C OMPOSITIO M ATHEMATICA
B. H ARTLEY
Subgroups of locally normal groups
Compositio Mathematica, tome 32, n
o2 (1976), p. 185-201
<http://www.numdam.org/item?id=CM_1976__32_2_185_0>
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185
SUBGROUPS OF LOCALLY NORMAL GROUPS
B.
Hartley
Noordhoff International Publishing
Printed in the Netherlands
1. Introduction
A group G is called a
locally
normal group, if each finite set of elements of G lies in a finite normalsubgroup
of G. If H is asubgroup
of alocally
normal group
G,
we writeCIGH
andLcl,H
for theconjugacy
class andlocal
conjugacy
classrespectively
of H in G. A number of authors haveproved
theoremsasserting that,
under varioushypotheses, CIGH
=Lcl,H
if andonly
ifCIGH
is finite(see [4], [6], [7], [8], [9], and [10]).
Most,
if notall,
of these theorems either havehypotheses
whichvisibly imply
that G is ahomomorphic image
of aresidually
finitelocally
normalgroup, or can be
readily
reduced to that caseby
arguments like that of[8]
Theorem 4.1. Let us forbrevity
write X for the class of all groups which arehomomorphic images
ofresidually
finitelocally
normalgroups. For
X-groups,
Gorëakov hasproved
thefollowing
result :THEOREM 1:
(Gorëakov [1])
Let H be asubgroup of an X-group G,
andsuppose that
IHIHGI
is aninfinite
cardinal oc. ThenICIGHI
= oc andILCIGHI
= 2(%.
Here we have written
HG
=nXEG
Hx for thelargest
normalsubgroup
of G contained in H; we also write HG for the normal closure of H in G.
It follows from Theorem 1
that,
ifIHIHGI
isinfinite,
thenCIGH =1= LclGH.
On the other
hand,
ifIHIH,1
isfinite,
thenHG/HG
isfinite,
and since theconjugates
and localconjugates
of H lie betweenHG
andHG, CI G H
andLcl,H
are finite. Infact,
it is easy to see thatthey
coincide(cf. [8]
Theorem3.1).
Thus Gorëakov’s Theorem has the immediateCOROLLARY :
If
G ~ Xand H ~ G,
thenCIG H
=Lcl GH if
andonly if
CIGH
isfinite.
and most of the other known results about the
equality
ofCIGH
andLclGT
can be deduced from this as we haveexplained.
We
give
in this paper an alternativeproof
of Gorëakov’s Theorem.In
fact,
Gorëakovactually
proves his theorem([1]
Theorem4)
under thehypothesis
that G is ahomomorphic image
of asubgroup
of a directproduct
of finite groups, but also proves([1]
Theorem2)
that every X-group is such an
image.
The outcome is Theorem 1 as we have stated it.However,
we shall notappeal
to Gorëakov’sdescription
ofX-groups,
but
proceed directly
to prove a result which on the face of it is moregeneral. Let
denote the class ofall locally
normal groups Lsatisfying
the
following
condition:for every infinite subset T of L.
Clearly,
the same class is obtained if the word ’subset’ isreplaced by
’normal
subgroup’. Also,
for asubgroup
H of alocally
normal groupG,
write ce =ce(H) = |H/HG|, 03B2
=03B2(H) = |ClGH| = |G : NG(H)|, 03B2
=03B2(H) =
|LclGH|.
We proveTHEOREM 1’: Let
G ~
andH ~
G.Suppose
that ce isinfinite.
Then03B2 = 03B1 and 03B2 = 203B1.
Theorem 1 follows from this and PROPOSITION 1 :
X ~ ,
which is immediate from Gorëakov’s result
[1]
thatX-groups
are homo-morphic images
ofsubgroups
of directproducts
of finite groups, but for which wegive
anindependent proof
in Section 2.In connection with Theorem
1’,
notice thefollowing straightforward
and well known
fact,
which can be establishedby
the considerations of[8]
Theorem 3.1:PROPOSITION 2 :
If one of 03B1, 03B2, 03B2
isfinite,
then all are, and in that case,03B2 = 03B2.
