R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
C ARLO C ASOLO
Subgroups of finite index in generalized T -groups
Rendiconti del Seminario Matematico della Università di Padova, tome 80 (1988), p. 265-277
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Subgroups T- Groups.
CARLO CASOLO
(*)
1. Introduction and main results.
The class of
T-groups
is the class of groups in which every sub- normalsubgroup
is normal. In[1]
we introduced and studied someclasses of
generalized T-groups.
We recall here the relevant defini- tions.Let G be a group, m a
positive integer,
then:1) (respectively
if each subnormalsubgroup
ofG has finite index
(resp.
index at mostm)
in its normalclosure;
2) (respectively
G EVm)
if each subnormalsubgroup
of Ghas a finite number of
conjugates (resp.
at most mconjugates)
in G(that
isJG : Na(g)
00, orrespectively JG : Na(.g) m)
for each H subnormal inG;
3 )
G E U(respectively
ifJ
is finite(resp.
for every H sn G.
We remind that He- and
denote, respectively,
the normalclosure and the normal core of H in G.
It is easy to check
(see [1]),
that U =T*,
and it is obvious that the class ofT-groups
is(properly)
contained in U.In this paper, we consider
subgroups
of finite index in groupsbelonging
to the classes above defined. Moreprecisely,
we aim tostudy
to which extentsubgroups
of finite index of a group in one(*)
Indirizzo dell’A.:Dipartimento
di Matematica e Informatica, Univer- sith di Udine, Via Zanon 6, 1-33100 Udine,Italy.
of these
classes, belong
to the same class. This is in thespirit
of arecent paper
by
H. Heineken and J. Lennox[2],
where the authors prove that asubgroup
H of finite index in aT-group
G is anUr-
group, the
integer
rdepending only
on the indexlG:HL;
; moreover, H is aT-group
if it contains some term of the derived series of G.In order to make shorter the statement of our first
Theorem,
we
give
thefollowing
definitions.Let H be a
subgroup
of the group G. We say that .H is almost normal(see [3,
page191])
in G ifIHG:HI
isfinite,
and almost sub- normal if .H has finite index in some subnormalsubgroup
ofG;
wesay further that H is
virtyally
normal in G ifNa(g)
has finite index inG,
andvirtually
subnormal if H is subnormal in asubgroup
offinite index in G.
THEOREM 1. Let G be a group; then the
following
areequivalent.
(i)
G E T*(that is,
every subnormalsubgroup of
G is almost nor-mal) ;
(ii)
every almost subnormalsubgroup of
G is almostnormal;
(iii)
everyvirtually
subnormalsubgroup of
G is almostnormal;
(iv)
the relationof
almostnormality
is transitive in G(we
meanthat whenever
G,
with K almost normal in H and H almost normal inG,
then K is almost normal inG).
Implication (i)
=>(iii) gives immediately:
COROLLARY 1. A
subgroup of finite
index in aT*-group
isagain
a
T*-group.
Also,
it will be easy to deduceCOROLLARY 2. A
subgroup of finite
index in aU-group
isagain aU-group.
An
analogous
result does not hold for the classV,
not even forthe class
U Ym .
We will show thisby
means of anexample,
whichmEN
is
essentially
taken from Heineken and Lennox[2]. Turning
to groups in which the index of every subnormalsubgroup
in its normal clo-sure is bounded
by
apositive integer,
we are able to prove:THEOREM 2. There exists a
function
d: ~T~T,
such thatif
G E
Tm
and H is asubgroup of G,
with then H EIt
easily
followsCOROLLARY 3. There exists a
function d: --~ ~T,
such thatif
G E
Um
and H is asubgroup of G,
withI G: H
n, then H EU(j(m,n).
An immediate consequence is
COROLLARY 4
(Heineken
and Lennox[2,
TheoremB]).
There ex-ists a
function f:
~ --~1~T,
such thatif
H is asubgroup of
index at mostn in a
T-group,
then H EUf(n).
(Indeed,
Heineken and Lennoxproved
ananalogous
of corol-lary
3 for a proper subclass ofU Um,
thatthey
call the class of X-groups).
MENIn the
hypotheses
of Theorem2,
if thesubgroup H
contains someterm of the derived series of
G,
then the bound d does notreally
de-pend
on the index of H in G. This is the content of our next result.THEOREM 3. There exists a
function
b : N such thatif
H isa
subgroup of finite
index in aTm-group G,
and H contains some termof
the derived seriesof G,
then H EIt will be evident from the
proof,
thatb (1 )
=1;
thus we havean alternative
proof
of another result of Heineken and Lennox[2,
The-orem
~~, namely:
COROLLARY 5. A
subgroup of finite
index in aT-group G,
whichcontains some term
of
the derived seriesof G,
is aT-group.
