I

ABELIAN SUBGROUPS OF p-GROUPS

SOO-SENG SIAH

B. S c . , NANYANG UNIVERSITY, I 9 6 6

A THESIS SUBMITTED I N

THE REQUIREMENTS

PARTIAL FULFILLMENT O F

FOR THE DEGREE OF

MASTER OF SCIENCE

I N THE DEPARTMENT

OF

MATHEMATICS

@ SOO-SENG SIAH 1 9 6 8

SIMON FRASER UNIVERSITY

NOVEMBER, 1 968

EXAMINING COMMITTEE APPROVAL

d- S e n i o r S u p e r v i s o r

Examining Committee

ABSTRACT

L

The main o b j e c t of t h i s t h e s i s i s t o review and d i s c u s s

some of t h e work done by J. L. A lpe r in , N . Blackburn and C .

Hobby on a b e l i a n subgroups of f i n i t e p-groups.

We f i r s t look a t some c o n d i t i o n s , under which

elements of t h e c e n t r a l i z e r of a n a b e l i a n normal si

t h e

~ b g r o ~

of a p-group G l i e i n A , i n terms of t h e exponent of A .

Then we cons ide r t h e problem of f i n d i n g a bound f o r t h e

number of g e n e r a t o r s of t h e subgroups of a p-group. We s e e

t h a t w i t h a g iven bound k on t h e number of g e n e r a t o r s of a l l

a b e l i a n normal subgroups of a p-group ( P > 2) G, we can

g i v e a n upper bound, i n terms of k, on t h e number of g e n e r a t o r s

o f a l l subgroups of G. I f k = 1, 2 , 3, t h i s p r o p e r t y i s

subgroup i n h e r i t e d i n t h e s ense t h a t i f k i s t h e bound on

t h e number of g e n e r a t o r s of a l l a b e l i a n normal subgroups

of a p-group ( p > 2 ) G , t h e n k i s a l s o t h e bound f o r t h e

number of g e n e r a t o r s of a l l a b e l i a n normal subgroups of H ,

f o r a l l subgroups H of G.

I n t r o d u c t i o n

S e c t i o n 1

S e c t i o n 2

S e c t i o n 3

S e c t i o n 11

S e c t i o n 5

B ib l iog raphy

Append x

TABLE UP COXTENTS

Page

Notat Lon acd Terminology I

Classical and fundamental results 4

C e n t m l i z e r s of a.be1ia.n normal subgroups of p-groups 7

'Depth 8nC r r iorn~al dep th of p-groups 19

I;ar.ge abe1ia.n subgroups o f p-groups 39

I wish 50 thank D r . J . L. Berggren who sugges ted t h e

topic and p a t i e n t l y a s s i s t e d me i n t he r e s e a r c h . I a l s o

wish t o thank D r . N . R . R e i l l y who has a s s i s t e d me d u r i n g

D r . Be rgg ren l s absence .

The suppor t r e c e i v e d f r o m D r . R . H a r r o p l s N . R . C .

Grant Number A-3024 i s a l s o a p p r e c i a t e d .

INTKODUCT ION

I

In t h e s t u d y of f i n i t c groups , many problems reduce t c

problems about t h e i r Syl.ow subgroups and, i n t h e s t u d y of

t h e l a t t e r , s h e l i a n subgroups a s w e l l a s a b e l i a n normal

subgroups p l a y a n impor ten t p s r t . Recent r e s u l t s conce rn ins

such s u b g r ~ u p s have been developed to some e x t e n t 5y J . I,.

A l p e r i n i n h i s remarkable papers ( s e e [ I 1, [ 2 ] and [j]) . A c l a s s i c a l theorem of W . Burils?.de ['I s t a t e s t h a t every

group G o f orde- p" has a n n b e l i z n riormal sabgr0u.p of o r d e r

m p w i t h m(m+1)22n. Hcwever, t h e c o n j e c t u r e t h a t G always .- t&] h a s a n a b e l i a n s u b g r ~ i ~ p of order p '- w.s demonstra ted to be

by A l n e r i n i n [2]. A l p e r i n has, 13 f a c t , c o n s t r t ~ - , t e d

a group of o r d e r p 3n+2 ( p ) 2) a l l of whose e b e l i e n sabgroups

. n-1-2 have o r d e r a t mas; p and a l s o 3 group of o r d e r z50 s l l

of whose abelS-en s ~ b g r o u p s h a - ~ e o rde r a t mos t 224. E x a c t l y

where, between Bu-rnsi.de Is and A l p e r i n l s r e s u l t s , t h e b e s t

p o s s i b l e one l i e s sti1:L rernains open.

The main o b j e c t of t h i s . t . he s i s i s t ,o r e v i e x and d i s c u s s

some of t h e r e c e n t work 6cn.e 5:. J. I,. Alpe r in , N. Blackburn

and C . Hobby.

The f i r s t two sec,Lons a r e of a n i n t r o d u c t o r y ~ i s t u r e ,

f o rmu la t i ng C e r ' h i t i o n s , termino1o;y and c m e r i n g sorrie

b a s j c r e s u l t s .

One of' t,he main r e s l ~ l t s i n Sec t ion 3 a s s e r t s t h a t i f

G is a p-group and M a subgroup, maxima!_ wSth :respect t o

be ing a b e l i a n , normal and exp(M) < pn, t h e n every element

of o rde r a t most pL' which c e n t r a l i z e s M l i e s i n M,

n u n l e s s p = 2 . Consider ing t h e p a r t i c u l a r case when G has

exponent p" we s e e t h a t t h e above s ta tement e s s e n t i a l l y

amounts t o s ay ing t h a t eve ry maximal a b e l i a n normal subgroup

of G i s s e l f - c e n t r a l i z i n g . If n = 1 and p > 2 , t hen w e

o b t a i n a u s e f u l consequence: every element of o rde r p which

l i e s i n C ( M ) i s i n M .

S e c t i o n 4 is devoted t o t h e i n v e s t i g a t i o n of t h e number

of g e n e r a t o r s of a b e l i a n and a b e l i a n normal subgroups of

p-groups. We in t roduce t h e terms 'dep th! and 'normal d e p t h '

of a p-group G t o denote m a x r m ( ~ ) 1 , where A ranges r e s p e c t i v e l y

over a l l a b e l i a n and a l l a b e l i a n normal subgroups of G, and

where m ( ~ ) denotes t h e mini-ma1 number of g e n e r a t o r s of A .

Our i n t e r e s t i n t h e s tudy of t h e normal depth i s s t i m u l a t e d

by t h e fo l l owing two c o n s i d e r a t i o n s . The f i r s t i s t h e f a c t

t h a t i n t h e s tudy of groups of odd order ( ~ e i t and Thompson,

[8]), t h e r e i s a fundamental d i v i s i o n of Sylow subgroups i n t o

two c l a s s e s : t hose which have ? -genera tor a b e l i a n normal

subgroups and t h o s e t h a t do n o t . The second r ea son i s t h a t

t h e l a r g e s t possibZe normal dep th p lays an important r o l e

i n t h e proof 01' t h e Thompson normal p-complement theorem

whicn g ives a c r i t e r i o n f o r t h e e x i s t e n c e of a

subgroup N of a f i n i t e group G such tha , t ( p , 1 N I ) = 1 and

r I G / N I = p ( f o r some p o s i t i v e i n t e g e r r ) i n terms of a

p-Sylow subgroup P, t h e c e n t r a l i z e r of t h e c e n t r e of- P and

( v i i )

t h e normal izer of t h e s o - c a l l e d Thompson subgroup of P.

For a d e t a i l e d t r ea tmen t of t h i s theorem, t h e r e a d e r i s

r e f e r r e d t o [ I 4 1 ( ~ a p t i l IV).

A remark.able r e s u l t ( s e e Theorem 4 .1 ) due t o J . G.

Thompson i n h i s unpubl ished work s t a t e s t h a t i f G i s a

p-group of odd o rde r such t h a t every a b e l i a n normal subgroup

has a t most k g e n e r a t o r s t hen every subgroup of G has a t

k+ 1 most ( ) g e n e r a t o r s . Whether o r no t ( "I) i s t h e b e s t

p o s s i b l e bound on t h e number of g e n e r a t o r s of subgroups of

G seems t o be unknown. I n t h e case k = 1 , t h e above

a s s e r t i o n reduces t o t h e c l a s s i c a l r e s u l t which says t ha t ,

i f a p-group ( p > 2) has t h e p r o p e r t y t>ha t every a b e l i a n

normal subgroup has one g e n e r a t o r t hen G i s i t s e l f a c y c l i c

group . I n t h i s s e c t i o n , we a l s o i n t roduce t h e p r o p e r t y Ak:

a p-group i s s a i d t o have p r o p e r t y A i f eve ry a b e l i a n k

normal subgroup can be gene ra t ed by k. e lements . N. Blackburn

[5 ] has complete ly determined t h e s t r u c t u r e of p-groups of

odd o rde r which have p r o p e r t y A2. We c l a s s i f y p-groups d

(p > 3 ) which have p r o p e r t y A2 by means of t h e t h e o r y of

r e g u l a r p-groups ( s e e Theorem 4.6), and deduce from t h i s

c l a s s i f i c a t i o n tha . t f o r p-groups of odd o rde r , p r o p e r t y

A 2 i s subgroup i n h e r i t e d . C . Hobby [ I I] has a l s o proved

t h a t f o r p-groups of odd o r d e r , p r o p e r t y A i s subgroup 3 i n h e r i t e d . The g e n e r a l problem about p r o p e r t y A k , f o r

k > 3, s t i l l l i e s open.

( v i i i )

The l a s t s e c t i o n f a l l s in t .0 t h r e e subsec t ions i n

which d i f f e r e n t t echn iques a r e involved i n t h e p r o o f s of

v a r i o u s s t a t e m e n t s . Apart from t h e theorems t h a t a r e

concerned, we l a y cons ide rab l e emphasis i n e x h i b i t i n g t h e

t echniques used i n t h e p r o o f s . For i n s t a n c e , i n showing

tha . t i f a p-group of odd o rde r has an a b e l i a n subgroup or

index p3 then it has a n a b e l i a n normal subgroup of index

p3, we use t h e Jordan canon ica l f orrn of a non-s ingula r l i n e a r

t r a n ~ f o r m a t ~ i o n . While doing s o , we r ega rd a n e lementary

a b e l i a n p-group H a s a v e c t o r space over Z t h e f i e l d of P'

i n t e g e r s modulo p , and e lements of AU~(H) a r e viewed a s

non-s ingu1a.r l i n e a r t r ans fo rma t ions of H. The appearance

of wreath p roduc ts i n t h i s s e c t i o n i s because of t h e i r

u s e f u l n e s s i n p rov id ing counterexamples t o c o n j e c t u r e s .

The l as t p a r t of S e c t i o n 5 i s concerned wi th a l t e r n a t i n g

forms and t , h e i r connec t ions w i t h and a p p l i c a t i o n s t o p-groups

o f c l a s s two.

S e c t i o n 1 Notat ion and Terminology

I

A f i n i t e p-group i s a f i n i t e group whose o r d e r i s a

power of t h e p r i m e , p . Throughout t h i s t h e s i s a p-group w i l l

mean a f i n i t e p-group. Nota t ion and terminology t h a t w i l l

a r i s e a r e b r i e f l y desc r ibed below. I n what fo l lows l e t

R, S denote any nonempty s u b s e t s and H any subgroup of a

group G.

Z+ s e t of n a t u r a l numbers.

z~ f i e l d of i n t e g e r s modulo p, p a prime.

'n c y c l i c group of o r d e r n .

1 i d e n t i t y element o r i d e n t i t y subgroup of a group

( u s u a l l y , t h i s w i l l be c l e a r from t h e c o n t e x t ) .

Z ( H ) c e n t r e of H.

c l ( H ) c l a s s of H ( t o be de f ined l a t e r ) .

I H 1 o r d e r of H ( 1x1: o r d e r of t h e element x ) .

minimal number of g e n e r a t o r s of H.

dep th of a p-group G ( t o be def ined i n S e c t i o n 4 ) .

normal depth of a p-group G ( t o be de f ined i n

S e c t i o n 4 ) .

c e n t r a l i z e r of S i n H.

normal ize r of S i n H.

When t h e r e i s no danger of confusion, we s e t

C ( S ) = c G ( s ) , N ( S ) = N G ( s ) .

subgroup gene ra t ed by S.

- 1 y x y, c a l l e d a conjuga te of x .

{X-lS X : B E S1

<sX : x s G>, t h e normal c l o s u r e of S i n G.

- 1 - 1 x y x y, a commutator.

H is a subgroup of G; < I means proper containment . H is a normal subgroup of G.

index of H in G.

automorphism group of H.

H is a characteristic subgroup of G.

Frattini subgroup of H, i.e., the intersection of

all maximal subgroups of H. k <x : x E G, Ix 1 < p >, where G is a p-group.

k <xP : x E G>, where G is a p-group.

We will write Q(G) = Cll (G) and U(G) = U1 (G).

wreath product of a group A by a group B (to be

defined in Section 5

exponent of H, i.e.,

of the orders of the

abelian and exp(~) =

).

the least common multiple

elements in H. If H is

p, then H is called elementary

abelian. We shall say an elementary abelian

group of order pn is of type (p, . . . , p) where p appears n times in the brackets.

0 The derived series of G is defined as : G = G,

G' = G(') = [G, GI, G ( k + l ) = [G(~), G(k)] for k 2 1 . The

n) least positive integer n satisfying G( = 1 will be called

the derived length of G. By the lower central series of G

we mean the invariant series G = G 2 Gg 2 ..., such that 1

Gp = [G, G] and G = [G,, G] for n 2 1. The upper n+ I

central series is the invariant series i=Zo(~)c~l(~)s...,

defined by the rule Zi(~) = n-I (z(G/~ (G) ) ) , where n i- 1

i s the canonical homomorphism ri : G - G/zi ( G ) , -

i = I 2, . . . . It i s well-kAown t h a t , i f G i s a p-group,

m ) t hen t h e r e e x i s t m, n E Z+ such t h a t G ( = 1 = Gn+, and

z,(G) = G. If n i s t h e l e a s t p o s i t i v e i n t e g e r such t h a t

Gn+ I = 1 , we say t h a t G has c l a s s n. A group G i s metacyl ic

(metabel ian) i f G possesses a c y c l i c ( a b e l i a n ) normal

subgroup N such t h a t G/N i s c y c l i c ( a b e l i a n ) . A p-group n n n n

pn G i s regular if a, b F G and n c Z+ imply (ab)' =aP bP cy ... c _

f o r some ci E (2, b>( ' ) . A group of order pn with n 2 3 ' 1

and c l ( G ) = n-I i s c a l l e d a p-group of maximal c l a s s . -. -- - - - . -

A p-group G i s c a l l e d s p e c i a l i f e i t h e r G i s elementary

abe l i an or G i s of c l a s s two and G ' = @ ( G ) = Z ( G ) i s

elementary abel ian ; i f , furthermore, G ' has order p, then

G i s c a l l e d ex t ra - spec ia l . A p-group G i s s a i d t o be

abso lu te ly r egu la r i f [G : = ( G ) 1 = p W ( G ) where W ( G ) < p.

