R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
P ETER V. H EGARTY
Minimal abelian automorphism groups of finite groups
Rendiconti del Seminario Matematico della Università di Padova, tome 94 (1995), p. 121-135
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of Finite Groups.
PETER V.
HEGARTY(*)
ABSTRACT - We determine the smallest odd-order Abelian group which occurs as
the
automorphism
group of a finite group.1. Introduction.
Finite
(non-cyclic)
groups whoseautomorphism
group is Abelianwere first studied
extensively by
G. A.Miller,
who wrote down in[8]
agroup of order 64 whose
automorphism
group is Abelian of order 128.Following
the author of[3],
I term a finite group G «miller» if Aut G is Abelian. Since Inn G is a normalsubgroup
of Aut G and Inn G ==
G/Z( G ),
a miller group isnilpotent
of class at most 2.Hence,
in any at-tempt
to characterize miller groups one can confine one’s attention to p- groups.By
a well-known result(see [2]),
theonly
Abelian miller groupsare the
cyclic
groups. In the non-Abelian case, the smallest miller 2- group is well-known to be theexample
constructed in[8].
In the oddprime
case, thequestion
of the smallest miller p-group took muchlonger
to resolve. It was tackledby Earnley [3]
andfinally
settled re-cently by Morigi [9].
She constructed a group of orderp 7
whose auto-morphism
group is Abelian of orderp 12,
where p is any oddprime,
andshowed that no smaller miller p-groups existed.
In this paper I propose to answer the natural
question running alongside
the issue of minimal millergroups-namely,
«What is the or-(*) Indirizzo
degli
AA.:Department
of Mathematics,University College,
Cork, Ireland.der of a smallest Abelian group which occurs as the
automorphism
group of a
finite, non-cyclic, p-group?». For p
=2,
the answer is G. A.Miller’s group
[8]
of order 128. This is borne outby
theclassification,
in
[5],
of all the groups whose orders divide128,
and which can occur asthe
automorphism
group of a finite group.For p odd,
it is natural toconjecture
that the smallest Abelian group with the desiredproperty
isthe one of order
p 12
inMorigi’s
paper. I shall prove that this is indeed the case.2. Notation and
terminology.
Most of the notation used is standard. All groups considered are
finite.
Cent G will denote the group of central
automorphisms
of a group G.A
purely
non-Abelian group(PN-group)
is one with no Abelian di- rect factor.d( G )
will denote the number of elements in a minimalgenerating system
for G.Gn
=1)
where n E N.Similarly,
Zp
denotes the field ofintegers
mod p. Anelementery
Abelianp-group G of rank n will be considered as a vector space of dimension n
over
Zp .
For a fixed of such a group, we shall asso-ciate to each n a E Aut G a matrix A = with entries in
Zp
such that" -
The
following piece
ofterminology
is non-standard: I shall call two groups G and Hhypomorphic
if andonly
ifThe set of all groups
hypomorphic
with G I shall term ahypomor- phism
class.3. Statement of theorem and
preliminary analysis.
It is our purpose to prove the
following
MAIN THEOREM 3.1. Let G be a
finite non-cyclic
p-group, podd,
for
which Aut G is Abelian. Then p 12 dividesAut
.Henceforth, then,
p denotes an oddprime,
G a finite p-group.If Aut G is Abelian then Aut G = Cent G
(see [3], 2.2),
and G is aPN-group ( [3], 2.3). Consequently,
Aut G is a p-group( [3], 2.4). Thus,
ifG is to contradict the
theorem,
Aut G must have orderp n
for some n %~ 11.
By Morigi’s result,
On the other handG ~ I
divides~ Aut G ~ I
when G is miller. Thusp’ I G I
Our first result allows us to eliminate
G ~ _ ~ 11,
and may be of in-dependent
interest. One may observe that the result isjust
aslight
im-provement
upon aspecial
case of that of Faudree[4],
that the order of every finite p-group of class 2 divides that of itsautomorphism
group.It is not
surprising, therefore,
that theproof
followsprecisely
the ap-proach
of Faudree. The notation for theproof
is takenentirely
from[4],
and henceforth I will assume the
familiarity
of the reader with that paper.LEMMA 3.2. Let G be a miller p-group, p odd. Then
I G I properly
divides
Aut
GI (1).
