R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
D EREK J. S. R OBINSON
J AMES W IEGOLD
Groups with boundedly finite automorphism classes
Rendiconti del Seminario Matematico della Università di Padova, tome 71 (1984), p. 273-286
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Groups Boundedly Automorphism
DEREK J. S. ROBINSON - JAMES WIEGOLD
(*)
1. Introduction.
Let G be a group and let .1~ be a
subgroup
of AutG,
the auto-morphism
group of G. Then 1~’ acts in a natural way on the group and on its set ofsubgroups.
Thus we mayspeak
of F-orbits of elementsand of
subgroups
of G. Inparticular
the Aut G-orbits are called auto-morphism
classes. Ourobject
here is toanalyze
the structure of groups.whose
automorphism
classes of elements orsubgroups
are finite with bounded order.Similar
problems
for Inn G-orbits were studiedby
B. H. Neumann almostthirty
years ago. In[4]
Neumannproved
that the groups withboundedly
finiteconjugacy
classes(or
InnG-orbits)
ofelements,
the so-called
BFO-groups,
areprecisely
the groups with finite derivedsubgroup (see
also[12]).
In asubsequent
paper[5]
Neumann wasable to show that the groups which have finite
conjugacy
classes ofsubgroups
areexactly
the centreby
finite groups, thatis,
the groups with finite innerautomorphism
group. It was later shownby
Eremin[2J
that this remains true if one restricts
only
theconjugacy
classes of abeliansubgroups.
RESULTS. Our first main result is a criterion for a group to have
boundedly
finiteautomorphism
classes of elements.(*)
Indirizzodegli
AA.: D. J. S. ROBINSON:Department
of Mathematics,University
of Illinois, Urbana, Illinois 61801, U.S.A.; J. WIEGOLD :Depart-
ment of Pure
Mathematics, University College,
Cardiff,Wales,
U.K.THEOREM 1. Let G be a group and let T denote the
torsion- subgroup of
the centreof
G. Then theautomorphism
classesof
elementsof
G areboundedly finite if
andonly if
T isfinite
and Aut G induc es afinite
group
of automorphisms
inGIT.
This result may be reformulated in such a way as to refer
only
to
T, GIT
and thecohomology
class d of the central extension T ~ G -~0
=GIT.
Recall that there is a natural left action of AutG
and a natural
right
action of Aut T on.Hr2(G, T). Using
these actionswe may restate Theorem 1 in the
following
form.COROLLARY 1. The group G has
boundedly finite automorphisms
classes
of elements if
andonly if
T andStAut õ(JAut T) ~
thestabilizer in Aut
G of
the Aut T-orbitcontaining L1,
are bothfinite.
In
general
theautomorphism
group can be infinite when the auto-morphism
classes of elements areboundedly
finite.In §
4 wegive
some
examples
of thisphenomenon:
we alsogive
necessary and suf- ficient conditions for theautomorphism
group to be finite.Turning
toautomorphism
classes ofsubgroups
we may state our conclusions in thefollowing
form.THEOREM 2. The
following properties of
a group G areequivalent.
(i)
Theautomorphism
classeso f subgroups of
G areboundedly finite.
(ii)
Theautomorphism
classeso f
abeliansubgroups of
G areboundedly finite.
(iii)
Either Aut G isfinite
or there is a directdecomposition
G = where
G1
is alocally cyclic
torsion group,G2
is a
finite
central extensionof
a directproduct of finitely
many groups
of type poo f or different primes
p, and01
andO2
do not contain elementsof
the sameprime
order.Thus the condition on
automorphism
classes ofsubgroups
comesquite
close toforcing
theautomorphism
group to befinite,
theonly obstacle,
so tospeak, being locally cyclic
torsion groups.Ultimately
Theorems 1 and 2depend
upon information about a group ofautomorphisms
F of anFC-group
G such that all r-orbits of G areboundedly
finite. This information is to be found in a paper of Baer[1], (p. 111).
We shallgive
a short andelementary proof
ofthe necessary facts. These can be summed up as follows.
PROPOSITION 1. Let r be a group
of automorphisms of
anFC-group
G. Then the r-orbits
of
G areboundedly finite if
andonly if
there isa
finite
F-invariant normalsubgroup
Eof
G such that .1~ induces afinite
group
of automorp hisms
inGIE.