It i s well-known t h a t an a b s o l u t e l y r egu la r p-group i s

regular . A p-group i s s a i d t o have property Ak i f every

abe l i an normal subgroup can be generated by k elements.

The c e n t r a l product of two f i n i t e groups H and K i s defined

a s fol lows:

Suppose A 2 z ( H ) , B 2 Z ( K ) and a : A - B i s an

isomorphism. Then H x K/[ (,, a (x- 1 )j : A 1 w i l l be

c a l l e d a c e n t r a l product of H by K, r e l a t i v e t o t h e p a i r

(A, B ) . I f A = z ( H ) and B = z ( K ) , then we s h a l l r e f e r t o

t h e c e n t r a l product of H by K, r e l a t i v e t o t h e p a i r (A, B )

simply as the c e n t r a l product of H by K.

Section 2 Classical and fundamental results I

2. I * - Classical- results

First of all, we shall summarize some classical results

showing the influence of the abelian subgroups on the global

structure of a p-group.

( 1 ) A p-group (p > 2) having property A1 is cyclic (see

[14]). his statement will be obtained as a corollary to Theorem 4.1.) The quaternion group of order 8 is an example

showing the necessity of p > 2.

(2) Among the non-abelian p-groups, those possessing a

cyclic subgroup of index p are characterized as follows [g]:

Let G be a non-abelian p-group of order pn which

contains a cyclic group of order p n- 2 and let a = l+p . n- 1 b a (i) If p is odd, G = -%, b : aP =bP=l, a =a >.

(ii) If p = 2 and n = 3, G is isomorphic to the dihedral

group or the quaternion group of order 8. 2n- I

(iii) If p = 2, and n > 3, then G = <a, b: a 2 b a =b =I, a =a >,

or G is a generalized quaternion group of order 2n.

Second, if G is a p-group how much information do we

know about the local structure. Needless to say, non-trivial

abelian normal subgroups always exist, e. g. the centre of G.

Since, however, it is often convenient to represent a

p-group as a permutation group on an abelian normal subgroup,

the "action" being conjugation, we would like to find abelian

subgroups which are non-central.

A classical result [7] states that if ( G I = pn then

there exists a maximal abelian normal subgroup M, / M / = pm,

such t h a t m ( w - 1 ) r 2n. This gives a lower bound f 01. m i n

t e r n s of n. Speaking informal ly we may conclude t h a t L

11 a b e l i a n normal subgroups of f a i r l y l a r g e " order e x i s t . O f

course, %he above theorem i s an exis tence theorem i n ' i t s

pures t form: it t e l l s us nothing about t h e s t r u c t u y e of M

o r where,. i n t h e l a t t i c e of subgroups of G, M l i e s . Also,

if t h e c l a s s of G i s given, say k, so t h a t Gk+l = 1, t hen

t h e r e l a t i o n G (i) G [10] shows t h a t G ( ~ ) = 1 , where

a 2 [log2(k+1 ) 1 t- 1. I n o t h e r words, G ( ~ - ' ) i s abelian,. A s

we shall. see i n Sec t i cn 3, if we impose t h e r e s t r i c t i o n

t h a t G be of maximal c l a s s , then we can say even more about

t h e exisLence of " l a rge" a b e l i a 2 normal subgroups. A f i n a l

remark i s t h a t ( c f . Theorern 3 . 4 ( i)) i f G has c l a s s k with

k 2 2i, t hen none of t h e maxima.l a b e l i a n narnal subgroups

of G i s contained i n 2. ( G ) ; t h e i r orders a r e a l l g r e a t e r 1

t han pi.

2.2. Related b a s i c r e s u l t s - Lis ted below a r e some known r e s u l t s we s h a l l use

f r equen t ly .

( 1 ) Every p-group of c l a s s l e s s than p i s r e g u l a r [ I 01.

( 2 ) Every p-group of o rde r a t most pP i s r e g u l a r [lo].

(3) In a r e g u l a r p-group G, I u ~ ( G ) I = CG: % ( G ) I [ I 01.

(4 j I n a p-group G, CZ~(G) ,G~I=I [I 0 I.

(5) If G has o rde r pn, then t h e order of a p-Sylnw n

subgroup of Aut(G) d iv ides p ( 2 ) [143.

(6) Let G be a group of order $ with p > 3. If G has

a t most pi-1 subgroups of order f o r some r i ~ ~ i t h

I s r l n - 2 , t h e a G i s r:letacyc!lric [ l h ] . I

(7 ) I n a r e g u l a r p-group G, [ u ( G ) , o ( G ) ~ = 1 c103.

We cl-ose thi .s s e c t i o n 5y giving without proof a w e l l -

known f a c t i n t h e theory of p-groups which we shall- make

use of quite often. T h i s f a c t reads a s fo i lows:

If G is a p-group and Ikf, N norxlal subgroups of G wri-kh

= pm, then f o r each k w i . t b Ggksrn, t h e r e e x i s t s a

H k norma.1 subgroup 11 of G s ~ ~ c l i t ha t P T ~ i ~ M and I /N 1 = p .

Section 3 Centralizers of abelian normal subgroups of --

p-groups I

We begin by proving a general proposition which, apart

from its future application, is of some interest as the

proof involved illustrates a general technique in proofs

concerning abelian normal subgroups.

Theorem 3.1. (i) If N is a normal subgroup of a p-group

2 G and if N has order p , then CG : C (N)] p.

(ii) If N is a cyclic normal subgroup of G,'

then C(N) 2 G I .

Proof. (i) If g c G, then since N 4 G, g induces by

con jugation an automorphism 6 of N. The mapping 6 : G-Aut (N) g

defined by ( g ) $ = ,dg is clearly a homomorphism of G into

~ u t (N) whose kernel is c (N). Hence c (N) d G and G/c (N) is

isomorphic to a subgroup of Aut (N). But, a p-Sylow subgroup

of Aut (N) has order p according to 2.2 (5). Thus, [G : C(N) ]

can be divisible only by 1 or p and the first part of the

assertion follows.

(ii) In the proof above we have just seen that

G/~(~) is isomorphic to a subgroup of Aut (N). Since N is

cyclic Aut (N) is abelian and so is G/c(N). Therefore

G' < c(N).

Remarks. (1) Theorem 3.l(i) no longer holds in case we

delete lnormalityl of N. As a matter of fact, N. Blackburn

[4] has shown that for odd p, any p-group of maximal class

contains a subgroup H of order p such that N(H) is elementary

abelian of type (p , p). Making use of this fact we conclude

e a s i l y t h a t f o r odd p any p-group of maximal c l a s s contains

subgroups of order p' which a r e s e l f - c e n t r a l i z i n g . 1

( 2 ) From p a r t (ii) of the above theorem, we may

ob ta in r e s u l t s on t h e p o s i t i o n of C ( N ) i n the l a t t i c e of

subgroups of G by r e s t r i c t i n g t h e s t r u c t u r e of N.

Lemma 3.2. Let G be any group and H, K two normal subgroups

with H r; K. Then K/H 2 z ( ~ / H ) i f arrl only i f CK, G] H.

< z ( G / ~ ) i f and only i f [ ~ k , ~ g ] = H fo r , a l l Proof. K/H -

k s K, g E G, i . e . , i f and only if [k, gl F H f o r a l l k E K,

g E G . The l a t t e r holds i f and only i f [K, GI 2 H.

Lemma 3.3. Let G be a p-group. If N 4 G and I N ( r pn, then

N s z ~ ( G ) .

Proof. Induct on n. For n = 1 , t h e a s s e r t i o n i s e v i d e n t l y

t r u e . Assume t h a t it i s v a l i d f o r k-I and l e t N - 4 G,

I N / = pk. Then N conta ins a normal subgroup M of G such t h a t

[N : M I = p. By induct ion hypothesis , M 2 Zk- ( G ) . Since

N / ~ 2 z ( ~ / ~ ) , EN, GI 2 M by Lemma 3.2. Hence

[N, G I 5 Zk-, ( G ) from which fol lows N 2 Z k ( G ) .

Theorem 3 . 4 . ( i) If M i s a maximal a b e l i a n normal subgroup

of a p-group G, then c ( M ) = M and i n p a r t i c u l a r , M i s a

maximal a b e l i a n subgroup of G. Moreover, i f 1 M 1 = then

G, M and G2, = 1.

(ii) If M I , M a r e maximal abe l i an normal 2 mi subgroups of a p-group G with e x p ( ~ ~ ) = p (i = I , 2 ) ,

then m2 5 2ml (and of course, ml 2 2m2).

Proof. ( i) Clearly, M 2 C ( M ) a s M i s abe l i an . Also,

G - C s ince M A G, c ( M ) 4 G. Se t C = c (b:), G = C = iM and

- - suppose M < C. Then > I and C 4 G , whence n z(<) > 1.

- - - - - - Let B C n Z ( G ) ::ilch t h a t B has order p. Then B 4 G

- and so A 4 G, wherc B i s tllc ?amplete inverse image of B i n

G. It i s evident t h a t B 2 C . -

Now, B = B/M has order p, so t h e r e e x i s t s b s B - M f o r

which B = < b, M >. Since b E C = c ( M ) , B i s abe l i an . We

have thus obtained an a b e l i a n normal subgroup B > M,

con t rad ic t ing t h e maximality of M. Hence C ( M ) = M.

Since ] M I = pm and M a G, M s z,(G) by Lemma 3.3. I n

If G has c l a s s c , then i t fol lows from I M 1 = pm and

Gm > Gm+l > ... > Gc+, = 1 t h a t c s 2m - 1 . Therefore GPm = 1 .

(ii) Let N = < M I , M2 >. Then

and s o Z ( N ) = MI n M2. Since [ M I , ~~1 < M n M2 = z ( N ) , 1

we conclude t h a t N has c l a s s a t most two. It fo l lows t h a t

Remark. It follows from t h e theorem above t h a t once we

know the order of a maximal a b e l i a n normal subgroup of G we

can bound both t h e c l a s s of G and t h e order of any o the r

maximal abe l i an normal subgroup of G.

More genera l ly , t h e r e s u l t of t h e f i r s t p a r t of Theorem

3 . 4 ( i ) i s s t i l l v a l i d provided G i s a n i l p o t e n t group

without necessa r i ly being a p-group.

Using t h e preceding r e s u l t s together with a f a c t i n

Scc t ion 2 w e can e a s i l y deduce t h e fol lowing well-known

c l a s s i c a i r e s u l t [7 3. ,

Theorem 3.5.(a) Every group of order pn n e c e s s a r i l y con ta ins --- a n a b e l i a n n o m a l subgroup of order pm, where m ( m f 1 ) 2 2n.

Furthermore, every group of order pn with n 2 4 con ta ins

an a b e l i a n normal subgroup of order p3. +

Proof. Let \G ( = pn. The a s s e r t i o n holds t r i v i a l l y i f G

i s abe l i an . Since G i s f i n i t e , it i s poss ib le t o p ick

a maximal a b e l i a n normal subgroup M of G. Ey Theorem 3 . 4 ( i ) ,

C ( M ) = M and f r o n t h e proof of Theorem 3 . l ( i ) we have t h a t

G/M i s isomorphic t o a subgroup of Aut ( M ) . Suppose M

has order pm. Then i n view of 2 .2 (5 ) , t h e order of a ( 3 )

p-Sylow snbgroup of A u t ( ~ ) d iv ides p . . Thus I = p n-m

( 3 ) a l s o d iv ides p , whence p n-m * P

(5) from which fol lows

m ( m + 1 ) 2 2n, a s a s s e r t e d .

For p-groups of maximal c'ass, t h e r e e x i s t a b e l i a n

normal subgroups of ' l a r g e s i z e r a s w i l l be j u s t i f i e d i n

a statement fol lowing these 'remarks.

We r e c a l l a p-group of maximal c l a s s i s a p-group of

o rde r pn and c l a s s n-I , where n 2 3. Evidently, such a

group i s non-abelian. Blackburn [4] has s tud ied such groups

i n g r e a t d e t a i l . A p-group of maximal c l a s s has t h e fol lowing

main p roper t i e s :

( I ) G / G ~ i s elementary abe l i an ( so G2 - @ ( G ) ) of order

p2; G : Gi+, 1 = p f o r 2sisn-1.

( 2 ) G has p + 1 maximal subgroups.

* See Appecdix,

k (3 ) G has a unique normal subgroup of index p ,

1

k = 2, ..., n.

We a l s o remark t h a t i n any p-group G, i f N 4 G, N Gi,

and i f N i s non-abelian, then z ( N ) has order a t l e a s t pi.

Indeed, N 6 z ~ ( G ) s ince [Gi, Z i ( G ) 1 = 1 by 2 . 2 ( 4 ) . Now

N f l z ~ ( G ) 4 z ( N ) . We have, f o r j $ i,

Iience, 1 < N fl Z 1 ( G ) < N n z ~ ( G ) < ... < N fl z ~ ( G ) and s o

pi s /N n z~(G)\ I z ( N ) I . This j u s t i f i e s our a s s e r t i o n .

With t h e preceding remark and t h e p r o p e r t i e s of p-groups

of maximal c l a s s , we e a s i l y deduce

Theorem 3.5.(b) I f G i s a p-group of maximal c l a s s and i f G

has order pn, then G possesses an a b e l i a n normal ( i n f a c t ,

t o be prec ise , a unique c h a r a c t e r i s t i c a b e l i a n ) subgroup of

m n+ 1 order p , where m 2 [-I. 2

Proof. I f n i s even, say n = 2k, then Gk i s abe l i an , f o r

otherwise, i t s cen t re z ( G k ) would have order a t l e a s t p k

k by t h e foregoing remark; but Gk a l s o has order p , hence

Gk i s abe l ian .

k+ I I f n i s odd, l e t n = 28 + 1 . Then Gk has order p .

k I f Gk i s non-abelian, I z ( G ~ ) 1 2 p by t h e remark p r e c e d h g

t h e theorem. Hence, [ G ~ : I Z(G~)] 2 p, a con t rad ic t ion .

Thus, G must be a b e l i a n . k

The uniqueness fol lows f ~ o m proper ty (3) of a p-group

of maximal c l a s s .

Renark. Although p-groups of ~naximal c l a s s possess a'celian

normal su'bgroups of f a i r l y l a r g e cxder, Theorem 3 5 (a)

does no t guarantee every p-group of order pn conta ins n -

a b e l i a n normal subgroups of order pC2'. I n f a c t , Alper in [3 1

has cons t ruc ted f o r every prime p 2 5, a group of order pn, 2

where a = 6p- -t p -I- I , a l l of whose mnaxim:&l a b e l i a n subgroups

2 have order a t most pm, where m = 3p . Before s t a t i n g t h e main r e s u l t i n this secti .cn we f i rs%

prove a c s e f u l lemma.

