SOME THEOREMS ABOUT THE
AUTOMORPHISM GROUP OF
CERTAIN p-GROUPS

Thesis for the Degree of Ph. D.
MICHKSAN STATE UNIVERSITY
JAMES E. VANDEVENTER
1970

 

LIBRARY

JHFS“

" Michigan Sum
University

 

This is to certify that the

thesis entitled

Some Theorems About the
Automorphism Group of
Certain p-Groups

presented by

James E. VanDeventer

has been accepted towards fulfillment
of the requirements for

EILD. degree inMathematics

(fled,

Major profesf) r(/

Date IM /¢7o

0-169

 

ABSTRACT

SOME THEOREMS ABOUT THE AUTOMORPHISM GROUP
OF CERTAIN p-GROUPS

By

James E. VanDeventer

There are several possible types of relations between a
finite group G and its group of automorphisms A(G). One rela-
tion is concerned with the size of G as compared to the size of
A(G). For example, there is a conjecture that \G‘ divides
‘A(G)‘ if G is a finite nonabelian p-group, where \G‘ denotes
the order of G. Another relation concerns which conditions on
G guarantee a certain type of automorphism of C. An example of
this type of relation is Gaschfitz's result that a finite nonabelian
p-group possesses an outer automorphism of p-power order. Still
a third type of relation concerns how knowledge about A(G) in-
fluences information about the structure of C. Thompson has shown
that if a finite group G possesses a fixed point free automorphism
of prime order, then G must be nilpotent. In this dissertation,
we consider several questions of this nature.

Let G be a p-group, p a prime. Otto has shown that if
G is abelian and \G‘ = pn, n > 2, then ‘G‘ divides ‘A(G)\ if,
and only if, G is not cyclic. He has also shown that if G is a
p-group with no abelian direct factors, of order pn and nilpotence

class m - 1, G/G2 is elementary abelian, and lGi/G for

i+ll = p

i = 2,3,...,m-1 where Gi is the ith member of the descending

James E. VanDeventer

central series, then ‘G‘ divides \A(G)‘. Faudree has shown that
if G is a finite nonabelian nilpotent class two p—group, then
\G\ divides \A(G)\.
A p-group G, p # 2, is said to satisfy the conditions W)
if:
1). G has nilpotence class at least three,
2). G has no abelian direct factors,
3). G' is cyclic and ‘G'\ = ph,
4). ‘G' n Z(G)‘ = pm where m < h,
5). \G/¢(G)\ = pr.
In this dissertation, we prove the following theorems:
Theorem 11.1: If G is a p-group satisfying the conditions
W), and Z(G) s G', then G is a central product of the form

ph ph-l-m tpm
l’lexl = X2 = 1’ CXz’xi] = X2 ’

G = AB where A.= <x
(t,p) = L> and B is a p-group of nilpotence class two.

Theorem 11.3: If G is a p-group satisfying the conditions
W), and Z(G) s G', then \G‘ divides \A(G)‘.

Theorem 11.4: If G is a p—group satisfying the conditions
W), Z(G) = Z*<8 A where 2* = Z(G) n G', and exp A 2 exp 2*, then
\G| divides \A(G)\.

Theorem 11.5: If G is a p-group satisfying the conditions
W), Z(G) = 2* 8>A where 2* = Z(G) D G', and h 2 2m, then ‘6‘
divides ‘A(G)‘.

For a p-group G with no abelian direct factors, one knows,
in a certain sense, the size of 1(6) and the size of the group of

central automorphisms AC(G). It would be useful to know what con-

ditions, if any, are necessary and/or sufficient that AC(G) s 1(G).

James E. VanDeventer

In connection with this problem, Sanders has shown that a p-group
G, p # 2, of nilpotence class exactly two has Ac(G) s 1(G) if,
and only if, G' = Z(G) is cyclic.

We prove the following theorems which are concerned with
the problem:

Theorem 11.2: If G is a p-group satisfying the conditions
W), and Z(G) s G', then AC(G) s 1(G).

Theorem 111.1: If G is a p-group, p # 2, with no abelian
direct factors and such that AC(G) s 1(G), then Z(G) s G'.

Theorem 111.2: If G is a p-group, p # 2, with no abelian
direct factors, AC(G) 3 1(6), and Z(G) is not cyclic, then
22(0) s ®(G).

It is also conjectured that if G is a p-group, p # 2,

with no abelian direct factors and such that AC(G) 3 1(6), then

Z(G) must be cyclic.

SOME THEOREMS ABOUT THE AUTOMORPHISM GROUP
OF CERTAIN p-GROUPS

By
'0
M“
James E. VanDeventer

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1970

_‘-
K"
,1

DEDI CATION

TO JOAN

ii

ACKNOWLEDGMENT

I want to thank Professor J.E. Adney for his guidance
throughout the preparation and writing of the dissertation. His
patience and encouragement has served as a source of inspiration

for the completion of this work.

iii

TABLE OF CONTENTS
Chapter Page
I INTRODUCTION 0 ....... 0 O I O O 0...... 0.. O. O O O. O O O O 000000 1

11 SOME SUFFICIENCY CONDITIONS FOR \G\ TO DIVIDE

|A(G)\ ............................................. 6

III SOME NECESSARY CONDITIONS FOR AC(G) S I(G) .. ...... 36
BIBILIOGRAH'IY 0.0.0....OOOOOOOOOOOOOOOOCCC0.0COOOOOOIIOOOOOOO 43

APPENDIX 0.00.00.00.00000000000000000000000000 ........ 00.... 44

iv

NOTATION

G is a finite group
G = <x1,x2,...,xr> means that G is generated by the set of
elements {x1,x2,...,xr} where xi 6 G.

<x .,xr‘E> denotes the group generated by the symbols

1,..

x .,xr subject to the relations Rh

1,..
H s G denotes H is a subgroup of G.

H A_G denotes H is a normal subgroup of G.

G = AB denotes G is a product of the two subgroups A and B.
(X

,x > s G denotes the subgroup of G generated by the set of

1 2

elements {x1,x2}.
‘x‘ denotes the order of the element x.
lG| denotes the order of G.

a|b denotes a divides b.

[x,y] = x' y x y

[X052] = [Emyli-ZJ

[H,K] = <[h,k]|h e H, k e K)

c' = G2 = [G,G]

Gk+1 = [Gk,G] for k 2 1 is the kth member of the decending
central series for G.

Z(G) denotes the center of G.

22(6) = <y E G|[x,y] E Z(G) for all x E G> is the second center

of G.

¢(G) denotes the Frattini subgroup of G.

A(G) denotes the automorphism group of G.

I(G) denotes the group of inner automorphsims of G.

AC(G) = {a E A(G)‘x-1xa E Z(G) for all x 6 G} is the group of
central automorphisms of G.

llG is the identity automorphism of G.

a‘H is the automorphism of H s G obtained by restricting
a E A(G) to H.

Hom(H,K) denotes the set of homomorphisms from the group H into
the group K.

For y E G, fly denotes the inner automorphism of G which is
obtained by conjugation by y.

G g H denotes g is a homomorphism from G to H.

CHAPTER I

Introduction

The question of possible relations between a group G and
its automorphism group A(G) has been considered in several papers.
One type of relation is concerned with how their sizes are related.
An example of this type of relation is the conjecture that the
order of a finite, nonabelian p-group divides the order of its
automorphism group. Another type of relation is concerned with how
knowledge about the group can guarantee the existence of certain
types of automorphisms. The result of Gaschfitz [4], which states
that a finite p-group possesses an outer automorphism of p-power
order, is concerned with this type of relation. A third type of
relation is concerned with how knowledge about the automorphism
group helps determine the structure of the group. An example of
this type of relation is Thompson's result [9], which states that
if a finite group has a fixed point free automorphism of prime
order, then the group is nilpotent.

As stated above, it is conjectured that if G is a finite,
nonabelian p-group, then ‘G‘ divides ‘A(G)‘. Since \I(G)\ =
‘G‘/‘Z(G)|, and for a finite p-group, p divides \Z(G)|, the con-
jecture is concerned with whether or not there exist sufficient
p-powered outer automorphisms of G. The result of Gaschfitz on the
existence of a p-power outer automorphism still leaves the question
of the size of the automorphism group unanswered.

1

The conjecture has been proven true for certain classes of
p-groups. In [3], Faudree has shown that if G is a finite, non-
abelian p-group of nilpotence class two, then ‘6‘ divides |A(G)\.
Otto has shown in [6] that if G is an abelian p-group of order
pn, n # 2, then \G‘ divides \A(G)\ if, and only if, G is not
cyclic. In the same paper he has also shown that if G is a p-
group of order p“, of nilpotence class m - 1, where 3 s m s n,

G/G is elementary abelian, and ‘Gi/G

2 = p for i = 2,...,m-1,

is
where G1 is the ith member of the descending central series of G,
then \G‘ divides ‘A(G)‘.

A p-group G is said to have no abelian direct factors if
G is not abelian and has no nontrivial abelian direct factors. In
the above mentioned paper by Otto, it is shown that if G is a
direct product P23 B of two subgroups P and B, where P is
abelian of order pr and B has no abelian direct factors, then
pr . ‘A(B)|p divides ‘A(G)‘, where ‘A(B)‘p denotes the highest
power of p dividing ‘A(B)‘. This result allows us to consider
only p-groups which have no abelian direct factors, that is, if the
conjecture is true for p-groups with no abelian direct factors, then
the conjecture is true for p-groups.

One very nice property is possessed by p-groups with no
abelian direct factors. Adney and Yen have shown in [1] that if
G is a p-group, p # 2, with no abelian direct factors, then AC(G),
the group of central automorphisms of G, is a p-group. A central
automorphism a of a group G is one for which g-1ga E Z(G) for
all g E G. In the paper by Adney and Yen, it has been shown that

if G is a p-group with no abelian direct factors, then

\AC(G)| = \Hom(G/G', Z(G))\. Also, if a e AC(G), then

f : x a x-lxa is a homomorphism of G into Z(G), and if

a
B e Hom(G/G', Z) and G g G/G', then v defined by g,Y = ggga
is such that y E AC(G). Thus for a p-group G, p # 2, with no
abelian direct factors, we know how to construct all possible
central automorphisms.
For notation purposes we make the following definition:
Let G be a finite p-group, p # 2. G is said to
satisfy the conditions (0) if:
1). G has no abelian direct factors,
2). G has nilpotence class at least three,
3). G' is cyclic and \G" = ph,
4). \Z(G) n G" = pm where m < h,
5). \G/cMGH = pr.
In Chapter II, we give several examples of groups satisfy-
ing the conditions 0”, and we prove the following theorems, which

concern the conjecture that ‘G‘ ‘ |A(G)‘.

Theorem 11.3: If G is a p-group satisfying the conditions

 

W), and Z(G) s G', then \G‘ divides ‘A(G)\.

Theorem 11.4: If G is a p-group satisfying the conditions

 

* * *
a”, Z(G) = Z 8>A where Z = Z(G) H G' and exp A 2 exp Z then

,

|c| divides \A(O)|.

Theorem 11.5: If G is a p-group satisfying the conditions

 

* *
w),2(c;) =z ®A where z =Z(G)nc' and h22m, then M
divides ‘A(G)‘.
Since a p-group always has an outer automorphism of p-power

order, we know that a Sylow p-subgroup of A(G) always properly

contains 1(G). The outer automorphism may or may not be a central
automorphism. For example, in [7], Sanders has shown that if G
is a p-group of nilpotence class two, then AC(G) s 1(G) if, and
only if, G' = Z(G) is cyclic. Thus there exist p-groups which
have all their outer automorphisms being noncentral automorphisms.
Regarding sufficient conditions on a group G to guarantee
AC(G) s 1(G), we prove in Chapter II the following theorem for
p-groups of class at least three, which extends Sanders' result in
a certain direction.

Theorem 11.2: Let G be a p-group, p # 2, with no abelian

 

direct factors, of nilpotence class at least three and such that
Z(G) s G', and G' is cyclic. Then Ac(G) S 1(G).

It is not known what conditions are necessary to place on
a p-group G to insure AC(G) s 1(G). In Chapter III, we obtain
some necessary conditions for AC(G) s 1(G). We first prove that
if Ac(G) s 1(G) for a finite p-group, p # 2, then one must have
G' 2 Z(G). This result is sufficient to give that not only does
\AC(G)\ = \Hom(c/c', 2)], but AC(G) E Hom(c/G', 2). This follows
from a condition in Sanders' work [7], which states: Let G be
a group with no abelian direct factors. Let R = {v(g): g 6 G,
v E Hom(G,Z)} and let B =f] (ker y). Then Hom(G,Z) E Ac(G)
if, and only if, R < B. SinZe Hom(G,Z) E Hom(G/G',Z), we have
the desired result. [See Appendix, 111].

