi < aour inductive hypothesisimplies that the expression above is an element ofLmodulo G4 and hence,since L contains G4, we find that g2k g1aL, as required. Hence, byinduction ona, our claim follows.

We have now chosen our generating set for G. We now exhibit a collectionof relations that these elements satisfy.


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For 1 i cand 1 j ri, letaijbe the smallest positive integer such

that gpaij

ij Gi+1 gi1 gi2 gij1. Note that every element gG maybe written uniquely in the form

g= ge1111 ge1212 ge1r11r1 ge2121 gecrc crc

where 0 ei j < paij. In the following, we will abbreviate this product tog

eijij 1 i c 1 j ri

Note that g Gi if and only ifeuv = 0 whenever u < i. This implies thatGi = p

cu=i

ruv=1 auv

In particular, sinceG = pm,c

u=1ru

v=1auv = m (5.8)

Note that (5.8) and the fact that the integers auvare positive implies thatfor anyi 1 2 c ,

cu=i+1

ruv=1

auv m r1 r2 ri (5.9)

For all integers iand jsuch that 1 i cand 1 j ri, define integers

bijuvby

gpaij

ij =

gbijuvuv 1 u c 1 v ru (5.10)

Similarly, for all i, jand k such that 1 i c, 1 j ri and 1 k r1,define the integers ci j k u vby

gij g1k= gcijkuvuv 1 u c 1 v ru (5.11)

Note that bijuv=0 whenever u < i and whenever u=i and v j.Moreover,ci jk u v = 0 wheneveru i.

We have defined a generating set for Gand a collection of relations thatthese generators satisfy. We now show that a subset of these relations (togetherwith the integersp,c,s,diandri) suffice to define the isomorphism class ofGuniquely.


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Theorem 5.6 The isomorphism class of the group G is determined by theintegersp, c,s, r1 r2 r c,d0 d1 ds together with the integers

aij 1 i c 1 j ribijuv 1 i c 1 j ri i u c 1 v ruc1jkuv 1 j < k r1 2 u c 1 v ruc2jkuv 1 k s dk1< j r2 3 u c 1 v rucijkuv 3 i c 1 j ri 1 k s i < u c 1 v ru

Proof: We will show that any group G of nilpotency class at most cgeneratedby the set of elements gij

1 i c 1 j riand satisfying the relations

(5.10) when 1 i c, 1 j ri and the relations (5.11) when i=1 and1 j < k r1; wheni = 2, 1 k sand dk1< j r2; and when 3 i c,1 j ri and 1 k shas order at most p

m. This is sufficient to prove thetheorem, for supposeLandMare groups that give rise to the same collectionof integers. Since L and Meach possess a generating set that satisfies theabove relations, both Land Mare isomorphic to quotients ofG. Moreover,Land Mhave orderpm, by(5.8). IfGhas order at most pm, we have that L

and Mmust in fact be isomorphic to G, and soLandMare isomorphic.To show thatGhas order at mostpm, letH Gbe the subgroup generated

by g11 g12 g1s. Define subgroupsH2 H3 H c by

Hi= guv i u c 1 v ruand define Hc+1= 1. For u 2, the relations we are given include thosedefining eachguv as a commutator and so we may write each element guv as

a commutator of lengthuin g11 g12 g1s. HenceG H H2 H3 Hc Hc+1= 1

Moreover, since G has nilpotency class at most c and the generators ofHcmay be represented as commutators of lengthc, we have that Hcis central inGand so in particular Hc is normal in H.

As the notation suggests, the subgroups Hi are the terms in the lowercentral series for H. However, we do not prove this. We do, however, wishto show that Hi/Hi+1 is central in H/Hi+1 for all isuch that i 2. Supposefirstly that i 3 and we have shown that Hi+1 is normal in H. (This iscertainly true when i = c 1, since Hc is normal even in G.) The relations(5.11) that we are given include those that imply that Hi/Hi+1 is centralisedby g11Hi+1 g12Hi+1 g1sHi+1, and so Hi/Hi+1 is central in H/Hi+1; inparticular Hi is normal in H. Induction on c inow shows that Hi/Hi+1 iscentral inH/Hi+1for i 3. It remains to show that H2/H3is central inH/H3.