Lemma 3.6. Let x and y be any two elements of a group G and

l e t z = [x, y]. If z c e n t r a l i z e s both x and g, then . .

(i) [xi, y j ] = z l J f o r a l l i n t e g e r s i, j;

i i ($ (ii) ( x y ) ' = x y Cy, X I f o r i 2 1 .

Proof. (i) z = [x, y ] implies xY = xz. Tnus,

i i i y = (& = ( X Z ) i = X 2 (x

s i n c e z c e n t r a l i z e s x, and

i 2 i r - x z

Contiriuing t h i s conjugat ion by y j-2 more times., we f i n a l l y

f o r j 2 0, while if j < 0, we can replace y by y-' i n t h e . .

above argument. Hence, [xi, y j ~ = z l J f o r a l l i n t e g e r s i, j

and t h i s proves t h e f i r s t p a r t of the lemma.

(ii) Induct on i, For i = 1, it i s obvious.

Then

++ ( ~ ~ ) ~ = ( x y ) ~ - ' ( x y ) = x i-I y i-I [y, XI ) x Y

Now, l e t t i n g i = 1 and j = i - I i n (i) we ge t

** yi-lx = xy i - 1 ~ 1 - i

S u b s t i t u t i n g (**) i n (*), we obta in

This completes t h e proof .

We a r e now i n a p o s i t i o n t o s t a t e t h e main theorem i n

t h i s s e c t i o n which i s due t o Alperin [ I 1. A s we s h a l l see

l a t e r , t h i s r e s u l t w i l l prove t o be very f r u i t f u l . The

proof we give here der ives c h i e f l y from an unpublished work

of Blackburn.

Theorem 3.7 [I 3. Let G be a p-group and M a maximal element

n -!- of t h e s e t C M : M I = I , M q G, e x ~ ( ~ ) 5 p 3, where n E Z . n

If x E C(M) and xP = 1 , t hen x E M, un less pn = 2..

n Proof. Let x c c ( M ) , xP = 1 and assume x , d M. Among a l l

t h e elements x s a t i s f y i n g t h e above condit ions, we may

pick an x i n such a way t h a t t h e subgroup X = < xg : g E G >

has smal les t poss ib le order . Let B = < x, M >; then B i s

abe l i an s ince M i s abe l i an and x c c ( M ) . Hence B has

exponent a t most pn. By t h e maximality of M, we see t h a t

B # G, whence [B, G ] $ B. ?"ne l a t t e r implies t h e r e e x i s t s

y c G f o r which [x, yl ,,d M.

Again, we may pick an element y i n such a manner t h a t

Y = < yg : g E G > has smal l e s t poss ib le order . S e t t i n g

z = [x, y ] we have z ,d M. Since x s C ( M ) and C ( M ) 4 G

it follows t h a t z = [x, y ] = x-Ixy c c ( M ) . Thus z s C ( M ) - M.

Furthermore, z = [x, y ] t: [x, G I < X. ( T O j u s t i f y t h e

l a s t i n e q u a l i t y we f i r s t note t h a t 1 < X 4 G, so t h e r e e x i s t s

W 4 G such t h a t W < X and [X : w 1 = p. It fol lows t h a t

X/W I z ( G / w ) , whence [x, G I I W < X by Lemma 3 . 2 . ) We have

< z g : g E G > ~ [x, G I

a s [x, G I 4 G, and

I < zg : g E G > 1 2 1 [x, G I I < 1x1. I n view of t h e choice of x, we must have zpn f 1 . Now, i f

[x, z ] M, then we can replace y by z . However,

z = Ex, y l E [G, Y 1 < Y, with CG, Y 1 4 G. This c o n t r a d i c t s

the choice of y and s o x z 3 E M.

Since x, z E c ( M ) , [x, z l E ~ ( x ) n ~ ( z ) . By

minimality of I x 1, xP E M and it i s the re fo re immediate t h a t

xp c ~ ( x ) f l ~ ( z ) , whence ExP, Z ] = I . By Lemma 3 . 6 ( i )

t h i s implies

* I = [x P , z] = rx, ZIP. I

On the o ther hand, we have

(by Lemma 3.6 (ii) ), n

n Together, (* ) and (**) imply p = 2. This e s t a b l i s h e s t h e

proof of t h e theorem.

Remark 1. I n case \ G I = pn i n the above theorem, t h e f i r s t

p a r t of Theorem 3 . 4 ( i ) fol lows a s a p a r t i c u l a r case of Theorem

Remark 2. To see why t h e case pn = 2 i s an except ion i n

t h e foregoing theorem, consider the d ihedra l group

of order 16. This group has nine elements of order two,

4 namely, t h e elements of t h e cose t < a > b toge the r with a ,

and has only one a b e l i a n normal subgroup of exponent two,

4 namely i t s cen t re Z ( D ) = < a >. Now, b ev iden t ly c e n t r a l i z e s

4 < a4 > and has order two, y e t b ,d < a >.

Alper in ls proof of t h e previous theorem makes use of

t h e fol lowing lemma which i s of independent i n t e r e s t .

Lemma 3.8 [I I. Let A be an abe l i an p-group and q an auto-

- 16 -

morphism of A with IT- 1 a d i v i s o r of pn. If q s t a b i l i z e s

Q,(A) ( i . . , q f i x e s Q,(A) elementwise), t hen .rl s t a b i l i z e s

A / ~ $ ( A ) except when pn = 2.

The proof of t h i s lemma was c a r r i e d out by induct ion

on the order of A. For d e t a i l , t he reader i s r e f e r r e d t o [ I 1.

Direc t consequences of Theorem 3.7 a r e the fol lowing.

Corol lary 3.9. Let G be a p-group with p odd. Let A be an

abe l i an normal subgroup of G such t h a t m ( A ) i s as l a r g e a s

poss ib le , where A ranges over a l l a b e l i a n normal subgroups

Corol lary 3.10. Let G be a p-group with p odd. Suppose

A 4 G and A < E, where E i s an elementary a b e l i a n subgroup

of G, then t h e r e e x i s t s an elementary a b e l i a n normal

subgroup B of G with [B : A ] = p.

Corol lary 3.11. Let G be a 2-group a l l of whose a b e l i a n

normal subgroups can be generated by two elements. I f

H C4 x C i s normal i n G, then every element of order 4 not exceeding 4 which c e n t r a l i z e s H l i e s i n H.

We conclude t h i s s e c t i o n by g iv ing an a p p l i c a t i o n of

Theorem 3.7. Other a p p l i c a t i o n s w i l l be found i n Sect ion 4.

The next r e s u l t i s due t o J. G. Thompson and the proof

presented here i s due t o N. Blackburn.

Theorem 3.12. Let p be an odd prime and G a p-group. If

each element of order p l i e s i n t h e cen t re of G, t h e n

m ( ~ ) ~ ( z ( G ) ) .

Proof. Induct on Ic - 1. The r e s u l t i s t r i v i a l i n case G I

has o rde r p. S e t B = R ( G ) . Then by hypothes i s ,

I3 = R ( z ( G ) ) z ( G ) , whence m ( ~ ) = ~ ( z ( G ) ) . Let m ( B ) = k

and suppose A/B i s a subgroup of' G/B maximal with r e s p e c t t o

b e i n g a b e l i a n , normal and exp(A/B) = p. S e t G = G/B

and = A/B. A s i s e lementary a b e l i a n , A '~BS(G).

Hence, c l ( A ) 5 2 and exp(A ' ) = p.

F i r s t , we show t h a t Ti = R ( z ( ~ ) ) . Let a E A, g E Gy

and b = [a, g]. Then, no rma l i t y of A imp l i e s b c A and -

exp(A) = p imp l i e s aP c B. We t h u s have -

( P ) by Lemma 3 . 6 ( i i ) . But p > 2, s o [by a ] = 1 and ( * )

reduces t o bP = 1, whence b c n ( G ) = B. Hence [A, G I C B

and s o A 5 ~ ( 2 ) by Lemma 3 .2 . Since A has exponent p, it i s

obvious t h a t A 5 ~ ( z ( G ) ) . On t h e o t h e r hand, i n view of

t h e choice of A, we conclude by t h e previous theorem t h a t

z . US, A = n(z(G) ) .

Next, we s h a l l show m ( G ) s k. Since A has c l a s s a t most

two and p i s odd, A i s r e g u l a r by 2 .2 ( 1 ) . A s = A/B i s

e lementary a b e l i a n , V ( A ) 5 B. Thus,

(+") I A / B I / A / u ( A ) I = I I I B I = P k

i n which t h e f i r s t e q u a l i t y fo l lows from 2 . 2 ( 3 ) . It fo l lows

t h a t m ( ~ ) r k. We may now a p p l y our i n d u c t i o n hypo thes i s

To complete t h e proof , we d i s t i n g u i s h between two cases

according t o whether R 5 @(G) or B 6 @(G). I n the f i r s t case,

m ( ~ ) = ~ ( G / @ ( G ) ) = ~ ( G / B ) 5 k

and we a r e done. As f o r the o the r case, @(G) being t h e

i n t e r s e c t i o n of a l l maximal subgroups of G, we can f i n d a

maximal subgroup M of G such t h a t B / M. Now,

Z ( M ) 2 M n Z ( G ) 2 M n B. Moreover,

M n~ = {X B M : XP = 1 1 = Q ( z ( M ) ) ,

which shows t h a t each element of M of order p l i e s i n z ( M ) .

Hence, once aga.in, we may apply t h e induct ion hypothesis t o

ob ta in

m ( ~ ) 2 ~ ( z ( M ) ) = ~ ( R ( z ( M ) ) ) = m ( ~ n M ) 2 k-1.

The proof i s now complete by not ing

m ( G ) m ( M ) + 1 k = ~ ( z ( G ) ) .

Sect ion 4 'Depth! and 'normal depth' of p-groups

Following D. or en stein ' [9 1 we introduce t h e terms

Idepth' and 'normal depthr of a p-group G t o denote the

maximum of m(~), where A ranges r e s p e c t i v e l y over a l l

a b e l i a n and a l l normal abe l i an subgroups of G. Thus, we

w r i t e d ( ~ ) , d , ( ~ ) f o r t h e depth, normal depth of G, r e spec t ive ly .

The motivation t o i n v e s t i g a t e p-groups with known normal

depth a r i s e s from t h e s tudy of groups of odd order C81 i n

which t h e r e i s a fundamental d i v i s i o n of Sylow subgroups i n t o

two c las ses : those which have 3-generator a b e l i a n normal

subgroups and those which do not . I n p rac t i ce , t h e l a t t e r

type of groups i s not easy t o dea l with (cf . [ I 1 ) . The main

purpose of t h i s s e c t i o n i s t o s tudy p-groups whose normal

depth i s small. Blackburn [5 1 has determined t h e s t r u c t u r e

of p-groups ( p > 2 ) whose normal depth i s two. Nevertheless,

2-groups having normal depth two a r e s t i l l f a r from being

handled.

To begin with, we s t a t e a very i n t e r e s t i n g theorem of

J. G. Thompson i n h i s unpublished work; t h e proof of t h i s

i s , again, due t o N. Blackburn, and it de,pends mainly on

Theorems 3.7 and 3.12 of t h e previous s e c t i o n .

Theorem 4 . 1 , Let p be an odd prime and G a p-group with

d , ( ~ ) r. Then every subgroup of G can be generated by

r+ 1 ( ) elements.

Remark. This g ives an upper bound f o r m ( ~ ) i n terms of t h e

normal depth of G, where H runs through a l l subgroups of G.

It i s n a t u r a l t o ask if ( r z l ) i s the b e s t poss ib le bound.

Whether or not t h i s i s the b&st poss ib le r e s u l t seems t o

be unknown.

Proof of Theorem 4 .1 . Without l o s s of g e n e r a l i t y , we may

assume t h a t t h e r e e x i s t s a maximal a b e l i a n normal subgroup

A of G with m ( A ) = r. (Fiere, we have assumed, without l o s s

of genera l i ty , d , ( ~ ) = r. ) Let B = n ( ~ ) and C = C ( B ) . Then obviously, B has exponent p and order pr. Applying

Corol lary 3.9, we see t h a t

Q ( c ) = Q ( c ( B ) ) = B.

Pick a subgroup S of G. It i s immediate t h a t S n C i s

normal i n S and t h a t < S, C > = SC i s a subgroup of G. We

have

S/S n c SC/C 5 G/C.

As i n the proof of Theorem 3. I (i), we know t h a t G/C i s

isomorphic t o a p-subgroup of A U ~ ( B ) . By 2.2(5), t he r

order of a p-Sylow subgroup of A u t ( ~ ) d iv ides p ( 2 ) , whence (5) [ s : s n c l r p .

We s h a l l show t h a t any subgroup of C can be generated

by r elements. Let D 5 C so t h a t D commutes with B

elementwise. I n p a r t i c u l a r , D 4 DB. We observe next t h a t

DB = D x B1 f o r some s u i t a b l e subgroup B1 i n DB. To see

t h i s , we f i r s t r e c a l l t h a t B i s elementary a b e l i a n of order

r p , s o it has a bas i s [ b l , . . ., b,], say. I f D n B = 1, w e

take B = B and a r e done because B 4 DB. Otherwise, we may 1

assume

D n B = al , ..., bs >.

Let B = < b 1 s+1 ' .. ., br >. ,Then evident ly , D n B1 = 1 ,

DB = DB, and s ince D commutes with B elementwise, B 1 4 DB. 1

Hence, DB = D x B, a s des i red . From t h i s , it fol lows t h a t

m ( D ) 5 ~ ( D B ) . As we wish t o demonstrate m ( D ) 5 r, we may

assume without l o s s of g e n e r a l i t y t h a t B I D C f o r we can

replace D by DB. Then B = s ~ ( c ) = R ( D ) . Since CD, B] = 1 ,

B 5 z ( D ) , whence R ( D ) s z ( D ) . That i s t o say, every element

of D of order p l i e s i n z ( D ) . Using Theorem 3.12, we conclude

m ( D ) ~ ( z ( D ) ) = ~ ( O ( Z ( D ) ) ) = m ( ~ ) = r.

It remains t o show m ( s fl C ) 5 r. However, t h i s i s

evident a s we have jus t demonstrated m ( D ) 2 r whenever D i s

a subgroup of C, so m ( s n C ) < r. The theorem i s proved

by not ing

An immediate consequence of t h e preceding theorem i s

the following well-known c l a s s i c a l r e s u l t .

Corol lary 4.2. Every p-group ( p > 2 ) having normal depth one

i s c y c l i c .

Theorem 4.3. Let p be an odd prime and G a p-group f o r which

d n ( ~ ) 2. Then d ( ~ ) 2 2.