If y E Z2(G), then [g,y] E Z(G) for all g E G. Thus
conjugation by an element in 22(G) induces a central automorphism

n

n
of G. Also, if "y E AC(G), where g y = y lgy, then g 1g y E Z(G)

for all g E G, and hence y E 22(G). Now if AC(G) s 1(G), we have

Ill

AC(G)

Ace)

22(G)/Z(G). Since AC(G) s 1(G) implies c' 2 Z(G), so

Ill

Hom(G/G', 2), we have 22(G)/Z(G) E AC(G) E Hom(G/G', 2).
This result is used to prove:

Theorem 111.2: Let G be a p-group, p # 2, with no
abelian direct factors, Ac(G) s 1(G) and Z(G) is not cyclic.
Then 22(6) s (5(6).

This theorem gives us information about the lattice of G
if Ac(G) S 1(G) and Z(G) is not cyclic. It is our conjecture
that if G is a p-group with AC(G) s 1(G), then Z(G) is cyclic.
So far, however, this has not been proven, but is a direction for

further research.

CHAPTER II

Some Sufficiency Conditions for ‘G‘ to Divide \A(G)\

Let G be a finite p-group, p # 2. G is said to satisfy
the conditions (U) if:
1). G has no abelian direct factors,

2). G has nilpotence class at least three,

3). G' is cyclic and ‘G" = ph,

4). ‘Z(C) O G" = pm where m.< h,
r
5). \G/¢(G)\ = p .
We shall now give several examples of p-groups which satisfy

the conditions (U).

h h+m

. A = p = p = =
Example 1). Let A <X1,X2‘X1 x2 1, [x2,x1]

m
x2p , m < h>. One can easily show by induction that (xlk)-1x2 x1k =
(1+p‘“)k . . .
x2 . U51ng this formula, we have the following subgroups of
m h
A. A' =‘Az = <x p > and thus \A" = ph. Z(A)k= <x2p > since
m+k m+k+l
(1+p‘“)P = 1+p +ep H

P '1 p = . .
(x1 ) x2 x x if, and only if, x2 x2

1 2
h+m m+k
ll)

2

if, and only if, p
pk A A m
x1 6 Z(A) if, and only if, x1p = 1. Since A' = <x2p >,
A k k 4 k A
x2 6 CA(A'), so [x2 ,y] = [x2,y] for all y E A. x2 6 Z(A)
. , k -l -l k -l k
if, and only if, (x2 ) x1 x2 = x1 , or [x2 ,x1] = 1 =
k m h+u1 h
[x2,x1]k = $2 p . But |x2‘ = p , so p \k., and we have 2m
>. Since ‘A = A', we have A = <[A,A ]> = <x >,
2 km 2 2

/~A _ P .
A,Ak]> - <x2 >. Since m < h and

, or k 2 h. But ‘x1‘ = ph, so
k

Z(A) = <x2p
A

and in general, Ak+1 = <[

h-{m A A
\x2‘ = p , A3 # <l> and the nilpotence class of A is at least
. A ph A A ‘A
three. Since Z(A) = <x2 >, we have Z(A) s A(A) and hence A
A A
has no abelian direct factors. We note that Z(A) s A'.
h+2m h+ni
Example 2): Let G = <x1,x2|x1p = x2p - 1,
m m k
_ p k -1 k __ <1+p >
[x2,x1] x2 , m < h>. One can Show that (x1 )m xle — x2
and as above, use this to obtain that: G' = <x2p >, \G'l = ph,
pkm ph ph ph ph
Gk+l = <x2 >, and Z(G) = (x1 , x2 > = <x1 > ®<x2 >.
[Z(G) n G'] = \<x2p >| = pm. Also Z(G) s ¢(c), so we have a can

have no abelian direct factors. Since m < h, G3 # <l> and G

. _ h+2m
has nilpotence class at least three. We note that Since [XI] = p

h+m
and ‘X2‘ = p , we have exp Z(G) 2 exp (Z(G) 0 G').
ph+s ph+m
Exampie 3): Let G = <X1,X2\X1 = x2 = l,
p , 1 s s < m <YL>. This example is similar to the

[Xz’xl] = x2
immediately above example. The difference is in this example,
h h

> ® <x2p

P

Z(G) = <x1 >, and we have exp (Z(G)) = exp (Z(G) n G') 2

h

exp <x1p >.

Example 4): Let G = x1,x2,x3,x4 h
[x3,x4]=x2p ,[xi,xj] = l
m
[x2,x1] xzp m
I P
One can Show G = <x >,
for l = 1,2 ; j = 3,4, m < 2
km
G = <x p >. Since mi< h, G # <1>, so G has nilpotence

3 h
p > s ¢(G), thus G has no abelian

class at least three. Z(G) = <x2

direct factors.

A A
G is a central product of the form AB, where A is the
m m m

group in the first example, and B = <x3,x4,c‘x3p = x4p = cp = 1,

[x3,x4] = c, [x3,c] = [x4,C] = 1>. We have 3' = Z(B) = <c>. A

h

and B are joined by amalgamating <x p > and <c>.

2

For the remainder of Chapter 11, we shall assume that [G
is a p-group, p # 2, which satisfies the conditions (U). Also,
for notational purposes, we let G' = <a>, G E G/C(G') and
G g G/G'.

Since G' is cyclic, we have G' s C(G'). Also,

G/C(G') Z A(G') which is cyclic since G' is a cyclic p-group
and p # 2. Thus G/C(G') = <§> is a cyclic p-group and we must

have if g E G is such that <gK> ='<)-<.>, then g 6 ¢(G).

- j(g)
g 18 81+S(g)p

Remark A: For g E G, let where

g:
(s(g),p) = l and j(g) 2 1, that is, we set j(g) = h if

g E C(G'). Then j(g) 2 m.

_ _ j
Proof: We note that g 1akg = (g 1ag)k = ak<1+Sp ) where
. . h m
s = s(g), j = 3(g)- Since lG'l = p . \Z(G) n G'\ = p and
h-m _1 h-m
G' =i<a>3 we have Z(G) n G' <ap >. Thus g ap g =

h- - - ' h- - ’
ap m(1+spj ph m+sph m+j p m Sph m+J

)=a

h-m+j h
(s,p) = 1, so ap = l, or p E 0 mod p . Thus

h-m+j 2 h, or j 2 m.

Remark B: Let g E G, s = s(g), j = j(g). Then for k 2 0,

' k h-m
k - 1 J
(g ) 1a gk = a( +Sp ) , and hence for all g E G, gp E C(G').

_ J 0
Proof: If k = 0, then (go) 1a g0 = a = a(1+Sp ) .

-1 +1
Assume the formula is true for k = n. Then (gn+1) a gn =

- l n - j n J j n
1a(1+8p ) g = (g lag)(1+sp ) = (a(1+sp ))(1+sp ) =

Hence, by induction, the formula is true for k 2 O.

h-m _ h-m j ph'm
) 1a 89 = a(1+Sp )

s'1((sn)'lagn)g = g
a(l+Spj)n+1

Let k = ph-m, then (gp

h-m-i-j h-m+j+l
a1+sp +BP = a since by Remark A, j 2 m, so h-m+j 2 h
h-m

h
and [G" = p . Thus gp E C(G')~

Remark C: For some g 6 G, j(g) = m.
Proof: Suppose not, then by Remark A, m g j(g) for all

g E G. Let j = min[j(g)‘g E G}, then m g j. For g F G, we have

- h-j h-j j(g) h-j h-j+j(g) h-j
3 lap g = ap (1+S(g)p ) = a? +S(g)p = ap since
h ph'j
j(g) 2 j and ‘G'I = p - Thus a E Z(G) n 6'. But
‘ap ‘ = pj, ‘Z(G) O G" = pm and m g j, a contradiction. Thus

for some g E G, j(g) = m.
h-m
Since gp E C(G') for all g E G, we have ‘G/C(G')\ s

ph-m. By Remark C, there exists g 6 G with j(g) = m. For this

- l
g we have g 1ag = a +sp where sl= s(g). By Remark B,

h-m-l h-m-l 'm' h-1 h
- p 1+sp +BP # a since

=3
h-
aSp # 1. Thus gp 4 C(G') and we have lc/C(G')i = p m.

- K
Remark D: If <gK> = <x>, where G a G/C(G'), then

1(a) = m.
. K - K - .
Proof: Since <g > = <x>, we may assume g = x. Since
h h-m-l
' 'm P I
‘<X>‘ = p , g 6 C(G ). Let j =hj(g], s = s(g), then we 1
_ j -m- _ h-m-l 1+3 j h-m-
have g 1ag = a1+Sp where j 2 m. (gp ) 1agp = a( p )p
h-m-1+j h-m+' h-m-1+j
1 J
= a 43p +Bp # a. Hence ap # 1, so h-m-l+j g h,

or j s m+l. But by Remark A, j 2 m. Thus m s j g m+l, and so
m = j.

Remark E: If g E G is such that <gK> = <§>, then there
exists y E G such that G' = <a> = <[y,g]>.

Proof: Suppose not. Then for all y E G, we must have

m
[3,g] E <ap>. Let a = [x1,x2]. Since g-lag = a1+Sp , where
m
1+8 -1 -l
(s,p) = l, we have a p = g ag = g le’leg =

-l -l -1 -1 -l -1 = -l -l -l -1
8 X1 gg X2 gg Xlgg ng X1 [X1 33])(2 [X2 ,gjx1[X1,g]X2[X2,g]

10

-l -l -1 -1
= X1 [X1 ,8]X2 [X2 :SJXIxztxlangxlagsx21Ix2:8]

-l -l -l '1 -l
= x1 [X1 :81x2 XIXZEXZ aglIxz ,g,X1XéKX1,8][X1:8,XZJIX2:g]

-1 -1 -1 -1 -1 -l -1
X1 X2 xlxztxl ,g][X1 ,g,X2 X1x2][X2 3g][X2 ,g,X1X2]

[x1,g][x1,g,x2][x2,g]
'1 '1 ‘l -l -l
= le’lele a81Ix1 9g’x2 XIX2][X2 ,g][x2 ,g,x1x2]

[Xlsgjtxlagsxzj[xzag]

But, by (10.2.1.2) of [5], [8b.c] = [a,c][a,c,b][b,c], so [a,c][b,c] =
-l . . . -l _
[ab,c][a,c,b] if GI is abelian. Hence [x1 ,g][x1,g] _

-l —l -l ~l -1 -l -1
[X1 xl’g%[x1 agaxl] and Ixz :g][xzag] = [X2 ngg][X2 3g’X2] ’
so all...Sp 1 1,g,x2]-1.

_ -l -l -1 - -1 -
- [Xl’x2][xl ’g,x2 Xlx2][:2 :g’XIXZJIXI :gsxl] [X2

m

1 m -
a +sp = g 1ag, we have
m

-l -l -l -l -l -l
<[x1 ,g,x2 x1X2][x2 ,g,x1x2][x1 ,g,x1] [x2 ,g,x2] > = <ap >.

However, since [y,g] E <ap> for all y E G, and by Remark A,
- j(W)
1an = ai>(1+s (W)p )

m+l
[ap w] E <ap >. Thus [x -1 g x -lx x ] [x ‘1 g x x ]
’ 1 ’ ’ 2 1 2 ’ 2 ’ ’ l 2 ’

-l

w , where j(w) 2 m, we have (ap)-1w-1apw =

m+l

[x1 ,g,x1]-1 and [x2”1,g,x2]-1 are all elements of <ap >,

and so we must have <ap

-l -l -1
> = ([XI :gsxz X1X2][X2 agaxlxz]
-1 '

1 m+l

[x1 ,g,x1]-1[x2-1,g,x1]— > S <ap >. But this is a contradiction.

Thus if g E G is such that <gK> = <§>, then there exists
y E G such that G' = <[y,g]>.

Lemma 11.1: If G is a p-group, p # 2, satisfying the
conditions W), then there exists a minimal generating set

{x1,c2,...,cr} for G with the properties <x K> = <x> and

1

ci 6 C(G') for i = 2,...,r. Furthermore, x may be chosen such

1
that |<x1> fl G'| is minimal.

11

Proof: By condition 5) of W), a minimal generating set

for G contains r-elements. Let x1 6 G be Such that <x1K> = <§>.

Then x1 4 ¢(G), so there exists a minimal generating set for G

containing x1, say G = <X1,x2,...,xr>. Without loss of generality,

K_- K_-w w-IK_
x1 ~ x. Let x2 — x . Then (x2(x1 ) ) —
K w -l K _ -w -w -1 _ —' w
x2 ((x1 ) ) - x (x ) 1, so we have x2(x1

we can aSSUme

)'1 e C(c') ker K.

