R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
J OHN L ENNOX
F EDERICO M ENEGAZZO H OWARD S MITH
J AMES W IEGOLD
Groups with finite automorphism classes of subgroups
Rendiconti del Seminario Matematico della Università di Padova, tome 79 (1988), p. 87-96
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Groups with Finite Automorphism Classes
of Subgroups.
JOHN LENNOX - FEDERICO MENEGAZZO HOWARD SMITH - JAMES
WIEGOLD (*)
1. Introduction.
The
automorphism
class of asubgroup
H of a group G is the orbita E Aut
G}
of H under the action of Aut G. One of the main aims of[7]
was toclassify
those groups for which theautomorphism
classes are
boundedly finite,
the answerbeing
as follows.THEOREM A. The
following properties of
a group G areequivalent.
(i)
Theautomorphism
classesof subgroups of
G areboundedly finite.
(ii)
Theautomorphism
classesof
abeliansubgroups of
G areboundedly finite.
(iii)
Either(a)
Aut G isfinite,
(*) Indirizzo
degli
AA.: J. LE-NNOX:Department
of Pure Mathematics,University College,
Cardiff CF11XL, Wales; F. MENEGAZZO: Istituto diAlgebra
e Geometria, Università di Padova, Via Belzoni 7, 1-35100 Padova
(Italia);
H. SMITH:
Department
of Mathematics,Michigan
StateUniversity,
EastLansing, Michigan,
USA; J. WIEGOLD:Department
of Pure Mathematics,University College,
Cardiff CFI IXL, Wales.The fourth author thanks the Istituto di
Algebra
e Geometria dell’Uni- versità di Padova for its warm and generoushospitality during
theperiod
that work on this paper was carried out, and
Consiglio
Nazionale delle Ri- cerche. for its financial support.Lavoro
eseguito
con il contributo del M.P.I.or
(b)
there is a directdecomposition
G =G1
XGa,
whereGz
is a
locally cyclic
torsion group,Ga
is afinite
centralextension
of
a directproduct of f initely
many groupsof type for different primes
p, andGl
andGa
donot contain elements
of
the sameprime
order.The
problem
ofdetermining
the groups in which the automor-phism
classes aremerely
finite was left open in[7 ].
We shall prove here that the two classes are the same :THEOREM B. The
following properties of
a group G areequivalent.
(i )
Theautomorphism
classesof subgroups of
G areboundedly finite.
(ii)
Theautomorphism
classesof
abeliansubgroups of
G arefinite.
(iii)
Either(a)
G is notperiodic
and Aut G isfinite,
or
(b)
G isperiodic
andof
the structure described in(iii) (b) of
Theorem A.To prove
this,
we needonly
show that a group with finite auto-morphism
classes of abeliansubgroups
is of thetype
mentioned in(iii)
of Theorem
B,
and thenapply
Theorem A.The
analogous
results forconjugacy
ofsubgroups
and of abeliansubgroups
are due to B. H. Neumann[6]
and Eremin[2];
groups with finite classes of abeliansubgroups
arejust
thecentre-by-finite
groups. We shall make much use of this result in our
proof,
whichis, surprisingly,
agood
deal harder than that of the Neumann-Eremin theorem.By
an abus delangage
we call the set (X E EndG}
of endo-morphic images
of asubgroup
H of a group G theendomorphism
class of
.g;
note thatendomorphism
classes need not bedisjoint.
Thefollowing
theorem is ananalogue
of Theorems A and B for endo-morphism
classes :THEOREM C. The
following properties o f
a group G areequivalent.
(i)
Theendomorphism
classesof subgroups of
G areboundedly finite.
(ii)
Theendomorphism
classesof
abeliansubgroups of
G arefinite.
(iii)
G is afinite
central extensionof
a directproduct of finitely
many groups
of type for different primes
p.89
Finally,
we have thefollowing
curious result:THEOREM D. Let G be a group in which the
endomorphism
classesof
elements areboundedly f inite.
Then G isfinite.