Proof. Pick a maximal a b e l i a n normal subgroup t o maximize

m ( ~ ) and we may assume without l o s s of g e n e r a l i t y t h a t

m ( ~ ) = 2 , f o r otherwise, G would be c y c l i c by Corol la ry 4.2

s o t h a t d ( ~ ) = 1 .

Suppose G conta ins an elementary a b e l i a n subgroup B

3 of order p . Let B1 = c ~ ( R ( I / ) ) ; then, s ince 0 ( ~ ) 4 G and

2 Q ( A ) has order p , we have [B : B1 1 p by Theorem 3. I (i).

2 Hence I B ~ 1 2 p . On t h e o the r hand, B1 5 0(A) by Corol lary

3.9, whence B1 = 0 ( ~ ) . Now, B c e n t r a l i z e s Q ( A ) and B $ Q ( A ) .

I n o ther words, t h e r e e x i s t s an element of order p which

c e n t r a l i z e s Q(A) and y e t t h i s element l i e s outs ide Q ( A ) . This

con t rad ic t s Corol lary 3.9, s o t h e proof i s complete.

Corol lary 4 .4 . Let G be a p-group ( p odd) f o r which d n ( c ) = 2.

Then d ( ~ ) 2 2 whenever H s G. I n p a r t i c u l a r , d,(H) 2 2

f o r a l l H 5 G.

We note t h a t while Theorem 4.1 gives an upper bound f o r

m ( ~ ) i n terms of t h e normal depth, d n ( ~ ) of G, t h e above

c o r o l l a r y gives a n upper bound f o r d(H) and d , ( ~ ) , where H

i s any subgroup of G, when d , ( ~ ) = 2. I n the s p e c i a l case

when d , ( ~ ) = 2, t h e c o r o l l a r y provides t h e b e t t e r bound f o r

d(H); from Theorem 4.1 we have d ( ~ ) 5 3, while from Corol la ry

4 .4 we g e t d ( ~ ) 5 2.

It i s a l s o c l e a r from t h e previous c o r o l l a r y t h a t , f o r

odd p, proper ty A2 i s subgroup i n h e r i t e d .

Theorem 4.5. Let p 3 and G a p-group f o r which d n ( c ) = 2.

Then e i t h e r G i s a b s o l u t e l y r e g u l a r o r G i s a p-group of

maximal c l a s s .

Proof. I n view of Theorem 1 . 1 [5], it s u f f i c e s t o show t h a t

any normal subgroup of exponent p i s of order l e s s than pP.

P Deny and suppose N 4 G, e x p ( ~ ) = p and IN 1 = p . By

Corol la ry 4.4., d , ( ~ ) 2. ~ h t then Theorem 3 . 5 ( a ) implies

t h a t N has an abe l i an normal subgroup M of order p3 s i n c e

p > 3. Since N has exponent p, e x p ( ~ ) = p. Thus, m ( ~ ) = 3,

contradicting t h e f a c t t h a t d , ( ~ ) 5 2 . Hence, t h e a s s e r t i o n

fol lows.

As w a s mentioned a t t h e beginning of t h i s s e c t i o n ,

p-groups of odd order whose normal depth i s two have been

completely determined (cf . [5 I). By adding t h e condi t ion

p > 3 we s h a l l now c l a s s i f y a l l p-groups G f o r which

d , ( ~ ) 5 2 by means of t h e theory of r egu la r p-groups.

Theorem 4.6. Let p > 3 and l e t G be a group of order pn with

d , ( ~ ) 2. Then G i s one of t h e fol lowing groups:

(i) G i s metacyclic. n- 2

p- p- p = [ X , Z ] = [ ~ (ii) G = < x, y, z : x -y - Z

I n t h i s case, G i s t h e c e n t r a l product of Z = < z > of order

pn- by t h e non-abelian group Y = < x, y > of order p3 and pn-3

exponent p, r e l a t i v e t o t h e p a i r ( < z >, < X Y Y > ( 1 ) ) .

Moreover, G i s t h e s p l i t t i n g extens ion of t h e a b e l i a n normal

subgroup < y, z > of type (pn-2, p ) by t h e automorphisrn

X x of order p : y = yz -pn-3 x , z = z. I n case n = 3, note

t h a t G i s non-abelian of order p3 and exponent p. n- 2

p- p- p = C Y , z] = 1 , (iii) G = < x, y, z : x -y -z

where n 2 4 and s = 1 o r s i s a quadra,tic non-residue modulo

p, I n t h i s case, t h e groups 'with the above defining r e l a t i o n s

together with s = 1 o r s = a quadra t ic non-residue (mod p )

a r e non-isomorphic, but both a r e t h e s p l i t t i n g extensions

of t h e a b e l i a n normal subgroup C ( G ' ) = < y, z > by t h e

automorphism x of order p:

yx = yz spn-3 x , z = yz.

Proof ( see t-141). For n = 3, G occurs i n (i) o r (ii) a s

mentioned above. Hence, consider n 2 4 i n the fol lowing

argument.

2 I f I R ( G ) I p , then G i s metacyclic by ( 6 ) i n Sect ion

2 .2 and s o G i s a group i n (1). Assume then \ R ( G ) 1 2 p5.

We s h a l l show f i r s t G i s regu la r . Since R ( G ) charG (and PO

R ( G ) 4 G ) O ( G ) cannot be a b e l i a n because of t h e hypothesis

t h a t dn (G) I 2. But O ( G ) contains a (necessa r i1y abel ian)

2 normal subgroup A of G whose order i s p . Set C = c ( A ) ;

then A = R ( C ) by Corol lary 3 .9 , whence R ( c ) < R ( G ) and hence

c < G. I n view of Theorem 3.1, [G : C] = p. As s ~ ( c )

has order p2 and p > 3, C i s metacyclic by Sect ion 2.2(6).

Therefore C/U(C) i s metacyclic of exponent p and i s thus of

order a t most p'. Now

[G : u ( G ) I 5 CG :'[I@)] s p 3

implies G i s a b s o l u t e l y r egu la r ( s ince p > 3 ) and hence

regular . It fol lows from (3) of Sec t ion 2.2 t h a t

( n ( ~ ) I . Thus, / O ( G ) I = p3. A s remarked e a r l i e r

R ( G ) i s non-abelian so z ( R ( G ) ) has order p. We claim t h a t

Q ( c ) # U ( C ) . Assume t h e c o n t r a r y . It fo l l ows from r e g u l a r i t y I

of G and 2.2(6) t h a t V ( G ) commutes wi th R ( C ) elementwise, s o

U ( G ) n Q ( G ) z ( R ( G ) )

and

which i s

n ( c ) <u(c) n Q ( G ) I U ( G ) n R ( G ) < z ( Q ( G ) ) ,

2 impossible s i n c e / R ( c ) 1 = p and I z ( Q ( G ) ) I = p. 2 Hence, Q ( C ) $ U ( C ) and t h i s imp l i e s , s i n c e 1 ~ ( c ) 1 = p , t h a t

U ( C ) con ta in s a s i n g l e subgroup of o rde r p . The re fo re ,

n-3 U ( C ) i s c y c l i c ( s e e [ I 43) and i s e v i d e n t l y of o rde r p . Mow, l e t z c C w i t h U ( C ) = < zP :, and l e t y E R ( c )

n-2 w i t h y n o t equa l t o a power of z . Then ( z / = p and

C = c y , z>. If R ( C ) # z ( c ) , t h e n t h e r e e x i s t s c c R ( C ) - Z ( C )

s o t h a t t h e subgroup < Z ( C ) , c > i s a n a b e l i a n normal subgroup

having 3 independent g e n e r a t o r s . Th is c o n t r a d i c t s d , ( ~ )

Hence R ( C ) 5 Z ( C ) and s o C = < y , z > i s a b e l i a n of t ype

s a t i s f i e s t h e r e l a t i o n s n-2

(R) xP = yP = zP = 1, [y, Z ] = 1 .

We s h a l l show t h a t G i s one of t h e groups i n (ii) o r pn-3

(iii) . R e c a l l t h a t u ( c ) = < zp >, t h u s 'ISnm3(c) = < z >

and Un-?(C) charC A G , whence 'CT ( c ) 4 G . S ince U ( c ) n-3 n- 3

has o rde r p , ru ( c ) 5 z ( G ) . The f a c t o r group R ( G ) / Y I - J ( ~ ) n-3

2 has o rde r p . By d e f i n i t i o n of Q ( G ) , we have

n-3 R ( G ) = < x, y, zP > and we know t h a t Q ( G ) i s non-abe l ian .

There fore ,

from which fo l lows

On t h e o the r hand, s i n c e xp 5 1 and G i s

1 = [ z , xP] = [z , x l P Y whence n-3

( * [z, X I E R ( C ) = < Y Y zP >.

r e g

A t t h i s p o i n t , we d i s t i ngu i - sh two p o s s i b i l i t i e s acco rd ing a s

x la

Cz, X I E un-+c) 0' [z . X I # u n w 3 ( c ) n- 3

I n t h e f i r s t c a se , [z, x] E U n W j ( c ) = < zP > imp l i e s n-3

( * [ z , X ] = zsp , i . e . , z X = z I fSpn-3 Y

f o r some int ,eger s . Since r # 0 (mod p ) , t h e r e always e x i s t s

a,n i n t e g e r t such that

( * s f rt - 0 (mod P ) .

Then w e s e e that,

( z y t ) X , , x ( Y x ) ~ = z yzrpn-3) ( b y ( ")and

= z n-3 t rtpn-3 l+sp y z ( s i n c e y commutes w i t h z ) ,

NOW, i f we r e p l a c e z by z = ( ~ y ~ ) ~ , t hen

( s i n c e n 2 4 ) , 1pn-3

s o yx = yz and t h e o t h e r r e l a t i o n s analogous t o ( R )

n-3 a r e xP = yP = z l P = 1 , [y, z ' ] = 1 . The re fo re ,

lp"-2 G = < X , y , Z' : xP = YP = z = [x, z l ] = [y, z'] = 1 ,

i s t h e r e q u i r e d group i n ( i i ) . n-3

I n t h e second c a s e , [z , XI ,d Un-3 ( c ) = < zp >, s o

by renaming we may s e t [z, x] = f o r some s u i t a b l e g

in C where I Y ~ = p. Then y ,d < z > and C = <y, z>=<y> x <z>, I

YX = yz rpn-3 , where r if 0 (mod p ) . Let t, be a quad ra t i c

non-res idue modulo p and l e t k be a n i n t e g e r such t h a t

2 2 k I- 1 (mod P ) o r k r 5 t (mod p ) . A s t r a i g h t forward

induc t ion w i l l show t h a t

k krp n-3 k n-3 Yx = yz , z = y z B 1 + ( 2 ) r p

k Set. a = x ,

t i o n w i l l y i e l d

Then a r o u t i n e computa-

where s i s equa l t ,o 1 o r t .

To sum up, we have demonstrated i n t h e second case t h a t n-2

~ = < a , b , c : a P = b P = c P = [b, c ] = 1 ,

where s = 1 o r s = t , a q u a d r a t i c non-res idue modulo p .

Namely, G appears i n (iii) . F i n a l l y , we show t h a t t h e s e groups determined by s a r e

non-isomorphic. To s e e t h i s , w e observe t h a t

G = < x, y, z > w i t h t h e d e f i n i n g r e l a t i o n s i n (iii)

t o g e t h e r w i t h s = 1 has t h e p r o p e r t y

where z E c ( G ' ) - G I and x E G - C ( G 1 ) . Consider now t h e

group G = < x , y , z > w i t h t h e d e f i n i n g r e l a t i o n s i n (iii)

t o g e t h e r w i th s = t , a qua.dra t ic non-res idue modulo p , and

suppose t h e r e a r e elements u, v , w i n G s a t i s f y i n g

pn-3 [u, v , v ] = u , u E C ( G 1 ) - G I , v E G - c ( G ' ) .

i j Then u = z y , v = wxl, where w E c ( G ' ) and i, j, 1 a r e some

i n t e g e r s . We n o t i c e tha t ,

T h i s completes t h e p roo f .

Remarks. ( 1 ) It fo l lows from Theorem 4 . 6 t h a t G ' i s a .be l ian

(and hence G i s metabe l ian) i f G i s a p-group ( P > 3) w i t h

d , ( ~ ) 2 . I n f a c t , a s a r e s u l t of Alackburnfs work [5],

G ' i s a b e l i a n even i f G i s a 3-group w i t h d,(G) 2 2. Thus,

we conclude that p-groups of odd o rde r which have normal

depth two have de r ived l e n g t h a l s o equa l t o two.

( 2 ) There i s no analogue f o r 2-groups which have

normal depth two. Recen t ly , Thompson [I51 remedied t h i s

s i t u a t i o n wi th t h e fo l l owing r e s u l t :

Let G be a non-abel ian 2-group f o r which d , ( ~ ) 2.

Suppose A i s a group of automorphisms of G of odd o rde r and

- 1 G = < g ~ ( g ) : E G y 7 € A >.

Then G i s isomorphic t o one of t h e fo l l owing :

( i) A qua t e rn ion group of o rde r 8.

(ii) The c e n t r a l p roduc t of a qua t e rn ion group of

o rde r 8 by a d i h e d r a l group of o r d e r 8.

(iii) A s p e c i a l group of o rde r 64 w i th e x a c t l y t h r e e

i n v o l u t i o n s ( a n i n v o l u t i o n i s a n element of o r d e r

two) , each of which i s c e n t r a l .

B. Huppert [ I41 showed t h a t i f a 2-group can be expressed

a s t h e product ( a group G i s t h e product of two subgroups H

and K i f $ E G imp l i e s g = hk f o r some h E H , k E K ) of two

c y c l i c groups, t h e n m ( G ' ) 5 2 . He a l s c showed that TOY any

odd prime p , a p-group t h a t can be expressed as t h e p roduc t ,

of two c y c l i c groups i s me tacyc l i c ; i n p a r t i c u l - a r , G I i s

c y c l i c . Using some r e s u l t s of Blackburn [6] we s h a l l show

a 3-group of maximal c l a s s ha s p r o p e r t y A, excep t when t h e L

4 group has o r d e r 3 .

Theorem 4.;[6]. I n a p-group G, G s z ( G 1 j provided m ( G 1 ) s 2. 3

p roo f . See [61.

C o r o l l a r y 4.8. I n a p-group G, i f m ( ~ ) s 2 and m ( G 1 ) 2,

t h e n G ( 2 ) = 1 ( and s o G i s meta .be l i sn) .

P roof . If g , h a r e g e n e r a t o r s o f G, t hen G I = < Gj, [ g , h ] >

and s o G'/G i s n y c l i c . By Theorem 4.7, 3 G3 5 z ( G ' ) and s o

G ( * ) = 1 ,

The above c o r o l l a r y f i n d s i t s a p p l i c a t i o n s i n what

f o l l o w s . F i r s t , H u p p e r t t s r e s u l t , r e f e r r e d t o above,

t o g e t h e r w i t h Coro l l a ry 4 .8 g i v e s r i s e t o

Theorem 4 .9 , If G i s a 2-group which can be expressed a s t h e -- produc t of two c y c l i c groups , t h e n m ( G 1 ) 5 2 and G I i s

abe l - ian . ( ~ o t e t h a t m ( G ) 2 . )

h he nex t application of Coro l l a ry 4.8 concerns 3-groups

of maximal c l a s s .