Thus x where c2 6 C(G'). G # <x ,xr> since

1,X3,ooo
there are r-elements in a minimal generating set for G.

w K -
2 = x1 c2 6 <x1,c2,x3,...,xr>, so G = <x1,c2,x3,...,xr>, x1 = x,

c2 6 C(G') and {x ,xr} is a minimal generating set

=XWC
2 1 2

X

1,c2,x3,...
for G.

K -
..,x > where x = x,
r

2,...,ck_1,xk,. 1
ci 6 C(G'), ci 4 ¢(G) for i = 2,...,k-l < r. Let x K = £3,

k
then (xk(Xls)-1)K = (xk)K((Xls)-1)K = £s(§3)'1 = 1. Hence we

-1

Now suppose G = <x1,c

S

s
have xk(x1 ) E C(G'), and xk = x1 ck where Ck é C(G')-

000 00. . " S
G # <x1,c2, ,ck_1,xk+1, ,xr>» Since there are r element

in a minimal generating set for G. xk = xlsck E <x

and we must therefore have C = <x ..,xr>. Thus

1c2,...,ck,xk+1,.

{x1,c2,...,ck,xk+1,...,xr} is a minimal generating set for G
with the properties le = x, ci 6 C(G') for i = 2,...,k s r.
Thus by induction, we have proven that there exists a minimal gen-

erating set for G of the form C = <x1,c2,...,cr> where

<x1K> =‘<x>, ci 6 C(G') for i = 2,...,r.

Since the only property of the element x1 that was used

in the above proof was that <x1K> = <x>, we may now choose x1

to be an element of G such that <x1K> = <§> and ‘<x1> O 3" =

min{‘<x>’n G'll<xK> = <§>}.

1’C2"°°’Ck’xk+l’°'

.,x

12

Corollary 11.1: In the notation of Lemma 11.1, c may be

 

2

chosen such that G' = <[C2,x1]>.

. K - .
Proof: By Remark E, Since <x1 > = <x>, there eXists y E G
tr nik
such that G' = <a> = <[y,x1]>. Let y = x1 ( H ci )a . Then

I
r n

n, r n,
c. 1>ak][( n c. 1>ak.x l
2 1 i=2 1 1

II :JWN

n 1:
i k
2ci )a ,xl].

= [(

i

":1"!

Using the facts [ab,c] = [a,c][a,c,b][b,c] and c1 E C(G'),
r 11. k r ni k
a e C(G'). we have [y.x J = [( n c. 1)a .x l = ( n [c.ix 1 )[a.x I -
l ._ 1 l ._ 1 l 1
1—2 1’2
' r Di k
Thus G = <[y,x1]> = <(.H2[Ci’xi] )[a,x1] > s

1:
m

P -
<[c2,x1],[c3,x1],...,[cr,x1],[a,x1]>, [a,x1] E <a > s $(G'), so we
have G' = <[c2,x1],[c3,x1],...,[cr,x1]>. Since G' is cyclic,
by choosing i such that \[ci,x1]\ = maxfl[cj,xl]flj = 2,...,r},
we have 6' = <[ci,x1]>. Now reindex, if necessary, to set i = 2,
and we have 6' = <[c2,x1]>.

1+t m
p where

-1
We now have that x1 [c2,x1]x1 - [c2,x1]
(t’p) = 1-

Lemma 11.2: Let G be a p-group, p f 2, satisfying the

 

conditions (U), and suppose C = <x1,c2,...,cr> where <x1K> = <x>,
ci 6 C(G') for i = 2,...,r, and G' = <[c2,x1]>. Then
[Ci’cj] e Z(G) n G' for i,j = 2,...,r.

lggggf: By (10.2.1.5)of [S], we have [x,y,z][y,z,x][z,x,y] =
[y,x][z,x][z,y]x[x,y][x,z]y[y,z]x[x,z][z,x]y = 1 since 6' is
abelian. Thus [x1,ci,cj][ci,cj,xl][cj,X1,ci] = 1. But
Ci’cj 6 C(G') gives [x1,ci,cj] = 1 = [Cj’xl’ci]° Thus
[ci,cj,x1] = 1, that is [ci,cj]-1x1-1[ci,cj]x1 = 1. Since

k
' =
[ci,cj] E G , [Ci’cj] [c2,x1] for some k. Thus

13

m

-l _ -l k = -l k _ k(1+tp )
x1 [ci,cj]x1 x1 [c2,x1] x1 (X1 [c2,x1]x1) [Ci’xl]
_ l+tp"' - l -1 _ - l 1+t:pm
_ [Ci’cj] Hence [Ci’cj] x1 [ci,cj]x1 — [ci,cj] [C1,Cj]

m

_ tp _ m _ m
_ [C1,Cj] - l, and we have \[CI’Cj1‘ s p . But ‘Z(G) O G" _ p

and since 6' is cyclic, we must have [ci,cj] E Z(G) n c' for
i,j = 2,...,r.

w
P

- 1+
Remark F: If g 1[c2,x1]g = [c2,x1] S where (s,p) = l

w k
(1+sp4) '1 for k 2 l.

(l+spw2-1
a] w -

 

k -1 k
and w 2 m, then (g ) y g = YIY:g]
SP

-1 -l -1
222213 3 ye = yy 3 ya = yfy.g] = yiy.
5P
Now assume the formula is true for k = n and let [y,g] = [c2,x1]u.

w n
n+1 -l n+1 -l n -l n -l (1+sp -1
Then (g ) y g = 8 ( 8 l y 8 )8 = 8 YIyigl w)

Sp
-1 -1 l+s -1 -l 1 w n-1
g yg(g [y,gls)( pm) = yfy.s](g [62»X118)u (+SEWIT‘
sp w n SP
X (1+Spw)u (1+SP 1 -1
i 1]

 

Wl'l

 

yiy.gjicz Spw
w w n+1 w w n+1
sp 1+5 -l-S (1+SP -
yiy.g]——;'[y.s]( p ) w 9* = yfy,gl %

sp Sp SP

 

1

Thus, by induction, the Remark is proven.

Lemma 11.3: If g E G is such that G' = <[g,x1]>, then

 

‘gg‘ 2 pm, where C Q G/G'.

r
-l = al+sp

Proof: Let g ag where (s,p) = l, r 2 m

m-l
m-l m-l r p
-1 1+ -
and G' = <a>. By Remark F, (gp ) Xlgp = X1EX1’g]( splr 1

 

m-l m m-l m-Ip

_ P +89 P _ P
- x1[x1,g] . Hence we have <[g ,x1]> — <[x1,g] >

pm-l
<[c2,x1] > since 6' =<a> = <[c2,x1]> = <[g,x1]>.

m-l m-l
p

p _ k
e s', so g — [czell
m

m
If \gg| g p , then g
m-l k p
for some k. Thus <[gp ,x1]> = <[[c2,x1] ,x1]> = <[c2,x1] >.

m
-l k _ k(1+tp )
But x1 [c2,x1] x1 - [c2,x1] , so we have

14

k ktpm k pm
[[CZ’X1] ,x1] = [c2,x1] and <[[c2,x1] ,x1]> s <[c2,xl] >
m-l m

$ <[c2,x1]p >, a contradiction. Thus we must have ‘ggl 2 p .

Lemma 11.4: Let G satisfy the conditions W), and

 

{x1,c2,...,cr} be as in the conclusion of Lemma 11.1. Also

suppose G' = <[c2,x1]>.

t h+m l Pm
I). If Z(G) s G , then ‘C2‘ = p and G = <c2 >.
* s
2). If Z(G) = z e.A where z = Z(G) n G' and
*
exp Z = exp 2 2 exp A, then \c2‘ = ph+m and
p2m-1 pill-1
<c2 > = <[c2,x1] >.

Proof: Consider the following conjugation relations:
m

l 2 = X1EX1’C2

tpm tpm tpm ‘1 th
] = x1[x1,c2] . (c2 ) ch2
m m
t t .
= cj[cj,c2 p ] = cj[cj,c2] p = cj for all j = 2,...,r by

Lemma 11.4.

-1 -l -1
Also [c2,x1] x1[c2,x1] = xlx1 [c2,x1] x1[c2,x1]

(1+tpm))-l

-1 -1
Xl(x1 [Cz’xl]xl) [CZ’XI] = x1([C2’Xl] [CZ’X1]

tpm -1 tpm -l

= xl([C2,x1] ) = x1[xl,c2] . [CZ’Xl] cj[C2,xl] = cj Since
cj E C(G') for j = 2,...,r.

Thus we have n = n where n denotes the

[C :X J m y
2 1 c tp
2
inner automorphism of G induced by conjugation by the element y.
m
Hence we have [c2,x1]z = c2tp where z E Z(G).
m
_ _ P
1). If Z(G) s G" then <[c2,x1]z> — <[c2,x1]> — <c2 >

and ‘[c2,x1]z| = ‘[c2,x1]\ = ‘CZP ‘ = ph’ so [c2] = Ph+m.

* 7':
2). If Z(G) = Z (8 A and exp Z(G) = exp Z 2 exp A,

m-l m-l m-l

then ([c2,x1]z)p = [c2,x1]p 2p 6 G'. Thus

m-l t 2m-l 2m-1
([c2,x1]Z)p = c2 p E G' and we have <c2p > s G'.

15

‘ th‘ = ‘C Wm] ‘ECZ ,x1]z‘ = , so ‘Cz‘ Thus we have
[<c2pzm 1>‘ = ph-m+1 and since G' is cyclic, we now have
2m-1 m-l
<c2 > = <[c2,x1]p ‘>.
Remark G: If y = xlcza, then for k 2 0 we have
——a——2 <<l+tp "‘-)k-1-‘“kcp >
yk = x1k(cza)k[c2,x1](tpm )

Proof: Let k = 0. Then y0 = l

_ a o
— x1 (c2 ) [c2,x

m
0
’1
r
ll

n+1
= x

n.

a a n a n
1 C2 [C2 3X1 1(C2 ) [C2

--a—2- «1+ch ) °-1>
(tpm )
11

“+1 n a

_ = n a n
Then y — y y XlCZ x1(c2 ) [c2,x

Assume the formula is true

—3-—(<1+tp‘““ > -1- -ntp m)

(tpm )2
1]

(1+tpm)n-l-ntpm)

‘—-——-((1+tpm)n-l-ntpm)

 

2
= n+1 a n+1 n a (th )
Xl (C C2 ) ECz’xl 3 [Cz’xl]
c2 6 C(G').
mn
By Remark F, [c2,x1n]a = [crxl]:1+tP ; - ,
th

n+1 =
Y X1

11+
=X

 

- m2<tpm (1+tp M)“ pm)

n+1 a n+1

(.2 > [cx1](t:m)

since

--—§((L+tpm ) n_1_ -ntp m)

[C2 ,X 1021))

a m——((l+t pm)n+1-l-(n+l)tpm)

a n+1 (tpm )2

 

11(C2 ) [C Cz’xl1

tion, we have proven the formula.

and by induc-

Lemma 11.5: Let G be a p-group, p # 2, satisfying the

 

conditions (u), and let {x1,c2,...,cr] be a minimal generating

set of G

satisfying the conclusions of Lemma 11.1.

Let x1 be

chosen such that [<x1> n G" is minimal, and Suppose

16

_ h
6' - <[c2,x1]>. Then ‘xll 2 p .

Also, <x1>11 G' = <1> if either of the following conditions

 

 

hold:
1). Z(G) sG',
* N k
2). Z(G) =z 99A where z =Z(G) n G',
*
exp Z(G) = exp Z 3 exp A and h 2 2m.
h-l
h-l h-l m p
. P -1 p _ L1+t21 -1
Proof. (x1 ) cle - c2[c2,x1] m
(It?)
h-l+8ph h-l
= c2[c2,x1]p # c2 since [c2,xl]p # 1. Thus
h-l
P h - l
x E C(<c >), hence \x ‘ 2 p . Also, Since c E C(G ), we
1 h 12 h 1 h 2h
have x p 4 G'. x p E Z(G) since (x p )-1g x p =
l 1 1 l
h
h h+l
1+t:mp - . h
8[8:X1]( pm) 1 = g[g,x1]P +Bp = g Since \G" = p .

tP
1f <x1>n (3' ye <1>, then <xl>fl G' s Z(G) n G'. Under

either of the conditions 1) or 2) of Lemma 11.5, we have
h+m, h
, so assume \XI‘ = p +g

pZm-l> =
2

h
exp Z(G) = pm. Thus p 3 \x1‘ 3 p

where O s g s m. Also we have, by Lemma 11.4, <c

m-l
<[c2,x1]p > 2 Z(G) 0 6' under condition 2) of Lemma 11.5, or

<c > 2 Z(G) n G' under condition 1) of Lemma 11.5. Let

2

ph+g-1 r m-l
x1 =‘% p where <E> = Z(G) H G'. Then for some a,
m-l ph+m-l

-1
where (oz,p) = 1, (er? ) = e20“

ph-l aph+m-g-l O[pm-g m ph-l h+m-l
x1 C2 E‘32”’C1]”“'u'r“2'((Ht!) )
(tp )

h-l h-l h-l
x1 é C(<c2>) implies yp é C(<c2>), so yp 4 G'.