This
generalizes
a theorem of Baer[1] stating
that a group is finite if itsendomorphism
set is finite. See[7]
for groups in which the auto-morphism
classes of elements areboundedly
finite. The boundedness condition in Theorem D isessential,
as aglance
at the directproduct
of
infinitely
many finite groups ofcoprime
orders shows.All our methods are
elementary
and notation standard. We usewithout further comment the fact that groups of the
type
under dis-cussion are
centre-by-finite
and thus have finite commutator sub- groups.Further,
if oc is anyendomorphism
of a group G into the centre of G such that a2 =0,
then the map 1+
a is anautomorphism
of G.[4]
is agood
reference for facts on abelian groups that we use.We thank Joachim Neubuser for
pointing
out a serious error in afirst version of Theorem C.
2. Proof of
Theorem
B : theperiodic
case.In this
section,
G is aperiodic
group in which the orbits of abeliansubgroups
arefinite,
and we show that G has the structure defined in(iii) ( b )
of Theorem A. Theproof
isaccomplished
in severalstages.
1)
G does not have a divisiblep-subgroup of
rank2, tor
anyprime
p.Suppose
that G contains asubgroup U
XV,
where U I"JTT ^~
Since G is
central-by-finite,
U and V arecentral;
and since G’ isfinite,
Thus
U V’G’ /TTG’
X.R/YG’
for some1~ c G,
and maps 99 such that
UVG’jVG’
-~V,
1give
rise to endo-morphisms
oc: G --~Z(G)
such that a2 = 0. Butsince there are
2No
different choices for thecomplement
of U in U XV,
this means that the Aut G-orbit of U is
infinite,
a contradiction.2)
The reducedpart of
everyp-subgroup of
the centreof
G is
finite.
If
not,
thenG jG’
has the sameproperty,
y and we can write- ~aG’> xX/G’,
whereaG’>
is a non-trivialcyclic
p-group: sinceoo,
Z(G)
must have an infiniteelementary
abelian sub-group
Y,
because a basicsubgroup
of an abelian p-group is finite if andonly
if it is a direct factor and thus the reducedpart
is finite.Any homomorphism
99: -Y, together
with the trivial mapXfG’ - 1, gives
rise to anendomorphism
a of G such that a2 =0,
and -- ~ac ~ (aG’ )~> ;
since Y isinfinite, a>
has infinite Aut G-orbit.A consequence of 1 and 2 is:
3) Every infinite Sylow p-subgroup of Z(G)
isof
theform
where
Fp
isfinite.
For every p > the
Sylow p-subgroups
of G are central. Weuse this fact to prove:
4)
T hesubgroup
S =Fp : p
> is a directfactor of
G.Every subgroup
is a direct factor of since it is finite and the reducedpart
of aSylow p-subgroup,
so thatSG’/G’
is a direct
factor
ofGIG’,
sayG/G’= XLIG’.
But then G =and so that and S is a
direct factor.
The next
step
isproved
in a very similar way to 4 and we omit theproof.
5)
For every p > divisiblep-subgroup of
G is a directfactor of
G.For our final
step,
note that S is the directproduct
of theF1J.
6)
Almost all theF 1J
arecyclic.
It is
enough
to show that almost all theF1J
with p >IG’I
arecyclic.
Suppose
thatinfinitely
many of them arenon-cyclic,
so thatF1J
hasa direct factor of the form
~a~ , b1J: = bp ~
=[a~ ,
=for p
lying
in an infinite set 11:. Consider anautomorphism 99
of G that extends the
automorphisms b~ --~ b~ ,
where
( ~,~ , p)
= 1. Theimage
of the abeliansubgroup A = ~a~ : p
E11:) under 99
is and it is clear that suitable choices of the2. give
rise toinfinitely
many AutG-images
of A.A
quick glance
now shows that G has the structurerequired,
andthe
proof
of theperiodic
case of Theorem B iscomplete.
91
3. Proof of
Theorem
B : thenonperiodic
case.Throughout
thissection,
G is anon-periodic
group in which the orbits of abeliansubgroups
are finite and T itsperiodic part (which
is a
subgroup
since G is anFO-group). Again
weproceed
in severalstages
to theproof
that Aut G is finite.1)
T has nosubgroups isomorphic
to aZpoo.