We mention that a
topological proof
of this result hasrecently
been
given by Schlichting [10].
RiF,mARiics:
(i)
It will be seen that theproof
of Theorem 1provides
bounds ---albeitcomplicated
ones-for the order of T and the index ofOAuto(GjT)
in terms of the least upper bound of the numbers of elements in anautomorphism
class.(ii)
leave open theproblem
ofdetermining
the groups in which theautomorphism
classes of elements(or subgroups)
are finite but
possibly
unbounded.2.
Automorphism
groups withboundedly
finite orbits.We
begin
with someelementary
facts.LEMMA 1. Let G be a group in which all
automorphism
classesof elements’have
at most n elements.T hen
(i)
Aut G isresidually
asubgroup of
thesymmetric
groupSn;
thus Aut G is
residually finite
and hasfinite exponent (ii)
the group G’ isfinite
andGjZ(G)
is centreby finite.
PROOF. The first statement is true because Aut G
permutes
theelements of an Aut G-orbit
gAut G
and Theremaining
state-ments follows from B. H. Neumann’s theorem on
BI’C-groups
and aresult of P. Hall
(see [7 ], 4.25).
Our immediate aim is to prove
Proposition
1. For this we shallrequire
a lemma whichalready
occurs in Baer’s paper.LEMMA 2. Let T be a group
of automorphisms of
a group G andassume that all r-orbits
of
G have at most n elements where n > 1. Let g be an elementof
G such that the r-orbitof
g hasexactly
n elements anddefine
N to be thesubgroup generated by
all(gx)Y
where x E G and y E 1~.I f
K =Cr(N),
then each K-orbitof GIN has
at most n -1 elements.PROOF.
Suppose
that on thecontrary
some x in (~ has=
n,. Thenclearly
so that
Cr(x)
=Cr(xN).
Now considerCr(gx).
If o~ ECr(gx),
thengx =
gaxa
and xa -= x mod N. Thus d ECr(xN)
=Cr(x) and g
=ga.
Therefore
Consequently
and thus
GYr(g)
=Cr(gx).
It follows that K =Cr(x)
and This
gives
the contradiction l = = nPROOF OF PROPOSITION 1. The
sufficiency
of the condition is clear.Assume that all r-orbits of G have at most n elements. If n then 7~=1. Thus we can assume that n > 1 and use induction on n.
Let g
be an element of G withIgrl
= n.Denote
by N
thesubgroup
of Ggenerated by
all(g~)y
where x E Gand ye 7~. Set K
equal
toCr(N).
Since .1~permutes
a finite set ofgenerators
of theF(J-group N,
we conclude that is finite.By
Lemma 2 and induction on n there is a finite K-invariant normal
subgroup
of such that K induces a finite group of auto-morphisms
in DefineSince
E’o
is afinitely generated FC-group, IX :FoI
is finite and soIF:FoJ ]
is finite.
Clearly .I’o - 1
where 0m ~ n ! ;
since[G,
and[Eo,
==
1,
we deduce that[G,
=1..As a torsionsubgroup
of thefinitely generated .F’C-group
the group[G, To]
is finite. Now define B to be then E is a finite I’-invariant normalsubgroup
ofG,
while
Cr(GfE) ) I
is finite sinceTo
~Or(GfE).
From
Proposition
1 we can derive a first criterion for a group to haveboundedly
finiteautomorphism
classes of elements.LEMMA 3. Let G be a group. Then the
automorphism
classesof
elements
of
G areboundedly finite i f
andonly i f
there is afinite
char-acteristic
subgroup
E such that Aut G induces afinite
groupo f
auto-morphisms in G/E.
This follows at once on
taking
.1~ to be Aut G inProposition
1.(Notice
that if we take l~’ to be Inn G andapply
Schur’s Theorem oncentral
by
finite groups we recover Neumann’s result that aBFC-group
has finite derived
subgroup.)
COROLLARY 2.
If
G is a group withboundedly finite automorphism
classes
of
elements and T is thetorsion-subgroup of
thecentre,
then Aut Ginduces a
finite
groupof auromorphism in
PROOF.
By
Lemma 3 there is a finite characteristicsubgroup
Ewhich is minimal
subject
to Aut Ginducing
a finite group of auto-morphisms
inG/E.