PROOF. Let G be a
counterexample.
Aut G is a p-group. Follow-ing [4],
Aut G has asubgroup
T whose order isgiven by
it follows that
Then Faudree constructs a
subgroup
U of Aut G and shows that( UT : T ) ~
in all cases.Thus, ( Aut [ G I
unless n % 2. But if n =2,
we stillget G I
unlessd(G)
=2,
whichimplies
that G’is
cyclic
i.e.: that n = 1.Hence we can assume that G’ is
cyclic,
andT ~ I
=I G / G’ I
in thiscase. We consider the same
automorphisms
Ji , ~’2, 7: I and i2 as did Fau-dree,
anddistinguish
3possible relationships
between thequantities ta
and
tb , namely
(1) The author has been able to prove this result also
for p
= 2. Theproof
is omitted, as it would be irrelevant to the purpose of this paper.Suppose I)
holds.Replace
b toget tb
= ml. Thus i 1 has or-der mod Cent
G,
so Aut G Abelian +M2 .
But then a I hasorder
kb /ml
modT,
soAut
withequality possible
if andonly
ifkb
=M2 .
A similaranalysis
shows that we must have or,having
order ml mod
T, and ka
=kb , ta
= 1.Consequently, a2, T)
is a p- group of order 7niI G a
contradiction!Suppose II)
holds.Replace
aby
b-rlZ a
toget ta
= ml . Then í 2 must lie in Cent Gso ka ~
Then a2 ,T) I
will bestrictly
divisibleby 1 G I unless kb
= ml , in which case ~1 E T andI ( a 2, T~ ~ - ~ G ~ .
. Sinceta
=ml , it
is clear that Cent Gproperly
contains(~2 , T)
unlessd(G) =
= 2. In this case, a non-central
autmorphism fixing ( G ’ , b ~
elementwise iseasily constructed, using
Lemma 3.7 below.Finally,
supposeIII) applies.
z 1 E Cent G soThen,
as withI),
weeasily
deduce that(Q1,
a 2,T) I
isstrictly
divisibleby
.This
completes
theproof
of the lemma.Hence,
we can assume that if G contradicts thetheorem,
~
1° . My approach
will be to eliminate allpossible hypomorphism
classes of groups
one-by-one.
For most ofthese, straightforward appli-
cations of well-known results
suffice,
and nocomplete proofs
will begiven.
Some individual classes causegreater difficulty
and will be dealt with in more detail. I willrequire
along
sequence of results from the literature.First,
I note an immediatecorollary
ofequation (1)
above.
LEMMA 3.3. Let G be a
counterexampte
to the main theorem. Thend(G’ ) ~ 3.
This follows
straight
fromequation (1).
Lemmas 3.4-3.8 are all well- known results:LEMMA 3.4
[10].
Let G be aPN-group, for
anyprime
p. Thenwhere
pk
is theexponent of G/G’ and,
in acyclic decomposition of G/G’,
there occur rifactors of
orderpi.
Recall that in a finite Abelian p-group
A,
theheight
of an element xis
given by
We now have
LEMMA 3.5
[1].
Let G be a class 2 p-group withDefine
where
pb
is theexponent of
G’. Alsodefine
where
pd
=min(exp Z,
expG/G’).
Then CentG is Abelianif
andonly if R(G)
=K(G)
and eitherwhere XI is chosen
from
among. In
particular, RIG’
iscyclic.
LEMMA 3.6. Let G be a
finite
p-group. Then Aut G is not Abelianif
any
of
thefollowing
holds:LEMMA 3.7
[6].
Let N be a normalsubgroup of
afinite
group G such thatGIN
iscyclic of
order n. WriteGIN
=(Ng).
