R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
M. J. T OMKINSON
FC-nilpotent groups and a Frattini-like subgroup
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FC-Nilpotent Groups and
aFrattini-like Subgroup.
M. J.
TOMKINSON(*)
1. Introduction.
In
[6],
we introduced asubgroup IA(G)
which was a variation on theFrattini
subgroup
and we showedthat,
for various classes of infinite groups,g(G)
had many of theproperties
of the Frattinisubgroup.
Forexample,
we showed that satisfiednilpotency
conditions and that if satisfies anilpotency
condition then so does G.Supersolubili- ty
conditions and finiteness conditions on were considered insubsequent
papers[7, 8].
In all of these earlier papers the main results were obtained within the class of
nilpotent-by-finite
groups and the class ofSo-groups.
Our aim here is to extend the
applicability
of our resultsby
consid-ering
the class ofsoluble-by-finite FC-nilpotent
groups. This is a very considerablegeneralization
of the class ofnilpotent-by-finite
groupsand
requires
methods closer to those used forSo-groups.
We recall the definitions of
fJ.(G).
Let U be asubgroup
of the group G and consider theproperly ascending
chainsfrom U to G. We define
m( U)
to be the least upper bound of thetypes
aof all such chains. Thus
m( U)
is a measure of how far thesubgroup
U isfrom G. A proper
subgroup
M of G is said to be amajor subgroup
ifm(U) = m(M)
whenever M ~ U G. Then we defineg(G)
to be the in-tersection of all
major subgroups
of G. If G isfinitely generated
thenthe
major subgroups
arejust
the maximalsubgroups
of G and sog(G)
coincides with the Frattini
subgroup p(G).
One basic result
[6,
Lemma2.3]
enables us to use the Frattiniargument.
(*) Indirizzo dell’A.:
Depart.
of Math., Univ. Gardens,Glasgow
G128QW,
Gran
Bretagna.
PROPOSITION 1.1.
Every
propersubgroup of
a group G is con-tained in a
rrzajor subgroup of
G.When we considered
nilpotent-by-finite
groups in[6,7],
we firstconsidered
abelian-by-finite
groups and reduced to the case in which the groups are alsoperiodic
and so have agood Sylow theory.
This en-abled us to use the Frattini
argument
in the usual way.The FC-centre of a group
G,
denotedFC(G),
is thesubgroup
con-sisting
of all elementshaving only finitely
manyconjugates
in G. Agroup G is
FC-nilpotent
if it has a finite normal seriessuch that each factor is FC-central in
G;
thatis,
5
FC(G/Gi _ 1 ).
It is clear that everynilpotent-by-finite
group is FC-nilpotent
and is alsosoluble-by-finite.
However there are many con- structions of solubleFC-nilpotent
groups which are notnilpotent-by-fi- nite ;
forexample,
solubleFC-groups
and also directproducts
of(nilpo-
tent of class
c)-by-(finite abelian)
groups.For these groups we will not
normally
be able to make use ofSylow theory
and instead we aim atobtaining splitting
theorems in abelian-by-hypercentral
orabelian-by-hypercyclic
groups. Thepoint
isthat, by Proposition 1.1,
acomplement
must be contained in amajor subgroup
and so, in
particular, p.(G)
can not have acomplement.
Thesplitting
theorems that we obtain are rather weak
requiring
verystrong
condi-tions on the groups considered. These appear as Lemma 2.3 and Lem-
ma 3.1. Our main results are Theorems 2.4 and 3.2 and can be summa- rized as.
THEOREM. Let G be a
soluble-by-finite FC-nilpotent
group and let NG.(a) If N /N n p.(G) is
ahypercentral
group, then so is N.(b) If
is ahypercyclic
group, then so is N.Putting
N =g(G) gives
theCOROLLARY.
If
G is asoluble-by-finite FC-nilpotent
groups thenIA(G)
is ahypercentral
group.2.
