R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
P ATRIZIA L ONGOBARDI M ERCEDE M AJ
A VINOAM M ANN A KBAR R HEMTULLA
Groups with many nilpotent subgroups
Rendiconti del Seminario Matematico della Università di Padova, tome 95 (1996), p. 143-152
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Groups with Many Nilpotent Subgroups.
PATRIZIA
LONGOBARDI (*) -
MERCEDEMAJ (*)
AVINOAM
MANN (**) (1) -
AKBARRHEMTULLA(***)
The
following
result wasproved by
B. H. Neumann[N]
and alsoby Faber,
Laver and McKenzie[FLM], answering
aquestion
of P. Erd6s:A group G has the
property.,
that eachinfinite
subsetof
G con-tains a
pair of commuting elements, if
andonly if
the centreZ(G) of
Ghas
finite
index.This
prompted
J. C. Lennox and J.Wiegold [Lw]
to consider groups in which every infinite subset contains apair
of elementsgenerating
anX-subgroup,
where X is somegiven property
of groups, e.g.nilpotence.
Observing
that the characterisation of such groups ispossibly
veryhard, they
restrict themselves tofinitely generated
soluble groups, and among thesethey
characterise the groups with the aboveproperty
when X is the class of
polycyclic,
ornilpotent,
orcoherent,
groups(su- persoluble
groups were handled laterby
J. R. J. Groves[G]).
A more
complex
variation on the theorem of Neumann and McKen-zie, involving
several infinitesubsets,
is discussedin [LMR],
whilein
[CLMR]
the authorsimpose
restrictions on finite subsets with morethan 2 elements. The
present
paper is in a similarspirit.
For conve-nience,
we say «of class k» when we mean «of class at most k». Our basichypothesis
on a group G is:(*) Indirizzo
degli
AA.:Dipartimento
di Matematica eApplicazioni
«R. Ca-cioppoli»,
Universitadegli
Studi diNapoli,
Via Cintia,Napoli
80126,Italy.
(**) Indirizzo dell’A.: Einstein Institute of Mathematics, Hebrew Universi-’
ty, Givat Ram, Jerusalem 91904, Israel.
(***) Indirizzo dell’A.:
Department
of Mathematics,University
of Alberta, Edmonton, Canada T6G 2G1.(1)
This author’s research was done in partduring
his visits to theUniversity
of Manitoba and to the Universita
degli
Studi diNapoli
«Federico II», the latter under a CNR grant.(Nk) Any infinite
subsetof
G contains a setof
k + 1elements,
whichgenerates
anilpotent subgroup of
class k.Or the
following
moregeneral assumption:
(N) Any infinite
subsetof
G contains afinite
subsetX,
such that(X)
is
nilpotent,
and class((X))
We refer to such groups as
(N)
or(Nk)
groups. If G is an(Nk)-group,
and X is an infinite subset of
G,
thenRamsey’s
Theorem shows that X contains an infinite subset Y such that all(k
+I)-tuples
of elements of Ygenerate
anilpotent subgroup
of class k. Then(Y)
itself isnilpotent
of class k. This
implies
that(Nk)-groups
are also More-over, let us denote
by (Nt)
theassumption:
anyinfinite
subset containsr elements
generating
anilpotent subgroup of
class k. Then is the weakest among thesehypotheses,
while ourassumption (Nk) =
is the
strongest, implying
notonly
allpropertis (Nt),
but also all prop- erties(Nr)
with 1 > k. Thisimposition
of aprecise quantitative
assump- tion makes itpossible
to avoid furtherassumptions,
such assolubility.
Indeed,
once we derive thekey
fact that(N)-groups
have a non-trivialFC-centre,
we are able to show thatthey
arehyperabelian by finite,
and thus avail ourselves of the
techniques
and resultsof[LWI.
Thisyields
asatisfactory
answer forfinitely generated
groups.THEOREM A. For a
finitely generated
group G thefollowing
areequivalent:
a)
G is an(Nk)-group.
b)
G is an extensionof
afinite
groupby
anilpotent
groupof
class k.
c) G : Zk (G) ~ I
isfinite.