We do not consider this situation
further,
but go on to consider the relation between a,fi
andj8
in ageneral locally
normal group, when ceis assumed infinite. Of course, if G is any
locally
normal group with centerZ,
thenG/Z
isresidually finite,
and so in a sense Gdeparts only
slightly
from residual finiteness. But thisdeparture
seems to make thebehaviour and discussion of a,
fi
andj8 quite
a bit morecomplicated.
Weobtain without much
difficulty:
PROPOSITION 3:
1 f a is infinite,
thenThis is
proved
in Section 2.Thus,
if we assume the Generalized ContinuumHypothesis,
then 03B1and
j8
mustsatisfy
one of theequations
03B1 =03B2,
oc =20, fi
=2a,
andj8
and03B2
mustsatisfy
one of theequations e= j8, 20, 22ft.
We shall showby
anexample (Section 4) that,
forarbitrary
infinite a, each of thesepossibilities
occurs. It seems at first rather
surprising that 03B2
can be aslarge
as22ft.
Next we come to the relation between a and
03B2.
This is muchcloser,
and is the source of our main theorem :
THEOREM 2:
If H is a subgroup of a locally
normal group G anda(H)
isinfinite,
thenff(H)
=203B1(H).
This is an immediate consequence of Theorem B of Tomkinson
[9]
inthe case a =
0,
and indeed we shall use his theorem in ourproof.
How-ever, we have not been able to extend Tomkinson’s methods to the
general
case, and have had to
proceed
much moreindirectly.
Let Z be the centreof G. An immediate reduction
using
Theorem 1 allows us to assume thatHZ « G. We next deal with the case when
HZ/Z
is central inG/Z.
Inthat case, we use some of the
mildly topological
methods and results of[3],
the argumentdepending ultimately
onknowing
the cardinal of thecompletion
of an abelian groupequipped
with a Hausdorfftopology
inwhich certain
subgroups
of finite index form a basisof neighborhoods
ofthe
identity -
an abelian cofinite group in theterminology
of[3].
Whenoccasion
demands,
we shall usefreely
the notation of[3].
The case whenHZIZ
is central inG/Z
seems to be the crucial one, and the rest of the argument deduces thegeneral
case from itby purely
group theoreticmeans. The
proof
of Theorem 2 isgiven
in Section5,
and that of Theorem l’ in Section 3. Section 2 contains somepreliminary results,
most ofwhich are in essence well known.
The author is
grateful
to theUniversity
of Wisconsin in Madison for its support andhospitality
while this work wasbeing
carried out.2.
Preliminary
resultsWe
begin by proving Propositions
1 and 3 of the Introduction.PROOF oF PROPOSITION 1: We have a
homomorphic image
G of aresidually
finitelocally
normal group, and have to show that G ~. By (1)
of theIntroduction,
we have to showthat G : CG(T)| ~ |T|
for eachinfinite subset T of G. Since G is
locally normal, 1 TI
=1 TGI,
and so wemay assume that T is a normal
subgroup
of G.We first consider the case when G itself is
residually
finite andlocally
normal. The set of all finite normal
subgroups
of G contained in T is a setof
cardinal 1 Tl.
Since G isresidually finite,
we maychoose,
for each such finite normalsubgroup F,
a normalsubgroup N(F)
ofG,
contained as asubgroup
of finite index inT,
such that F nN(F)
= 1. We may then choose a finite normalsubgroup A(F)
ofG,
contained in T, such thatA(F)N(F)
= T.For
each g
EG,
letF(g) = gG
n T, wheregG
is the normal closure ofg>
inG,
and let6(g)
denote theautomorphism
induced onA(F(g)) by conjugation
with g. The map0: g - (F(g), 03C3(g))
has asimage
a set ofcardinal at
most |T|, since,
for each of thepossible subgroups F(g),
thereare
only finitely
manypossibilities
for6(g).
Butif g, g’
E G and9(g)
=6(g’), then F(g)
=F(g’), N(F(g))
=N(F(g’)), A(F(g))
=A(F(g’)),
andu(g) = u(g’).
Now
[gG, N(F(g)) ~ F(g)
nN(F(g))
=1,
and so gand g’
centralizeN(F(g))
=N(F(g’)).