Hypercentral
groups.We
begin
with adecomposition
result forlocally
finite modules. Thesignificance
of this result is that if A is a normalperiodic
abelian sub-203
group contained in the FC-centre of a group G then each finite set cf elements of A is contained in a finite normal
subgroup
ofG;
thus A canbe considered as a
locally
finiteZ(G/A)-module.
LEMMA 2.1. Let G be a
locally nilpotent (locally supersoluble)
group and let A be a
locally finite
ZG-module. Then A has adecomposition
where every irreducible
ZG-factor of A +
is trivial(cyclic)
and no irre-ducible
ZG-factor of A +
is trivial(cyclic).
Thisdecomposition
isunique.
PROOF. Let
Fi ,
i EI,
be the finite submodules of A so that A = Since is a finitenilpotent (supersoluble)
group, theiEI
module
Fi
has aunique decomposition Fi
=Fi+
61Fi-,
where the irre- ducible ZG-factors ofFi+
are trivial(cyclic)
and no irreducible ZG-fac- tor ofFi-
is trivial(cyclic) [2].
It is clear that if thenFi+ 5 F~+
and
Fi~ 5 F, .
Thus the submodules AE Fi+
andA - - ~ Fi-
havethe
required properties.
Recall that if A is a normal
subgroup
ofE,
then E is said tosplit conjugately
over A if A has acomplement
K in E(that is,
A.K = E andA rl K =
1)
and any twocomplements
to A in E areconjugate
in E.B efore
proving
oursplitting
theorem forabelian-by-hypercentral
groups it is useful to record the
following
useful result which appearsas Lemma 3 in
[5].
PROPOSITION 2.2. Let E be a group
containing subgroups
A ~ Hsuch that A is a normal abelian
subgroup
and H is an ascendant sub-group.
If CA (H)
= 1 and Hsplits conjugately
overA,
then each com-plement
to A in H is contained in aunique complement
to A inE,
andE
splits conjugately
over A.We have
only
been able to obtain asplitting
theorem for extensions of alocally
finite ZG-module Aby
ahypercentral
group G under verystrong restrictions,
but thefollowing
result will be sufficient for ourmain purpose.
°
LEMMA 2.3. Let G be a
hypercentral
group and A alocally finite
ZG-module which is also a p-group.
Suppose
that G has a normal sub- group Hoffinite
indexprime
to p such that each irreducibleZG-factor of
A is H-trivial but not G-trivial.Then any extension E
of
Aby
Gsplits conjugately
over A.PROOF. We prove the result
by
induction onI GIH I
and suppose first thatG/H
iscyclic
ofprime order q #
p.In the extension E of A
by G,
let M be the extension of Aby
H.Then E =
M(x)and x
e E can be chosen either to have infinite order orto have order
qn.
If V is some irreducible ZG-factor of A then
[V, G]
= V but= 0. Since E =
M(x),
it follows thatSince A is a
locally
finitemodule,
we see that[A, x]
= A andCA (x)
= 0.Let L =
A(r) 5 E;
since x either has infinite order or has orderq n
p, it is clear that
~x~
is acomplement
to A in L. Let(y)
be asecond
complement
to A inL;
then(y) = (xa),
for some a E A. But there is an element b E A such thata -1
=[b, x]
and so xac =b -1 xb.
Thus(y)
= Lsplits conjugately
over A.Since
E/A
ishypercentral,
L is ascendant in E and so,by Proposi-
tion
2.2,
Esplits conjugately
over A.Now suppose that
E/M
=G/H
is not ofprime
order and choose asubgroup X/M
ofprime
order in the centre ofE/M.
SinceX/A
is ahy- percentral
group, the ZX-module A has adecomposition
A =AX 61 Ax
such that every irreducible ZX-factor of
AX
is X-trivial and no irre- ducible ZX-factor ofAX
is ZX-trivial. SinceAX
andAi
areZG-submodules of A. _
First consider X =
X/AX .