17

) = le, so since
h+u

<x1K> = <x>, we have <yK> = <x>. Now suppose that yp E G'

K a: =

h+u h+u h+u+m-g m-g
where 0 g u s g-l. Then yp = x p Czap [C2:X1]a m 2

(tp )
h+u h+u+m-g
((l+tpm)p -1-tph+‘“*“). If 2(0) 3 c', then ezC"p e c'

m
* *
since 6' = <c2p >. If Z(G) = 2 <3 A, exp Z 2 exp A and
2m-1

h 2 2m, then <c2p > s G' and also h+u+m—g 2 h 2 2m-l, so

a h+u+m~g h+u
c2 p E G'. In either case, since yp E G', we must also

h+u

have x1p E 6'. Thus ‘<y>-n G" g ‘<X1>.n G", a contradiction

to the minimality of ‘<x1>i1 G". Thus <x1>41 G'

<l>.

Theorem 11.1: Let G be a p-group, p # 2, satisfying the

 

conditions (U), and suppose Z(G) S G'. Then G is a central pro-
h h+m

duct of the form C = AB where ‘A = <x1,c2lx1p = c2p = 1,

[c2,x1] = cztpm> and B is a p-group of nilpotence class two
having a cyclic center.

Egggg: The proof consists of three steps. First we find
a new minimal generating set for G consisting of {x1,c2,g3,...,gr},
where x1,c2 are as before, but gi E 22(G) for i = 3,...,r.
Next we analyse <x1,c2> s G. Thirdly, we write G as the desired

central product.

Step 1: Let G be a p-group, p # 2, satisfying the con-

ditions (U) and also such that Z(G) s 0'. By Lemma 11.1,

18

G = <x1,c2,...,cr> where <x K> = <§> and ci 6 C(G'). By

1
Corollary 11.1, G' = <[c2,x1]>. By Lemma 11.4, we have |c2\ = p
m

and G' = <[c2,x1]> = <c2p >. By Lemma 11.5 and the assumption

‘<x1> H G" is minimal, we have \x1‘ = ph and <x1>ffi G’ = <1>,
h

him

since x p e Z(G) s c' and <x

1 >nG'=<1>.

1
For i 2 3, we wish to show we can replace each C1 in our
minimal generating set by a g1, where g, 6 22(6). Since
1
[Cj’ci] E Z(G) n G' for i,j 2 2, if [x1,ci] E Z(G), we have
ci E 22(G). Thus suppose [Xl’ci] 4 Z(G). Z(G) $ 6' and G'
cyclic imply that Z(G) s <[x1,ci]>.
[X C ]X [X C 1-1 = x x -1[x c ]x [x c ]-1
1’ i l 1’ i 1 l 1’ i l l’ i
= x (x -l[c x ka )[x c 3-1 where [x c ] = [c x ]k
1 1 2’ 1 1 1’ i 1’ i 2, 1

_ k(1+tpm) -1
’ X1EC2’X1] [Xl’ci] =

tpm
x1[x1,ci] .

-1 _ . . .

[XI’CiJCj[X1,Ci] cj for J 2 2 Since Cj E C(G ).

m m m m
tp ) 1c,c tp

tpm)-1X tpm _ t
i

P _ tP
Also (ci 1ci — x1[x1,ci ] — x1[x1,ci] xci

m m

tp tP

= c, c, c, = c. c, c, = c, b Lemma 11.2.

J[ J, l 1 Ji 3’ 11 J y m
-1 t

, so [x1,ci] z = C1 p where

Thus n -1 = n
[Xl’ci] tpm
c.
1 pm
2 E Z(G). But Z(G) ;. <[xl,ci]>, so <Fx1,ci]> = <ci > s <ci>.
ph-m-l
Moreover, Z(G) g <[c2,x1] > s <[x1,ci]> fl <[c2,x1]>

S <c.>l1 <c >.
1 2

Let Hi = <C2’Ci>' H1 is two-generated since 6' g Hi

and cj,ci are in a minimal generating set for G. Hi s Z(G) = 2

since [c2,ci] e Z(G). Let Hi 3 Hi/Z. Hi/Z is abelian and two-

m
. . t
generated Since Z s G' and Since ci p E <[x1,ci]> s G', we have

19

m
Y 11 h+m ’Y__h . I: P
\ci ‘ s p because \Ci‘ 3 p . \cz | - p Since G <c2 >.
Y Y . . . = y
Thus ‘cz ‘ 2 ‘Ci ‘, so if Hi/Z is cyclic, Hi/Z <c2 >. Thus
yk _ k ,

= (c2 ) for some k, or C1 — c2 2. But 2(a) g G 3 (C2)
then implies ci 6 <c2>. Thus G = <x1,c2,...,ci_1,ci+1,...,cr>,

a contradiction to the minimality of the generating set {x1,c2,...,cr}.

Since Hi/Z is two-generated and c2Y is an element of

highest order in Hi/Z, we have that Hi/Z = <c Y>®<b>, and

N

<c2Y>11 <b> = 1. Let gi 6 Hi be such that giY = b. Then

Hi/Z = <c Y> x><giy> and <c2> n <gi> 3 Z. TY —(g 1Y)w (c 2y)S for

2
w s u
some w,s, so 01 = gi c2 2 where z E Z(G) s G s <c2>, so
= w u f

ci 8i c2 or some w,u.

G ¢i<x1,c2,...,ci_1,ci+1,...,cr>, but Since
c = w u E c c c c > we have

i 31 C2 <X1’ 2"°°’ i-l’gi’ 1+1""’ r ’
G = <x1,c2,. ...,ci _1,gi ,ci+1, ...,cr>.
_ . . = a
gi 6 Hi - <C2,:i>d implies gi c1 c2 for some a,d.
d

Thus [cj ,g1]= [cj ,ci 2d] = [c J,c2 d][cj ,ci a][c J,cia ,c2 ]

= [cj,c2 d][cj ,c a] e Z(G) for j = 2,...,r. Also, gi e C(G')
since c2,ci E C(G )
[x1,gi] e Z(G) and hence gi e 22(0), for if [x1,gi] é Z(G)

we have:

-1 _ -1 -1 _ tp
[Xl’g11XIEX1’gil — X1 X1 [xl’gilxltxl’gi] ‘ X1EX1’giJ

m
. - t
[x1,gi]cj[x1,gi] = Cj for J = 2,...,r. (gitp ) lxlgi p
m m m m
t t . tp -l tp
x1[x1,gi p j = x1[x1,gi] p Since gi E C(G'). (gi ) ngi
tpm tpm
= cj[cj,gi ] = chCj,gi] = cj Since [Cj’gi] E Z(G). m

-1
Thus we have n _1 fl tpm’ or [Xl’gi] z = 8
[x1.gi] gi

where z E Z(G). But [x1,gi] é Z(G) and we must have

20

tpm pm ph-m-l
<[x1,gi]> =<gi > =<gi >. Also, [c2,x1] E <[x1,gi]>,
h-m-l h-m--l
so [c ,x ]p E <g >. But [c ,x ]p E Z(G) on the one
2 l i h-%-11
hand, and on the other [c2,x1]p E <c2> fl <gi>. However,

<c2> fl <gi> s Z(G), thus we have a contradiction. Thus
[x1,gi] E Z(G), and we replace C1 by gi E 22(G) in our minimal
generating set for G.
Proceeding in this manner, we can write G in the form
C = <X12C23g3:°°°agr> Where (X K) = <32), C2 6 C(G'),

1
m
'= = p '=
G <[c2,x1]> <c2 > and g1 E 22(G), i 3,...,r.
S 20 A— A ‘ '= p A
tep . Let A — <x1,c2>. A A_G Since G <c

,« ph ph+m tpm
A = <xl,c2\xl = .2 = 1, [c2,X1]= c2 , (M)

II
!—‘
V
K
('7’

k A
w = up where (u,p) = 1. x1w E 2(3) if, and only if x1 E Z(A),
k k k

. . . p ‘1 p _ p
that is, if and only if, (x1 ) cle — c2. But (x1 )

k
= gi+t222p -1 _
C2EC2’X1J m ’ C2
tP
k+1 k

k
11p +fip - l, or we must have R 2 h. But x — 1 if

l+t m p -
if, and only if, [c2,x1]£ pal 1

= [c2,x
k 2 h, so <x1> n Z(X) = 1.

1 -1 w -1

w A . . w -
c2 E Z(A) if, and only if (c2 ) x1 c2 = x1 , or

[c2w,x1] = l which is true if, and only if ph‘w.
d e .I\ d e -1 d e e -l e
If x1 c2 E Z(A), then (x1 c2 ) x1 x1 c2 = (c2 ) xlc2 — x1,
e 4 d e e -l d A d A
or c2 E Z(A). Then (x1 c2 )(c2 ) = :1 E Z(A). But x1 E Z(A)

P
2

k A
Letting w = u p where (u,p) = l, we have x1w E 22(A)
k A
if, and only if, x1p E 22(A), if, and only if,
c -1(x pk).1c x
2 l 2 1

A
implies xld = 1, so we have Z(A) =‘<c >.

k
m P
A . -1 gl+tp_) -1 A
E Z(A), or equivalently c2 c2[c2,x1] m E Z(A).

tp
k k+1 k+m k+m+l
+ + ‘A
= [c2,x11p 8p = czp 8p 6 Z(A)

P

 

k
](l+tpm)p -1
l m
tP

But {c2,x

21

if, and only if k+m 2 h, or k 2 h-m.
d 4 . . d A .
c2 E 22(A) if, and only if, [c2,x1] E Z(A), which is

dpm 5' d .A .
E Z(A). Thus c E Z(A) if, and only

equivalent with c 2

2
if, ph-m‘d.
w d AA -1 w d -1 w d A
If x1 c2 .E22(A), then x1 (x1 c2) x1 x1 c2 E Z(A), or
- d - h
x 1(c ) 1x C d E 2(2) and we have c d E Z (A). Thus
1 2 1 2 2 2 h h
w d d -1 w ‘2 h z ,2 _ p "m p 'm
(X1 c2 )(c2 ) - x1 E ZZ( ). T us 2( ) — <X1 ,c2 >.

lHom(fi/A', Z(fi»| = p2m since we have ‘2 is a

\AC(A)\

p-group, p # 2, with no abelian direct factors. Also if X X K/A',

then since <x1>;1 G' = <x1> 0‘2} = <1>, we have \xlyl = ph,
‘CZY‘ = pm, and <x1Y> fl <c2Y> = <1>, so 2/3' = <X1Y> 8><C2Y>.
h-m h-m h
A A _ p p p _ 2m
‘ZZ(A)/Z(A)\ - l<xl ,c2 >'/<C2 >‘ - p . Thus we
A A A A A A A
have \Acm)‘ = ‘Z2(A)/Z(A)l = View) 0 1(A)|. Thus AC(A) s I(A).

Step 3: We now have G = <x1,c2,33,...,gr> where
gi 6 22(0), 1 = 3,...,r, and X = <x1,c2> A G such that AC@) 3 1(2).
We note that in Step 2, we have Shown that 3' = G' and 2(3) = Z(G).

Since gi E 22(6), we have E AC(K) s 1(3). Thus there exists

n A
gilA
It is known that if T is a

A
. E A Such that n A = n A.
yi gilA yi‘A

group, B s T and for x,y E T we have fix‘B = ny‘B’ then
‘1 . __ - A
xy .E CT(B). Since “gilfi — nyi‘fi, we have giyi E CG(A), or

that = d whe d EC K S' A A d

gi yi i re i G( ). ince 22(A) s ¢(A) an
A A
A A_G, we have by 7.3.17 of [8] that 22(A) s ¢(G). Thus yi E ¢(G).
If (11 E ¢(G), then gi E ¢(G), a contradiction to gi being in a
minimal generating set for G. Thus di E ¢(G).
G # <x1,c2,g3,...,gi_1,gi+1,...,gr>, but Since yi E <x1,c2>,

= d .0. .0. .