Theautomorphisms
induced on Tby
g andg’
aretherefore
completely
determinedby
their restrictions toA(F(g))
=A(F(g’)),
that
is, by 03C3(g)
=03C3(g’).
Henceg’g-1
ECG(T).
Thus 0 isinjective
on anyset of coset
representatives
ofCG(T)
inG,
and|G: CG(T)| ~ |T|.
To deal with the
general
case, suppose G =L/M,
where L is aresidually
finite
locally
normal group and M i L. Then T =SM/M
for somenormal
subgroup
S of L such that|S| = |T|. By
what we havejust
seen,|L: C| ~ |S|,
where C =CL(S).
SinceCM/M
centralizes T and has index atmost |S| = |T|
inG,
wehave G : CG(T) ~ |T|,
asrequired.
PROOF oF PROPOSITION 3: We have a
subgroup
H of alocally
normalgroup
G,
with oc =IH/HGI,f3
=|ClGH|, 03B2
=|LclGH|,
and have to assumece infinite.
(i)
ce~
203B2. To seethis,
let S be a transversal toNG(H)
in G. Then|S|
=fi.
Let SES. Then as G islocally
normal andCH(s) ~
H nHS,
wehave
|H: H ~ Hs|
oo. Hence H n HS contains a normalsubgroup H(s) of H, of finite
index inH,
andcontaining HG’ Clearly HG
=n,.s H(s),
and so
HIHG
can be embedded in thecomplete
directproduct
ofthe 03B2
finite groups
H/H(s).
Hence|H/HG| ~ 203B2.
(ii) fi ~ 03B2 ~ 203B1 ~ 22P. Trivially, 03B2 ~ fl.
Since G islocally
normal and|H/HG|
= a, we have|HG/HG|
= ce. Since localconjugates
of H lie be-tween
HG
andHG, 03B2 ~
203B1 The lastinequality
is immediate from(i).
The
remaining
results in this section are technical facts needed for theproofs
of Theorems l’and 2.LEMMA 2.1
(Gorcakov [1]):
Let G be alocally
normal group, and sup- pose that{N03BB:
À EAI
is a setof normal subgroups of finite
indexof
G suchthat
nAEA NÂ
= 1.Suppose
thatlAI
is aninfinite
cardinal oc. Then|G: Z(G)| ~
oc.PROOF : The set of all intersections of
finitely
manyNA
may be indexedby
a set of cardinal oc, and so we may assume that the intersection of anytwo of the
NA
contains a third. For each À EA,
letThen |H03BB: H03BB
nN03BB|
oo,as IG: NÀI
00.Clearly HA
nN.
is central inN..
We have G =NÀA
for some finite set A. Now the centralizer of A inH03BB
nN.
isclearly
central inG,
and has finite index inH03BB
nN.. Thus,
if Z =
Z(G), JH.: HA
nZI
oo, and soIHÀZ/ZI
ce. Now everyelement g
of G lies in a finite normalsubgroup
B ofG,
and we haveB n
NÀ
=1,
and hence[B, Nj
=1,
for some 03BB E A.Thus g
EHA.
It fol-lows that
G/Z
is the union of the finite normalsubgroups H03BBZ/Z(03BB
eA),
and
so |G/Z| ~
03B1.LEMMA 2.2: Let G be a
locally
normal group, let N be aninfinite
normalsubgroup of G of cardinal
Il, and let C =CG(N).
Then there exists a normalsubgroup
Dof
G such that C ~ D,[D, G] ~ C, [D, N] ~ Z(G),
and|G:D| ~ 03B1.
PROOF : The set of all finite normal
subgroups
of G contained in N is a set of cardinal a.Clearly
C =n CG(F),
whereF ranges
over all suchfinite normal
subgroups. Thus,
ifB/C
is the centre ofG/C,
Lemma 2.1gives
(1) IG: B| ~ 03B1.
On the other
hand,
if Z =Z(G),
thenG/Z
is well known to beresidually
finite.
By Proposition 1,
ifA/Z
=CGIZ(NZIZ),
wehave IG: A| ~
Il. LetD = A n B. From
(1), IG: D| ~
oc, and[D, G] ~ [B, G] ~ C, [D, N] ~ [A, N] ~ Z(G).
The
following
result was usedby Gorcakov [1].