Moreover,
G is an(N)-group if
andonly if
it is an(Nk)-group, for
some k..
Part
b) implies
that in an(N)-group
G the elements of finite ordergenerate
alocally
finitesubgroup T,
and thatG/T
is torsionfree. For these factors of G we haveTHEOREM B.
a)
A torsionfree (N)-group
ishypercentral, b)
A torsionfree (Nk)-group
isnilpotent of
class k.c)
A torsion(N)-group
ishypercentral by finite.
145
Along
the way we derive thefollowing
characterisation for FC- groups :THEOREM C. An
FC-group satisfies (Nk), for
somek, if
andonly if
it contains a
subgroup
Hof fznite
index suchthat, for
some n, any two elementsof
Hgenerate
anilpotent subgroup of
class n.If
we choose 1and n as minimal
possible,
thenk -
nI
~ 1.Moreover,
in an FC-group which is an
(N)-group
the termof
the upper central series hasfznite
index.The authors are
grateful
to the referee for his manyhelpful
remarks.
The
following
result will be used time andagain:
NEUMANN’S COVERING LEMMA.
If
a group G is the unionof finitely
many
subgroups,
then someof
thesesubgroups
havefinite indices,
and G isalready
the unionof
thesefinite
indexsubgroups.
For the
proof,
see e.g.[T2, 2.2].
We start our
proofs by noting
a very useful fact.PROPOSITION 1. Let G
satisfy (N).
Then the FC-centreof
G is nottrivial
PROOF. We consider subsets A of G such that if B is a subset of
A,
andI B I
=k,
then(B)
is not anilpotent
group of class less than k.By (N),
thereexists,
among suchsets,
one which is finite and maximal. Let C be this maximalset,
andlet x o
C. Then C U contains a subset B such that(B)
isnilpotent
of class less thank,
where k = Then B == D U
{x},
for some subset D of C. But thenD ~ I
= k -1,
and(D)
is not anilpotent
group of class less than k -1,
so there exists a commutator w ofweight k -
1 in the elementsof D,
such that w ~ 1. But[w, x]
=1,
sox commutes with w. Since C is finite and k £ the number of
possi-
bilities for w is
finite,
and G is coveredby
the centralisers of these ele-ments,
hence one of these centralisers has a finiteindex,
so that the cor-responding w
is an FC element.COROLLARY 2. An
(N)-group
isFC-hypercentral.
Aninfinite (N)-
group has an
infinite
FC-centre.PROPOSITION 3.
If G is
an(Nk)-group,
theI G : Fk (G) I
isfinite,
where
Fk(G) is the
k-th FC-centreof G.
PROOF.
Suppose
thatI
is infinite. We will construct aninfinite sequence
(x2)
such thatAssume that we have
already
found xl , ... , xnsatisfying
these re-straints. We consider the
subgroups
for any indices
i( 1) i(2)
...i( t ) ~
n. The restraints1),
... ,I~ )
im-ply
that thesesubgroups
are each of infiniteindex,
as isF~ ( G ),
so wecan find an element x which
belongs
neither to one of thesesubgroups
nor to
Fk (G),
and then take Xn + 1 = x . Thiscompletes
the construction of the sequence(xn ),
whichobviously
contradicts theassumption ( Nk ).
PROPOSITION 4. An
(N)-group
ishyperabelian by finite.
PROOF. We may assume that G is
infinite,
and it suffices to showthat G contains a non-trivial normal abelian
subgroup. Suppose
no suchnormal abelian
subgroup exists,
and let x be anon-identity
elementwith
only finitely
manyconjugates,
let N =(X)G
be the normal sub- groupgenerated by
theseconjugates,
and let C =CG (N).
ThenG :
C ~I
is
finite,
and N fl C =Z(N)
is a normal abeliansubgroup
of G. Hence N n C = 1. Therefore N isfinite,
andhaving
a trivial centre, it is cer-tainly
notnilpotent.