Let x eZ(N)
such that
gn
=(gx)n .
Then the map a : G - Ggiven by
can be extended to an
automorphism of
G.LEMMA 3.8
[7]. Suppose
thefinite
group Gsplits
over an Abeliannormal
subgroup
A. Then G has anautomorphism of
order 2 which in-verts A elementwise.
4. Proof of main theorem.
Let G be a
counterexample.
Wealready
knowNow most of the
hypomorphism
classes of groups of these orders can be eliminatedby using
Lemmas 3.2-3.8 above.Obviously,
the number of classes involved is far toolarge
for detailedproofs
to begiven
here. De-tails may be obtained from the author if
required.
The
analysis
revealed a small number ofclasses,
or collections of similarclasses,
which were not amenable to suchstraightforward
treat-ment. I now
give
a list of these:I shall eliminate the classes
individually
in a series of four lemmas.Each of the groups listed will be shown to have a non-central automor-
phism. My principal
tool will be thefollowing powerful criterion,
due toEarnley [3], 3.2,
for groups withhomocyclic
centralquotient -
a prop-erty possessed by
all the groups above-to possess a non-centralautomorphism.
LEMMA 4.1. Consider the extension 1 ~ Z -~ G ~
G/Z -~
1 whereG is a p-group and
G/Z is
a directproduct of n ( n ~ 2) copies of Cpt for
some
fixed
t. Let T :G/Z -~ ZIZPT
be thehomomorphism given by (Zx) T
=Zpt xpt . Also let [, ] : G/Z
xG/Z~Z
begiven by (Zx, Zy) [, ]
=- [x, y].
Now let a be in Aut(G/Z)
and~3
be in Aut Z.Then G has an
automorphism inducing
a onG/Z
andfl
on Zif
andonLy if
thefollowing
twodiagrams
commute:We now
begin
the process of elimination.LEMMA 4.2. Let G be a member
of
Class I. Then G has a non-cen-tral
autorrtorphism.
PROOF. Let G be a
counterexample.
Writewhere
apn, bp,
cP and dP are all in G’ . Cent G is Abelian so,by
Lemma3.5,
a can be chosen to haveorder p"’
and so thatZ( G ) _ ~ a p , G’.
Clearly, p 2 .
First suppose that a may also be chosen so thatCG ( a ) BZ
meetsGp .
Then[a, b]
= bP = 1 WLOG. We claim that c and dcan be chosen to commute. Choose both
arbitrarily
tobegin
with.[c, d] ~
1by assumption.
ButCG (b) d)
c Z= ~
as otherwiseCG (b)
would be a maximalsubgroup
of G and a non-centralautomorphism
ofG could be constructed
by
Lemma 3.7. Thus G ’ -([b, d], [c, d])
and ourclaim follows
easily. Indeed,
we can also assume that cP = 1 WLOG. Butif we could also choose d of order p, then G would
split
over the normalAbelian
subgroup (Z,
a,b ~
and have a(non-central) automorphism
oforder
2, by
Lemma 3.8. So we can take it thatx (dP ).
Setzi and dP = z2 for convenience. There exist
i, j, k, 1,
m, n inZp
such that
Consider the matrices
with entries in
ZP .
Let a andfl
be theautomorphisms
ofG/Z
and Z as-sociated with M and
N,
and withrespect
to the bases{Za, Zb, Zc, Zd}
and
f Zl,
ofG/Z
and G’respectively.
Then one mayverify that, by
Lemma
4.1,
there exists anautomorphism
of Ginducing
a andf3 provid-
ed that
But
(i, j ) ~ (0, 0)
as otherwiseCG (c)
would be maximal in G and anon-central
automorphism
of G could be constructedusing
Lemma 3.7.Similarly, CG ( b )
is not maximal inG,
so det . So choose 0and a
(unique)
solution( y, 3)
toequation (12),
and hence a non-centralautomorphism
ofG,
isguaranteed
to exist.We may therefore assume that a cannot be chosen so that
CG (a)BZ
meets
Gp .