Theorem 4.10[6]. Every 3-group of ma.,ximal c l a s s i s me tabe l i an .

P roo f . Le t / G I = 3n and c l ( ~ ) = n-I where n 2 3 . Then

2 G / P ( G ) i s e lementary a b e l i a n of o rder 3 and s o m ( G ) = 2.

We s h a l l show m ( G 1 ) I 2. Deny; t h e n G ' / H ( G ' ) i s e lementary

3 a b e l i a n of o rde r a t l e a s t 3 . Let N 4 G such t h a t

G ' > N 2 @ ( G I ) and [ G I : N] = 33. Then G/N has o r d e r 3 5

and s o c ~ ( G / N ) 2 4, whence ( G / N ) , = 1 . Obviously, G, 5 N. J J

5 Since G has maxima.1 c l a s s , [G : G 1 = 3 . Also, s i n c e - 5 3 [G : N ] = 3 , it fo l lows t h a t G = N. Hence G/N ha s c l a s s

5 f o u r and ( G / N ) ~ = G /N has o rde r 3 3 . (Note t h a t G2/N i s

e lementary a b e l i a n . ) The fo l l owing argument shows t h e

e x i s t e n c e of such a group l e a d s t o a c o n t r a d i c t i o n .

Suppose H i s a group w i t h t h e above p r o p e r t i e s . S ince

2 H2/H4 has o rde r 3 and i s a normal subgroup of H/H4, t h e

c e n t r a l i z e r C / H ~ of H 2 / ~ 4 ha s index a t most 3 i n H / H ~ , by

Theorem 3.1. If [H/H4 : c /H4] = 1 , t h e n [H2, H ] g H4,

which i s imposs ib le . Thus [ H / H ~ : c /H4] = 3.

P ick g E H - C , g l E C - H2. Then H = < g, Q1 >.

Let g2 = kl , 81, g3 = [g2, g l . Then H2 = < H3, g2 > and

g2 # H 3 . Also, by t ,he cho ice of g , g3 E H 3 - H4. Next,

s i n c e g E C , [ g l , g2] E H4. Now,We have 1

- I = k 3 3 g2 1 = 1 ( s i n c e [H3, H2] g H = I ) . 5

Since H 3 = < H 4 , g3 > t h e fo rego ing argument immediately

imp l i e s g l E c ( H 3 ) , s o C c ( H 3 ) . But C has index 3 and

3 FI ~ ( 1 3 ) ~ hence C = C ( H ) . It i s easy t o s e e g E H2. 3 3 ' 3 Furthermore, we a s s e r t t h a t g E Hh. For , if g3 E H - H4,

3 t hen H = <g 3 , H >, which immediately impl ies g E c ( H ~ ) , 4 whence H 5 z ( H ) , a c o n t r a d i c t i o n . On t h e o t h e r hand, i f 3 D-) E H2 - H3 o , t h e n a similar reason ing shows g E C , a g a i n

a con t , r ad i t i on . Hence g3 E H4 = Z ( H ) . But t h e n a n

e lementary c a , l c u l a t i o n shows A

IT IT 2

D-31 = D D- "IT " - 1 = Cg, , , 0 0 2 - 82 3" a3 2" "3 g = [g3, g] ,

s i n c e H2 i s e lementary a b e l i a n . This i s a c o n t r a d i c t i o n

s i n c e g cannot commute w i th bo th g e n e r a t o r s of H. Thus 3 m ( ~ ' ) < 2 and t h e theorem fo l lows from Coro l l a ry 4.8.

As a r e s u l t of t h e p reced ing theorem we o b t a i n

C o r o l l a r y 4.11. A 3-group of ma.ximal c l a s s has p r o p e r t y A 2 ,

4 except when t h e group has o rde r 3 .

Proof. Let [ G I = 3n and c l ( G ) = n-1 w i th n 2 3. I f n = 3 ,

t h e n e v i d e n t l y , every a b e l i a n normal subgroup has a t most

two g e n e r a t o r s . Consider t hen n 2 5. From t h e proof of t h e

p rev ious theorem we conclude t h a t any subgroup of G ' has a t

most two g e n e r a t o r s . By (3) preced ing Theorem 3 . 5 ( b ) , t h i s

t akes c a r e of any a b e l i a n normal subgroup of index a t l e a s t

2 3 . It remains t o cons ide r t h e ca,se of a n a b e l i a n maximal

subgroup of G .

Suppose M i s a n a b e l i a n maximal subgroup of G and

m ( ~ ) r 3. Then n ( ~ ) i s a n e lementary a b e l i a n normal

3 subgroup of G of o r d e r a t l e a s t 3 . Hence R ( M ) c o n t a i n s a

3 subgroup N such t h a t N 4 G and 1 r ~ l = 3 . It i s c l e a r t h a t

m ( ~ ) = 3. However, by remark ( 3 ) preced ing Theorem 3 . 5 ( b ) ,

we have A = G n- 3 ( r e c a l l n 2 5) and s o m ( ~ , - ~ ) 2 3,

a c o n t r a d i c t i o n .

With r ega rd t o t h e c o r o l l a r y above, we now e x h i b i t t h e

e x i s t e n c e of a group of o rde r p 4 ( p > 2 ) and c l a s s 3

which has a n e lementary a b e l i a n maximal subgroup of o r d e r

3. P 2

Let G = < a, b : aP = bP - b c - 1 , a = a c , a = a , 1 +P

Then Z ( G ) = < aP > = G 3' G2 = < G [a , b ] > = c aP, 3 '

Hence G ha s c l a s s 3. Let M = < aP , b , c >. Then, M

o r d e r p3 and s o i s maximal i n G . It i s c l e a r t h a t M

c >.

has

i s

e lementary a b e l i a n .

From Coro l l a ry 4.2 and Corolla,ry 4.4, we have a l r e a d y

seen t h a t i f a p-group of odd o r d e r has p r o p e r t y A ?

( r e s p e c t i v e l y A ~ ) , t h e n every subgroup has p r o p e r t y A1

( r e s p e c t i v e l y A~). It i s n a t u r a l t o pu t forward t h e

ques t i on : I f a p-group ( p > 2) has p r o p e r t y Ak, k 2 3,

does eve ry subgroup i n h e r i t t h i s p rope r ty? For k = 3,

Hobby [ I 1 ] g ives a p o s i t i v e answer t o t h i s q u e s t i o n a s

w i l l be shown i n t h e nex t theorem. Never the less , t h e gene ra l

problem s t i l l remains open.

The proof of t h e l e m a below i l l u s t r a t e s a u s e f u l

dev ice i n u s ing e lementa ry a b e l i a n subgroups of p-groups t o

o b t a i n groups of l i n e a r t r ans fo rma t ions .

Lemma 4.12. Let A be a n e lementa ry abe1ia.n normal subgroup

3 of o rde r p i n a p-group G . Suppose N 4 G and [A : N] = p.

2 I

Then [ c ( N ) : c ( A ) ] s p .

Proof . A s i n t h e proof of Theorem 3 . l ( i ) we s e e t h a t G / C ( A )

i s isomorphic t o a p-subgroup of Aut(A) . Wri t ing A as a n

a d d i t i v e group and u s i n g t h e f a c t t h a t A i s e lementary a b e l i a n ,

we may r ega rd A as a Z -module, i . e . , a v e c t o r space over P

z~ . Thus we may a l s o r ega rd elements of A U ~ ( A ) as

be ing non-s ingula r Z - l i n e a r t r ans fo rma t ions of A onto A . P

On t h e o t h e r hand, every non-s ingula r l i n e a r t r ans fo rma t ion

of A i s c l e a r l y an automorphism of t h e ( a d d i t i v e ) group A .

Hence, A U ~ ( A ) i s isomorphic t o t h e group of non-s ingula r

l i n e a r t r ans fo rma t ions of A .

3 By 2 . 2 ( 5 ) , we s e e t h a t G / C ( A ) has o r d e r a t most p . 3 Assume [ c ( N ) : c ( A ) ] = p . Then C ( N ) = G and s o N I z ( G ) .

Let f x l , x2] be a b a s i s f o r N and [ x l , x2, x ] a b a s i s f o r 3 A where N and A a r e regarded as v e c t o r spaces over Z .

F P

If g e G, a E A , t h e mapping qD : a a" i s e a s i l y 'a

seen t o be a n automorphism of A ; whi le g qF i s a homo- "

morphism of G i n t o Aut(A) . Indeed, f o r a l l x E A , g l , g2 E 6,

we have

g l g 2 g 2 ~ l k l , g2] D F

NOW, x = x = x O2"I s i n c e A i s normal of o rde r

p3 and c o n t a i n s a c e n t r a l subgroup of index p , i . e . ,

A I ' Z 2 ( G ) and hence [A, G2] = 1 by 2.214) ; i n p a r t i c u l a r ,

Hence, g -+ qD is indeed a homomorphism of G into Aut(A) o

whose kernel is c(A). Further, qD(xl) = xl, q,(x2) = x2 0 0

as XI, X2 are central. Since [x3, g] E [A, GI 6 N,

[x3, g] = ixl + jx2 for some integers i, j modulo p; and 0-

since Ex3, g] = - x3 + xjo, it follows that

Let [q,] denote the matrix representing qD with respect 0 3

to the basis {xl, x2, x 1 for A. Then 3

is a homomorphism of G into the group of 3 x 3 non-singular

matrices with entries in Z whose kernel is c(A). Evidently, P

2 there can be at most p such matrices, since i, j E Z . P

Hence, G/C(A) is of order at most p2 and so the assumption

that [c(N) : c(A)] = p3 is fa,lse. This proves the lemma.

Theorem 4.13[11]. Let p be an odd prime and G a p-group

with d,(~) 3. Then d,(~) 5 3 for all H 2 G.

Proof. Deny and let G be a minimal counterexample. In

view of Corollary 4.4 we may assume without, loss of gener-

ality that dn(G) = 3. Suppose A is an elementary a.belian

3 normal subgroup of G of order p . By the induction hypothesis

we can assume that there exists a maxima,l subgroup M of G

such that d,(~) $ 3. Thus, M must contain an elementary 4 abelian normal subgroup E of order p . We shall carny out

the proof by demonstrating that there is an element of order

p which c e n t r a l i z e s A b u t does not l i e i n A .

Using t h e f a c t / A M I = f ' '7 we s e e t h a t A fl M has

2 orde r p3 o r p and s o we may f i n d a normal subgroup N of

2 G such t h a t N 5 A n M and N has o r d e r p . S e t C = c ( N ) .

Then C ha s index a t most p by Theorem 3.1 ( i ) . Again, u s i n g

3 I E C / = -wl wenote I E n C ] 2 p . Hence, we may assume,

wi thout l o s s of g e n e r a l i t y , t ,hat N < E, f o r i f n o t , we could

p i c k a subgroup E l , s ay , from t h e subgroup gene ra t ed by E n C

and N such t h a t N < E1 4 M and El i s e lementary a b e l i a n

4 of o rde r p . With Coro l l a ry 3 .9 a p p l i e d t o A , we conclude

t h a t a l l t h e elements of o r d e r p l y i n g i n C ( A ) a r e i n A .

S ince E n C ( A ) i s e lementary a b e l i a n , t h i s imp l i e s n c ( A ) / s p3

If E n C ( A ) has o rde r p3 t hen A = E n C ( A ) and s o by

Coro l l a ry 3.10 we conclude t h a t A < B, where B i s a n

4 elementJary a b e l i a n normal subgroup of G of o rde r p . This

c o n t r a d i c t s t h e hypo thes i s dn(G) = 3. Consequently,

2 N = E n c ( A ) , whence [C : c ( A ) ] 2 p . However, by t h e

2 preced ing lemma, [C : c ( A ) ] p . It fo l lows tha t

C = < E, C ( A ) > = E C ( A ) . Thus, i f e E E, t hen eg E C , 0.

s o eO = e c f o r some e l E E, c E c ( A ) . S ince E i s a.n 1

elementary a b e l i a n normal subgroup of M, Eg i s a l s o a n

ele.mentary a .be l i an normal subgroup of M of t h e same o r d e r . 0'

Hence t h e de,r ived group of EG = <[EO : g E G]> i s con ta ined

G i n t h e c e n t r e of EG, whence c l ( E ) 2. Furthermore, EG

G has exponent p . For EG i s r e g u l a r , s i n c e p i s odd and c l ( E ) 52.

Thus, s i n c e E" i s generat ,ed by elements or' o r d e r p i t fo l lows

G F t h a t e x p ( ~ ) = p. There fore , from em = e l c , i . e . ,

- 1 '3-

e , e " = c , a l s o fo l l ows cp = 1, whence c E A by Coro l l a ry

3 .9 . I n b r i e f , we have showh eg E E A f o r a l l e E E, g E G ,

s o EA -4 G.

Now, from our p rev ious work, N -s E n A 5 E n C ( A ) = N ,

whence N = E n A . This imp l i e s [EA : El = p s i n c e

I E/E n A ( = (EA/A I = p2 imp l i e s EA has o r d e r p5. Also,

G E # G impl ies E~ > E , and EA 4 G impl ies EA 2 E . IT

Hence EA = EG = EE" f o r some g E G. It fo l l ows that E n E~

has o rde r p3 and c e n t r a l i z e s E as w e l l as E ~ , s o

E fl E' 5 z(EA);: G. Hence t h e r e i s a n element x E E n J j g

( s o x must be of o rde r p ) such t h a t x E C ( A ) - A . By

Coro l l a ry 3 .9 , t h i s i s a c o n t r a d i c t i o n and s o t h e theorem

i s proved.

Remark. We have remarked e a r l i e r t h a t i n a p-group ( p > 2 )

G f o r which d , ( ~ ) 2 , t h e de r ived group of G i s a b e l i a n and

s o G ( 2 ) = 1 . Never the less , t h e r e e x i s t p-groups ( p > 2 ) f o r

which d ( G ) = 3 and i n which t h e der ived s e r i e s is a r b i t r a r i l y n

l ong LIZ?].

Theorem 4.14[11]. Let p be a n odd prime and G a p-group f o r

which every a b e l i a n normal subgroup of G con ta ined i n Gn

can be gene ra t ed by n e lements . Then m ( ~ , ) 5 n .

Remark. I n case n = 1 , t h e above theorem says that i f every

a b e l i a n normal subgroup of G i s c y c l i c t h e n G i s c y c l i c , a s

was seen i n Coro l l a ry 4 .2 .