WriteN1
=N, CI
=C,
and in C find an element y withfinitely
manyconjugates.
Then y has alsoonly finitely
many con-jugates
in G. LetN2 = ~ y~G .
In the same way, we see thatN2
is a finitenon-nilpotent
group.Moreover, N2
cC,
so that andN2 generate
theirdirect
product
in G. WriteC2
=CG(NIN2),
and continue the processby choosing
an FC element inC2 ,
etc. We thus find in G an infinite directproduct NI
xN2 x
... XNk X
... of finitenon-nilpotent
groups.Being finite, Ni
contains apair
ofelements xi and yi
whichgenerate
a non-nilpotent subgroup.
Now let z1 = xl , z2 = y1 x2 , Z3 = yl y2 x3 , ... , zn == yl y2 ... yn -1 xn , .... If
i j , then zi and zj project
on the elements x2and yi
ofNi ,
hence(zj,
is notnilpotent,
so that theset (zj)
violates
(N).
147
At this
point acquaintance with [LW]
is recommended.First,
forcompleteness,
weperform
one of thatpaper’s «finger
exercises».PROPOSITION 5. Let G be a
finitely generated hyperabelian
group in which everyinfinite
subset contains apair of
elementsgenerating
apolycyclic
group. Then G ispolycyclic.
PROOF. If we
change «hyperabelian»
to«soluble»,
this is TheoremB of
[LW].
To prove thepresent
version we assume that G violates theProposition.
Sincepolycyclic
groups arefinitely presented,
we can finda normal
subgroup
N such thatG/N
is notpolycyclic,
but all proper fac- tor groups ofG/N
arepolycyclic.
We mayreplace
Gby G/N.
Since G ishyperabelian,
it contains a proper normal abeliansubgroup
A. ThenG/A
ispolycyclic,
and now theproof
of Theorem Bof [LW] applies.
THEOREM 6. A
finitely generated
group Gsatisfies (N) if
andonly if it
isfinite by nilpotent,
and in that case itsatisfies ( Nk), for
some k.PROOF. Let G be a
finitely generated (N)-group. By Proposition 4,
G contains ahyperabelian subgroup
H of finite index. Then H is alsofinitely generated,
soProposition
5 shows that it ispolycyclic,
and G it-self satisfies the maximum condition. If G is not finite
by nilpotent,
wefind in it a normal
subgroup
N such thatG/N
is not finiteby nilpotent,
but all proper factor groups of it are finite
by nilpotent. Replace
Gby G/N. Naturally
G isinfinite,
so theproof
ofProposition
4 shows that G contains a normal abeliansubgroup.
Nowagain
aproof
from[LW],
thistime of Theorem
A, applies
to show that G is finiteby nilpotent,
acontradiction.
Conversely,
if G is finiteby nilpotent (not necessarily finitely
gen-erated),
it is well known that some termZk ( G )
of the upper central series of G has finite index in G[R, 4.25],
and it is immediate that then G satisfies(Nk).
COROLLARY 7. In an
(N)-group
the setof
elementsof finite
orderforms
alocally finite subgroup.
Thus an
(N)-group
is an extension of alocally
finite groupby
a tor-sion free one. For the
top
factor we can now proveparts a)
andb)
of Theorem B.COROLLARY 8. A torsion
free
group Gsatisfying (N)
ishyper-
central.
PROOF. Theorem 6 shows that G is
locally nilpotent,
andCorollary
1 to
[R, 4.38]
and ourCorollary
2 show that it ishypercentral.
REMARK. The fact that a
locally nilpotent (N)-group
ishyper-
central will be used on several occasions below.
COROLLARY 9. A torsion
free (Nk)-group
isnilpotent of
class k.PROOF. Since G is
locally nilpotent, Proposition
3 combinedwith
[R, Corollary
2 to4.38]
shows thatG : Zk (G) I
is finite. Then G is atorsion free
nilpotent
group, and in such a group all factor groupsG/Zi ( G )
are also torsion free. Therefore G =Zk ( G ).