It will suffice to show that Fix g inG;
then L =
G»)
is characteristic with finite index in G. Hence Aut G induces a finite group ofautomorphisms
inOIL n
E. ThusEL by minimality
of E andEZ(G).
COROLLARY 3. The
automorphism
groupof
a group withboundedly finite automorphism
clas3esof
elements is abelianby finite.
The
point
here is that thesubgroup 1-’o
whichappeared
in theproof
of
Proposition
1 is abelian since it satisfiesTo]
= 1.3. The
torsion-subgroup
of the centre.In this section we shall prove Theorem 1. The
sufficiency
of thecondition in the theorem is clear. Thus in view of
Corollary
2 all thatremains to be done is to show that if G is a group with
boundedly
finite
automorphism
classes ofelements,
thetorsion-subgroup
of thecentre, T,
is finite. We shall establish this in a series of fivesteps.
Let C denote the centre of G and set
Q equal
to Write d for thecohomology
class of the extension C ~ G -~Q.
(i)
There is apositive integer
e such thateH2(Q, C)
= 0.By
theUniversal Coefficients Theorem
Here
Qab = Q/Q’
andM(Q)
is themultiplicator
ofQ.
NowQ ~ Inn G,
so we may deduce from Lemma 1
that Q
has finiteexponent.
HenceEgt (Qab, C)
iscertainly
bounded.Next
QIZ(Q)
is finiteby
Lemma 1(ii).
It is now clear from the .six termhomology
sequence for the extensionZ(Q) ~
thatM(Q)
is bounded([11],
p.105).
Therefore Hom(M(Q), C)
isbounded,
and we deduce that
H2(Q, C)
isbounded,
asrequired.
(ii)
Thesubgroup
T is reduced. Otherwise we can write.0 =
C1 X C2
where01
is ap°°-group
for someprime
p. Then themappings
0, F-+ 9 C21-+ c2, 1(ei
EC,),
determine anautomorphism
yof C. Now
4y
= L1 because e4 = 0. Hence y extends to an auto-morphism
ofG;
however thisautomorphism
has infiniteorder,
incontradiction to Lemma 1.
(iii) If p is
anyprime,
thep-component T~ of
T hasfinite exponent.
Suppose
that 0 =01
XC,
where01
is acyclic
group of orderp I.
Theassignments
01 H 7 C2 H 0,,(ei
EOi),
determine anautomorphism
of C whose order is
pl-m-2
or 1 if I m+ 2;
here prn is thep-share
of e. This
automorphism
extends to anautomorphism
ofG,
so itsorder is
c n ! by
Lemma 1. Hencep ~-2 c e(n !),
and there is an upper bound for 1. It follows from(ii)
and well-known facts about basic.subgroups
of abelian p-groups([3], VI)
thatT~
has finiteexponent.
(iv) If p is
anyprime,
thep-component T~
isfinite.
The groupQ
is centre
by
finite with finiteexponent by
Lemma 1. From this it is -easy to show thatwhere
Qo
is finite andQ1
is abelian. WriteQo = Go/C
andQ, = G1/O.
Suppose
first thatQ,,
and henceQab ,
has an element of order p.Since Hom
(Qab, T~)
isisomorphic
with asubgroup
of AutG,
the groupT1J
cannot have rankgreater
than n; this is because anautomorphism
class cannot contain n
+
1 elements.By (iii)
the groupT1J
is finite.Now assume that
Q1
is ap’-group.
We can writeGo
= XC with.X a
finitely generated subgroup.
Now/X:X n 0/
isfinite,
so Cis
finitely generated.
SinceT1J
has finiteexponents,
there aredecomposi-
tions where and
To
isfinite.
Combining
theseequations
we obtainThe next
step
is to prove thatPI
is also a direct factor of G. Toaccomplish
this we consider theLyndon-Hochschild-Serre spectral
sequence for the extension
Now our extension restricts to a
split
extensionTi w Go -
and so comes from the kernel of the restriction.Thus we need
only
consider the termsE2~
andE2°
in thespectral
sequence. In fact both of these vanish.
In the first
place
because
Ti
is a p-group andQ,
is an abelianp’-group-here
it is relevantthat
Tl
is a trivialQ1-module.