Proof . Suppose G i s a minimal counterexample. Then @ ( G n ) i s

n o n - t r i v i a l as Gn i s non-abel ian . S ince @(G,) cha r Gn

char G y +(G,) n Z ( G ) > I . 'Let A 2 @(G,) n Z ( G ) and \ A / = p;

t h e n A 4 G. It fo l lows from A 2 @ ( G ) 2 + ( G ) t h a t n

m ( ( G / A ) ~ ) = m ( G n ) . I n view of our hypothes i s (G/A), must

c o n t a i n a n e lementary a b e l i a n normal subgroup B/A of G/A

w i t h I B / A / = pn", f o r o the rwi se , every a b e l i a n subgroup of

(G/A) , normal i n G/A would have a t most n g e n e r a t o r s and

s o t h e i nduc t ion hypo thes i s would app ly t o G/A.

Denote t h e preimage of B/A i n G by B. Then B 4 G,

and ( B ( = pn'2. It i s ea sy t o s e e B' 2 A 2 Z ( G ) . Hence,

B has c l a s s a t most two and s i n c e p i s odd, B i s r e g u l a r ,

whence e x p ( ~ ( B ) ) = p. Since B/A has exponent p , u ( ~ ) I: A .

By r e g u l a r i t y of B, B/U ( B ) ha s t h e same o rde r as s ~ ( B ) ,

whence / R ( B ) 1 2 pn'l . A s R ( B ) char B 4 G imp l i e s

R ( B ) u G , O ( B ) 2 C such t h a t U ( C ) = 1 , (c( = pnC1, where

C a G. Let N 4 G such t h a t [C : N] = p. By Lemma 3.3,

N z , (G) . Also, [ z , ( G ) , Gn] = 1 ( b y 2 . 2 ( 4 ) ) , whence

[N, Gn] = 1 . S ince N < Gn, t h i s imp l i e s N s z(G, ) . A s

N c C R ( B ) r Gn, C i s a b e l i a n . We have t hus ob t a ined

a n e lementary a b e l i a n normal subgroup of G whose o rde r i s

pn+', and which i s con ta ined i n Gn. Hence t h e a s s e r t i o n

fo l lows by no t ing t h i s c o n t r a d i c t ion .

On t h e b a s i s of t h i s r e s u l t , we now deduce

Coro l l a ry 4.15[111, Suppose t h e hypo thes i s i n Theorem 4.14

ho lds and Gn has exponent p r . Let t be a p o s i t i v e i n t e g e r

t. s a t i s f y i n g n i 2 . Then G ( t t d = 7 .

Proof. We f i r s t show t h a t u(G,) = P ( G ) . Since n

$(G,) = 7 J ( G n ) GI,, i t s u f f i c e s t o show ?J(G,) GI,. We

can t h e r e f o r e assume ?I(G,) = 1 and show G I n = 1 . But t h e n

it i s s u f f i c i e n t t o cons ider t h e case where 1 G f n 1 = p. By

t h e preceding theorem m(G,) n , s o G n / P ( G n ) has o rde r i pn,

i . e . , Gn/G; has o rder a t most pn ( s i n c e we assumed U (G,) = 1 ) . Thus, Gn has o rder a t most pn". Let N < Gn such t h a t

N G , [Gn : N] = p. Then N has o rde r a t most pn , whence

N 2 Z ( G ) by Lemma 3.3, and [N, G,] = 1 by 2 . 2 ( 4 ) . The n

l a t t e r impl ies N 2 z(G,) and s o Gn i s a b e l i a n , i . e . ,

G 'n = 1 . Hence 7.3 (G,) = $(G,) . It fo l lows from [I3 t h a t

1 B u t G ( t ) 5 G Gn s o G ( r + t ) = 1 'n 2 t

S e c t i o n 5 Large a b e l i a n subgroups of p-groups

I

Q. Abel ian subgroups of index p k

It has been remarked i n S e c t i o n 3 t h a t a p-group does

no t always pos se s s a b e l i a n subgroups of v e r y ' b i g s i z e 1

r e l a t i v e t o t h e o rde r of t h e l a r g e r group. Suppose G i s a

k p-group and H a n a b e l i a n subgroup wi th [G : H ] = p . Does

k t h i s imply t h e r e e x i s t s K w i t h K 4 G, K ' = 1 and [G : K] = p ?

There i s no reason t o expect it does though it i s t r u e f o r

k = 1 , 2, and k = 3 ( i f p r 2 ) . For k = 1 , it i s no s u r p r i s e

s i n c e H 4 G, whi le f o r k = 2, t h e proof i s s t r a i g h t - f o r w a r d .

However, f o r k = 3 and p > 2, t h e r e s u l t c e r t a i n l y i s a

s u r p r i s e and t h e proof i s d e f i n i t e l y n o n - t r i v i a l .

Theorem 5.1. If G i s a p-group wi th a n a b e l i a n subgroup of

2 2 index p , t h e n G has a n a b e l i a n normal subgroup of index p .

Proof . Let A be an a b e l i a n subgroup of index p2 i n G. Then

A i s con ta ined i n a maximal subgroup M of G. If A i s t h e

on ly a b e l i a n maximal subgroup of M, t h e n A i s c h a r a c t e r i s t i c

i n M and hence i s normal i n G.

We may t h e r e f o r e suppose A 1 i s a n a b e l i a n maximal

subgroup of M such t h a t A # A 1 . Then c A , A 1 > = AA1 = M.

S ince A and A1 a r e bo th a b e l i a n , A fl A 1 5 z ( M ) . The r e v e r s e

i n c l u s i o n i s obvious 8s A and A1 a r e c e r t a i n l y maximal

a b e l i a n subgroups of M . It fo l l ows t h a t A n A , = z ( M ) .

Hence, we can choose a normal subgroup N of G such t h a t

Z ( M ) < N < M. Obviously, N i s a b e l i a n and [M : N] = p . The I

a s s e r t i o n fo l lows immediately.

Coro l l a ry 5.2. Let G be a non-abel ian p-group wi th an

a b e l i a n subgroup of index p . Then G pos se s se s e i t h e r 1

o r p + 1 a b e l i a n subgroups of index p (of cou r se , a l l of

t h e s e subgroups must be normal) , where i n t h e l a t t e r c a s e ,

2 , , [G : z ( G ) ] = p .

Proof . If G c o n t a i n s more t han I a b e l i a n maximal subgroup,

we have j u s t seen i n t h e proof of Theorem 5.1 t h a t

2 [G : z ( G ) ] = p . We observe t h a t t h e r e i s a one-one

correspondence between a b e l i a n maximal subgroups of G and

t h e proper subgroups of G / Z ( G ) . Since G/Z(G) i s non-cycl ic

2 of o rde r p , t h e number of a b e l i a n maximal subgroups i s

p r e c i s e l y p + 1 .

Remarks. ( I ) The above c o r o l l a r y g i v e s us t h e in?ormation

of t h e ! s i z e T of t h e c e n t r e provided t h e r e i s more t h a n one

a b e l i a n subgroup of index p .

( 2 ) With r ega rd t o Theorem 5.1 and a remark

preced ing i t , it i s worthwhile t o p o i n t ou t t h a t t h e c a s e s

2 of a b e l i a n subgroups of index p and p a r e ve ry s p e c i a l and

t h e p roo f s i n t h e s e ca se s g i v e no i n d i c a t i o n t h a t t h e r e s u l t

3 should be t r u e f o r a b e l i a n subgroups of index p . Alpe r in [2] has proved t h e corresponding r e s u l t f o r

a b e l i a n subgroups of index p7, where p i s odd. The proof

i t s e l f i s f a r more d i f f i c u l t and complicated t h a n t h a t of

Theorem 5 .1 , and t h e t echnique involved deserves i t s own

expos i-t i o n . ,

Theorem 5.3[2]. Let p be a n odd prime and G a p-group.

3 I f G con t a in s an a b e l i a n subgroup of index p , t h e n i t has

3 an a b e l i a n normal subgroup of index p .

Proof . Let N be an a b e l i a n subgroup of index p3 and M a

2 maximal subgroup c o n t a i n i n g N w i t h [M : N] = p . By Theorem

2 5.1, M has an a b e l i a n normal subgroup A of index p . Since

M i s maximal i n G , t h e r e e x i s t s g E G - M such t h a t G = <M, g>.

If AO' = A , t h e r e i s no th ing t o prove; s o assume A~ # A .

Then s i n c e [G : N ( A ) ] = [G : M ] = p , A has p con juga t e s .

,i Let i E [0 , 1 , ..., p - I ] and de f ine Ai = AD . Fol lowing

t h i s d e f i n i t i o n , A. = A and A i # A if i $ j. ( ~ o t i c e t h a t j

a l l A i a r e a b e l i a n normal subgroups 01' M . ) We a l s o observe

t h a t f o r any i, j w i t h i # j ,

2 [ A i : A i MA] = [ A A : A . ] < [ M : A . ] = p . J i j J J

The proof of t h e theorem f a l l s i n t o t ,he d i s c u s s i o n of two

major cases : ( I ) [Ai : Ai n A .] = p and [A : A . n A .] = p J j n 1 J

2 f o r a l l i j, i # j (11) A . I : A . 1 n A . ] = p f o r some i, j.

J I n case ( I ) , we d i s t i n g u i s h between two p o s s i b i l i t i e s ,

namely A2 < A O A l o r A 2 4 A o A 1 . Suppose A2 < A O A 1 ; t h e n we

now show A i < A O A l f o r a l l i. Induct on i and assume

D-J- Then Ai = Am

A O A 1 Hence A . < A O A l f o r a l l i. Now, t h e normal c lo su re , '

I.

G G G A of A s a t i s f i e s A 4 AOA1 , S O A = A O A 1 . A s t a n d a r d

G . argument, a s i n t h e proof of Theorem 5.1 , on Z(A ) completes

t h e p roof .

Next, suppose A2 4 AOA; This t o g e t h e r w i t h

2 AOA1/AOA1 fl A, we have [AOA1 :AOAlnA,]=[(~oAl)~2:~2]=[~:~2]=p . Also, [AOA1 : A l l = [A1 : A1 M2] = p , whence

2 [AOAl : A1 fl A2] = p . Now,

2 [AOAl : AOAl n A2] = p ,

2 [AOAl : A1 fl A2] = p and A1 il A:, g AOAl Tl A2 imply

A1 fl A, = A A n A,. Replacing A, by A. i n t h e above 0 1

argument y i e l d s A. n A2 = ADAl fl A2. The l a s t two equa t ions

imply A. fl A, = A, n A, n A,. Again, a s i n t h e proof of

Theorem 5.1, A. Tl A 1 fl A2 ,< Z(A~A,A~) = Z(M) , s o

CM : z(M)] 6 [M : A. Tl A,] = p3. Thus, t h e r e e x i s t s B 4 G

such t h a t M > B 2 Z(M) and [M : B] = p 2 . Such a B i s

n e c e s s a r i l y a b e l i a n , complet ing t .he p roof .

2 In case (II), i . e . , [Ai : Ai fl A .] = p f o r some i and J

-i j, conugat ing t , h i s equa t ion by & we may assume i = 0 and

'2 get, [A : A fl A;] = p , where t h e "j" appea r ing h e r e i s n o t J

2 t h e same a s above. This imp l i e s [AA : A.] = p , whence

j J M = AA. because [M : A .I = p2 and AA. < M. Hence

J J J

A n Aj 4 z(M). If A fl A < z(M), t hen we may assume 3 3 [M : Z(M) ] = p and choose a normal subgroup B of G f o r which

M > B > Z(M) and [B : z(M)] = p. There fore , B s e r v e s a s

t h e d e s i r e d a b e l i a n subg-oup. Suppose t h e n Z(M) = A fl A s o ;i

2 t h a t A/Z(M) has o r d e r p .

Consider 'irst t h e c a s e when A/Z(M) i s c y c l i c , say - ,J A/Z(M) = < a Z(M) >. Then A ./z(M) = < ao Z(M) > which i s

J a l s o c y c l i c . S ince M = AA. and A, A a r e normal subgroups 01'

J j

M of i cdcx pz, M' < A n A , = z(H), whence cl(M) < 2. Since J

2 p > 2, M i.s regul-ar and as, A/z(M) has o r d e r p , aP E z(x). 0. J

Therefori i , l e t t i n g b = a" and u s i n g r e g u l a r i t y of I4 we no te 2 2

1 = [ a P , b ] = [a , b l P = LaP, bP].

Thus, H = < z(M), aP , bP > i s a b e l i a n . However,

M/Z(M) 2 (A/z(M)) x (A ./z(M)), t h e f a c t c r s beLng bo th c y c l i c 3

2 of o r d e r p , s o II/'Z(M) = R(M/z(M)) and Ii cha r M 4 G imp l i e s

- 1 H 4 0. Yurthermore, [a : fij = p3 and s o H i s t h e d t s i r e d

subgroup.

It remafns t o c o n s i d e r t he case when A/Z(M) i s e lementary

a b e l i a n . Then A,/z(M) i s also e1emc;ntary a b e l i a n and J

M/Z(M) i s isomorphic t o t h e d i r e c t product of t h e s e two

subgroups . The re fo re , we may r ega rd &I/Z(M) a s a 4-dimensionzl Lf

v e c t c r space over Z P *

The mapping m -, mC, where :lz E: M, i s

a n autornorphism o f M and i t induces a n automorphism g* ~n

M/Z(M) : g*(m z(M)) = mgz(M). We n o t i c e t h a t t h e induced

au.tornorphism has o r d e r p s i n c e gP E M and cl(~) = 2. This may

be regard-ed a s a l i n e a r t r a w f o r m a t i o n T on M/z(M). Le t

[TI = [a i j ] be a 4 x 4 m a t r i x over Z such t h a t T' = 1. P

Then t h e minimal p o ~ g n o m i a l f o r [TI d i v i d e s xP-l = (x-I) P

and s o a l l t h e character!-st ?"? va lues of [TI ccinci .de w i th 'I.

Hence t h e Jordan c a n o n i c a l form f o r T has a l l t h e e n t r i e s 1

down t h e main d i agon3 l . Four p o s s i b i l - i t i e s f o r t h e Jord-an

forms o f T a r i s e :

( 1 ) one 2-dimensional and two I -di.mensional b locks ,

(2) two 2-dimensional b locks ,

( 3 ) one 3-dimensional and one I -dimensional b l o c k ,

(4) one 4-dimensional b lock provided p > 3.

S ince A 4 G, t h e case of f o u r 1 -dimensional b locks cannot;

happen. For t h e f i r s t two ca se s t h e r e a d e r may r e f e r t o [2].

We s h a l l work ou t and d i s c u s s i n d e t a i l t h e remaining two

ca se s . ( ~ e c a l l t h a t M/Z(M) i s e lementary a b e l i a n of o r d e r

p4 s o t h a t it has f o u r b a s i s elements .)