We include aproof
for
completeness.
Notice that ityields
an alternativeproof
ofProposition
1.
LEMMA 2.3: Let G be a
residually finite locally
normal group, and let N be a normalsubgroup of G of infinite
cardinal oc. Then there exists a normalsubgroup
Mof G
suchthat IG: M| ~
oc and M n N = 1.PROOF : For each
non-identity
elementof N,
choose a normalsubgroup
of finite index of G which does not contain that
element,
and let L be theintersection of the
subgroups
so obtained.By
Lemma2.1,
ifS/L
is thecentre
of G/L, |G: SI
~ a.Passing
toG/L,
we may assumethat G : Z| ~
a, where Z =Z(G).
Let M be a
subgroup
of Z maximalsubject
to the condition M n N n Z- 1. Then M «
G,
and if we can show thatthen G : M| ~ 03B1,
and N n M = 1.Write U =
Z/M,
and let V be the naturalimage
of N n Z in U. Then|V| ~
a, and every non-trivialsubgroup
of U intersects Vnon-trivially.
Thus,
for eachprime
p, thesubgroup
of elements of orderdividing
p in U liesin v
and so has cardinal at most a. TheSylow p-subgroup
of Utherefore also has cardinal at most oc, and
hence ! |U| ~ 03B1, establishing (2).
Finally
we recall thefollowing
facts aboutlocally
inner automor-phisms :
LEMMA 2.4.
(Stonehewer [5]):
Let G be alocally
normal group, letA ~
G and B « G. Then(i) Every locally
innerautomorphism of A
can be extended to alocally
inner
automorphism of
G.(ii) Every locally
innerautomorphism of G/B
isnaturally
inducedby
alocally
innerautomorphism of
G.An alternative
proof can
begiven by viewing
the groupof locally
innerautomorphisms
of alocally
normal group as thecompletion
of the group of innerautomorphisms
in its natural cofinitetopology,
as in[3].
However,
since this seems to be acomplication
rather than asimplifica-
tion at present, we shall not pursue it further.
3. Proof of Theorem l’
Let
G ~ and H ~
G. Thehypothesis
is that oc =!N/NJ
isinfinite,
and we have to show
that 03B2
= oc andj6
= 21X.We have H =
HGS
for some set S of cardinal oc, and the definitionoff
gives |G : CG(S)| ~
03B1. SinceCG(S) ~ NG(H), 03B2 = |G: NG(H)| ~
(X. On theother
hand,
G =NG(H) T
for some subset T of cardinal03B2.
Since G ~9), IG: CG(T)| ~ 03B2,
and sinceHG ~ CH(T),
oc= |H: HG| ~ 03B2. Thus,
we haveproved
thatFrom
Proposition 3,
stated in theIntroduction,
we havefi
~ 203B1. To establish the reverseinequality,
we consider familiesof finite normal
subgroups N03BB
ofG, satisfying
the conditions(4)
For each À EA,
there exists an element x AENA
such that(H ~ N03BB)x03BB ~ H ~ N03BB.
It follows
easily
from these conditions that theN.
are all distinct. The collection of all such families may bepartially
orderedby
set-theoreticinclusion,
and Zorn’s Lemma allows us to choose a maximalfamily,
say
(2).
We shall show that
|039B| ~
oc. If this is not so, thenwriting
N =03A003BB~039B NA,
we
have |N|
a. SinceG ~ , IG: CI
oc, where C =C,(N).
Now ifH n C a
C,
then03B2(H
nC)
a, andhence, by (1) applied
to H nC,
weobtain H
n C :(H
nC)G|
a. SinceHG
n C aG,
we conclude thatIH
n C:HG
nCI
oc, and sinceJH:
H nCI
oc, thatJH: HG
nCI
oc.Finally,
weobtain H : HGI
oc, a contradiction.Therefore,
H n C is notnormal in C. Let y be an element of C and h an element of H n C such
that hy ~
H nC,
and let M be a finite normalsubgroup
of G containedin C and
containing h, y>.
Then(H
nM)y ~
H nM,
and M centralizes eachN.
since M ~ C =CG(N). Thus,
we mayadjoin
M to(1)
to obtaina
larger
suchfamily,
a contradiction.We have shown that
|039B| ~
a. It is nowconvenient, though
notstrictly
necessary, to introduce a
completion
G of G in any cofinitetopology
ofG
(see [3]).