Each irreducible ZX-factor of A isnontrivial,
but each irreducible ZM-factor of A is M-trivial. If V is anirreducible ZX-factor of
A,
then0;
butCv (M)
is a ZX-sub-module and so is
equal
to V. Therefore each ireducible ZX-factor ofA
is M-trivial. SinceI
it followsby
induction that Xsplits conjugately
over A. and = 0 so,by Proposition 2.2,
Esplits conjugately
over A. Thatis,
E contains asubgroup K
such thatAK = E and A n K =
AX
and all suchsubgroups
areconjugate.
Now K is an extension of
AX by
G. Each irreducible ZK-factor ofAX
is nontrivial. Each irreducible ZX-factor ofAX
is X-trivial and hence each irreducible ZK-factor ofAy
is(K f1 X)-trivial.
SinceI
it followsby
induction thatKsplits conjugately
over
AX
and it now followseasily
that Esplits conjugately
over A.THEOREM 2.4. Let G be a
soluble-by-finite FC-nilpotent
group and let N G. Then N is a huvercentral orouv if and onlu ifNlN f1
205
PROOF. The
hypotheses
on G ensure that it has a finite normal seriessuch that
G/Gn
is finiteand,
for i =1,
..., n,Gi /Gi _ 1 is
an abelian nor-mal
subgroup
contained inWe suppose that
N/N
f1 is ahypercentral
group and prove that N ishypercentral.
Since N fl f1 N nli(G)
isfinite,
the usualFrattini
argument
shows that it isnilpotent.
Thus there is a finite series of normalsubgroups
of Gsuch that is abelian and is contained in
By
induc-tion on the
length
of this series we may assume that G has an abelian normalsubgroup
A ~ N flp.(G)
such thatN/A
ishypercentral
andA 5
FC(G).
Note first that it is sufficient to prove that each G-chief factor of A is central in N.
For,
since A 5FC(G),
A has anascending
series of G-admissible
subgroups
each factor of which isfinitely generated.
Thiscan be refined so that the factors are either finite or torsion-free.
Any
finite factor is
hypercentrally
embedded in N. Also a free abelian G-ad- missiblesubgroup
F of rank n is contained in the n-th term of the upper central series ofN,
since for eachprime
p, and so[F,N, ... , N] 5
nFp=1.Suppose
then that there is a G-chief factorU/V
with U 5 A andU/V
not central in N. ThenU/V
is anelementary
abelian p-group forsome
prime
p. Let M be maximal withrespect
to MG,
M f1 U = V.Then
MU/M
is theunique
minimal normalsubgroup
ofG/M.
We re-place
Gby GIM, U by UM/M
and Aby AM/M ~
NM ng(G) M/M %
~
nM/M
n Thus U is contained in every nontrivial normal sub- group of G. We claim that A is a p-group. It is clear that the torsionsubgroup
of A is a p-group and if A is not torsion then it contains an el- ement x of infinite order. Since AFC(G),
the normal closure F =x>G
is an infinite
finitely generate
abelian group. There is aninteger k
suchthat
F k
is torsion-free. Thus U = 1 andF~
a Gcontrary
to ourcondition on U. Hence A is a p-group, as claimed.
In
particular,
A is alocally
finite ZG-module and also a ZN-module.By
Lemma2.1,
A has adecomposition
A =A’ 61 AN
whereA’
andAN
are
ZN-submodules,
each irreducible ZN-factor ofA’
is N-trivial andno irreducible ZN-factor of
AN
is N-trivial. Since NG, AN
andAN
are ZG-submodules
and,
since U is contained in every nonzero ZG-sub- moduleof A,one
of them must be zero. But U is not central in N and so= 0 and hence
U 1 AN .
Therefore A =AN
and no irreducible ZN-factor of A is N-trivial.Now let C =
CN (U);
sinceOp (G/CG (U))
is trivial it follows thatN/C
is a fmite
nilpotent p’-group.
Let F be any nontrivial finite normal sub- group of G contained inA;
thenN/CN (F)
is finite andnilpotent.