I n case ( 3 ) , we may choose f o u r d i s t i n c t e lements

y l , ..., yq which, t o g e t h e r w i t h Z ( M ) g e n e r a t e M such t h a t D 0

Y; = Y ~ Y ~ , Y$ = Y ~ Y ~ , Y: = 93, yf = 94 (mod z ( M ) )

( A t t h i s p o i n t , we remark t h a t s i n c e we s h a l l be i n t e r e s t e d

i n computing commutators Cyi, y j ] and y , Y lg and s i n c e M j

has c l a s s 2, it c l e a r l y does no t ma t t e r whether t h e s e

e q u a l i t i e s ho ld a b s o l u t e l y o r only modulo Z ( M ) . Hence, wi th

t h e s e cau t iona ry remarks having been made, we s h a l l t r e a t

these equali-Lies a s h o l d i n g without, a.ny q u a l i f i c a t i o n . )

Denote Y ~ Z ( M ) by x i , i = l , . . ., 4. Now, t h e subspace

W = < x 2, X3' xq > of M/Z(M) has dimension 3 and i s i n v a r i a n t ; -

t h e subspace A of M / Z ( M ) corresponding t o A/Z(M) has -

dimension 2. Since A 4 G, A 4 W and we conclude t h a t - - - A I l W has dimension 1 . Let a E A fl W , b t h e o t h e r b a s i s

k

element f o r A and l e t a , b be t h e corresponding e lements i n

i j k. A. Then we may assume a = y2 y j yq , b = y,y21yjmy4n, where

i, j, k , 1, m , n a r e s u i t a b l e i n t e g e r s modulo p . By t a k i n g

c o n j u g a t e s of t h e s e b a s i s elements we may assume, i n

m ' n ' t h a t b = y1y3 yq . @ If i = o ( p ) , t h e n a = y

and s o by renaming we may c a l l y, = b and y = a . 3 I n t h i s

s i t u a t i o n we may t h u s assume A = < y , , y3, z ( M ) >. ' Theref o r e ,

1 = [ y l , y ] and con juga t i ng t h i s equa t i on twice by g , we 3

It f o l l o w s t h a t < y2, y3, Z ( M ) > i s t h e d e s i r e d group.

@ Suppose i # 0 ( p ) ; t h e n by t a k i n g powers we may

j k assume a = y y y4 . We r e c a l l t h a t b = y y my n 2 3 1 3 4 . C l e a r l y ,

we can choose b as our new y l , and we may choose our new y4

j k t o be y y 3 4 . Thus, a = y2y4 wh i l e b = y , , and s o

1 = [ y , , y2y4]. Conjugat ing t h i s by g we g e t

1 = [y1y2, Y ~ Y ~ Y ~ I = L Y ~ , Y ~ I [ Y ~ , Y ~ I C Y ~ , Y ~ I C Y ~ ~ Y ~ I C Y ~ ~ y41

= C Y , 9 y31[y2. Y ~ I C Y ~ y41*

i n view of t h e r e l a t i o n [ y , , y21 [y I , y41 = 1 . Conjugat ing

a g a i n by g we obtai-n, ana logous ly ,

I = [Y, 9 y3][y2, Y ~ ~ [ Y ~ , y41

- 1 Tee., Cy3, Y2Y4 1 = 1.

- 1 But < yj , Y2Y4 , Z ( M ) > i s normal i n G and , a s t h e above

shows, i s a b e l i a n . Th is completes t h e proof w i t h one b lock

of dimension 3 and one of dimension 1 .

I n c a s e ( 4 ) , we may choose f o u r d i s t i n c t e l e m e ~ t s

y l , y2, y3, y4 which, t o g e t h e r w i t h z ( M ) , g e n e r a t e M such IT F 0-

t h a t y l " = Y 1 Y 2 ~ y2" = Y g Y y gjo = YJY4, yqg = Yq* Adopt

t h e same n o t a t i o n and use t h e same argument i n ( 3 ) . Then

i j k a = y2 y3 y4 , b = y y y , where i, j, k , m, n a r e i n t e g e r s 1 7m qn

modulo p . a Suppose i I o ( P ) ; t h e n by renaming we may

assume a = y (however, t o assume t h i s , we must a l s o assume 3 l m n

b = y1y2 y3 94 ) . Our fundamenta l commutator r e l a t i o n i s now

1 n [y,, y1y2 yq ] = 1 . Conjugating t h i s by g we o b t a i n 1

which becomes, u s i n g our fundamental r e l a t i o n ,

Once aga in , we con juga t e t h i s by g t o g e t , a f t e r s i m p l i f i c a t i o n

by us ing t h i s second r e l a t i o n ,

1 [ Y ~ Y y 2 3 [ y 4 ~ Y31[y4~ y 3 I [ y 4 ~ Y j 1 = 1

2 Again, con juga t ing by g and s i m p l i f y i n g we o b t a i n [y4, y3] = 1.

Therefore < YJ, y4, Z(M) > i s a n a b e l i a n normal subgroup of

t h e d e s i r e d index. @ Suppose i f o ( p ) ; t h e n wi thout

J k l o s s oT g e n e r a l i t y , a = y2yJ y4 , and we can s u i t a b l y rename

n our g e n e r a t o r s such t h a t a = y and b = yly3n1y4 , Thus our 2

fundamental r e l a t i o n i s [y2; y y my n ] = 1 . 2 3 4 Conjugat ing

2 consecu t ive ly f o u r t imes by g , we conclude t h a t [y3, y4] = 1 ,

whence [y y4] = 1 ( s i n c e p > 2 ) . Hence < z ( M ) , yj, y4 7 3 ' i s t h e d e s i r e d a b e l i a n normal subgroup. The proof of t h e

theorem i s now complete.

Remark. When p = 2, Theorem 5 .3 i s f a l s e . We s h a l l now

c o n s t r u c t a group of o r d e r p9 which has a b e l i a n subgroups of

6 6 order 2 b u t k ~ s no a b e l i a n norm81 subgroups of o rde r 2 . Let H = < x

2 2 2 2 1 7 X29 Y ? , y2 : X I =X2 = Y I =y2\.=l, c ~ ( H ) = 2,

Moreover, H has a n automorphi-sm 5 of o rde r 2: I

5 ( x 1 ) x 2 5 (x2 ) X I 5 ( y 1 ) = 3 C(y2) = 91. - 1 Now, l e t G = < H , 5 : x5 = 5 x 5 Tor a l l x E H > . Then G

has o rde r 2' and c o n t a i n s two a b e l i a n subgroups

6 < z ( H ) , x l , y1 2, < z ( H ) , x2 , g2 > of o r d e r 2 which a r e no t

normal. We a s s e r t t h a t G has no a b e l i a n normal subgroups of

6 orde r 2 . F i r s t , suppose A i s a n a b e l i a n normal subgroup of G and

] A ] = p6, A H. Obviously A 2 z ( H ) , f o r o therwise H would

have a n a b e l i a n normal subgroup A Z ( H ) of o rde r a t l e a s t

27, which it does n o t . So, we must have

€1 6 1 Y 1 w i th a l l exponents i n t h e s e t l o , 11. Let S=xl x2 Y1 Y2 ,

q = x &2 62 Y2 O2 x2 y1 y2 . Since A i s a b e l i a n , [s, q] = 1 , and

u s ing t,he f a c t c l ( ~ ) = 2 t o work ou t t h i s commutator and

c o l l e c t terms we g e t

This g i v e s 4 congruences t o be s a t i s f i e d :

( 3 a l Y 2 + Y l b 2 0 ( 2 ) ; (4) Y 1 0 2 f O 1 y 2 0 ( 2 ) .

Consider and c2. We no te t h a t

(* ) (S ince we may s u b s t i t u t e sqc f o r e i t h e r 5 o r q,

where c E Z ( H \ = H I )

we may assume e i t h e r (i) = E~ = 0 o r ('5.1) = 0, c2 = 1 .

Suppose t hen E , = c2 = 0. If y l = y2 = 0, t hen A = <H', 5 , 6 1 w i t h 5 = x2 y2 , 7 = x 62 9 2 y2 . Now, one of 5, q must

- invo lve x2 and t h e o t h e r y2. By symmetry, t h e n 5 = x y U 1

2 2 ' 62 q = x2 y2. W e observe t h a t no t bo th = 1 and b2 = 1,

f o r t h e n 5 = q. Suppose ol = 0. Then, cons i d e r i n g i f

necessary Sq f o r q, we may assume b 2 = 0 . Hence 5 = x 2' q = y2. But t h e n SL = x , and x , ,d A , s o A 4 G, a c o n t r a d i t i o n .

S i m i l a r l y , if 62 - 0, we a g a i n o b t a i n a c o n t r a d i c t i o n . It

fo l l ows t h a t no t bo th y,, y2 a r e 0. By ( * ) , and renaming

i f necessary , we may assume no t bo th yl = 1 and y2 = 13 so

we may assume y, = 1 , y2 = 0. Then ( 4 ) imp l i e s o2 = 0 and

(3) impl ies 62 = 0. But t h e n E~ = y 2 = a = b 2 = 0, i . e . ,

5 7 = I, a c o n t r a d i c t i o n , f o r i t says ] A ] 4 2 . This completes

t h e d i s c u s s i o n of case (i) . Case (ii) i s ana lysed

s i m i l a r l y , and a g a i n it y i e l d s a c o n t r a d i c t i o n .

b r Thus, we may suppose t h a t i f ] A I = 2 , A = 1 alnd

A 4 G y t hen A gf H. S ince H i s maximal. i n G y we have

5 I A n H I = 2 a n d A = < ch , A n H > f o r some h E H. If I

A n H # H ' t h e n < A fl H, H' > i s a normal ( i n G, s i n c e

A f l H 4 G, ~ ' 4 G ) a b e l i a n ( s i n c e bo th A n H and H ' = Z ( H ) 6 a r e a b e l i a n ) subgroup of H of o rde r 2 , c o n t r a r y t o our

s u p p o s i t i o n . Thus, ch commutes with each element of H '

I ( s i n c e H A n H ) , and s i n c e h E H and H' = z ( H ) , we have

t h a t c commutes w i th each element of H'. However, t h i s i s

impossible , s i n c e , e . g . 5 does no t commute w i t h [x, , yp]e H ' .

Th is j u s t i f i e s our a s s e r t ion .

As regard t o Vaeorerrz 5.3, one ml-ght ask:

a p-group ( p > 2 ) , tihSch has an a b e l i a n subgrc

If G i s

)up of inde k p , k > 3 , i s it t r u e t h a t G has 3n abe l i an normal subgroup

of t h e same icdex? The answer of t h i s i s i n t h e negat ive ,

I n h i s rmpubl.ished work, Alperin cons t ruc ted a group a s

fol lows :

Let H be a group of exponent p > 2 and c l a s s two such

t h a t H = < xi, .. ., xp, y l , . . ., Y~

: [xi, y i l = l , i = i , . . . , p >. /

ThenH==<[x,, x j ] , Cyi., y j ] , [xi, yj] : i f j, i, j= 1 ,..., p>

2 2 has order pp -2p and H has order pa . Let

z z - CI = < H, z : x i = xi+, , yi - yi+! ( i mod p ) , zP=1 >.

2 G has a b e l i a n subgroups of order p 2p -2p+2 , but it does not

have an a b e l i a n normal subgrmp of t h i s order .

If we impose a condi t ion on a p-group G by requ i r ing t h a t

'r' G' have order p , t hen Theorem 5.3 can be genera l ized .

Theorem 5.4. Let G be a p-group f o r which G' has order p2.

k If G has an a b e l i a n subgroup of index p , then t h e r e e x i s t s

an a b e l i a n normal subgroup of t h e same index.

P roo f . Let A be an a b e l i a n subgroup of G whose index Ls

k p . We may assume without l o s s of g e n e r a l i t y t h a t Z(G) < A

2 and t h a t I A fl G ' I =I Z ( G ) fl G ' I = p. Since G ' has order p ,

[G : C ( G ' ) I d p by Theorem 3 .1 ( i ) ; and s k c e G t 4 z ( G ) , I

[ G : c ( G ' ) ] = p. Now, A 4 c ( G f ) f o r otherwise, < A , G >

would be a b e l i a n and normal. Hence, A c ( G ' ) = G and

from t - b = 1 G I , it fd l lows that A n c ( G ' ) h a s index

k+ 1 P

Eviden t ly , A c ( G ' ) c e n t r a l i z e s bo th A and G ' , s o

< G I , A n c ( G ' ) > i s a b e l i a n . Also, < G', A n c ( G ' ) >

k p r o p e r l y conta , ins A i l c ( G ' ) and s o has index a t most p . It i s c l e a r t h a t < G I , A n c ( G ' ) > i s normal. There fore

t h e t,heorem fo l lows .

It i s easy t o s e e that if G i s a p-group w i t h c l ( G ) = 2

and if G has an a b e l i a n subgroup of index pk t h e n G ha s a n

a b e l i a n normal subgroup of t h e same index. For , by i n d u c t i o n

on k we may assume t h a t t h e g iven subgroup A c o n t a i n s Z(G) . But, s i n c e c l ( ~ ) = 2 , t h i s means A 2 G ' and t h e r e f o r e A i s

normal. I n p a z t i c u l a r , every s p e c i a l o r e x t r a - s p e c i a l

p-group has a n a b e l i a n normal subgroup of index pk whenever

k i t has a n a b e l i a n subgroup of index p .

5.2. Wreath Products

I n t h i s subsec t ion , we s h a l l prove a few s t a t emen t s

concerning a b e l i a n subgroups i n t h e wreath product, of f i n i t e

groups . The wreat,h product has become a n i n c r e a s i n g l y

v a l u a b l e t o o l i n t h e c o n s t r u c t i o n a,nd i n v e s t i g a . t i o n of bo th

f i n i t e and i n f i n i t e groups . Among o t h e r t h i n g s , t h e \

importance of wreath p roduc t s i s that it p rov ides counte r -

examples t,o c o n j e c t u r e s . The wreath product Z 1 Z shows P P

t , ha t we cannot bound t h e c l a s s of a n i l p o t e n t g roup 'by

bounding i t s de r ived l e n g t h . , A s we a r e i n t e r e s t e d i n on ly

f i n i t e groups, we w i l l be concerned w i t h wreath p roduc ts of

f i n i t e groups.

Le t A and B be f i n i t e groups. Le t K = IT Ab, t h e d i r e c t b EB

produc t of I B I cop ie s of' A , indexed by e lements of B. We

can t h i n k of K a s t h e group of' a l l f u n c t i o n s from B t o A

w i t h m u l t i p l i c a t i o n de f ined componentwise; i. e . , if

f , g E K, t hen f g i s a n element of K de f ined by

( f g ) ( b ) = f ( b ) g ( b ) f o r a l l b E B. If f E K, b E B,

b we d e f i n e f b E K by f ( x ) = f ( x b - ' ) f o r a l l x E B. We may

i d e n t i f y B i n a n a t u r a l way w i t h a group of automorphisms of

K. Indeed, t h e group of automorphisms of K g iven by f -+ f b

f o r a l l f E K, i s isomorphic t o B. The wreath p roduc t W

of A and B, w r i t t e n symbo l i ca l l y W = A[ B, i s now de f ined a s

t h e s p l i t t i n g ex t ens ion of K by B, viewed a s a subgroup of

A U ~ ( K ) ; namely: W = < ( b , f ) r b E B, f E K > i n which b2

m u l t i p l i c a t i o n i s def ined by ( b l , f , ) (b2 , f 2 ) = ( b l b 2 , f f 2 )

f o r a l l b , , bg E B, f l , f 2 E K. With u s u a l convent ions ,

K a W , B 6 W and K n B = 1 . An element w of W ca.n t h e r e f o r e

be w r i t t e n in iq ,ue ly i n t h e form bf w i th b E B and f E K.