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Assume, as an inductive hypothesis, that

H2/H3 is centralised by g11H3 g12H3 g1a1H3

This hypothesis is true when 2 a 3, for since d0= d1= 0 it is impliedby the relations (5.11). For our inductive step, it is sufficient to show thatg2 g1aH3 for 1 r2. This is implied by our commutator relationswhen > da1, and so we may assume that 1 da1. One of the givenrelations (5.11) is the equality

g2= g1x g1y

where 1 x < y < a, by our choice (5.7) of the generatorsg2j. We claim thatfor any i1 2 r 2, g2i g1y g1aH3. When i > da1, this followsfrom our given commutator relations. But when i da1, the commutatorrelation (5.7) shows that g2i=g1u g1v where 1 u < v a 1. By ourinductive hypothesis, g1uH3 and g1vH3 centralise H2/H3, and so g1u g1vH3centralises H2/H3. Since one of the given commutator relations expressesg1y g1aas a product of elements inH2, this implies that g2i g1y g1a

H3

for all values ofiand so our claim follows. Nowg11ag2g1a= g11ag1x g1yg1a

= g11ag1xg1a g11ag1yg1a= g1xg1x g1a g1yg1y g1a

Since g1x g1aH2, this element commutes with g1x and g1y modulo H3by our inductive hypothesis and commutes with g

1y g

1amodulo H

3by the

claim we proved above. Hence,

g11ag2g1a= g1x g1yg1y g1amod H3

But g1y g1aH2, and so commutes modulo H3 with g1x and g1y by ourinductive hypothesis. Therefore

g11ag2g1a

=g1x g1y

=g2modH3

sog1acentralisesH2moduloH3, as required. Thus, by induction on a,H2/H3is central in H/H3.

The given commutator relations directly imply that H/H2 is abelian, thatHis normal inGand thatG/His abelian, hence the quotients of consecutiveterms in the sequence

G H H2 H3 Hc 1


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are abelian. But now the relations (5.10) imply that

Hi/Hi+1 p

rij=1 aij

H/H2 psj=1 a1j andG/H p

r1j=s+1 a1j

HenceG pc

i=1ri

j=1 aij = pm, and so the theorem follows.

5.5 Proof of the upper boundThis section completes the enumeration ofp-groups by providing an upperbound on the number of isomorphism classes of groups of order pm.

Theorem 5.7 Letpbe a prime number. Then

fpm p 227 m

3+Om5/2

Theorems4.5and5.7combine to produce the following theorem.

Theorem 5.8 Letpbe a prime number. Then

fpm = p 227 m3+Om5/2

Before embarking on proving Theorem5.7, we give the following elementary

lemma.

Lemma 5.9 Letnbe a positive integer. Then there are exactly 2n1 orderedpartitions ofn.

Proof: An ordered partition ofnis produced by taking the expression

1+1+ 1+ + 1 = nand grouping the terms of this sum by replacing some of the + signs by + . There are n 1 plus signs in this expression, and so there are 2n1ways of adding brackets. So the number of ordered partitions ofnis 2n1.

Corollary 5.10 Letnbe a positive integer. The numberpnof(unordered)partitions ofnis at most2n1.


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5.5 Proof of the upper bound 41

Proof of Theorem 5.7: Let a prime p and a positive integer m be fixed.The previous section showed that the isomorphism class of a group of orderpm is determined by a certain collection of integers, namely the integerscsri diaijbijuv and cijkuv. We will provide an upperbound on the number of choices for these integers. We start by showingthat there are relatively few choices for the integers csri diaij andbijuv, and so we need to concentrate on finding a good upper bound forthe number of choices for the integers ci j k u vto provide a good upperbound on the number ofp-groups of orderpm.

The sum = r1 + r2 + + rc is such that 1 m. The integersr1 r2 r c form an ordered partition of the integer with c parts, andso once is fixed the number of choices for c and r1 r2 r c is at mostequal to the number of ordered partitions of . Lemma5.9shows that thenumber of ordered partitions of is 21. Since 1 m, there are atmost

m=1 2

1 choices for c and r1 r2 r c, which is less than 2m. The

integers d2 d1 d3 d2 ds ds1form an ordered partition ofr2, sincethe integers di are strictly increasing. Moreover, since d0= d1= 0, the inte-gers di and s are determined by this partition. So, again using Lemma5.9,

the number of choices for s and d0 d1 ds is at most 2r21 2m1. Theintegers aijare positive and sum to m, and so the number of choices foraij is at most the number of ordered partitions ofm. Hence there are atmost 2m1 choices for the integers aij. Finally, to count the number ofchoices for the integersbijuv, note that 0 bi j u v < pauv and sothere are at most pauv choices for each integer bijuv. When i and jare fixed, the number of choices for the integers bijuvis at most

cu=i

ruv=1

pauv = pc

u=iru

v=1 auv pc

u=1ru

v=1 auv = pm

Thus there are at most pm choices for the integers bijuvwhen iand jare fixed. But there are at most mchoices for the integers iand j, for once ihas been chosen there are ri choices for j, and

ci=1 ri m. So there are at

mostpm2

choices for the integers bijuv.In summary, the number of choices for the integers c s ri diaij and

bijuvis bounded above by

2m2m12m1pm2

and this is much less than our error term ofpOm5/2. Hence to prove the upper

bound we require, it suffices to provide a good upper bound on the numberof choices for the integers ci j k u v once the integers scri diaijand bijuvhave been chosen.