IfC/CN (F)
is not a p-group, then there is a finite normalp’-subgroup Q/CN(F)
ofG/CN(F)
such thatBy Fitting’s Lemma [3,
Theorem5.2.3], F = [F, Q] x CF (Q).
SinceQ 4 G,
both[F, Q]
andCF (Q)
are normal in Gand, by
the condition onU,
one of them must be theidentity subgroup.
ButQ ~
C =CN(U),
soU ~
CF (Q).
Hence[F, Q]
= 1 andQ
=CN (F).
ThusC/CN (F)
is a finitep-group.
It follows that F is
hypercentrally
embedded in C and so every N- chief factor of F is centralizedby
C. This is true for all F and so every irreducible ZN-factor of the module A is C-trivial. Thus A satisfies the conditions of Lemma 2.3(for
G =N/A
and H =C/A)
and so Nsplits conjugately
over A.Now N a G and since every irreducible ZN-factor of A is nontriv-
ial, Proposition
2.2 shows that Gsplits conjugately
over A. But this iscontrary
to A ;g(G)
and theproof
iscomplete.
3.
Hypercyclic
groups.We
begin
with asplitting
theorem for extensions of alocally
finiteZG-module A
by
ahypercyclic
group G which willplay a
similar role to Lemma 2.3 for thehypercentral
case. We need evenstronger
restric-tions for this result but
again
these conditions arise in a natural way in theproof
of our main theorem.Recall that a ZH-module V is
H-polycyclic
if it has a finite series ofZH-submodules such that each factor is a
cyclic
group.LEMMA 3.1. Let G be a
hypercyclic
group and A alocally finite
ZG-module which is also a p-group.
Suppose
that G has a normal sub-group H
of finite
indexprime
to pcontaining
G’ and suppose that each irreducibleZG-factor of
A is G’-trivial andH-polycyclic
but notcyclic.
Then any extension E
of
Aby
Gsplits conjugately
over APROOF. We prove the result
by
induction onI
and suppose first thatG/H
iscyclic
ofprime
p.207
In the extension E of A
by G,
let M be the extension of Aby H
sothat
= so that
M/M*
andE/E*
are thelargest
factor groups of ex-ponent dividing p -
1 ofM/E’A
andE/E’A, respectively.
If p = 2 wetake
M * = E ’ AM 2
andE * = E ’ AE 2.
In both casesE * ~ E ’ AE 2.
Since each irreducible ZG-factor of A is
H-polycyclic
and G’-trivial it isM*-trivial,
since it is notcyclic
it is notE*-trivial;
hence E * _> M *. In E =EgM *
there is an element x of q-power order such that E =M(5).
Thus has
exponent dividing p - 1 (or 2,
if p =2)
and so(x) ; E
*.Thus
E*/M*
is nontrivialcyclic
of q-power order. There is an elementx E E* such that
E * = M * ~x~
and x either has infinite order or has or-der a power of q.
If V is an irreducible ZG-factor of
A,
thenand it follows that A =
[A, x]
andCA (x)
= 0. It is clear that(x)
is acomplement
to A inA(x).
If(y)
is a secondcomplement
then(y) =
=
(xa)
for some a e A. There is a b e A such that =[b, x]
and so xa =and
(y) = b -1 (x) b.
ThereforeA(x) splits conjugately
overA.
By taking
anascending
normalcyclic
series forE/A
and intersect-ing
withE’A/A
we obtain anascending
serieswith
E« + 1 /E«
either infinitecyclic
orcyclic
ofprime
order and% AE ~ . We prove
by
induction on « that(x)
is containedin a
unique complement I~«
to A inE« ~x~
and thatE« ~x~ splits conju- gately
over A.Case
(i):
a - 1 exists.Then
E« _ 1 ~x~
has aunique complement containing ~x~.
Ifis infinite
cyclic
orcyclic
of order p, then iscyclic
of orderdividing
2 orp -1
and so E * ~CE (E« /E« _ 1 ).
Sincex E E*, E« _ 1 ~x~ a E« ~x~.