Let A
= ~a1, a2 , ...)
be such asubgroup, with ai
=1,
= aifor
i > 1;
asbefore, A
is central.If
(G/T)p =1= GIT,
take x E G such that(GIT)P.
For k >1,
setXk/T = (G/T)pk;
thenxXk
has maximal orderpk
inGiXk
I so thatC-lXk
= X for somesubgroup Yk
of G. If ak is the endo-morphism
of G definedby
= IY’k
==17
then 1 Aut Gand - but this
gives
the contradiction thatx~
has in-finitely
manyimages
under Aut G.Thus we may assume that is
p-divisible.
Let be a com-plement
of thequasicyclic subgroup AG’ /G’
in then G =AX,
X = is
finite, XI(T
nX)
isp-divisible
and there is a sub- group with T nX
Y suchthat X/Y
=...)
gzA,
the
isomorphism being given by ai
==(ci
for all i. For ap-adic integer
a, I let ya , be theautomorphism
of G that is 1 on A and suchthat for x E X. Then and
X Ya n A = X n A. We claim that if
fl. Indeed,
ifXYlX = XYP we have
Xya, so
that Xv- n A =am~,
andso
pm(a - fl)
= 0. But then ot= fl,
so our claim isjustified
and~XY : y
EE Aut G}
isinfinite;
this is notyet
acontradiction,
since X is non-abelian if G is.
However,
a very easyargument
now shows that the abeliansubgroup X~,
n =IG:
is such that(Xn)a
ifLx
~,
and this is therequired
contradiction.2) Every Sylow p-subgroup of
T isfinite.
Clearly,
since there are no divisiblep-subgroups by 1,
it isenough
to show that there are no infinite
elementary
abelianp-subgroups.
By
way ofcontradiction,
assume that E is one. Since G’ isfinite,
the
P-component TpIG’
ofTIG’
is notdivisible,
so that it has acyclic
direct factor
aG’>
of orderpk > 1,
and of course we may choose a to be of orderph,
WriteGIG’=:
Then the inter-section
Eo
= E r1Z(G)
r1 B is stillinfinite,
since B r1Z(G)
is of finiteindex;
for each theidentity
map of B extends to an auto-morphism q?,,
of Gsending a
to az, andagain
we have the contradiction that the Aut G-orbit of(a)
is infinite.3)
T isfinite.
Let m be the set of
primes p
for whichZ(G)
has ap-element.
Ourtask is to show that co is
finite,
so we assume that it isinfinite,
andwithout loss of
generality
that the index n ofZ(G)
in G’~ iscoprime
top, if pEW.
CASE 1. There exist an element g
of infinite
ord er and aninfinite
subset
Wo ç
(JJ snchthat, for
every p E wo, gfails
to beinfinitely p-divisible.
Enumerate the elements of coo in some way, coo = 9 P2
...~,
and for each Pi E coo
define ki
to be thelargest integer
such thatx~~;
= gfor some x E G. Without loss of
generality
we may assume that g EZ(G).
Set yo = g, and suppose thaty;il
- yo . Thep2-characteristic,
of y, is
again k2,
and we choose y. such that - y1. Proceed in this way; once yo , yl , ... , yr have beenchosen,
thepr+1-characteristic of Yr
iskr+1,
and we choose such thaty’+,
= yr, where s is short forBy construction,
thesubgroup Y Yo 7 Yi 1,- 7 Yr 9
islocally cyclic
andtorsion-free,
andmoreover yi 0
Wi for i > 0.Now let ti be an element of order pi in
Z(G)
such that~ti>
Gvl ~# YG?’ = if one
exists,
and let (pi be the usualautomorphism
induced from a
homomorphism GIG",
-~ti>
with kernel acomplement
of
containing
If no such tiexists,
then G =(Ui)
Xwhere
(Ui)
is thep;-part
ofG,
which is of course theSylow pi-subgroup
ofZ(G)
since In that case, let ggi be theidentity
onK;
and a non-trivial power on(Ui) (which
must existexcept
in the caseZa,
and we canignore this).