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We need to choose the integers cijkuv when i=1 and 1 j 1. When i, jand kare fixed, the number of choicesfor the integers ci j k u vis at most

c

u=i+1ru

v=1 pauv

pmr1r2ri

by (5.9). When i 3, there are sri choices for j and k and so there are atmost

psrimr1r2ri

choices for the integersci j k u v. Wheni = 2, there arer2 dk1choicesfor jonce kis fixed. Since the sequence d1 d2 ds is strictly increasing


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5.5 Proof of the upper bound 43

and non-negative, we have that dk1 k 2 whenever k 2. Hence, sinced0= 0, we find that the number of choices for jandkis

sr2 s

k=1dk1= sr2

sk=2

dk1

sr2 s

k=2k 2

= sr2

s 12

and so there are at mostpsr2

s12mr1r2

choices for the integers c2jkuv.Thus the number of choices for the integers ci j k u vis at most pto

the powerM, where

M=Npr1 r2 s +

r12

m r1 r2 +

sr2

s 1

2 m r1 r2

+c

i=3srim r1 r2 ri

Nowrim r1 ri ri

j=1m r1 ri1 jand soc

i=3srim r1 r2 ri s

ci=3

rij=1

m r1 ri1 j

= sr

3+r

4++r

ck=1

m r1 r2 k

smr1r2

k=1m r1 r2 k

= s

m r1 r22

So we find that

MNpr1 r2 s +1

2r21 m r1 r2

+

s 1r2 1

2s 12

m r1 r2

+12

s 1m r1 r22 + Om2

(5.14)


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We need to show that this upper bound for Mis at most 227 m3 +Om5/2. We

consider two cases, depending on the size ofr1. Suppose that r1 610 m. By

the upper bound (5.6),

M1

2r21 m r1 s 1 +

1

2r1r2s 1

+

s 1r2 1

2s 12

m r1 r2

+12

s 1m r1 r22 + Om5/2

Settingx = r1/m,y = r2/m,z = m r1 r2/mand u = s 1/m, we needto show that the function Axyzudefined by

1

2x2z + y u +1

2xyu +

uy 1

2u2

z +1

2uz2

is at most 227 whenx y z uare non-negative and satisfyx +y +z = 1,u yand x 610 . This may be shown using standard techniquessee LemmaA.1in AppendixAfor details.

Now suppose thatr1 610 m. The bounds (5.5) and (5.14) combine to showthat

M1

2r21 m r1 s 1 +

s 1r2

1

2s 12

m r1 r2

+12

s 1m r1 r22 + Om8/3

As before, definex=

r1/m,y

=r

2/m,z

=m

r

1 r

2/mandu

=s

1/m

and let Bxyzube defined by

Bxyzu =12

x2z + y u +

uy 12

u2

z +12

uz2

It is not difficult to show that the maximum value ofBxyzu on theregion defined by x 0, y 0, z 0, u 0, x + y + z=1, u minxyand x 610 is strictly less than

227 (see LemmaA.2in AppendixA). But this

implies thatM m3

+Om8/3

2

27 m3

+Om5/2

whenr1 6

10 m. We haveshown thatM 227 m3 +Om5/2whatever the value ofr1, and so the theorem

follows.


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III

Pybers theorem


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6

Some more preliminaries

This chapter contains material that we will use in the proof of Pybers upperbound on the number of soluble groups of ordern. Section6.1contains resultson Hall subgroups and Sylow systems of soluble groups and Section 6.2deals with the Fitting subgroup of a soluble group. Section6.3concerns itselfwith primitivity in permutation and linear groups, and also contains a resultbounding the number of generators of a permutation group.

6.1 Hall subgroups and Sylow systems

LetGbe a finite group. A subgroup HofGis said to be aHall subgroupifHandG Hare coprime. Let be a set of prime numbers. A subgroupHofGis a Hall -subgroupifHis a number (a product of primes in) andG His a number (a product of primes not in ).Theorem 6.1 Let G be a soluble group of order p11 pkk . Let r be aninteger such that1 r kand let= p1 p2 pr. Then:

(i) The group Ghas a Hall -subgroup(of orderp11 prr ).(ii) Any two Hall -subgroups are conjugate in G.