ThusG
is a compacttopological
groupcontaining
G as anormal dense
subgroup,
and theautomorphisms
induced on Gby
G areprecisely
itslocally
innerautomorphisms.
Let M be any subset ofA,
and for each finite subset 0 ofA,
letG(03A6)
denote the set of all elements g EG
such thatThen
G(03A6)
contains the element03A003BB~M~03A6
XA and so is not empty; further-more it is an intersection of cosets of the closed
subgroups N a(N ¡
nH) (2
~03A6),
and so is a closed subsetofG.
The intersectionof any
finite number of setsG(03A6) clearly
contains another such set, and asG
is compact, there exists an element x = xm in their intersection. WriteHm
= HX. ThenHM
nNA
= H" nNÂ = (H
nN¡)X
isequal
to(H
nN¡)xÂ
or H nNA according
as 2 E Mor  0
M. Therefore thesubgroups Hm
are alldistinct.
By
Theorem 5.5 and Lemma 3.1 of[3], they
are alllocally conjugate
to H in G. Since|039B| ~
a, there are at least 2CX of them.Thus, 03B2 ~ 203B1,
and theproof
of Theorem l’ iscomplete.
4. An
example
We notice first that the
possibility
a =oc(H)
=03B2(H), fl(H)
= 2P(H)certainly
occurs forarbitrary
infinite oc. Forexample,
take G to be thedirect
product
of acopies
of thesymmetric
group ofdegree
3 and H tobe a
Sylow 2-subgroup
of G.The
example illustrating
the otherpossibilities
isessentially
a versionof the ’infinite
extra-special p-group’
first introducedby
P. Hall in[2]
for p = 2 and
presented
in a different way forarbitrary
pby
Tomkinson[8].
It will be convenient for us togive
a thirddescription
of these groups.Let
A, B,
C be three additive abelian groups, and suppose we aregiven
a bilinear map of A x B into C. In
anticipation
of what is to come, wedenote the map
by (a, b)
~[a, b].
We define abinary operation
on theset A x B x C
by
A routine verification shows that A x B x C becomes a group with
identity (o, 0, 0).
The setCo
of all elements(0, 0, c)
with c ~ C is a centralsubgroup
of G with abelian factor group. Thus G is
nilpotent
of class at most 2.We now assume that the bilinear map is
non-degenerate,
in the sensethat,
for aE A,
b EB,
Then it is easy to
verify
thatCo
isprecisely
the centre of G. We nowidentify A,
B and C withsubgroups
of G via the maps a -(a, 0, 0),
b -
(0, b, 0),
c ~(0, 0, c) respectively.
Then[a, b]
becomes the com-mutator of a and
b,
and G = ABC. We have[G, AG] ~ A
n C = 1 sinceG/C
isabelian,
and soAG ~
A nZ(G)
= A n C = 1.Arguing similarly
with
B,
we haveThe
subgroup
AC isabelian,
and soNG(A) ~
AC. Howeverand
by (2),
we obtain B nNG(A)
= 1. Since G =ABC,
we obtainNG(A)
= AC.Arguing similarly
with Bgives
In
general
G will not be anFC-group.
Since G = ABC and C iscentral,
G will be an
FC-group
if andonly
if each element of A ~ B has centralizer of finite index. In terms of the bilinear map, this says that(5) If a
EA, b
EB,
then there existsubgroups X,
Yof finite
index in Aand B
respectively,
such that[X, b] = [a, Y] =
0.This will
certainly
be satisfied if C isfinite,
since for fixed a EA,
the map b ~[a, b]
is ahomomorphism
of B intoC,
andsimilarly
with the roles of A and Binterchanged.
We now introduce
particular
choices forA, B,
C. Let be anarbitrary
infinite
cardinal,
let p be aprime,
and let A be anelementary
abelianp-group of rank 03BB. Let C =
Zp
be an additive group of order p, andB = A* = Hom
(A, Zp). Then,
if we write[a, b]
for theimage
of a E Aunder b E
B, (1), (2)
and(5) hold,
and so we obtain anFC-group
G asabove. Since G is a p-group it is
locally normal,
ofnilpotency
class 2 andwith centre of order p.