In both cases, and and this shows that our sub- group Y has infinite Aut G-orbit.CASE 2.
Every
elemento f infinite
order isinfinitely p-divisible for
almost all p.
Assume first that
GIT
has infinite rank. Then there is a countableindependent
subset ofGIT
which we may indexby
elements of and we can suppose that each x~ is inZ(G).
For eachPEW, choose ap E
8fJ(G),
theSylow p-subgroup
ofG,
of maximum93
order,
and aninteger sp
such 1 and p, where the con- gruences are modulo theexponent
of Note thatSi)(G)
is a directfactor of G since
S~(G)
~1 G’= 1(p
does not divide the index ofZ(G)
in G and therefore does not divide
IG’I [8])
and is a directfactor of Let
C~
be acomplement
of inG,
and define anautomorphism çp,
of G that is theidentity
onOil
and the powerautomorphism
on~S~(G).
With X = p Eo))y
we haveX,
and thus X has infinite Aut G-orbit.
Th,aC,
we may assume, thatGIT
has finite rank and isp-divisible
for almost all
primes
in ro. Choose aprime
so thatq ( IG’I.
as
before,
andG/T
isq-divisible.
Then G =Si)(G) X B,
where B is
q-divisible, Z(B)
isq-divisible
and has no q-torsion, by
1.(Recall
that n = If r is the orderof q
mod n, then themapping
a : B - Bgiven by ba
= bqr is anautomorphism
of B
(this because qr =1
modand
a induces anautomorphism
on
Z(B) ),
whichclearly
extends to anautomorphism
of G.Every
element of infinite order in B has
infinitely
manyimages
under4 )
every PEW,IG: Gpl is finite.
If is
infinite,
thenIG: GpO’ Qi)1
isinfinite, where Q,
is theSylow p-subgroup
ofZ(G).
Take anysubgroup
N ofindex p
con-taining
andlet 99
be ahomomorphism
of G intoZ(G)
withkernel N. Write Then
and,
since there are
infinitely many N,
there must beinfinitely
many X n N sinceIG:XI I
andIG:NI I
are(boundedly)
finite. Thus there areinfinitely
manyWe come now to the final
stage
of theproof.
Let 1~ be a transversal forZ(G)
inG;
if K is anysubgroup
of Gcontaining .R,
then- Z(G).
Set 1~ = Aut G.For a finite subset .I1 of
G,
we define the closureH(F)
of .F as fol-lows :
Clearly H(F)
is a characteristicsubgroup, G/.H(.h’)
istorsion-free,
and
Z(.8’(.F)) c Z(G).
From here on we assume that Aut G is infinite and find a contradiction.5)
.For everyfinite
subset Fof G, Cr(H(F))
hasf inite
index in.1~,
and
Cr(B’(I’) ) r1 Cr(G/.H’(.F’) )
isfinite.
Clearly, S2(.F) : _ ~ f y :
~J u T is a finiteT-invariant set. Thus the kernel g of the restriction T-+
Sym (Q(F))
has finite
index; E
is the centraliser in1~
of)I E lJ
T. To establish the first claim we need to show thatIX: I
is finite.Every
element cx of .g centralises.H’(.h)/T
and
.L,
so it induces ahomomorphism
T r1Z(L), namely
it is
important
here to notice thatZ(L)
sinceHowever, Cr(H(F))
isjust
the kernel of this map, and thus has finite index since itsimage
is in T and therefore finite.Next, Cr(H(F))
r1Or(GjH(F))
isisomorphic
toHom (G/H(.F’), Z(H(F)))Hom ( G/.g(.F), Z(G))
via the obvious map. There is nohomomorphism
withnon-periodic image,
elsethere would be elements
a E G,
0 EZ(H(F))
of infinite order andy E
Cr(.g(.I’) )
r1Cr( G/H(.F’) )
such that av = ae, and then = would be an infinite set ofimages
ofa~.