Next considerof course
which is a bounded abelian p-group. The action of
G1
on .J3 is thefollowing;
if 0 EH and g
EG1,
then 09 sends0153TI
inGo fT
to(x[x, g-11] T,,)
0.Now
[Go ,
and since isa p’-group,
so is[Go ,
Therefore[Go,
ismapped by
0 to theidentity.
It follows that 8D = 8 aDdG1
actstrivially
onH,
so that H is a trivialQ,-module.
HenceH1(Q1, H)
=Hom (Q1, H)
= 0.Consequently E21 -
0 and thus Gsplits
overTi.
This means thatT1
is a direct factor of G. SinceT1
has finite
exponent,
we conclude thatT1,
and henceTp,
must befinite.
(v)
Thesubgroup
T isfinite.
Let p be aprime greater
than1 + n ! and assume that
Now Q
hasexponent
at mostn ! ,
so it must be a
p’-group.
In additionT,
is a direct factor of Cby (iii).
we may argue as before thatTp
is a direct factor of Gby
ap-pealing
to thespectral
sequence for the extensionQ.
Thus G has a
cyclic
direct factor of some orderpk >
1. However thisimplies
thatp -1 c r~ ! . By
this contradictionif p > 1 + n!.
It now follows from
(iv)
that T is finite.PROOF OF COROLLARY 1. Recall that T is the
torsion-subgroup
ofthe centre of G and G =
GIT.
We know that G hasboundedly
finiteautomorphism
classes of elements if andonly
if both T andare finite. Let x E Aut
G;
then x E I if andonly
if there is a r in Aut T such that xJ == J T([11],
p.22) .
This isprecisely
the condition for x to leave the orbit JAutT fixed as a set. Hence I =4. Some groups with infinite
automorphism
group.We
begin
with a result which tells usexactly
when a group withboundedly
finiteautomorphism
classes of elements has finite auto-morphism
group.PROPOSITION 2. Let G be a group with
boundedly finite
automor-phism
classesof
elements. Let C =Z(G), Q
=GIC,
and let T be thetorsion-subgroup of C;
write0
=C/T .
Then Aut G isfinite i f
andonly it
and arefinite
where t =PROOF.
By
Theorem 1 theimage
of the canonicalhomomorphism
is
finite,
while its kernel isclearly isomorphic
with .H =Hom((G/T)ab, T) .
The
given
conditionsimply
that.g,
and hence AutG,
is finite. Theconverse is easy.
REMARK. It is not difficult to show that if Aut G is
infinite,
it willbe uncountable : in this connection see
[6]
and[9].
The relation between finiteness ofautomorphism
classes andcompactness
of the automor-phism
group is discussed in[6],
a work in which ourinvestigations
hadtheir
origin.
LEMMA 4. Let p be any
prime
and let F be atorsion- f ree
abeliangroup such that Aut F is
finite
buth’/.F’9
isinfinite.
Then G =is a group with
boundedly finite automorphism
classesof
elements but hasinfinite automorphism
group.This follows at once from
Proposition
2. Such an abelian group~’ can be constructed in the
following
way. Let psy ... and ql, q27 ...be two sequences of distinct
primes.
Let v2,...I
be a basis fora rational vector
space Tr
ofcountably
infinite dimension. Define F to be thesubgroup
of Vgenerated by
all elements of the formwhere m eZ and
i = 1, 2, ....
It isstraightforward
to prove thatlaut Fl
=2,
while it is clear that is infinite.REMARK. The form of this
example
istypical
of an abelian group withboundedly
finiteautomorphism
classes. Indeedby
Theorem 1such a group is
always
a directproduct
F X T whereI TI
isfinite,
Fis torsion-free and Aut .F is finite.
A more
challenging problem
is to find a torsion group withboundedly
finite
automorphism
classes of elements which has infiniteautomorphism
group.
CONSTRUCTION. If p is any odd
prime,
let G be the group withgenerators
and the
following
relations:It is clear that G is an infinite
nilpotent
p-group of class 2 andexponent p2.
Furthermore C =Z(G) = a>
Xb~
andQ
=GIC
areelementary
abelian: the latter has a basis =1, 2, ...}.
Concern-ing
this group we shall prove thefollowing.
PROPOSITION 3. The least upper bound
of
the ordersof
the auto-morphism
classesof
elementsof
Gequals (p ( p
Aut Gis
infinite.
PROOF. We shall
merely
sketch theproof, omitting
most of thecomputations.