Now
|B|
=2’. Thus,
from(3)
and(4),
we obtainThus all
pairs (oc, fi)
allowedby Proposition
3 and the Generalized ContinuumHypothesis
can occur.We next calculate
03B2(A)
and03B2(B).
Now AC ~ G and)AC) =
03BB. Since AC contains every localconjugate
ofA, 03B2(A) ~ 2’.
But03B2(A)
=2À, by (6).
HenceIt is in fact not difficult to see that every
complement
to C in AC isconjugate
to A inG,
so that the stronger resultC1GA
=LclGA
holds(cf. [8]
p.210). However,
this does not concern us here.Now let A** = Hom
(A*, Zp),
and let L be the group obtainedby
theabove construction from the
triple (A**, A*, Zp).
Then A can be em-bedded in A** in the usual way via the map x
(cf. [3] § 3),
whichassigns
to an element a E A the element i ~
r(a) (TEA*)
ofA**,
and the map(a, b, c)
-(x(a), b, c) (a
EA,
b E B =A*,
c ~C) clearly
embeds G as anormal
subgroup
of L. SinceIA**l
=22À,
the considerations used above show that|ClLB|
=22À.
We claim that the elements of L induceby
con-jugation locally
innerautomorphisms
ofG;
from this it will follow thatso that all
pairs (fi, fl)
allowedby Proposition
3 and the Generalized ContinuumHypothesis
can occur.Let D be the
subgroup
of Lcorresponding
to A**. Since the set oflocally
innerautomorphisms
of G is a group and L =DG,
it will follow that L induceslocally
innerautomorphisms
on G if we can show that Ddoes so. Let
F1
andF2
be finitesubgroups
of A and Brespectively,
andF =
F 1 F2 C.
Then F is a finite normalsubgroup
ofG,
and every finite set of elements of G lies in such an F. Let d ~ D. Then d centralizesF 1 C,
and commutation with d induces a
homomorphism
ç ofF2
into C. Wehave to show that 9 can be induced
by
commutation with an elementa E A. Then a and d induce the same
automorphism
on F.Now in the bilinear map
notation, F2
is a finitesubgroup
of A*. LetE be its annihilator in A and let
E1
be the annihilator of E in A*. ThenE1
isnaturally isomorphic
to Hom(A/E, Zp),
andF2
is asubgroup
ofE1
which has zero annihilator inA/E. By
theduality theory
ofimite
abelian groups
(or
of finite dimensional vectorspaces), F2
=E1-,
andevery element of Hom
(F2, Zp)
isnaturally
inducedby
an element ofA/E,
thatis, by
somex(a)
with a E A. Inparticular,
ç is soinduced,
and ais the
required
element.We have thus shown that L induces
locally
innerautomorphisms
onG. In
fact,
it is not hard to see that L induces on G its full group oflocally
inner
automorphisms, though
we shall notrequire
that fact here.5. Proof of Theorem 2
Throughout
thissection,
H denotes asubgroup
of alocally
normalgroup
G,
a =IHIHGI
isinfinite, and g
=|LclGH|.
We have to show thatfl
= 203B1. Webegin by stating
thefollowing
immediate consequence of Lemma 2.4:LEMMA 5.1 : Let B be a
subgroup of a locally
normal groupA,
X ~A,
The next lemma establishes a
special
case of Theorem2,
and in factprovides
thekey
to thegeneral
situation.LEMMA 5.2:
If [H, G, G]
=1, then e
= 203B1.PROOF : Let L be the group of
locally
innerautomorphisms
of G andA the group of inner
automorphisms
of G. We may take the set of central- izersCL(F),
whereF ranges
over the finite normalsubgroups
ofG,
as a basis ofneighborhoods
of 1inducing
on L the structure of a Hausdorfftopological
group. Then L is a cofinite group in the sense of[3],
andby [3], Proposition 5.6,
L iscompact
and A is dense in it. In otherwords,
L is a
completion
of A in thetopology
which the latter inherits from L.Now in
proving
Lemma5.2,
we may assumeby
Lemma 5.1(iii),
thatHG
= 1.Writing
Z =Z(G),
we then have H n Z =1,
and soNow every inner
automorphism
of G is trivial onHZ/Z
andZ,
and hence so is everylocally
innerautomorphism.