81 yi i E <X1,C2,g3, ’gi-l’di’gi+l’ ,gr> Thus we replace

gi by di' Proceeding in this manner, we can write

22

A
G = <X1,c2,d3,...,dr> where di E CG(A) for i = 3,...,r.
A A
Set B = <d3,...,dr>, A = <x1,c2>’A_G. Then G = AB,

B s 06(3) so ‘3 AG, B A.G. A.n B s Z(A) and we have that G is

 

a central product of the desired form since Z(B) s Z(G) which is
cyclic.
" ph'h“ ph
m Corollary 11.2: A = <x1,x2‘x2 = x1 = l, [XZ’XI]
= xgp , (t,p) = l, m.< h> is a group such that all its central

automorphisms are inner automorphisms.

lgggggz Step 2 of Theorem II.1.

Theorem 11.2: Let G be a p-group, p # 2, with Z(G) g G'
and G' is cyclic. Then AC(G) s 1(G).

lggggg: Since Z(G) g G', G has no abelian direct factors.
6' cyclic and Z(G) g 6' yield G has nilpotence class at least
three, and Z(G) n G' = Z(G) s G', thus G satisfies the conditions

W), and also the hypothesis of Theorem II.l. Thus we have G =‘RB

h h+m

A
as a central product where A = <x1,c2‘x1p = c2p = 1, [CZ’XI]

m
t A h
= C2 P 9 (t3?) = 1: m < h), C. = A. = <C2p >’ ‘6" = P :

h
A
Z(G) = Z(A) = <c2p >, ‘Z(G)\ = pm, and Z(G) s G'. G a central
product of ‘A and B gives A.n B s Z(G) s G'. Using the results

of Step 2 of Theorem 1, and the fact that B s CG(A), we have

 

 

ph-m ph-m
22(6) =i<x1 , c2 , B>. Thus
ph-m ph-m
22(6) = |oc1 ,cz >HB‘ = 3m B
Z(G) Ph-m ph-m pm G H B
\Z(G)H<x1 ,c2 > 0 8|
h h-m A h-m
since Z(G) = «:2p > s <c2 > and A n B s Z(G) n B s.<<:2p >

nBsG'nB.

23

A
I A BC. I A I I A A
Let yG 6 ET-n ET_ , then yG e A/G , but G s A so y e A.

Also yG' e BG'/G', thus y 6 BG'. Hence y E'Acw BG' = G'(A n B) = G'.

' - A BC. _ '-
Thus yG = l and we have ET'fl GT_ - 1. Thus
A A
c A B ' s A B
a?“ = GT®C_;_'_ = 'G'T® E67 . ‘AC(G)‘ = ‘Hom(G/G', Z(G))\

= \uoucii/c', z(G))HHom(qu—G-r , Z(GM = szlHomqfigT, Z(GM

= 2m B
p Bnc

A
because B is of nilpotence class two and B centralizes A.

since if b E B, then [y,b] E Z(G) for all y E C

m m
But [y,bp 3 = [y,b]p = 1 since |2(c)\ = pm. Thus

m

 

exp HgG' 5 pm since bp E Z(G) s G' for all b E B. Thus
Hom (E3573 Z(G)) - 3561' We now have \ZZ(G)/Z(G)‘

= \AC(G) n I(G)| = \AC(G)\, or AC(G) s 1(G).
Theorem 11.3: If G is a p-group, p # 2, Satisfying the
conditions u)) and Z(G) s G', then ‘G‘ divides ‘A(G)\.

Proof: By Theorem 11.1, G =‘AB where B is a p-group
A ph ph+m
of nilgotence class two, A = <x1,c2‘x1 = c2 = l, [c2,x1]
t A
= c2 p , (t,p) = l> and Atfl B s Z(A) = Z(G) = <c >
h-m

- P
- <[c2,x1] >.

Define ¢ on

>>
G
\<
X
I
X
I—'
O
N
ll
'0
0

v
'0
I
”U
I
H
A
D
N
is»
v
v D
II
A
X
H
'U
D
V
'0

E, _ . ”
] 2 - 1 Since x E Z2(A).

II
A
><
H
V
O
N
r"'1
O
N
H
X
H

m

¢ ¢» = P = ¢ tP
[c ’Xl ] [x1 c2,X1] [c2,x1] and (c2 )
h-m m h m h-m

tp p m
C2 [c2,x1

tpm(tpm-1l _ tp
3 2 “ C2 '

II
A
>4
|"-‘
O
N
ll
H

ab A . ab®_ a
A" For x1 c2 E A, define (x1 c2 ) — (x1 )
a b c d a+c b+d b a
(*1 C2 xi c2 )fi = (X1 C2 [C2 ’Xl 1>¢

a+ b+d b
= (X1 C)¢(C2 )®[c2 ,xla] since $‘A’ = I‘A,.

Thus ¢ A = 1‘

A

24

a+c ph-m

h-m
1 (x1 c2) (b+d) b a+c-Pp (b+d)C (b+c)

a
[C2 ’XI 1 ’ X1 2

h-m
b hm bfihl
[C2 ’X1CJEC2’X1P 3i 2% Tl °

h-m h-m

b C p d
c2) x1 (x1 c

2)

_ - h-
ph m bgb-l) X c dph m dEC x p m]
1 2 1 X1 C2 2’ 1 2

3(Cp

a b c d
(X1 C2 )¢(X1 C2 )¢ = X 1

1
= x a+bph-mC b[ X
1 2 C2’ 1

_ .a+c+bph'm b b c dph‘m d ph-nbgb-ll-l-dgd-ll
x1 c2 [c2 ,x1 1x1 c2 [c2,x1 ] 2

h-m h-m h-tm
_ a+c+(b+d)p b b dp d b c p b(b-1L+d gd-Q
x1 c2 [c2 ,x1 ]c2 [c2 ’Xl ][c2,x1 ] 2
= a+c+(b+d)ph’m b+d[ b c ph’m bgb-1)+d(d-1)+2bd

x1 C2 C2 ’Xl 1ECz’xl 1 2

 

h-m h-m
_ a+c+p (b+d) b+d b c p
X1 C2 [C2 ’X1 1C2’X1 3

 

(b+d) £b+d - ll .
2

Thus ¢ e A(A) and $13} = lifi" so @1AnB = 1‘313.

* 7‘: 9: 7':
' o a = m ¢ = w {b = .
Define ¢ on G by. X1 x1 , c2 c2 , di di

* * .
Thus ¢ ‘3 = 1‘B’ and ¢ ran = QYQWB = 1‘AflB Since
811 B S Z(G) = Z(A). Hence ¢* E A(G).

k h-m h-m
(¢*) _ kp - - _ ¢* _ ¢ _
c2 x1 c2 Since 1f nil 1, c2 — c2 — x1 c2.

n
Assume true for k = n. Then c2(®*) = (02(Qk) )®*
h-m
* *
np )¢ ‘ x np c ¢ = x (n+1)p c . Using this formula,
1 2 l 2
h

pm
C(¢*) =XpC=C

m 2 " 1 2 2’ 2 1 2 2’
w)" = d (so "
i

we see C

x x1, = di. Thus pm“¢*| and \®*‘ ;.pm ,

. k
hence |¢*‘ = pm, (¢”) E 1(G) for k = 1,2,...,pm-1, because if

k h-m
* -
(¢*)k E 1(G), then c2(® ) = x kp c = y 1c2y for some y E G.

h-m l 2

Thus x kp = [c2,y] E G'. But <X

1 >{1 G' = l, a contradiction.

1

25

‘<¢*>I(G)\ = pm T%%%YT = pm l§#'= \G‘, and we have ‘6‘
P

*
divides \<¢ >I(G)\, so ‘6‘ divides \A(G)‘.

Theorem 11.4: If G is a p-group, p # 2, satisfying the

 

* *
conditions (u), Z(G) = 2 ex A where z = Z(G) n G', and
*
exp A 2 exp Z , then \G‘ divides ‘A(G)‘.
Proof: By Lemma 11.1 and Corollary 11.1, we have

K _
G = <x1,c2,...,cr> where <x1 > = <x>, ci E C(G') and

G' = <[cz,xl]>. By Lemma 11.5 and its proof, we know \x1\ 2 ph
h- h-l
x1p E CG(<c2>), so le E G'.

and
, A ' A A A'YA'
Con81der K = <x1,C2>. c s K so K A_G. Let K ~ K/G .
We wish to show R/G' is a two-generated abelian group of type
a b
(p ,p ) where a 2 h, b 2 m.

Clearly ‘R/G' is abelian and at most two-generated. If

fi/G' is cyclic, then fi/G' = <x1Y> or fi/G' = <c2Y>. If

A/I- Y h .. w I I
K G - <x1 >, t en c2 — x1 g where g E G . But G s @(G),

so G = <x1,c3,...,cr,G'> = <X1’C3"'°’Cr>’ a contradiction to the

assumption that a minimal generating set for G has r-elements.

Y

2 >. Thus fi/G' is a two-

A similar argument shows fi/G' # <c
generated abelian group.

If \x1y\ 2 ‘CZY‘, x1Y may be chosen as one generator of
E/G', so ‘E/G' = <x1y><&«<y>. ‘Now let g E B be such that gY = <y>.
Then R/G' = <x1Y><®-<gy>. c2Y E fi/G', so c2Y = (xly)r(gY)S, or
c2 = xlrgsu, where u E G'. Since G' = <[c2,x1]>, we have
6' = <[X1rgsu,x1]>. [xlrg8u,x1] = [Xlrgs,x1][x1rgs,x1,u][u,x1]
= [xlrgs,x1][u,x1]. But u E 6' implies [u,x1] E G'(p),thus

r s
G' = <[x1 g ,X1]>.

26

r s r r s S 5
[X1 g ,X1] = [x1 ’XIJEXI ,x1,g ][g ,x1] = [g ,xl]. Thus
3 . s s
G' = <[gs,x1]> s <[g,x1] > s <[g,x1]> = G' Since [g ,x1] = [g,x1] w

k
where w is a product of elements of the form [g ,x1,g], and

'(p).

hence w E G
G' = <[g,x1]> implies, by Lemma 3, ‘gy‘ 2 pm. Thus if

‘lel 2 ‘c2Y\, we have ‘R/G' = <x1Y><®«<gY>, and \xly‘ 2 ph,

“ b
‘gY‘ 2 pm. Hence K/G' is of type (pa,p ) where a 2 h, b 2 m.

Now suppose K/G' = <c2Y> 8><y>. \CZY‘ 2 pm since in

Lemma 4, we have shown [c2,x1]z = c2tp where z E Z(G), and

hence \c2y| 2 pm. We wish to show \§\ 2 ph.

Y ' - A A
H =1<CZ ”My? = 119—911 =1 “9' 1
y y y <c2,G'>/G' '

c > c >
< 2 < 2

It is sufficient to show that if x1u E <c ,G'>, then

2

1,c2> X <c2y>ka> i <y>, we have (xlu)YJ

= ((le)u)J = (leJ)u = l if, and only if, x1U 2,
If x1u E <c2,G'>, then for some k,n we have x1u =

k n , ,. . u
c2 EC ,xl] . G s CG(<c2>) implies x1 E CG(<c2>). But

h . .
p ‘u, Since in <x

E <c G'>.

P u . . h
x1 E CG(<c2>), thus x1 E CG(<c2>) implies p ‘u.
In either case, we have ’fi/G' is a two-generated abelian

group of type (pa,pb) where a 2 h, b 2 m. Now suppose

* * G
exp A 2 exp Z and Z(G) = Z €>A. ‘I(G)\ = TE%E%TI’
\AC(G)\ = \Hom (G/G',A)|lHom (c/c',z*)\, \Ac(c) n 1(c)\ s

‘Hom (G/G',Z*)\ since if a E AC(G) n 1(G), x'lxa E Z(G) for

all x E G, and since a = ny, x-lxa -1Xn

- *
thus x lxa E Z(G) n G' = Z . Thus a induces a homomorphism

= x y E G' for all X E G,

from G/G' into 2*, and hence ‘AC(G) fl I(G)\ s ‘Hom(G/G',Z*)\.

27

\Hom(G/G',A)‘\Hom(G/G',Z*)“Zig)l

 

‘Ac(G)I(G)‘ = {AC(G) n I(G)\ 2
2(2) |Hom(G/G',A)| = \Gllnom(G/G',A)|/|z*|\A|.

Let a e A be such that [al = exp A. Let c 5 G/G'.
Then ‘agl = la‘ since if (ag)k = l, then (ak)g = l, or
k . * k * g
a E G n Z(G) = z . Thus a e z n A = <1> and ‘a \ = {a\.
Thus there exists y E G/G' with the property y is in a minimal
generating set for G/G' and \§| 2 exp A.