Thus we may choose a and b so that[ a, b]
= 1 and G’ == x
(b P ). Consequently,
we can choose c and d both to have order p.It follows that and must both be contained in N =
=
(Z, b,
c,d).
From this weeasily
deduce that either[c, d]
= 1 orZ(N) properly
containsZ ( G ).
In the former case, Gsplits
over the Abelian normalsubgroup (Z,
a,b)
and has anautomorphism
of order 2. In the latter case, a non-centralautomorphism
iseasily
constructedusing
3.7.
This
completes
theproof
of Lemma 4.2.We continue
immediately
toLEMMA 4.3. Let G be a member
of
Class II. Then G has a non-cen-tral
automorphism.
PROOF. Let G be a
counterexample.
Writewhere
a p n + 1,
bP and cP are all in G ’ . Cent G is Abelian so,by 3.5,
wemust have
a ~ I _ ~ n + 1
andZ ( G )
= G ’ If b and c could be chosen tocommute,
then A =(G’, b, c)
would be Abelian withG/A =
and so G would have a non-central
automorphism by
3.7. It follows that[a, b]
= 1 WLOG. If b could be chosen to have order p, then a non-cen-tral
automorphism fixing
B =(Z,
a,b~
elementwise could be construct- edusing
3.7. If c could be chosen of order p, then G wouldsplit
over Band have an
automorphism
of order2, by
3.8. Thus we can take it that G’ =(bP)
x(cup).
Set zl = = cP and Z3 = aP . There existi, j, k, 1
inZp
such thatLet a be the map on
G/Z
associated with the matrixrelative to the basis
{Za, Zb, Zc}.
Let be the map definedby
One may
verify
that the conditions of Lemma 4.1 are satisfied pro-vided the
following equations
hold:.... I r
Since det
I
, one canreadily
check that a solution(y, d, E, ~) ~
~
(1, 0, 0, 1)
to theseequations
exists in all cases. Furthermore we canchoose our solution to
satisfy
y # 0and ~ ~ 0,
thusguaranteeing
that aand
P
define(non-trivial) automorphisms
ofG/Z
and Zrespectively,
and hence the existence of a non-central
automorphism
of G.This
completes
theproof
of Lemma 4.3.Next we have
LEMMA 4.4. Let G be a member
of Class
III. Then G has a non-cen-tral
automor~phism.
PROOF. Let G be a
counterexample.
Writewhere bP and cP are all in G ’ . Cent G is Abelian so we
must, by 3.5,
have
a I
WLOG. LetZ = z1, z2 , z3
with and G’ _-
(Zl,
z2 ,WLOG,
there existk, 1
inZp
such thatI
distinguish
two cases,according
to whether[a, G] n(Z3)
is trivialor not.
So first suppose that
[a, G] n (z3 ~ _ ~ 1 ~.
Then there is no loss ofgenerality
inassuming
that[a, b]
= 2:1,[ a, c]
= z2 and[b, c]
=zl’-’.
.Let a be the
automorphism
ofG/Z
associated with the matrixrelative to the basis
IZa, Zb, Zc}. Let fl
be theautomorphism
of Z de-fined
by
One verifies
easily
that the conditionsimposed by
Lemma 4.1 re-duce to the matrix
equation
But
fl
is anautomorphism
of Z for any choice of E and0.
Hence a sol- ution(y, d,
E,~) ~ (0, 0, 0, 0)
toequation (21)
isguaranteed,
and G hasa non-central
automorphism.
Now
secondly
suppose that[ac, G]
is non-trivial. In this case, there is no loss ofgenerality
inassuming
that[a, b]
=[ac, c]
= z2 and[ b, c]
= zl . Let a be theautomorphism
ofG/Z
associated with the matrixrelative to the basis
{Za, Zb, Zc}. Let fl
be theautomorphism
of Z de-fined
by
One
easily
verifies that the conditionsimposed by
Lemma 4.1 re-duce to the
following
3equations
in the 3 unknowns a,0,
yNotice that the first two
imply
the third whenfl
~ 0. But there obvi-ously
exists a solution(a, ~8, y)
to the first twoequation
for whicha ~
0, p
~ 0. Hence G has a non-centralautomorphism.