AlsoCA (K«) ~ CA (x)
= 1 and so,by Proposi-
tion
2.2,
there is aunique complement K«
to A inE« ~x) containing
and
E« (x) splits conjugately
over A.If
1 has prime
order r # p, then(r) [
= r and itfollows from a result of Gaschftz
[4, Hauptsatz 1.17.4]
thatE« ~x~ splits
conjugately
iver A and there is aunique complement
to A inE« (x)
con-taining ~x~.
Case
(ii):
a a limit ordinal.For
each p
«, letl~a
be theunique complement
to A in whichcontains x. then
and = U =
1,
so thatKa
is acomplement
to A inE« (x)
and
clearly Ka
contains x. IfLa
is a second suchcomplement
thenLa
nis a
complement
to A in and so = ThereforeLa
= U(La
= UKp
=Ka ,
asrequired.
This
completes
the inductivestep
and we haveshown,
inparticular,
that
E’A(x) splits conjugately
over A. But andCA (x)
= 1so,
by Proposition 2.2,
Esplits conjugately
over A.Now suppose that
E/M
=G/H
is not ofprime
order and choose asubgroup X/M
ofprime
order. SinceX/A
ishypercyclic,
A has a de-composition
A =AI 61 Ai
such that every irreducible ZX-factor ofA
is
cyclic
but no irreducible ZX-factorof Ai
iscyclic.
SinceX a G, A and AX-
are ZG-submodules of A._
First consider X =
X/AX .
Each irreducible ZX-factor of A= A/Af
is
noncyclic
but since the irreducible ZG-factors are G’-trivial so arethe irreducible ZX-factors. Since
X ’ ; E ’ A,
the irreducible ZX-fac- tors of A are X’-trivial.Also,
since every irreducible ZH-factor of A iscyclic,
the irreducible ZX-factors of A areM-polycyclic.
SinceI
it follows that Xsplits conjugately
over A. Since Ahas no X-trivial
factors, CA (X )
= 1 and follows fromProposition
2.2 that Esplits conjugately
over A. Let K = be acomplement
so that AK = E and A fl K =AX
and K is an extension ofAX by
G.Each irreducible ZX-factor of
AX
iscyclic
and so the irreducible ZK-factorsof Af
are(K n X)-polycyclic.
SinceI I GIH I
itfollows that K
splits conjugately
overAX
and it iseasily
deduced thatE
splits conjugately
over A.THEOREM 3.2. Let G be ac
soluble-by-finite FC-nilpotent
group and let N G. Then N ishypercyclic if
acndonly if
ishypercyclic.
PROOF. As in the
proof
of Theorem2.4,
we may assume that G hasan abelian normal
subgroup
A such that A 5N/A
ishyper- cyclic
and A ~FC(G).
We show that N ishypercyclic.
Again
we note that it is sufficient to consider the G-chief factors ofA,
this timeshowing
thatthey
areN-polycyclic. For,
A has an ascend-ing
series of G-admissiblesubgroups
in which the factors are finite or209
free abelian of finite rank. Each finite factor can be
replaced by
a finiteseries of N-admissible
subgroups
in which the factors arecyclic.
If F isfree abelian of finite rank with then
G/CG (F)
is finite and weconsider the
split
extension H of Fby
G =G/CG (F).
For anypositive integer
n,H/Fn
is finite and each H-chief factor ofF/F"
isS-poly- cyclic,
where S K H is the extension of Fby
N =NCG (F).
ThusS/Fn
issupersoluble.
Hence all finitehomomorphic images
of thefinitely generated abelian-by-finite
group S aresupersoluble.
It followsfrom
[1]
that ,S issupersoluble
and so F has a finite series of N-admissi- blesubgroups
withcyclic
factors. Hence A has anascending
series ofN-admissible
subgroups
withcyclic
factors and so N ishypercyclic,
asrequired.
Suppose
then that there is a G-chief factorU/V
with U ~ A andU/V
notN-polycyclic.