COROLLARY 10. The class
( Nk )
isproperly
contained in PROOF OF THEOREM A. Theimplication a)==>b)
followsby
combin-ing
Theorem 6 andCorollary 9, c)
~a)
isclear,
and theequivalence
ofb)
and
c)
is a combinationof [R,
4.24 andCorollary
2 to4.21].
We consider
locally
finite groups next. In view ofProposition 1,
it is of interest to look atFC-groups.
If G is anFC-group,
thenG/Z(G)
is aresidually
finite torsion group. In such groups theproperty ( Nk )
turnsout to be
closely
related to aproperty
sometimes denotedby ( 2
-k), namely:
thesubgroup generated by
any two elementsof
G isnilpotent of
class at most k. With thisnotation,
wehave (2)
THEOREM dl. Let G be a
residually finite
torsionFC-group.
ThenG is an
if
andonly if
G contains a( 2
-k) subgroup off-
nite index.
PROOF. Let G be an
( Nk )-group,
and suppose it does not contain a(2 2013>~) subgroup
of finite index. Then we can find elements xl , yl , such that is notnilpotent
of class 1~. The normal closureN, =
=
(x1, yl )G
isfinite,
therefore there exists a normal finite indexsubgroup MI satisfying M1
= 1. We can find inMI
anotherpair
X2, Y2, such that(X2, Y2)
is notnilpotent
of classk,
andproceeding
in the same waywe
repeat
the contradiction in theproof
ofProposition
4.Let G be a
(2
-1~)
group. We will proveby
induction on r that G sat-isfies
(Nk).
Let X be an infinite subset ofG,
let x EX,
and let M be a fi-(2)
Ouroriginal proof
of Theorem 11 and Theorem C relied on a structure theorem of M. J. Tomkinson forFC-groups
[Tl, 2.24]. The presentelementary
proofs
are due to the referee.149
nite index normal
subgroup satisfying
M fl(X)G
= 1. There is a cosetMy
for whichMy
fl X isinfinite,
andby
induction there are elements ml y, ... , that(m, y,
is of class k. Let L ==
(x,
... , ThenLM/M = ~x, y) M/M
is of classk,
as isL(x)G
so L itself has class k.To conclude the
proof
we show that if G contains an( Nk)-subgroup
H of finite
index,
then G is an( Nk )-group.
We first find a normal finitesubgroup
F such that G =HF,
and then a finite index normalsubgroup
M H such that M n F = 1. Let X be an infinite subset of
G,
andfind y
so that X flMy
is infinite.Write y = hf, h
EH, f E F.
ThenX fl My
contains elementsml hf,
such that..., is of class k. Let L =
~ml hf, ... , Then LF/F
is of class k and
LM/M
iscyclic,
so L is of class k.PROOF OF THEOREM C. The claims about
(Nk)-groups
followby
ap-plying
Theorem 11 toG/Z( G ).
Now let G be an(N)-group
that is an FC-group. To show that the ro-th centre has a finite
index,
it suffices toshow that for
G/Z(G),
so we assume also that G is torsion and residual-ly
finite. We first show that G is(locally nilpotent)-by-finite.
Ifnot,
wefind elements x1, such that
yi)
is notnilpotent.
ThenNI =
=
(Xl’
isfinite,
and there exists a normalsubgroup
of finite indexM
such that
N1
flmi
=1, MI
contains two elementsgenerating
a non-nilpotent subgroup,
etc.Let H be a
locally nilpotent subgroup
of finite index.By [R, 4.38],
His its own ro-th centre. There exists a finite normal
subgroup
N suchthat G =
HN,
and it follows thatZn (H)
flCG (N) Zn (G).
This con-cludes the
proof.
We can now prove Theorem B
c).