- A I . , a b

Since K/G is a Subgroup of G/G of type (p ,p ),
where a 2 h, b 2 m, by Theorem 5.5.7 of [8], we know G/G' is at
least two-generated, say by the above y and g, and is such that

1§1 2 exp A, \g| 2 pm. Thus \Hom(G/G',A)| 2 1A1pm = \A‘\Z*‘.

Hence we have ‘G‘ divides ‘AC(G)I(G)‘, so ‘6‘ divides ‘A(G)‘.

Theorem 11.5: If G is a p-group satisfying the conditions

 

* *
(U), Z(G) = Z <8 A where Z = Z(G) D G' and h 2 2m, then \G‘

divides ‘A(G)|.

*
Proof: If exp A 2 exp 2 , we have C satisfies the hypothesis

of Theorem 4, thus ‘6‘ divides ‘A(G)‘. Thus we assume exp A ; exp 2

Then we have exp Z = pm.

By Lemma 11.1 and Corollary 11.1, we may assume

K -
G = <x1,c2,c3,...,cr> where <x > = <x>, ci E C(G') for

l

i = 2,...,r and G' = <[c2,x1]>. By Lemma 11.4, we have ‘c2\ = ph+m
m-l

2m-1
and <C2p > = <[C2,x1]p >. By Lemma 11.5, ‘XI‘ 2 ph and

<x1> n G' = (1>.

Now, since h 2 2m, we have h-m 2 m 2 m-l, thus
m-l m
<[c x]p > p = *
a 3 <[C ,X ] > Z . Hence we know that <c > 2
22m11 2 l 2
‘ A

p > 2 z

<C:2

*

28

By Lemma 11.2, [C1,Cj] E Z(G) n 6', thus if [X1,Cl] E Z(G),
ci E 22(6). If ci E 22(G), we replace ci, for i = 3,...,r, by
di’ where di 6 22(6) to obtain a new minimal generating set for

K ..
G of the form {x1,c2,d3,...,dr} where <x > = <x>, c E C(G'),

1
(11 E 22(6), 1 = 3,...,r and G' = ([cz,x1]>. To obtain di’ we

2

proceed as follows:
Since [x1,ci] E Z(G) n G', we have

-1 -l -1 -l k '1
[xi’cilxl[xi’ci] = X1X1 [Xi’cilxi[xi’ci] = X1x1 [Cz’xil XiEXi’Ci]

_ k _ k(1+tpm) -1 _ tpm
where [xl’ci] — [c2,x1] ,— x1[c2,x1] [x1,ci] - XlEXI’Ci] .

-l . .
[xl’ciJCjEXI’Ci] = c. for J = 2,...,r Since cj E C(G').

J
tm-l tIn tm tm,
(ci p ) xlci p x1[x1,ci p 3 = x1[x1,ci] p Since ci E C(G).

m m
tp -l tp
(ci ) cjci

tpm tpm
= c, c. c, c. c, c, = c, since
1[ 1’ 1 1 J[ J’ 1] J

pm. Thus we have n ‘ n

[xl’CiJ-l C tp

* *
[Cj’ci] E Z and exp 2

m

-l _ _ tp .
or [Xl’ci] z — [ci,x1]z — ci , where z E Z(G). Since

m
tp
[x1,ci] E Z(G), ci 4 Z(G).
J
Let Hi = <c2,ci,Z(G)> and consider Hi ~ Hi/Z(G). Since

H.’ = <[c2,ci]>, Hi' 3 Z(G) and Hi/Z(G) is abelian. c J # l

i 2

J _ 2m-1

since c2 = 1 implies c2 E Z(G), thus c2p E Z(G), thus
m-l
p . . J - . _ -

<[c2,x1] > s Z(G), a contradiction. ci # 1 Since ci — l

m
implies citp E Z(G) which is impossible.

1f Hi/Z(G) is cyclic, then either Hi/Z(G) = <c J> or

2
_ J _ J _ a
Hi/Z(G) — <ci >. If iii/Z(G) — <c >, then ci — c 21, where

2 2

2i 6 Z(G). G # <x1,c2,...,ci_1,ci+1,...,cr>, but Ci = C 2, =

C 0.. C Z C o.
“1’ 2’ ’ i-1’ i’ i+1’

Since ci e Z(G), zi 6 22(6). 1f iii/Z(G) = <ciJ>, then

°’Cr> = G. Thus replace C1 by z

e
c2 = Cibz Where 2 E Z(G). If (b,p) = 1, then c1 = C2Zi Where

29

<c2J>. Thus we are in the above case.

21 e Z(G) and Hi/Z(G)
, _ _ b

If p‘b, then Since G' - <[c2,x1]>, we have 6' — <[ci z,x1]>.

But [cibz,x1] = [ci,x1]b E G'(p) since p‘b. But this is

impossible, thus we must have (p,b) = 1. Thus if Hi/Z(G) is

cyclic, we can replace ci by di’ where (11 E Z2(G), in a minimal

generating set for G which contains x and C

1 2°
If Hi/Z(G) is two-generated, then Since ‘c2J\ = pm and
J m J _
‘Ci ‘ s p , we may assume that Hi/Z(G) = <c2 > 8><b>. Let
8- G H. be Such that g,J = b. Then H /Z(G) = <c J> ga<g.J>,
1 i i 1 2 i
and <c2>r1 <gi> S Z(G). Since gi E Hi = <c2,ci,Z(G)>, we have
J J k J b k b
e c c' . L t = , = .
gi. - ( ) 8 Ci (C2 ) (gi ) Then Ci C2 gi z, whtre
' b _ R J J b
2 E Z(G). Set (11 = gi 2, so ci — c2 di. We note d1 = (gi ) ,
so <di>67 <c2> s Z(G). G ¥ <x1,c2,...,ci_1,ci+1,...,cr>, but
k .
C1 = C2 di E <X1,c2,...,ci_1,di,ci+1,...,cr> = G. Thus (11 is
in a minimal generating set for G which contains x1 and c2.

. _ k -1 u
Since di (c2 ) Ci’ we have [Cj’di] E Z(G) H G for
j = 2,...,r. If [x1,di] e Z(G) n c', then d. e 22(0). Thus
m-

t P
Suppose [Xl’di] E Z(G) n G , then [c2,x1] E <[x1,di]>. Also

-1 -1 -1
d =
[Xl’ i3XIEXI’di] X1X1 [Xl’diJXlEXI’di]

-1 k -1 k _
xlx1 [c2,x1] XIEXl’di] where [c2,x1] - [Xl’di] ,

._ XIECZ’XI] [X1,d1] — X1[X1,di] LX1,di] — XIEXl’di]
-1 tpm -1 tpm
m m m m
tp tp _ tp _1 tp
xltxladi ] = X1[X1:di] Since di 6 C(G'), (di ) dei =

7" m

m

tp . >1: .
c, c. d, = c, Since c, d, E Z and ex Z = .
J[ J, 1] J E J, 1] p p

tp:n
c.[C. d.
J J’ 1 J

m
Thus n = n m: so [x ,d,] 12 = d,tp where
-l tp l i i
[X ,d.] d.
1 i i

tm J th .
z e Z(G). Thus [di,x1]z = di P , and [di,x1] = (di 9 ) . SlnCO

30

pm-l * pm-l
[c2,x1] E <[x1,di]> = ([di,x1]> and Z g <[c2,x1] >, we
m-l
- p J J J - . .
now have 1 # ([c2,x1] ) E <c2 >'fl <di > = 1. But this is

impossible, thus we must have [x1,di] E Z(G) n G'. We now replace

c. by di so we may assume G = <x1,c2,d

...,d > where
1 r

3’

K - - h _ '
<x1 > - <x>, ‘X1‘ 2 p , G' — <[c2,x1]>, <x1>i7 G

<1>, c2 E C(G')

and di E 22(6) for i = 3,...,r.

Set A = <x1,c2>, B = <d3,...,dr>. Then A A G since
G' S'A and we have G = AB, A.fl B s A11 2 (G).
2
~ - 2m-l 2m-l _
Let A R A/G'. Since <c2p > s G', we have (c2p )p = l,
p 2111-]. , ' _ p _ ' _ h+S
so \cz ‘ s p . Since <x1>;1 G — <1>, \xl ‘ — \x1\ — p
where 0 s s s m. h 2 2m implies \xlp‘ 2 ‘Czpl’ thus x1p is

an element of A/G' of highest order, So we may assume

A/G' = <X p>'89<b>. Let g E A be Such that g0 =‘b. Then

1
A/G' = <xlp>€®‘<gp>.
c2p = (le)a(gp)b so c2 = xlagbw, where w E G'. Since

a b
G' = <[c2,x1]>, we have G' = <[x1 g w,x1]>.

a b _ a b a b
[X1 g WSXl] — [X1 g )X1][x1 g ,X1,W][W,X1]

a a b b a b b
~ [X1 ,X1][X1 ,X1,g JEg 3X1][X1 g ’Xl’WJEW’Xl] _ [g 3X1][w3x1]'
w E G' implies [w x ] E G'(p) thus 0' = <[gb x ]> s /[g x ]>
3 1 3 « 3 1 ‘ ’ 1
since [gb,X1] = [g,x1]bu where u is a product of elements of

k
the form [g ,x1,g] which are in G'(p).

By Lemma 11.3, we have
‘gp‘ 2 pm. Thus A/G' is abelian of type (pa,pb) where a 2 h,
b 2 m. Thus by 5.5.7 of [8], we have G/G' contains a subgroup

of type (pa,pb) where a 2 h, b 2 m.

We may assume that c p = (xlp)a(gp), or that c

x ag w
3
2 1 h+s-m
fi, (:8— p

where w E G'. Now define ¢ on x1,g by x1 = x1, g“ — x1 g.

2

31

h+s-m

Then, Since G' = <[g,x1]>, we have [g$,x1®] = [xlp g,x1] =
ph+S-m -1 -1 ph-f-s-m ‘1 -1
(X1 8) X1 X1 g X1 = g x1 g x1 = [g,x1], so
¢ pm ph+s-m pm ph+s pm ph+s-m pm
m- = Haw Also, <g> = (.1 s> = x. g ml 1
{Em ) m h+s-m *
2 1 = gp since xlp E 22(6) and exp Z = pm.

2 a b
If y E A, y can be written in the form y = x1 g u where

u E G'. Define ¢ on ‘K by y¢ = (xla)¢(gb)®u. @ is well-

defined, for if xla(1)gb(1)u1 = xla(2)gb(2)u2, where ui E G'
for i = 1,2, then Xla(l)-a(2) = gb(2)u2u1-1(gb(1))-1. Thus
(X1a(1)-a(2))p = (gb(2)-b(1))p = 1 since K/G' = <xlp> Q9 <gp>.
Hence Xla(1)-a(2) E 6', but <x1>r1 G' = <1>, so we have
xla(1) = x2a(2). Thus gb(1)ul = gb(2)u . But then we have
gb(1)-b(2) = uzul-l, s (gb(1).b(2))p = l. ‘gp\ 2 pm implies

m

l - 2 1 -b 2
’50 (gb<>b<>)o=gb<> m:

m pm a p
P \(b(1)-b(2))- But (8 )” = g

uzul-l, or (gb(1))¢((gb(2))¢)-1 = uzul-1 which implies

(8b(1))¢u1 = (gb(2))®u2 Thus ¢ is well-defined.

For g1,g2 E G', we have:

a b c d ¢ _ a b c c d a
(x1 9. 81 X1 8 g2) — (x1 g x1 glig1,x1 1g g2)
a+c b b c d c c d
(X1 g [g 3X1 lglg [gl,x1 thl’xl ’g 1g2)¢

a+c b b c d d c c d
(X1 8 [3 :X1 18 gIEgl’g ][g1’X1][-g1’x1 3g 182)“?5

_ a+c b+d b c b c d d c c d ®
— (x1 g [g ,xl Jig ’X1 ,g ]g1[gl,g][g1,x1][g1,xl ,g 182)
h+s-m

= xla+c(xlp g)b+dw where

S

_ b c b c d d C c d
— [g 5X1 ][g ,X]. 9g1g1[g13g ][g13X1 ][gl’xl 9g ngg

a+c+(b+d)ph+s'm b+d ph+s'“‘ grid) (b+d-1)
= x1 g [g,x1 ] 2 w.