This
completes
theproof
of Lemma 4.4.I now turn to the final and most
complicated
case.LEMMA 4.5. Let G be a member
of
Class IV. Then G has a non-cen-tral
automorphism.
PROOF.
Clearly, (G : Gp Z) ~ p 2 ,
and(G : CG (x)) ~ ~ro 2
for all x E G.The case in which for all x E G is that which causes the most
dificulty,
and we assume this to be the case inwhat
follows,
until otherwise indicated. Writewhere
Gp = ~G’ , b,
c,d).
Cent G is Abelian so,by 3.5,
a can be chosenso that
I a [ _ ~ n + 1
and Z =(aP, G’.
We must haveZ/Zp
= xx but will find it necessary not to assume that e p E G’ . The fol-
lowing
two assertions areeasily
verified:(i)
G has no Abeliansubgroup
ofindex p2.
(ii)
Let x1 EGBZ.
Let X2 ECG (x1 )B~Z,
Let X3 E(x2 )BCG (Xl
Then
CG (CG (Xl), x3>-therwise stated, (CG (xl), CG (x2), CG (x3))
= G.We divide the
analysis
into 2parts, according
to whetherZ(Gp)BZ
isempty
or not(the
non-centralautomorphism
wefinally
construct willbe
slightly
different in the twocases).
So first suppose that
Z(Gp )
c Z. It is easy to see that for some g notin We claim that a has this
property
WLOG.
Suppose
not. Thenif g
has theproperty
we must havegP
E G ’ .Let
[g, b] _ [g, c]
= 1. Then[ b, c] ~
1by
assertion(i)
so[ b, d]
= 1 WLOG.By
assertion(ii),
a can be chosen so that[c, a]
= 1. Let x EE
CG (d)BZ(a,
c,g). Clearly
x exists. But x cannot be chosen to lie in(c, g) by
assertion(ii). Therefore,
we canreplace a by x
and we have[a, c] =
=
[a, d]
=1,
thusproving
our claim.By
similarreasoning
it is easy to de- ducethat,
for anappropriate
choice of a,b,
c, d and e, thefollowing
commutation relations hold:
Let
G’=(ZI)X(Z2)
where Now there exist m, n, q, r,s, t
inZp
such thatLet a:
G/Z ~ G/Z
be themapping
associated with the matrixrelative to the basis
Zb, Zc, Zd, Ze}.
Let be the map-ping
definedby
a and
~i
defineautomorphisms
of theirrespective
groupsprovided
y # 0. The conditions
imposed by
Lemma 4.1reduce,
as may be verifiedby
thereader,
to thefollowing
set ofequations:
for the seven variables y, ... , v. If det set y = 1.
Otherwise,
set y =
2,
say. In either case, a solution(v, 3 ) # ( o, 0)
to(28)
is guaran- teed. NowCG ( d )
is not maximal inG,
so det Thus whenwe substitute 6 into
(26),
the existence of a solution(I, E)
is guaran- teed.Similarly, CG ( b )
is not maximal inG,
so det , andI I
when we substitute v into
(29)
the existence of a solution(Ø, p)
isguaranteed.
Hence
(26)-(28)
have a solutionaccording
to which a is a non-trivialautomorphism
ofG/Z,
and we conclude that G has a non-central auto-morphism
in this case.Secondly,
suppose thatZ(Gp ) ~ Z. Gp
is notAbelian, by
assertion(i),
soZ ( Gp ) _ ~ G ’ , b)
WLOG. A series of routine calculations lead us to conclude that a, c, d and e may be chosen so that thefollowing
commu-tation relations hold:
Let zl and z2 be defined as before. Then there exist
i, j , k, l,
m, n, q, r,s, t
inZp
such thatLet a
and B represent exactly
the samemappings
ofG/Z
and Z re-spectively
as before. Onceagain
a and~8
defineautomorphisms
of theirrespective
groupsprovided
y # 0. Thistime,
the conditionsimposed by
Lemma 4.1 reduce to the
following, slightly different,
set of equa- tions :Now one reasons in
precisely
the same manner asbefore,
to con- clude that G possesses a non-centralautomorphism.