ThenU/V
is anelementary
abelian p-group forsome
prime
p. As in Theorem2.4,
we let M be maximal withrespect
toMG,MQ!7=V
and considerG/M.
Then we may assume that U is contained in every nontrivial normalsubgroup
of G and A is ap-group.
In
particular
A is alocally
finite ZG-module and also a ZN-module.Since
N/A
ishypercyclic,
A has adecomposition
A =A; 61 AN
whereAN
areZN-submodules,
each irreducible ZN-factorof A;
iscyclic
and no irreducible ZN-factor of
AN
iscyclic.
Since N aG, AN
andAN
are ZG-submodules
and,
since U is contained in every ZG-submodule ofA,
one of them must be zero. But U is notN-polycyclic
and soU ~
Therefore A =AN
and no irreducible ZN-factor of A iscyclic.
The
subgroup [ U, N’ ]
is normal in G and so is either 1 or U. Consid-er first the case in which
[ U, N’ ]
= U. SinceN’A/A
is ahypercentral
group, when considered as a
ZN’-module,
A has adecomposition
A == A +
61 A -,
where every irreducible ZN’ -factor of A + is N’-trivial andno irreducible ZN’-factor of A + is N’ -trivial. Since N’ a
G,
A + and A -are ZG-submodules and one of them must be zero. Since
[ U, N’ ]
=U,
there are nontrivial irreducible ZN’-factors in U and so A- # 0. There- fore A =
A - ;
thatis,
no irreducible ZN’-factor of A is N’-trivial.Let H =
CAN, (U); then,
as aZ(H/A)-module,
A has adecomposition
A =
AJ 61 A - ,
where every irreducible ZH-factor ofA9
is H-trivialand no irreducible ZH-factor of
An
is H-trivial.Again
since H aG, A9
and
An
are ZG-submodulesand,
since U ~A +,
it follows that A == A9 .
Thatis,
every irreducible ZH-factor of A is H-trivial. If V is someirreducible
Z(AN’)-factor
ofA,
thenCv (H ) ~
0 and soCv(H)
= V. Thatis,
V is H-trivial. Thus theZ(AN’ /A)-module
A satisfies the conditions of Lemma 2.3 and so AN’splits conjugately
over A. Since G andCA (N’)
=1,
Gsplits conjugately
overA, contrary
to A ~Therefore we may assume that
[ U, N’ ]
= 1.By considering
the de-composition
of A as aZN’-module,
A = A +EÐ A -
in which each irre- ducible ZN’-factor of A + is N’ -trivial and no irreducible ZN’-factor of A - is N’-trivial we see that U ~ A + and so A = A +. Thatis,
every ir- reducible ZN’-factor of A is N’-trivial.Let C = then C > N’ and so
N/C
is a finite abelian group.Also
Or (GICG (U))
is trivial and soOp (N/C)
is trivial. ThusN/C
is a fi-nite abelian
p’-group.
Consider A as a ZC-module. SinceC/A
ishyper- cyclic,
A has adecomposition
A =A~ 61 Ac ,
where every irreducible ZC-factor ofAC
iscyclic
and no irreducible ZC-factor ofAc
iscyclic.
Again
C is normal in G and soA~
andAc
are ZG-submodules of A.Since U ~ it follows that A =
A~
and every irreducible ZC-factor of A iscyclic.
Hence every irreducible ZN-factor of A isC-polycyclic.
Also if V is an irreducible ZN-factor of A then 0 and so V is N’-trivial. Therefore the A satisfies the conditions of Lemma 3.1 and so N
splits conjugately
over A. Since N G andwe see that G
splits conjugately
overA, contrary
toA 5
tA(G).
This
completes
theproof.
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Abh. Math. Sem. Univ.Hamburg,
23 (1957), pp. 11-28.[2] R. BAER, Durch Formationen bestimmte
Zerlegungen
von Normalteilern endlicherGruppen,
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Manoscritto pervenuto in redazione l’l ottobre 1991.