We c a l l K t h e base subgroup of W and Ab t h e f a c t o r s of K.

Lemma 5.5. I n W = A I B , C ( B ) = Z ( B ) x D, where

D = {f E K t f ( b ) = f(1) f o r a l l b E B]. 1

Proof . F i r s t , we n o t e t h a t i f c E C (B) t hen t h e r e e x i s t s

un ique ly b c B, f E K such that c = b f . I f b , E B, t h e n

b l b l b l I I bf = ( b f ) = b f a n d b E B , f E K . B y t h e u n i q u e n e s s

b l I

I of b and f we conclude b = b, f = f . Thus b E Z ( B ) . 1 Now f = f f o r a l l b l E B impl ies f(bl) = f ( 1 ) f o r a l l

b , ~ B. So, i f c E C ( B ) t h e n c = bf', where b E z(B)and

f E D = [f E K : f ( b ) = f ( l ) f o r a l l b E B] .

Conversely, if f E D , t h e n

1 -1 1 f ( b ) = f ( b b l ) = f ( 1 ) = f ( b ) , i . e . , f = f

f o r a l l b , E B. Hence, i f b E Z ( B ) and f E D, it i s

c l e a r t h a t b f E C ( B ) . It fo l lows t h a t C ( B ) = Z ( B ) D . By

what we have shown above, [D, B] = I . The re fo re , t h e

product, i s d i r e c t , i . e . , C ( B ) = Z ( B ) x D .

Theorem 5.6. ( i) Let A , B be f i n i t e groups w i t h 1 B I a n odd

i n t e g e r . Then a l l a b e l i a n normal subgroups of W = A 1 B a r e

con ta ined i n t h e base subgroup K of W.

( i f ) If H i s a a a.be1ia.n subgroup of A \ Z P

and is no t conta.ined i n t h e base subgroup K of A 1 Z t h e n P'

\ H I ,( ap , a be ing t ,he o r d e r of t h e l a r g e s t a b e l i a n subgroup

of A . Moreover, i f A ha s a n a b e l i a n subgroup A 1 of o rde r a ,

t hen t h e r e

B has o rde r

Lemma 5.5.

Proof . ( i)

x i s t s a n a b e l i a n subgroup H, of A 1 Z w i t h P

= ap .

( iii) I n W = A B, any a b e l i a n subgroup c o n t a i n i n g

a.t most I Z ( B ) 1 I D I , where D i s de f ined a s i n

Assume t h e a s s e r t i o n i s f a l s e . Then t h e r e

e x i s t s a n a.be1ia.n normal subgroup N of W such t h a t t h e r e i s

a n x E N - K. We can w r i t e , x = bf w i th b E B, f E K. We

s h a l l now show t h a t t h e r e e x i s t s a n a b e l i a n normal subgroup

of A 1 < b 3 c o n t a i n i n g a n element of t h e form bg wi th b a s

above.

F i r s t , we no t e t h a t , s i n c e N i s a b e l i a n normal, x

commutes w i t h a l l i t s con juga tes and t 'hey g e n e r a t e a n

a b e l i a n normal subgroup. Thus, x commutes w i t h a l l i t s

con juga tes i n any homomorphic image of a subgroup con t ra in ing

x. Next, we s h a l l show t h e r e e x i s t s a subgroup of W

con t a in ing x and having a homomorphic image isomorphic t o

A 1 < b >. To do t h i s , we decompose K i n t o K1 and K2, i. e . ,

K = K1 x K2, where K1 i s genera ted by a f a c t o r o f K t o g e t h e r

w i t h a l l i t s con juga tes under powers of x and K2 t h e p roduc t

o f t h e remaining f a c t o r s of K. Then K 2 4 < K, x >, and

< K, x > A b > To s e e t h e l a t t e r , w r i t e x = bh

w i t h 1 # b € B, h E K. Moreover, we assume t h a t t h e index ing

of t h e components of t h e base subgroup i s such t h a t t h e f i r s t

n o f them a r e indexed by t h e n elements i n < b >. F i n a l l y ,

s i n c e < K, x > obvious ly c o n t a i n s K, x I x . . . x I as a

subgroup we may assume t h a t h has only t ,he t , r i v i a l element.

a s i t s f i r s t n components ( f o r we can a p p a r e n t l y a d j u s t h

s o t h a t i t does by u s i n g elements from < K , x >) . Bearing

t h i s i n mind, cons ide r t h e mapping /

( b h ) i k ,... knk - bi k l . . . k n

from < K, x > t o A i < b >, where k l , ..., k, a r e from t h e

f i r s t n components of t h e base subgroup and kt E K2; Now,

i j I ( (bh) k l . . . k$ ') ( (bh) C , . . . C ~ C ) =(m) i + J ( k l . . .kn) (bh) j' el ..A kc n I , I?

where ci E K and k" E K2. ' Since, by whmt we s a i d e a r l i e r ,

h E c(K!) a n d s ince , by d e f i n i t i o n o f K1, b E X ( K ~ ) , we

conclude tha.t

(bh) i + j ( k l . ' .k n ) (bh) c l . . .cnkffc '=(bh) ( k . . k b j c 1 . . .c,k"cf,

b J 0 which (by our ru le ) i s ma~ped t o bi'j(kl . . . kn) C , . . n* But t h i s i s p r e c i s e l y t h e product of t h e imzges of

j 4 (bh)kk ,... knkf and (bh) c l . . . c n c . Hence, t h e napping a s

def ined ezr l i . e r i s a homomorphism whose ke rne l i s K2.

Si-nce the m y p i n g i s clear2.y onto, we ccnclude t h a t ,

Hence we may assuxe t h a t bf belongs t o an a b e l i a n

normal. subgroup of A 1 ( b >. L e t t i n g 1 bl = m we may w r i t e

f = f, ... l where f i b-longs t o t h e i t h f a c t o ~ of K. Uefi-ne m

where % belongs t3 the mth f a c t o r of H. Then

-1 bf - k, b f l = g g - - 1 b f - 1 l f - (d = g g l - D-% where

- - F b If - and g1 = g1 - Z1 I f 1

g l a r e elements of the f i r s t

f a c t o r of K. Since bf i s an element of an a b e l i a n norna>l

subgroup, ( ( b f ) g ) b f = (bf)'. But ( t ~ f ) ~ = b f [ b f , g] , s o 0. bf - 1 -1 bf - 1

( ( b f ) " ) = b f g l g a n d (bf g l g ) = b f g , g, whence -1 bf - 1 b f ( g l g ) = bf g1 g . Hence cance l l ing bf' we have

- 1 f - 1 (g2 g ; ) = g, g where g l ' , and g2 = g I h belong t o t h e

f i r s t and t h e second f a c t o r s of K r e s p e c t i v e l y . Therefore, - 1 f 2 fl - 1

(g2 ) ( g l ' ) = 91 g; t h i s i s a c o n t r a d i c t i c n s i n c e t h e

l e f t -hand s i d e has a non- iden t i ty - t e r n i n the second f'acto?

of K, whereas t h e r i gh t -hand s i d e does n o t , u n l e s s m = 2 ,

[Reca l l t h a t g is a non-identity eleixnt in the mtth f a c t o r of' K. 1

(ii) Since H $ K, t h e r e e x i s t s h E X - K.

S e t h = b f , b E Z f E K. Consider C ( b f ) and suppose P' K

t h a t g - o l . . . $ E K and t h a t g commutes w i t h b f . Then -

kl *op f - i.e., f c l f . . . g p - l - gl...gp with

"P " obvious n o t a t i o n . Hence, - f f g - , g2 = g1 , "., CT =

f "P "p 1

and s o g, g iven t h a t it commutes w i t h b f , i s determin2d b y

any one of i t s c o o r d i n a t e s and f . Now HK = A x Z = W, s o P

H/H' fl K W/K, a group of o r d e r p . Thus, H = < H n K, bf > and c l e a r l y , ~ ( b f ) 2 H n K. By our p rev ious remark concerning

a t y p i c a l g E I: n K we s e e t h a t IX fl XI( a , where a i s t h e

o rde r o f t h e l w g e s t a b e l i a n subgroup of A ( i . e . , of any one

of t h e f a c t o r s of K) . Since H = < H n K, bf ), it fo l l ows

t h a t !HI < ap.

To prove t h e second p a r t o f (ii) , l e t A l be a n a b e i i a n

subgroup of A t h a t has o r d e r a , and l e t

D = ( f E K : f ( 1 ) = f ( b ) E A , f o r a l l b E Z 1. 1 P '

Then c l e a r l y , Dl i s a b e i i a n and cornmutes w i t h Z i. e . , P '

D l x Z is a b e l i a n and has o r d e r ap . P

( i i i ) Th is i s a n immediate consequence of Lemma

5.5.

5.3. A l t e r n a t i n s Forms

I n t h i s s h o r t s u b s e c t i o n , we s h a l l f i rs t prove a

s t a t emen t concerned w i t h forms and v e c t o r spaces and t h e n

a p p l y it t o p-groups of c l a s s two. For t h i s purpose , w e

s h a l l assume, wi thout p r o o f , t h e fo l lowing known f a c t i n

l i n e a r a l g e b r a .

Let V be a n n-dimensional v e c t o r space over a f i e l d F.

We c a l l a b i l i n e a r f u n c t i o n a l f of V i n t o F a n a l t e r n a t i n g

form o n V i f f ( v , v ) = o f o r a l l v E V. If n i s odd, t h e r e - e x i s t s w E V, w i th w # o, such t h a t f ( v , w ) = o f o r a l l

v E V.

Theorem 5.7[2]. If f , g a r e a l t e r n a t i n g forms on a n

n-dimensional v e c t o r space V, t h e n V has a subspace W of

1 dimension [$n+l)] on which f and g bo th v a n i s h .

Proof . Induc t on n. It s u f f i c e s t o prove t h a t t h e a s s e r t i o n

ho lds f o r n a p o s i t i v e odd i n t e g e r , f o r t h e n we can conclude

by app ly ing t h e i nduc t ion hypo thes i s . Thus, l e t n = 2k-1

where k E z'. Since n i s odd, by a remark preced ing t h e

theorem, t h e r e e x i s t s v E V w i t h v # o such t h a t f ( w , v ) = o

f o r a l l w E: V. As g ( . , v ) i s a l i n e a r f u n c t i o n a l o n V , V

c o n t a i n s a n (n-2)-dimensional subspace W w i t h v # W such t h a t

~ ( w , V ) = o f o r a l l w E W . By induc t ion hypo thes i s , W ha s a 0

subspace U of d.imension k-I such t h a t f , g b o t h v a n i s h on U.

Now, l e t T be t h e k-dimensional subspace spanned by v and

U. Then f , g bo th v a n i s h on T, and t h e theorem f o l l o w s .

The prev ious r e s u l t can be a p p l i e d t o prove t h e fo l l owing .

Theorem 5.8[2]. Suppose G ha s o rde r pn, s a t i s f y i n g 2 k

e x p ( ~ ) = p, c l ( G ) = 2, J G ' I = ~ , I z ( G ) \ = p .

Then G c o n t a i n s a b e l i a n subgroups of o rde r p j where

1 I

j 2 k + [?(n-k+l)] .

Proof . S ince c l ( G ) = 2, G ' ,( Z ( G ) and G / Z ( G ) i s element,ary

n-k a b e l i a n of o r d e r p . We may t h u s r e g a r d G/Z(G) a s an

(n-k)-dimensional v e c t o r space over Z . A s G ' i s e lementary P

2 a b e l i a n ( s i n c e exp(G) = p and G ' h a s o rde r p ) , G ' = < c f , cD > 0

f o r some c f , c, E G . I f x, y c G , t,hen 0

where f ( x , y ) and g ( x , y ) depend s o l e l y on t h e c o s e t s of Z ( G )

c o n t a i n i n g x, y and a r e determined only modulo p . ,Thus ,

we may r ega rd f , g a s f u n c t i o n s from G / Z ( G ) x G / Z ( G ) i n t o Z P *

Elementazy c a l c u l a t i o n s show

f ( x , y , + ~ ~ ) = f ( x , y , ) + f (x, ~ 2 ) ' f (x , i y ) = i f (x, Y) 9

f o r a l l c o s e t s of Z ( G ) and i a n i n t e g e r modulo p . Also,

f ( x , x) = o f o r a l l x E G . There fore f i s an a l t e r n a t i n g

form on G / z ( G ) . Likewise, g i s a n a l t e r n a t i n g form on

G / Z ( G )

Applying Theorem 5.7 , we may l e t H / Z ( G ) be a

1 [-i (n-k+l)] -dimensional subspace of G / Z ( G ) such t h a t f and

g bo th van i sh on H / z ( G ) . Hence H i s a b e l i a n . Obviously, 1

H has o rde r pa, w i t h a = k c lp(n-k+l ) 1, from which t h e

a s s e r t i o n f o l l o w s .

Remark. The preced ing r e s u l t s mot iva te t h e ques t i on :

Suppose t h e r e a r e more t h a n two a l t e r n a t i n g forms o r G ' h a s

order g r e a t e r t h a n -p2, what w i l l t h e consequence be?

T e n t a t i v e l y , it seems t h a t t h e prev ious r e s u l t s do not

ho ld when t h e hypo thes i s i s weakened, making t h i s an

i n t e r e s t i n g ques t ion t o end wi th .

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Appendix -

I

Proof of t h e second p a r t of Theorem 3 . 5 ( a ) . From t h e fS--st

p a r t of Theorem 3 . 5 ( a ) , when n24, we observe t h a t G has a n

a b e l i a n normal subgroup of order pm with m 2 3 . We s h a l l cow

show tha.t G indeed has an a b e l i a n normal subgroup of order

Since G has order pn and n d , it i s c l e a r t h a t G has

a n ' a b e l i a n normal subgroup N of order pz. In view of

Theorem 3.1 (i) ,' t h e c e n t r a l i z e r , c ( N ) , of N has index a t

most p . By a remark on page 6, t h e r e i s a normal subgroup

M of G such t h a t N < + M < C ( N ) and lM/N1 = p . Hence t h e r e

e x i s t s x E M-N such t h a t M =< N, x >. Since N i s a b e l i a n

and x c e n t r a l i z e s N , it follows t h a t M i s a b e l i a n , Therefore

M i s a subgroup of t h e des i red order .