 

32

a b ® c d ¢ 3 ph+s-m c ph+s-m
(x1 g g1)'(x1 g g2) = x1 (x1 g) glx1 (xl g) 82

_ dgd-lz
h+s-m h+s-m Elk-ll h+s-m h+s-m
-x a+bp b[ ‘x p 1 2 x c dp d[ x p 1 2 g
1 g g’1 gllxl g g’l 2

m bgb-llfigd-ll

h+S-m h+s-m d h+s- 2

a+bp b c dp p
X1 g glxl x1 g gzig,xl ]

h+s-m
E 22(6)

since x

l

bgb-1)+d(d-11
2

 

a+b

h+s-m h+s-m h+s-m
= X p bX C [g X Clx d
1 g 1 g1 1’ 1 1

d p
P 8 gztgaxl J

ggb-iwdm—i)

 

a+c+bph+s-m h+S-m h+s-m

b b c dp c d p 2
= X1 g [g ’Xl 1X1 glcgl’xl 1g g2[g1X1 ]

_ x a+c+bph+sum b dph+S-mE b c d[ x c E c d

1 3 X1 8 ,x1]g1g gl’ 11 gl’xl ’3 1g2
h+s_m bgb-1)+dgd-Q
p 1 2

 

[3,x1

h+s-m h+s-m
a+c+ b+d b d b c d d

c c d ph-+3-'m bib-llgdld-ll
[g1,xl ][g1,X1 ,g nglg,xl ]

a+c+(b+d) ph+s ”m

b b c d d c c d
— x1 g [g ,x1 18 g1[g1,g Jigl,x1][g1,x1 ,g ]g2
h+s-m 99'9“ (d'll b h+s-m

2 dp
[g,X1p ] [g ,xl ]

a+c+(b+d) ph+S “m b+d

_ b c b c d d
- X1 g [3 ’X1 1118 ,XI ,gjg1181,8 ]
c c d ph+s-m £b+dléb+d-ll
[gl’xl 1E813x1 ,8 1g2[gaxl J

Qfi) (b+d - 1)
a+c+(b+d)ph+s 'm b+d ph+s 'm 2
= X1 g WEg:X1 3

Thus ¢ E A(K).

33

k k h+s-m
gq5 - x1 p g for k 2 l. The formula is true for
n+1 n
k = 1. Thus assume it is true for k = n. Then g® = (gq5 )® =
h+s-m h+s-m h+s-m h+s-m
(x “P g)¢ = X “P X p g = X (n+1)p
l 1 1 1
. ¢Pm pm pmph+s-m
Since x1 = x1 and g¢ = x1 g = g, we have
m pm"1 ph+s-l m-l
‘¢‘\P . g® = x1 g # g, so ‘¢\ > p and we have
m
lal = P -
Because c2 = xlag w where w E G', we have c26 =
a h+s-m ph+s-m pm ® a pm ph+s-m pm
x1 X1 g w = x1 c2. (c2 ) = (cz’) = (x1 c2) =
Ph+S pm ph+s-m pm pm
X1 C2 [C2:X1 ] = c2 . Thus we have ¢\ pm =
<C >
2
1 m 3 (D = 1 and C5 ' 1 ' ThUS
‘<c2p > ‘<xl> ‘<x1> G . ‘6
¢ m = 1
P I m
‘ql’cz ,G > ‘01,C2p ’6')

Let y E K n B, then y E 22(6). Suppose y = xlrczsgl,

r S
where gl 6 c'. [x1,y] E Z(G), so [x1,y] = ix1,x1 c2 g1] =

-l -l s -l l

r - s -1 s -1 5
x1 g1 (c2 ) (x1 )

r .—
X1X1 C2 g1 ’ X1 (C2 g1) X1C2 g1 '
[x1,c23g1] E Z(G). Moreover, since [c2,c28g1] E Z(G) and

S
idi.c2 g1] e Z(G), we have c Sgl e 22(c). Thus

2

r s s -l r r .
x1 c2 g1(C2 g1) ' — x1 E 22(G). But x1 E 22(6) if, and only
h-
if, p m‘r using Remark F, the fact ‘Z(G) n G" = pm and that
tpm

di E Z2(G). Also, Since [c2,x1] = c
kpm
have g1 = c2 2

2 z where z E Z(G), we

1, where 21 E Z(G). Since czsg1 E 22(0),

s+kpm s+kpm
C2 2 2

b
only if, [c2 ,x1] E Z(G), or equivalently, [c2,x1]b E Z(G). But

b
. ' d
21 E 22(G), or c E 22(G) But c E 22(C) If: an

[c2,xl]b E Z(G), if, and only if, ph-m\b. Thus ph-m‘(s+kpm), so

34

- h-
np m = s+kpm, or up m - kpm = 3. But h 2 2m implies h-m 2 m,
h- -

so np m - kpm = pm(nph 2m_ k) = s, or pm\s.

~ h-m m
Thus we have 22(G) fl A.s <x1p ,czp ,G'> = D. But
m
D s <x1,c2p ,G'>, so we have ¢D = 1‘D and hence $[ZDB = llZflB.

In [2], Demana has proven the following theorem:

Theorem 1.2: Let G = AB where A A_G. Let a E A(A),
B E A(B). A necessary and sufficient condition that there exists
y E A(G) such that y‘A = a and y‘B = B is:

1)‘ a‘mB=B‘AnB3

2). n a = anbe on A where b E B.

b
. . * . 'k @7?
Defining ¢ in G by ¢ [A = $‘X and di’ = di’ we
*
have ¢ E A(G) since:

‘1:

*
1* <0 \an = <1>th = 1mg '“' e ‘3’

>1

2). nb¢ = ¢nb on where b E B. This is because,

"13¢

for b E B, b E 22(6), a (b-lab)¢

(aia.b3>¢ = a¢[a,b],

and a¢nb = b-1a¢b = a¢[a¢,b]. Let a = xlvgugl’ where g1 E G'°

- _ _ u u-l
The a¢ - v( ph+s m u = X v uph+s m‘UE ph+s m _L7__l
n — x1 x1 g) gl 1 x1 g g,x1 ] gl
h+S-m h+S-m M h+s-m h+S-m 533—11
= x up [ x p ] 2 v u = x up 1 x p 2 a
1 g’ ]_ X1 3 gl 1 g: 1 J .
uph+s-m ph-Fs-m 3931;12-
Thus [a¢,b] = [X1 [g,x1 ] a,b]
uph+S-m Ph+S-migs%;ll uph-H".m ph-PS-m E£%;ll
= [X1 [g9x1 ] ’bJEXI [g’xl ] ,b,a][a,b]
h+s-m
= [a,b] since b E Z2(G),[g,xlp ] E Z(G) and h+s-m 2 m.

Since ‘¢\ = pm, we have ‘¢*‘ = pmo

35

Now consider LAC(G)I(G)\. \I(G)\ = \GL/\z(G)\,
\Ac(G)\ = \Hom(G/G',z*)\\Hom(c/c',A)\, and \AC(G) n I(G)\ s

‘Hom(G/G',Z*)‘. Thus

11(G)l\AC (9)1 [Hom(G/G' ,A)LLHom(G/G' ,2 *)L_
‘Ac (G)I(G)‘ \A (c) n 1(G)]

 

LHom(G/G' ,2 *)L

igLLHom(G/G ,Aji .
‘2 HA!

2
type (p8 ,p b) where a 2 h, b 2 m, we have ‘Hom(G/G',A)\ 2 \A‘ .

Thus LAC (c)1(c)L 2 lEllél__ lELLéL= .L_Ll_l

\Z HA\ 12 l Pm

Since \<xl> n Z(G)‘ = p8 and .<x1>11 G' = <1>, we have

Since .A/G' is a Subgroup of G/G' of

|Al 2 pS. Thus LAC(G)I(G)‘ 2 \G\/pm-S.
Now consider ‘AC(G)I(G) <¢*>‘. We claim (¢*)k E AC(G)I(G)
m-s

*
for k = 1,2,...,p - 1. For suppose (¢ )k E AC(G)I(G). Then

k
(¢*)k = any where a E AC(G) and y e c. g(¢*) = 3(e) =

kph+S -m an n _1
x1 g = g y = (g ga) y = y g y ga where gaE Z(G). Then
- h+s-
we have x ph+S m = [y g-ljg or x kp mg.1 = [y g-l]
1 , or, 1 a , .
x1 ga x1 (ga — Since ‘ 1 p
m -1 pm -' l I
and exp Z(G) = p . Thus Ly,g ] = 1, or [y,g ] E Z(G) n G .
_1 k h+S-m h'].
Thus [y,g ]ga e Z(G) and we have x1 9 e Z(G). xlp é Z(G)
h - - -
so we have p ‘kph+S m, or pm S‘k. But for k = 1,2,...,pm S - l,
pm-s does not divide k.

Thus \1<G>Ae<0> <¢*>\ 2 pm‘S\I<c>Ac<c>\ = pm'slcl/pm‘s =

LG\ and we have ‘6‘ divides LA(G)|.

CHAPTER III

Some Necessary Conditions for AC(G) S 1(G)

We shall now assume that G is a p-group, p # 2, having
no abelian direct factors. For such a group G, there are two
p-subgroups of A(G) which are of interest, namely AC(G) and
1(G). They are of interest because we know their sizes and, in
a certain sense, how one obtains the automorphisms in them. A
question arises as to what, if any, are necessary and/or
sufficient conditions that AC(G) s 1(G).

As was mentioned in the introduction, Sanders has shown
that all the central automorphisms of a nilpotence class two group
G are inner automorphisms if, and only if, G' = Z(G) is cyclic.
We shall give a different proof than Sanders gave of this fact
at the end of this chapter. Also, Theorem 11.2 of Chapter 11
gives us one sufficiency condition for AC(G) s 1(G), that being
G is a p-group, p # 2 with Z(G) s G' and G' is cyclic.

The assumption that AC(G) s 1(G) should somehow place a
restriction on what the lattice of subgroups for G can look
like. We shall prove that if AC(G) s 1(G), then Z(G) s G'.
Furthermore, if Z(G) is not cyclic, we shall Show that
22(6) S $(G). It is our conjecture that if Ac(G) s 1(G) for a
finite p-group G, p f 2, then not only must Z(G) 3 6', but Z(G)

must be cyclic.

36

37

Lemma 111.1: Let G be a p-group with no abelian direct

 

factors. Let K A'G with G' s K and let G E G/K. If
B E Hom(G/K,Z(G)), then for a defined by ga = g ggB, a E AC(G).

Proof: Since G/K E c/G'/K/G', there exists y: G/G' « G/K

such that ng = gg.
G
p Q
\
I
(”G 21—29 c/x —__B__.> 2(a).
Then ga = g g’:,8 = g ngB = g gD(VB). ya 6 Hom(G/G',Z(G)), thus

by the proof of Theorem 1 of [I], a E AC(G).

Remark: A necessary condition for a p-group G, p # 2,
with no abelian direct factor to have AC(G) s 1(G) is that
Z(G) s ¢(G).

2593f: Suppose AC(G) s 1(G), but Z(G) é $(G). Let
z E Z(G) be such that z E @(G), and let {2,x1,...,xr} be a
minimal generating set for G containing z. Let a E Z(G) H @(G)

be such that \a‘ = p.
n -

Suppose G B G/®(G) = <g> 8><xl> 8L..®><xr>, where z = z,
x,‘n = ; for i = l,...,r. Define ° g———-~ <a> by (E)8 = a,

1 i ' ¢(G)
G .
x. = 1 for i = l,...,r. a E Hom(mruGD. Let or be defined

by ga = g gnB. Then by Lemma 111.1, a E AC(G) and we have

a NB

2 = z z = za ¢ 2 since a # 1. Thus a E 1(G) since a\Z(G) ¢ 1LZ(C).

But this is a contradiction to AC(G) s 1(G). Thus we must have

Z(G) s ¢(G).

38

Theorem 111.1: Let G be a p-group with no abelian direct
factors and such that AC(G) s 1(G). Then Z(G) s G'.

Egggfz By the above remark, we know that Z(G) s @(G).
Suppose that Z(G) £ 6' and let y E Z(G) be such that y E G'.
We consider two cases.

Case 1). exp G/G' 2 exp Z(G). Let {x1,...,xr} be a
minimal generating set for G such that in G S G/G' =

U, Q =
<a1><3...br<ar>, xi a1 and \a1\ 2 exp Z(G).

Define B: G/G' a <y> by 318 = y, ai8 = 1 for i = 2,...,r.

Since ‘y‘ s exp Z s \al‘, we have B E Hom(G/G',Z(G)). Defining

a by ga = g gQB, we have a E AC(G) s 1(G). Thus there exists

h E G such that a = nh. Since xla = x1 xlgB = le’ we have

h-lxlh — le’ or Lxl,h] = y E G'. But this is a contradiction
since y E G'. Thus Case 1) cannot happen and we must have
exp G/G' $ exp Z(G).