(iii) Any-subgroup ofGis contained in a Hall -subgroup.

Proof: The proof is by induction on G. The theorem is trivial when G = 1,so we may assume thatG > 1 and that the theorem is true for all groups oforder less thanG. We must show that the group Gsatisfies properties (i),(ii) and (iii) of the theorem.

Let Mbe a minimal normal subgroup ofG. SinceGis soluble,M = pfor some prime p dividing n. Applying the inductive hypothesis to G/Mwe find that G/Mhas a Hall -subgroupK/M. Further, all such subgroups

47


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48 Some more preliminaries

are conjugate in G/Mand any -subgroup ofG/M is contained in a Hall-subgroup ofG/M. The proof is now divided into two cases.

The first case is whenpbelongs to. In this caseKis a Hall-subgroup

ofG, and so property (i) of the theorem follows. IfHis any Hall-subgroupof G then H contains M (since HM is a -group). Moreover, H/M isa Hall -subgroup of G/M and so our inductive hypothesis implies thatgM1H/MgM = K/Mfor some gM G/M. But this means that K=g1Hg. Thus we have shown that property (ii) follows. Finally, suppose thatL is a -subgroup of G. Then LM/M is a -subgroup of G/M and sois contained in a Hall -subgroup H/M of G/M. But then H is a Hall

-subgroup ofG, andL H. Thus (iii) follows, and we have established theinductive step in this case.The second case is when p does not belong to . ThenK/MandMare

coprime. The SchurZassenhaus theorem implies that there is a complementH forM inKand any two complements are conjugate in K. (We will provethis in Chapter7, Corollary7.17.) But thenH = K/Mand so His a Hall-subgroup ofG, and so property (i) follows. LetH1be any Hall-subgroupof G. To establish property (ii), it is enough to show that there exists aconjugate H2ofH1that lies in K. (For then H2is a complement ofMin K, andso the SchurZassenhaus theorem implies thatH2andHare conjugate.) Now,H1M/M is a Hall -subgroup of G/M. By the inductive hypothesis thereexistsg1inG such thatg1M

1H1M/Mg1M = K/M. Thusg11 H1g1 Kand so property (ii) follows. It remains to prove property (iii). Let L be a-subgroup ofG. NowLM/Mis contained in some Hall -subgroupK1/MofG/M (so L K1). Since all Hall -subgroups ofG/Mare conjugate in

G/M,K1andKare conjugate inG. So there exists a conjugateL1ofLsuchthat L1 K. We have that K= HMand so

L1M= L1M HM= L1M HMNowL1andL1MHare complements forMinL1Mand therefore are con-

jugate by the SchurZassenhaus theorem. ThusL1is conjugate to a subgroupofH. ThusL is contained in a Hall -subgroup ofG (namely an appropriate

conjugate ofH). So we have established property (iii) in this case, and so theinductive step follows here also.The theorem now follows by induction onG.

LetGbe a group of order p11 p22 pkk . ASylow system inGis a family

P1 P2 P kwherePiis a Sylowpi-subgroup andPiPj=PjPifor alli j1 2 k . Wheneveri=j, we find thatPiPjis a subgroup and its order isp

ii p

jj .Further,if p1 p2 pk then pi

Piis a Hall -subgroup ofG.


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6.1 Hall subgroups and Sylow systems 49

Theorem 6.2 LetGbe a soluble group of orderp11 p22 pkk . Then

(i) Ghas a Sylow system;

(ii) any two Sylow systems are conjugate in G;(iii) ifH GandQ1 Qk is a Sylow system forHthen there is a SylowsystemP1 P k forGsuch thatQi= H Pi for all i 1 2 k .

Proof: For eachi 1 2 k , definei= p1 p2 pk \ pi. We saythat a subgroup His aHall pi-complementifHis a Hall i-subgroup. Let bethesetofallSylowsystemsin G andlet ibe the set of all Hall pi-complements

in Gfor 1 i k. We claim that there is a bijection between the set ofSylow systems P1 P2 P kofG and the set 1 2 kof sequencesG1 G2 Gk where each Gi is a Hall pi-complement. For, given a Sylowsystem P1 P2 P k, we may define Gi=

j=i Pjfor all i1 2 k .

The remark after the definition of a Sylow system shows that each Gi is aHallpi-complement. In the reverse direction, suppose thatG1 G2 Gkaresuch that eachGjis a Hall pi-complement. The index ofGjisp

jj and hence

(since these indices are coprime) we find that for any subset S

1 2 k

the index ofjS Gj isjS pjj and sojS Gj =jS p

Documents

[Simon R. Blackburn, Peter M. Neumann, Geetha Venk-Enumeration of Finite Groups