For any g in G we have and of courseIG: =
p. The latterproperty
characterizes Xl in thefollowing
sense.
(i) If IG: Go(g)1
= p, then g Exl, 0).
HenceXl’ 0)
is charac- teristic in C.Write g = xil
...X:k
mod C where Ui EZ_~
and letk > 1. If h E
0,,(g),
we can write h =XII
...r[$§’ with
vic- 4
at theexpense of
deleting
elements which are known to commute with g.Then 1 =
[g, h] yields
Now use the
defining
relations to obtain two linearequations
forvi , ... , the final terms are
Since
u~~ 0
and occursonly
in the secondequation,
this alinearly independent system.
But thisimplies
that a con-tradiction.
(ii)
Thesubgroups a>
andb>
are characteristic in G. Since(in)
Thesubgrotip
X3, ..., X2i-1,C~
is characteristic in Gfor
all i > 0.
Clearly 01
=C~ ) _
~3 , ~4 , ... ,C~
is character-istic ;
alsoZ(Ol) === a). By
theargument
of(i)
we havecharacteristic in
01
and hence in G. Thus x3,0)
is characteristic in G. The same method can beapplied
toC3
=Oa«Xl’
X3,0»)
etc.In what follows a is an
arbitrary automorphism
of G. Then wecan write
with Considerations of order show that
ail = 0
if i > 1.Also
‘~2=_1 = ~32$’1,8 /~b9i-l,a ~~. vV2i i ~,9’~-1
mod 0 lf 2 > oby (iii).
(iv)
For eachi >
0 thesubgroup C)
is characteristic and=
u‘/a=j~s ...
... mod Cfor j ---1 2, ...~20131.
This isproved by
induction on i : the cases i = 1 or 2 areclear,
y so 2..By applying
(X to[X2i,
andusing
thedefining
relations one obtains(X2i,2i = 0 for j =1, 2, ’"?~2013~’
One can thenapply
a to[X2i, x2i+1]
to show that
for j
i. It follows thatC~
is char-acteristic in G.
Finally
from[x2i, X2i+l]
one obtains thatfor j
=1, 2, ... , 2 -1.
Thiscompletes
the induction.So far we know that
By applying
a to the commutators[x2, x,], [X2’ X6], [X2’ X8]2
..., 7 weobtain a2, = a25 = _ ... = 0 = a43 = = aS3 = .... We can treat x~, xa etc. in a similar fashion. Thus we conclude that
(v) 0)
is characteristic in G. From this it isstraight--
forward to
identify
Aut G.(vi)
There is asplit
exact sequence.It is now easy to deduce
Proposition
3.REMARK.
Proposition
3 is still valid if p =2,
asslight change
in the
argument
show.5.
Automorphism
classes ofsubgroups.
In this section we shall prove Theorem 2. Of course it is clear that condition
(i) implies
condition(ii).
The maindifficulty
in theproof
is to show that(ii) implies (iii).
This will beaccomplished
inseveral
steps.
(a)
We assume that noautomorphism
class of abeliansubgroups
of G has more than k elements where k is some
positive integer.
WriteC =
Z’(G)
andQ
=G/C,
and let T be thetorsion-subgroup
of C.By
the theorem of Eremin
quoted
in the introductionQ
isfinite,
withorder q
let us say. If L1 is thecohomology
class of the central extension.0» G
-~9~
thenq4
= 0.(b)
The group T is the directproduct of
alocally cyclic
group L .and a groupof finite exponent
E. We consider a directdecomposition
C =
PI xP2 X 00
where the p-groupsPi
andP2
arecyclic
or oftype poo
and~P2~.
Let 0 E Hom(Pi , P2);
then theassignments
x Hx H x,
(x E P2 X Co),
determine anautomorphism y
of C.Since
q4
=0,
we have4 y = L1,
sothat y
extends to anautomorphism of G,
and in consequence there are at most ksubgroups
of the form Nowimplies
thatq0
=q0’. Hence q
Hom(PI, P2)
has order at most 1~. This tells us at once that there is at most one
subgroups
oftype poo
for eachprime
p.Now suppose that
P1
iscyclic
of orderpl
and writepm
for thep-share
of q. Then and soptqk
for allprimes
p. Hence t = 0 for almost all p. It follows from well-known facts about basicsubgroups ([3], VI)
that thep-component
0 islocally cyelic
for almost
all p.