ThusHom
(G/H(F), Z(H(F)))
is in fact HomZ(H(F))
nT),
whichis finite
by
3 and 4.6)
For everyfinite
subset Fof G,
there exist acyclic subgroup C/B’(F) Z(G), and y
ECr(H(F))
such that6
If
not,
then = forall g
E G(since
some power of every element iscentral)
and thus theautomorphism
group inducedby Cr(H(F))
onG/.g’(.I’)
would bejust (-1); however,
that is im-possible
since nCr( G/.H’(.F’) ))
is infinite.Next,
we defineinductively
a sequence yl , y2 , ... of elements of r and a sequence a2, ... of elements ofZ(G)
as follows :yi is any non-trivial element of .1~
(other
than -1 if G isabelian);
al is any element of
Z(G)
such that(al);
if and a1, ac2 , ... , an have been
chosen,
use 6 to choosesuch that
95
Assume that there is an
element g E
T suchthat ;
iwhich contradicts the choice of an+m . Thus
We are now
ready
for the finalstep. Let i, j
beintegers
such that1 i j .
Thenby
7. On the otherhand, ai
E X’’s nB’(a1,
... ,ai),
whereasari £ ~
0
...,ai) T by
our choice of yi . Thusai) =1=
Xvl n() H(a1,
..., so thatXY~ ~
and X hasinfinitely
manyimages
under 1~.
4. Proofs of
Theorems
C and D.To prove Theorem
C,
we note that a group G in which the endo-morphism
classes of abeliansubgroups
are finite must have one of thestructures described in
(iii)
of Theorem B. Since it is clear that G cannot be the directproduct
ofinfinitely
many nontrivial groups, if it isperiodic
it must be of thetype required
in Theorem C.To show that G is
periodic, proceed
like this. With n: = , the transfer map to the centre isjust g - gn [5, 10.1.3].
It followsthat every
endomorphism
of the abelian group Gn extends to one of G.If G has an element of infinite
order,
so doesGn ;
if a is such anelement,
the
endomorphism
class ofa~
under the powers of theendomorphism
x H x2 of
Gn,
and thus of its extension to anendomorphism
ofG,
isinfinite.
Conversely,
suppose that G is a finite central extension of a directproduct
C of groups oftype
for differentprimes
p, let .g be a sub- group of G and oc anendomorphism
of G. ThenHa CIC
is one of bound-edly finitely
manysubgroups
ofG/C,
andB’a/B’a
r1C;
but .Ha r1
C ~ (H r1 C)a,
and since C isfully
invariant and a directproduct
of different(H
nC)a
is one ofboundedly
many sub- groups. Thus is one ofboundedly
manysubgroups,
asrequired.
Finally,
we prove Theorem D. Let G be a group in which theendomorphism
classes have cardinal at most k(finite).
Inparticular,
for any elements x, y of G there is an
integer r k
such0(y),
and it follows that
G/Z(G)
has finiteexponent
ml, say.Further,
G’is finite since the
conjugacy
classes of elements areboundedly
finite[6],
of order m2 say. Let m be any
positive integer
divisibleby
m1 and m2.For all x, we have for some so
that
(xy)~$ =
and the map x --~x~$
is anendomorphism
of G.It follows that G has finite
exponent.
Now every
periodic
group with finite commutatorsubgroup
islocally finite,
and thus it has an infinite abeliansubgroup
if it is in- finite. One could refer to[3]
for thisresult, though
it is very easy to prove itdirectly
in our case.Suppose
that our group G is infinite and let A be an infinite abeliansubgroup.
Then A contains an infiniteelementary
abelianp-subgroup
P say, whence must have a non-trivial
cyclic
direct factor of p-power order. For each a EP,
the map extends to an
endomorphism
ofG,
and thus x has infiniteendomorphism
class. This contradiction proves that G isfinite,
asrequired.
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Groups
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classesof conjugate
abeliansubgroups,
Mat. Sb., 47
(1959),
pp. 45-54.[3]
P. HALL - C. R. KULATILAKA, A propertyof locally finite
groups, J. London Math. Soc., 39(1964),
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L. FUCHS,Infinite
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New York and London, 1970.[5]
B. HUPPERT, EndlicheGruppen
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York, 1967.[6] B. H. NEUMANN,
Groups
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Groups
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J.WIEGOLD, Multiplicators
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