Thus[H, L] ~ Z, [Z, L]
= 1.
Therefore,
forb E L,
the map03C8(b):h ~ [h, b]
=h - 1 hbis
a homo-morphism
of H intoZ, and 1/1
is ahomomorphism
of L into the additive group H* = Hom(H, Z), consisting
of all abstract group homomor-phisms
of H into Z.Let U denote the
image
of1/1,
and for each finitesubgroup
E ofH,
letIf F is a finite normal
subgroup
of Gcontaining E,
then F isL-invariant,
L/CL(F)
isfinite,
and03C8(CL(F) ~ E~.
Therefore El has finite index in U.The
subgroups E~
form aseparating
filter base ofsubgroups
of finiteindex in
U,
and induce on it a cofinitetopology
in the sense of[3].
Thefact that
03A8(CL(F)) ~
E~ as above showsthat 03C8
is continuous with respectto the
topologies
of L and U. Since L is compact and U isHausdorff, 03C8
is a closed map, and so maps the closure L of A onto the closure T of T =03C8(A).
Thus U = T We haveFurthermore,
T is compact,being
the continuousimage
of the compact space L. Hence T is acompletion
of the cofinite group T(cf. [3]),
in thetopology
induced on Tby
TLet
b E L, h
E H.Then hb
=h[h, b]
= h ·(03C8(b))(h),
and it followseasily that,
forb,
b’ EL, Hb
= Hb’ if andonly
if03C8(b) = 03C8(b’). Thus,
from(2),
We now need to assemble
enough
information about T to be able to read off the cardinal of itscompletion
T from[3] Corollary
3.6.Now
by assumption, HG
= 1.Therefore,
if 1 ~ h EH,
there exists anelement a E A such
that h 0
Ha. Since ha = h ·(03C8(a))(h),
this means that(03C8(a))(h) ~
1. Thereforein the sense that if 1 ~ h
E H,
then there exists an element T E T such that03C4(h) ~
1.We also have :
For let a be an element of A such that
03C8(03B1)
= i, and let a be an element of Ginducing
the innerautomorphism
a.Then,
if hE H, (03C8(03B1))(h) = [h, a] = [h, a].
But[H, a]
iscertainly
finite as G islocally
normal.We need to recall that
Now since T is a
completion
of the cofinite groupT,
we obtain from[3] Corollary
3.6that 1 TI
=20",
where a is the cardinal of the setof open
subgroups
of T. In view of(3),
theproof
of the lemma may becompleted
by showing
that 6 = a. We now deduce this from(4), (5)
and(6).
Since each open
subgroup
of T has finiteindex,
the number of opensubgroups
of T isequal
to the number ofsubgroups
in any basis ofmeighborhoods
of 0 in T. We therefore calculate the number ofsubgroups
ET = {03C4 ~ T : 03C4(E)
=01 =
T nEb,
as E ranges over the finite sub- groups of H ; these form a basis ofneighborhoods
of 0 in T.For each h ~
H,
letx(h)
be the element i- i(h) (i E T)
of T * = Hom( T, Z).
Then x is a
homomorphism
of H into the additive group T*. If 1 ~ h EH,
thenby (4),
there exists T e T such that03C4(h) ~
1. Thus(~(h))(03C4) ~ 1,
and~(h) ~
0.Hence x
isinjective,
and allows us toidentify
H with asubgroup
of T*. Then
for each finite
E ~
H.For each finite
E ~ H,
defineH
being
identified with asubgroup
of T * as above.We claim that
X(E)
is finite. Infact, E j
is asubgroup
of finite index ofT,
and so T =E j
+D,
for some finitesubgroup
D of T.Applying (5)
to theelements of D in turn, we find that there exists a finite
subgroup Di
of Zsuch that
~(D) ~ D 1
for all ç~ H ~
T*. Since any element ofX(E)
isclearly
determinedby
its effect onD,
restriction to Dgives
aninjective
map of
X(E)
into the finite group Hom(D, Dl). Thus, X(E)
is indeedfinite.