MoreoverOf)
is at worst the directproduct
of a group of finiteexponent
and alocally cyclic
group. This establishes(b).
(c) I f G is a
torsion group, then 0 =01 X G2
whereG,
andG2
.are as described in the statement
of
the theorem. SinceQ
isfinite,
we-can write G = XC where X is
finitely generated
and therefore finite.’There are direct
decompositions
E =E0 X E1
and L =L0 X L1
whereEo
isfinite, La
satisfies the minimal condition and X nX Lo .
Then G = XG =
(XEoLo) Obviously E1
inherits the prop-~erties of
G,
soEl
isfinite, being
an abelian group of finiteexponent.
It is now clear that G has a direct
decomposition
of thetype
claimed.From this
point
on we shall suppose that G is not a torsion group._
(d)
Thesubgroup
T isfinite.
We observe first that the group~’ = C/T
cannot bep-divisible
for anyprime
p. For suppose that this is the case.By (b)
we can write C = for somesubgroup
D which is
evidently p-divisible.
Moreover themapping
x H x~ isan
automorphism
of D. There is apositive integer I
such thatLet y be the
automorphism
of C determinedby
theassignments x H (x
ED), x 1-+
x,(x E C~).
Then4y
=.1,
and so y extends to anautomorphism
of G. Let x be an element of infinite .order in D. Then there areonly finitely
manysubgroups
of the form=
1, 2,
... , which isplainly
absurd.We conclude that
CpHl
for all Now suppose thatCp
contains an element of order
p t.
Let 8 E Hom(OfCptT, C~) ;
then themapping
xT),10
is anautomorphism
of C which extends toan
automorphism
y of G.Writing
C = asbefore,
we concludethat there are at most k
subgroups
of the form DY. Now if Dy -DY’, then,
in the obviousnotation, q8
andq6’
take the same values onOfOptT, that is, qO = q8’.
It follows that q Hom(CICPT, Cp)
has order at most k. Hence
p t-m k
and where pm is thep-share
of q. Hence T has finite
exponent,
and we may write C = F X T withF torsion-free. Also G = XC where X is a
finitely generated subgroup.
Hence we have T =
To
XTl
whereTo
is finite and X r1C c To
X F.Then G =
(XTo.l4’)
XT,.
It follows thatT,,
and henceT,
is finite.(e)
Horn(GfT, T)
isfinite.
Let 0 E Hom(GfT, T) ;
then the map-ping x H x(xT)0
is anautomorphism
y of G.Writing
C = F X T asin
(d),
we can be sure that there areonly finitely
manysubgroups
FY.From this it follows that Hom
(OfT, T)
is finite.( f )
Aut G isfinite.
Let 1-’ denote the group ofautomorphisms
of C induced
by
AutG,
and let x E C. There are at most ksubgroups
with y in 1~. If x has infinite
order,
there are therefore not morethan 2k
possibilities
forxY;
it follows that the T-orbits of C =CIT
are
boundedly
finite. This allows us to deduce fromProposition
1that
T,
and hence AutG,
induces a finite group ofautomorphisms
in C.Define I~ to be the group of
automorphisms
of G whichoperate trivially
onT,
C andQ.
Then~Aut
is finiteby (d)
and theprevious
paragraph.
It is easy to see that .~ mustoperate trivially
onHence K -- Hom
(GfT, T),
which is finiteby (e).
Therefore Aut G isfinite,
and we have shown that(ii) implies (iii).
Final
step.
Tocomplete
theproof
of Theorem 2 we must show that(iii) implies (i).
Consider a group G = withG,
andG2
as described in the theorem. It is clear that we can find a characteristic
locally cyclic subgroup N
such thatNZ(G)
andGIN
isfinite,
oforder n say. Let H be an
arbitrary subgroup
of G. Then H r1 N ischaracteristic in
G,
so we can assume that N= 1. Hence H is finite of orderdividing
n. Let r = thenIf y
E themapping [x, y]
is ahomomorphism
from .H onto[H, y]
since[H,
Hence[H, y]
has orderdividing n
andso it is contained in
N[n] _ {x
ButN[n]
has finite order r, say, and it is clear that there are at most rnpossibilities
for HY. Hence the Aut G-orbitcontaining
H has at mostrn(n !)
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I. I. EREMIN,Groups
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