We need a lemma first.LEMMA 12. Let G be an
(N)-group
whose derivedsubgroup
G’ is a1l’ -group, for
some setof primes
1l. Then each a-elementof
G is anFC-element.
PROOF. Let x be a
.7r-element,
and suppose that x hasinfinitely
many
conjugates,
Then two of theseconjugates,
say xi and xj,generate
anilpotent
groupH,
which is then a ,r-group. Thenn G’ =
1,
so xi contradiction.PROOF OF THEOREM B
c).
We first prove the theorem in thespecial
case that G’ is
hypercentral.
We claim thefollowing:
If Jr
is a setof primes,
let 0 : =0~ ( G )
and L : =L~ ( G )
be theLargest normal1l-subgroup of
G and thesubgroup generated by
all yr-elements
of G, respectively.
ThenL/O
isfinite.
PROOF. Let H =
G/0.
Since H’ ishypercentral,
it has a normalHall
a-subgroup,
soby
the definition of 0 we see that H’ is aBy
Lemma12,
the r-elements of H are in its FC-centre K. Since K is alsohypercentral by finite, by
TheoremC,
and itshypercentre
is a.7r’-group,
by
definition of0,
we see that alljr-subgroups
of H are finite andof bounded order. We note also that H is
hypercentral by abelian,
so afinite
subgroup
of H is soluble. Let T be a maximalr-subgroup
of H.Since T consists of
FC-elements,
it is contained in a finite normal sub- group N. Let x be a n-element of H. Then T is a Hallsubgroup of (T, x),
so x is
conjugate
to some element ofT,
hence xE N,
andL/0 ~
N.Having
establishedthis,
let S be theproduct
of all the normalSylow subgroups
ofG,
andlet Q
be the set ofprimes
for which there is no nor-mal
Sylow subgroup.
Then S is contained in the Hirsch-Plotkin radical R ofG,
which is ahypercentral
group. Theprevious claim, applied
toeach
prime individually,
shows that theSylow subgroups
ofG/R
arefinite,
so ifQ
isfinite,
so isG/R.
We thus assume thatQ
is infinite.Let q e
Q,
and writeLl
for thesubgroup Lq
defined above. ThenLl
is not
hypercentral, by
the choice of q(note
that1),
and1l(L1),
theset of
primes occuring
as orders of elements inL1,
is finite.Suppose
that we have
already
foundnon-hypercentral
normalsubgroups
L1,
... ,Ln ,
whichgenerate
their directproduct
inG,
and such that ==
;r(Li
x ... xLn )
= is finite. L et K =L.,, ( G ), U
=On’ (G).
ThenK/ U is
finite. Let P = ;r’ - Then all P-elements of G lie in U. In par-
ticular,
we can find aprime r e Q
nP,
soLr ~ U,
andLr
is nothyper- central,
so we can choose =Lr .
We have thus an infinite direct
product
ofnon-hypercentral
sub-groups
Li .
ThenLi
is notlocally nilpotent
either.Being locally finite,
this means that we can find in
Li
apair
of elementsgenerating
a non-nilpotent subgroup,
and this leadsagain
to the same contradiction as inProposition
4.Now remove the
assumption
that G’ ishypercentral,
and let G beany
(N)-torsion
group. Let N be the Hirsch-Plotkin radical ofG,
whichis
hypercentral,
and suppose thatI
is infinite. ThenCorollary
2and Theorem C show that G contains a normal
subgroup M,
such thatM/N
is an infinitehypercentral
group. ThenM/N
contains an infinitenormal abelian
subgroup A/N.
Thus A’ ishypercentral,
so thespecial
case
implies
that A has a normalhypercentral subgroup B
of finite in-dex. Since B is subnormal in
G,
we have B ~N, contradicting
the in-finiteness of
A/N.
This proves our theorem.151
Note that if G is a
hypercentral (Nk)-group (for
some1~)
then itshy- percentral height
is at mostwk, by Proposition
3 and[R, 4.38].
REMARKS.