Case 2). Suppose exp G/G' g exp Z(G). Let {x1,...,xr}
be a minimal generating set for G with the property that in
G g G/G' = <al><@...%a<ar>, xig = ai. Let 2 E Z(G) be such that
\z‘ = exp Z(G) = pt. Since exp Z(G) 2.exp G/G', we have

‘ai‘ g pt for i = l,...,r.

r n.
y E G' implies yg # I, thus yg = U ai l where for some
n, _ n _ i=1
1, a1 1 # 1, say 31 1 # 1. Let ‘a1\ = pS where O < s < t.
- I B _ Pt-8 6 -
Define B: G/G q <z> by a1 - z , ai = l for l # l.
t-S S t

s s
(a1p )8 = (e115)p (z p )p = zp = 1, thus a e Hom(G/G',Z(G)).
Defining a by g0 = g ggB, we have a E AC(C).
t-S
r n, n p n
y = y y' - y( H a. 1)B = y z 1 . Since 81 1 # l, ps‘f n1.

39

Hence z 1 # 1, so ya # y. But if a E 1(G), then Since
y E Z(G), ya = y. Since AC(G) s 1(G), we again have a contradiction
and case 2) cannot happen.

Thus we must have Z(G) s G'.

Theorem 111.2: Let G be a p-group, p # 2, with no abelian

 

direct factors satisfying the following conditions:
1). AC(G) s 1(G),
2). Z(G) is not cyclic.
Then 22(G) s @(G).
Proof: Suppose 22 = Z2(G) i @(G). Let y E 22 be such

that y E ¢(G), and let {y,x2,...,xr} be a minimal generating set

for G containing y. Let Z(G) = <w1> $><w

2> ®...® <Wq> where
mi
LwiL — p , m1 2 m2 2...2 mq'
no - - r - - n--
5111313088 G ——-—<y>a<x>®...®<x> where y —y,
$(G) 2 m._1r
‘ J
x.n = x. for i = 2,...,r. Set 2. = w.p . Then we have for
l 1 m,-1 J J
1 L L PJ L df G (())
= a a , z = w. = . Now e inc .1"‘—’-* 3 Z G
J q 1 ‘1 " SJ em) 1
_ej _ Bj
by y = zj, xi = l for i = 2,...,r. If we define aj by
aj “B.
8 = g g J, then by Lemma 111.1, we have aj E Ac(G) S 1(G).

Hence there exists gj E Z2, j = l,...,q with the property that

oz. oz,
Let M = <x2,...,xr,@>. Since (yp) J = (y J)P = (”J-)1)

P P P
= z, = wehmm ag =1 .
y J y 3 JLM LM

a. = 1 and a. inner im 1 that M s C < ,> . Since
JLM LM J p y G< gJ )
M is a maximal subgroup of G, we must either have CG(<gj>) = M
or C (<g.>) = G. If c (< ,>) = G then , e Z(G), thus
G J o6 g1 ’ g1
j

a. = fl . = 1 . But
J g1 \G y

must have CG(<gj>) = M.

= Y Zj # y, thus gj 4 Z(G). Hence we

 

40

\G : M‘ = p = the number of conjugates of gj.

y = y J = y zj, we have

-1

-l
gj y g. = Y 2.: or y 8.Y = g.Z.

J J J J

2 oz)“1 - ( y'1> - z 2
y gj y y gj y gj j 3

p-l p-l -1 p-l
. = .2.
y gJ(y ) 8] J

Since szL = p, we have the p conjugates of g,
J

the elements , ,z_ g,z, ... ,z.
gJ’ gJ J’ J J ’ ’gJ J

are precisely

Since

By Theorem 111.1, we have, since AC(G) S 1(G), that

Z(G) s G'. Thus Hom(G/G',Z(G)) g AC(G). Since 5153- is r-
G _ , _ ui
generated, 67-— <al>'®L<az><8...6><ar>. Let LaiL — p and let
G 5 G/G'.
Q i . I v
. f f , th M .
gJ ¢ , or 1 gJ E G , en gj E G H 22 < Z(ZZ)
mg. a,
Thus, since y E 22, y J = yzj = y J # y. Thus let
s s s t t
E _ 1 2 r _ 1 2
g1 a1 a2 ...ar and g2 — a1 a2
Since szL = p, we have LBjL = p and thus LajL = p.
P P P P
a.) = . = . = 1 so we have 2 G s G'. Th 3
( J (ngJ) ngJ LG gj E ( ) u
ps pt pt
Q 9 _ p Q _ - ps1 p82 a r l 2
(gj ) (gj ) 1, so 81 a2 ... r 1
u,-l ,-1
Hence we have Si 5 0 mod p 1 and t. E 0 mod p 1
1
Because y E 22(G), fly E AC(G). Since ny-l - gjzj,

y-1 E Z(G), 80 y E Z(G) and ny # ILG. Moreover, since

AC(G) 5 Hom(G/G',Z(G)), there exists y E Hom(G/G',Z(G)) such

that (gjg)Y = z, for j = l,...,q, and in particular, for

J

j = 1,2. Suppose,wo10g , y is given by

i

h.
l

W

2

Then

pt

41

r r
b s z ,s, m -l
E Y 1_1 i 1 i=1 i i p 1
(g1 ) “wl w2 = z1"wl ’
r r
iilbiti 'Elhiti mz-l
(gZC)Y = w1 w21 = 22 = wzp and we have:
r m -l m r m2
2 b s, a p 1 mod p 1, 2 h,s, E 0 mod p ,
. 1 l i 1
i=1 i=1
r m1 r mZ-l m2
2 b.t. E 0 mod p , z h.t. a p mod p
. 1 i , 1 1
i=1 i=1
h b
If Lw L = Lw L define y* by a Y* = w iw i Since
1 2 ’ i l 2 ° '
*
y e Hom(G/G',Z(G)), y e Hom(G/G',Z(G)).
r r r
2 hisi Z bisi Z bisi
C» W = 1:1 i=1 = i=1 ‘ =
(gl ) w1 w2 w2 # l Sinus LwlL szL

’

* *
implies m1 = m2. Defining a by ga = g gQY* we have

*
o e AC(G) s 1(G). Thus there exists x e G such that o* = n .

f b'S' E b s X
“X _ 0* gy* _ i=1 1 1 i=1 i i ,
g1 - g1 = g1 g1 — glw2 $4 g1. Thus glw2 is
conjugate to g1. But this is a contradiction since the conjugates
-1
Of g1 are glaglzla°-°ngzlp .
r m2-l m
Now Suppose LwlL 2 szL. Since 2 h t, 2 p mod p 2’

m
for some k we must have hktk ¥ 0 mod p 2. For one Such k,

m -m
5 1 2 hk 6
define 6 on G/G' by ak = (wlp ) , ai = 1 for 1 ¥ k.
ml-mZ m b h

Since Lw p L = p 2 = Lw L and g Y = w kw we have

1 2 k 1 2 ’
5 e Hom(G/G',Z(G)). _

m 'mz h t h t pm1 m2
r t l
' k k
(sf)6 = < 2 81 1>8 = (wlp > k k = W1 r 1

i=1
m

'k \
since hktk i 0 mod p 2. Now define 6 by g = g g . Since

 

42

*
o E Hom(G/G',Z(G)), we have 5 e AC(G) s 1(G). Thus there exists

x
x E G such that 6 = nx.

m-
l 2
n h t P
x 6* Q6 k k
= = = W h
g2 g2 g2 g2 g2 1 1 g T ”S
m -m
hktkp 1
gzw1 is conjugate to g2. But this is a contradiction
. . -1
Since the conjugates of g2 are g2,g222,...,g222p .

Thus we must have 22(G) S ¢(G).
We now prove a result of Sanders.
Theorem: Let G be a p-group, p ¢ 2, of nilpotence class
two. Then AC(G) s 1(G) if, and only if, Z(G) = G' is cyclic.
‘Egggf: Let G' = Z(G) be cyclic. Then G has no abelian
direct factors and we know LAC(G)L = LHom(G/G',Z(G))L.
exp G/G' = exp G/Z(G) S exp Z(G) Since G' = Z(G) and G = 22(0).
Thus, because Z(G) is cyclic, we have Hom(G/G',Z(G)) E G/G'.
Thus LAC(G)L = LG/G'L = LG/Z(G)L = L1(G)L and AC(G) s 1(G).
Now Suppose AC(G) S 1(G). G has no abelian direct factors,
for suppose G = P'E B where P = <x> is cyclic. Let 2 E Z(B)
be such that LzL = p. For g E G, g = xkb where b E B. Define
a on G by g0 = (xz)kb. Since G is a direct product of P
and B and z E Z(B), a is a well-defined automorphism of G.
8-1ga = (Xkb)-l(xz)kb = 2k E Z(B) S Z(G), thus a E AC(G). But
xa = xz # x, so 3 E 1(G), contradiction to AC(G) S 1(G).
Since G has no abelian direct factors and AC(G) g 1(G),
by Theorem 111.1, Z(G) S G'. But G of nilpotence class two
implies G' s Z(G). Thus 6' = Z(G). By Theorem 111.2, if Z(G)
is not cyclic, then 22(G) S $(G). But 22(G) = G, thus 22(6) 5 3(0)

is impossible. Hence Z(G) = G' is cyclic.

BIB LIOGRAPHY

10.

BIBLIOGRAPHY

J.E. Adney and Ti Yen, Automorphisms of a p-group, Illinois
J, Math. 9 (1965), 137-143.

Franklin Demana, Some theorems on extending automorphisms,
Ph.D. Thesis, Michigan State University (1966).

R. Faudree, A Note on the Automorphism Group of a p-group,
Proc. Amer. Math. Soc. 19 (1968), 1379-1382.

 

W. Gaschfitz, Nichtabelsche p-Gruppen besitzen afissere p-
Automorphisms, J, Algebra 4 (1966), 1-2.

M. Hall, The Theory 9f Groups, (New York: The Macmillan
Company, 1959).

 

A.D. Otto, Central Automorphisms of a Finite p-Group, Trans.
Amer. Math. Soc. 125 (1966), 280-287.

 

P.R. Sanders, Some results in the theory of automorphisms of
finite groups, M.Sc. Thesis, University of London, 1967.

W.R. Scott, Group Theory, (Englewood Cliffs, New Jersey:
Prentice Hall, Inc., 1964).

J. Thompson, Finite groups with fixed-point free automorphiSms
of prime order, Proc. Natl. Acad. Sci. U.S. 42 (1959), 578-581.

 

 

H.J. Zassenhaus, The Theory of Groups, (New York: Chelsea
Publishing Company, 1958).

 

43

APPENDIX

APPENDIX

k
I. For p a prime, p ¢ 2, (1+tpm)p E 1+tpm+kmod pm+k+1

where (t,p) = l, m 2 1, k 2 l.

The proof is by an easy induction argument.

II. If c E CG(G'), k 2 1, then [x,ck] = [x,c]k.
Proof: [x,c1] - [x,c]. Assume true for k = n 2 1.
Then Lx,cn+1] = [x,cnc] I [x,cn]Lx,chx,c,cn] = [x,c]an,c]

+
- [x,c]n 1 since 0 e C(G').

111. An automorphism a of a group G is called central
if x-1xa E Z(G) for all x E G. The set of all central auto-
morphisms of 6 forms a group called Ac(G)‘

If a E AC(G), then fa: x a x-1xa is a homomorphism of
G into Z(G). If f E Hom (6,2), then af: x « xf(x) defines
an endomorphism of G. 9f is an automorphism if, and only if,
f(x) f x-1 for every x E G, x i l.

Adney and Yen show in [1] the following:

Theorem: For a group G with no abelian direct factors,
the correspondence a a fa defines a l-l mapping of AC(G) onto
Hom (6,2).

Corollary: If G is a p-group, p i 2, with no abelian
direct factors, than AC(G) is also a p-group.

Hom (G,Z(G)) 3 Ram (c/c',2(c)) since if y e Hom (c,z<c)),

c' g ker y.

44

 

45

Sanders, in [7], shows the following:

Let G be a p-group with no abelian direct factors. Let

R = {y(g)Lg E G, y E Hom (G,Z)} and B = /A\ ker y.

Then Hom (6,2) 3 AC(G) if, and only if, R < B.

£529£= If Ya’YB E Hom (G,Z). then (Yav8)(g) = va(g)ys(g).
Let 0,8 E AC(G) be such that a correSponds to ya, 8 corresponds
to YB° (ae)(g) = a(B(g)) = s(gva(g)) = a(s)a(vB(g))
= gya(g)y8(g)ya(Ya(g)). 08 thus corresponds to YaYB if, and

only if, ya(y8(g)) = l, or R < B.

 

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