We have now dealt
entirely
with the case in which( G : Gp ) _
=
( G : CG (x))
=p~
forall r g Z(G). Next,
we continue to assume that( G : Gp Z)
=p 2 ,
but also that there exists x such that( G : CG ( x ))
_ ~ . Ifx could be chosen to lie in
Gp ,
then we couldeasily
construct a non-cen-tral
automorphism using
3.7.Keeping
the same notation forG/G’
as inequation (24),
I claim that for anappropriate
choice of a,b,
c, d and e, e p EG ’ , ( G : C~ ( a ))
= p and thefollowing
commutation relations hold:In what follows I am
assuming
that eP E G ’ . I prove the claim in a num-ber of
stages.
Step
1.Z ( Gp ) c Z ( G ). Suppose
thecontrary. Gp
isclearly
non-Abelian, by 3.7,
so letZ ( G~ ) _ ~ G ’ , b). x (as
definedabove)
lies outsideGp .
Then we can choose c and d so thatCG (x) _ (Z,
c,d,
x,y)
for someGp Z.
ThenCG (g )
is maximal in G for some ge (c, d ~~G ’ ,
and G has anon-central
automorphism by
3.7-contradiction!Step
2.Suppose Gp
i.e.: that x can be chosen so that[ x, b ]
=- [x, c] = [x, d]
= 1. If xp EG’ then G/~Gp , x) = Cp~ ,
so G has a non-cen-tral
automorphism, by 3.7,
unless n =1,
in which case a and e are inter-changeable.
This means we can choose x for a. Now[b, c]
= 1WLOG,
whence (a, b, c)
is Abelian. There is some g inCG ( e ) ~Z~ a, b, c),
but gcannot lie
in (b, c) by
3.7.Thus,
wereplace
aby g
to obtain[ a, b ]
==
[a, c]
=[ b, c]
=[ a, e]
= 1. But now it is clear that d can also be chosen to commutewith e,
and the claim is established in this case.Step
3. We must haveGp
for some choice of x.Suppose
not.For a
given x
we can still choose b and c so that[x, b] _ [x, c]
= 1. If[b, c] = 1, proceed
as inStep
~. Thus[b, d] = 1
WLOG. Let y EE
x).
Routine calculations show that y and c can be chosen tocommute,
whence A =(Z, x,
c,y )
is anormal, Abelian, complement-
ed
subgroup
of G and G has anautomorphism
of order 2by
3.8-contra- diction ! Our claimregarding equation (36)
is now established in full.Now write with and
z2 = ep .
There existi, j , k, 1
inZp
such thatLet a be the
automorphism
ofG/Z
associated with the matrixrelative to the basis
{Za, Zb, Zc, Zd, Ze }.
Let¡3
be theidentity
map onZ. The conditions
imposed by
Lemma 4.1 arereadily
checked to reduceto the matrix
equation
for the four unknowns y,
d,
E,0.
The abovesystem
is underdeter-mined,
thusguaranteeing
the existence of a non-trivial solution(y, d,
E,~) ~ (0, 0, 0, 0)
andconsequently
of a non-central automor-phism
of G.We have now shown that Lemma 4.5 is true when
(G : GpZ)
=p 2 .
One
proceeds
inexactly
the same way as above when one assumes that(G : Gp Z) = p
or thatG = Gp Z.
Infact,
theargument simplifies
inplaces but,
in anyevent,
I do not think it necessary to go into any fur- ther detail.Hence,
theproof
of Lemma4.5,
andconsequently
that of the maintheorem,
iscomplete.
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