1)
As seems natural for aproblem posed by
P.Erd6s,
the paper
[N, FLM] apply Ramsey’s
Theorem. But this result washardly
used in this paper, nor in[LW].
We wish topoint
out that its usein
[N]
and[FLM]
is also not essential.Thus, Ramsey’s
Theorem is usedin these papers to prove that if a group G has the
property
that each in- finite subset contains acommuting pair,
then G is anFC-group.
Weprove this
alternatively
as follows. Let H be the FC-centre of G. Choosea subset A which is maximal with
respect
tohaving
anempty
intersec-tion with
H,
and notcontaining
acommuting pair.
Then A isfinite,
andeach element of G
belongs
either to H or to a centraliser of one of the elements of A. Since these centralisers have infiniteindices,
it followsthat G = H
(3).
Note that we have
replaced Ramsey’s
Theoremby
Neumann’s cov-ering
lemma. On the otherhand,
in[Tl, 7.3]
this lemma isproved by
ap-plying Ramsey’s
Theorem(which
was not used in theoriginal proof).
2)
A variation on theproblem
of this paper, alsosuggested by [LMR],
is to ask in(N) only
that some commutator ofweight [ X I
inthe elements of X is the
identity.
We were able to handleonly
astronger assumption
(N*)
For any infinite sequence xl , x2 , ... , ... of distinct elements of G there exist indicesi(l) i(2) ... i(k),
such thatXi(2), ... ,
2’Ik>]
= 1.It turns out that at least our
preliminary
results hold also under this weakerassumption.
We indicate the necessarychanges
in theproofs.
PROPOSITION 1*. A group G
satisfying (N*)
has a non trivialFC-centre.
PROOF. We assume that G is
infinite,
and well order itaccording
tothe least ordinal of
cardinality G ~ .
We denote the order relationby
« ». We can find a finite subset X of G maximal with
respect
to theproperty:
if Xl x2 ... Xk are elementsof X,
then[x1, x2 , ... , xk] ~
~ 1. Let z be the
largest
element of X. It then follows as in theproof
ofProposition
1 that each element x > z centralises one offinitely
many commutators in the elements of X.(3)
The referee made thefollowing
comment at thispoint:
«This argument is known to a number ofpeople
but has notappeared
inprint...
it came to me at theend of a chain in which the earliest link that I know of is Passman».
If G is
countable,
there areonly finitely
many elements which pre- cede z. Then G is the union of thesefinitely
many elements andfinitely
many centralisers.
Considering
the individual elements as cosets of1,
we see that
they
can beomitted,
G is the union offinitely
many cen-tralisers,
and one of these centralisers has a finite index.Suppose
thatG is uncountable. Then the elements
preceding z generate
asubgroup
H of
cardinality
smaller thanI G I
and thereforeG :
HI
is infinite. Thegroup G
being
the union of H andfinitely
manycentralisers, again H
can be omitted.
(We
see that thisproposition
can beproved
also under thefollowing
weaker condition: we first well order G as in the
proof,
and then in(N*)
consideronly
sequences which are orderedby
thisordering.)
PROPOSITION 4*. A group
sactisfying (N*)
ishyperabelian by finite.
PROOF. We
repeat
theproof
ofProposition 4, only choosing
xi, yidifferently. As Ni
is finite and notnilpotent,
it is not anEngel
group ei-ther,
so wechoose xi and yi
tosatisfy [xi ,
... ,1,
for any number of occurences of yi. With this choice theproof
ofProposition
4applies.
REFERENCES
[CLMR] M. CURZIO - P. LONGOBARDI - M. MAJ - A. RHEMTULLA,
Groups
withmany rewritable
products,
Proc. Amer. Math. Soc., 115 (1992), pp.931-934.
[FLM] V. FABER - R. LAVER - R. McKENZIE,
Coverings of
groupsby abelian subgroups,
Can. J. Math., 30 (1978), pp. 933-945.[G] J. R. J. GROVES, A
conjecture of Lennox
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