Now from
(7), E j
is the intersection of the kernels of the members ofX(E),
and so it followsthat,
for finitesubgroups El
andE2 of H, (El)T
=(E~2)T
if andonly
ifX(E1)
=X(E2). Further, E ~ X(E),
and soU X(E)
=
H,
as E ranges over all finitesubgroups
of H. Since|H|
= a and eachX(E)
isfinite,
the number of setsX(E)
is a and hence the number of sub- groupsET
is a, since these are inbijective correspondence
with theX(E).
The
proof
of Lemma 5.2 is nowcomplete.
The next three lemmas
provide
further steps towards thegeneral
caseof Theorem 2. The first is
perhaps
somewhatunexpected.
LEMMA 5.3:
If |G: HI
a,then e
= 203B1.PROOF :
By Proposition 3,
all we need to show is03B2 ~
203B1. We may assume,by
Lemma 5.1(iii),
thatHG
= 1. Let Z be the centre of G. ThenH n Z =
1,
and soIHZIZI = IHI
= a. LetU/Z
=(HZ/Z)G/z’
IfIHZ/UI
= a,
then,
asG/Z
isresidually finite,
Theorem 1 and Lemma 5.1(iii) give
03B2 ~
2". We may therefore assume that1HZ: Ul
a.Then |U/Z|
= a,IH
nUl
= a and(H
nU)G
=1;
also|LclGH| ~ |LclG(H
nU)| by
Lemma5.1
(ii),
andclearly IG:
H~ U|
a.Hence, considering H
n U insteadof
H,
we may assume thatLet B be a normal
subgroup
of G of cardinal y a such that G =HB,
and let C =CG(B). By
Lemma2.2,
there exists a normalsubgroup
D ofG such that
[D, G] C ~
Dand 1 G: D| ~
y a. Let K = H n D. ThenIKI
= a andKG
= 1.Furthermore,
from(8) [K, G] ~
C n HZ =(H
nC)Z
asZ ~
C. But H n C =CH(B) ~ HG
= 1. Hence[K, G] ~
Z.Lemma 5.2 now
gives |LclGK|
=2eB
and Lemma 5.1(ii) gives 03B2 = ILclaHI
~
2’,
asrequired.
LEMMA 5.4: Let X be a normal
subgroup of index
ocofG,
let K = H nX,
and supposeIK/Kxl
oc.Then ff
= 2a.PROOF : Let L =
KX.
Consider the chain ofsubgroups H ~ HGK ~ HGL ~ HG.
We have|H: HGK| ~ |H: K| ~ IG: XI
a, andIHGK: HaLl
~ |K : LI
oc. It follows that|L:
L nHal = |HGL: HG| = 03B1,
and sinceLG ~ HG n L,
that|L: LGI
=03B3 ~
03B1.There exists a normal
subgroup
B ofG,
of cardinal a, such that G = XB. Let E = LB. Then as L aX, LG
=~b~B Lb = LE.
ThusIL/LEI
= y, whilelE: LI
y.By
Lemma5.3, ILcl E LI
= 2y.Hence, by
Lemma 5.1(i), |LclGL| ~ 203B3 ~
22.But if ç is a
locally
innerautomorphism
ofG,
then ç leaves Xinvariant,
and so LI’ =
(Hep
nX)x.
Therefore distinct localconjugates
of L lie indistinct local
conjugates
ofH, and fl
=|LclGH| ~
2’. The reverse inclu-sion
resulting
fromProposition 3,
the lemma isproved.
LEMMA 5.5:
Suppose
that H is a p-group and that everyp’-element of
G normalizes H. Then
ff
= 2a.PROOF : Let
Q
be thesubgroup
of Ggenerated by
thep’-elements
ofG,
and let P be anySylow p-subgroup
of Gcontaining
H. ThenQ a
G andG =
QP. Thus, H.
=Hp,
and since(by Proposition 3)
all we have toestablish is
that 03B2 ~ 2a,
Lemma 5.1(i)
allows us toreplace
Gby
P andassume that G is a p-group.
Let Z be the centre of G.
Then,
as in the first part of theproof of
Lemma5.3,
we may assume thatHG
=1,
so that H n Z =1,
and that HZ « G.Now in the case a =