2. Proofs and related results.
PROOF OF THEOREM 1.
(i)
=>(ii).
Let H be an almost subnor- malsubgroup
of G ET*;
thenIHn:HI I
isfinite,
for some termgn
ofthe normal closure series of g in G.
Now, H, sn G,
whenceIH,,: H. I
is finite. But also
Hf
= and so =(
is fi-nite and H is almost normal in G.
(ii)
~(iii).
Let with H subnormal in L and E of finite index in G. We argueby
induction on the defect n of H in L. Letfirstly
SinceBG:L!
isfinite, La
has finite index inG,
andH r1
La-aLavG; hence,
inparticular,
H r1La
is almost subnormal in G. If W == we have therefore oo, andNow, ~
is finite andNa(HW»L,
that isoo.
By
Dicman Lemma(see [5, 14.5.7.]), (HW/W)alw
is finite. In
particular,
is finite. Because === is
finite,
we have that isfinite,
whenceis
finite,
and H is almost normal in G.Let now n >
1,
and T =gL; then, by
the case discussedabove, I
is finite.Now,
g isvirtually
subnormal in TO and the defect of .H~ in T is n -1.By
inductivehypothesis
and the fact that con-dition
(ii)
isobviously
inheritedby
normalsubgroups,
we have thatIHpG:H/
is finite. ButHpG
is subnormal inG;
thus .b~ is almost sub- normal inG,
and so H is almost normal in G.(iii)
~(i).
Obvious.(i)
=>(iv).
LetKHG,
with both andIHG:H/
finite.Then KH is
virtually
normal in Since(i)
=>(iii)
and(i)
isclearly
inherited
by
normalsubgroups,
KH is almost normal in.HG,
that isKH has finite index in =
K HG.
It follows that K has finite index inKHG. Now, KRG
is subnormal in Gand,
since(i)
~(ii),
Kis almost normal in G.
(iv)
=>(i).
Obvious.PROOF OF COROLLARY 2. Let H be a
subgroup
of finite index in G EU,
and let 8 be a subnormalsubgroup
of H. Then 8 isvirtually
subnormal in
G; thus,
since U sT*,
is finite. Let L =(S),~a,
then .L sn G
and,
inparticular, IL:LGI
is finite,. But18G:L/
is alsofinite,
so18°:LGI
is finite.Now, LG gives
A for-tiori, /8H:SHI
is finite. This holds for any thusproving
that.8~ is
a !7-group. a
We observe
that,
ingeneral,
asubgroup
of aT-group
need notbelong
to T*(nor
toV).
Let D be a directproduct
ofinfinitely
manycopies
of the additive group of therationals,
and a E Aut(D)
be theinversion map on D Then the natural semidirect
product G
==
D z ~a~
is aT-group.
Let A be a freesubgroup
of infinite rank ofD,
then thesubgroup A, 0153)
of G does notbelong
to VU
T*.Corollary
2 can beslightly improved, namely:
PROPOSITION 1. Let G he a. group; the are,
(i) G E U;
(ii) if KHG
and both and arefinite,
thenfinite.
PROOF,
(ii)
=>(i)
is obvious.(i)
=>(ii).
LetKHG,
with andfinite,
andlet G E U.
Now, by
the sameargument
used in theproof
of Corol-lary 2, IKHG:KH.1
is finite. ButKH;
and.gHG
are both subnormal in G. Since G EU,
we have oo and oo, andso is finite.
By contrast,
the class V is not closed undersubgroups
of finiteindex.
Indeed,
there exist groups, in which every subnormal sub- group has a bounded finite number ofconjugates,
that admit sub-groups of finite index which are not
V-groups.
Anexample
is thegroup constructed
by
Heineken and Lennox in[2].
Wereport
aslightly simplified
version of it.EXAMPLE. Let G be the group
generated by
a,b,
ei,di (i EN), subject
to thefollowing
relations:The G =
AB,
where A =di ; i
=1, 2, ... ~ a G
is anelementary
abelian
2-group, B
=a, b)
-S3
and A n B = 1. If Ssn G,
oneeasily
checks that either or~~(~Ly~.
Hence A normalizesevery subnormal
subgroup
ofG;
since =IBI
=6,
thisyields
G E
V6.
Let H =A, a~ ;
thenG : H
= 3 and H isnilpotent. Now, a~
is subnormal of defect 2 ing,
but =ci di; i
==
1, 2, ...)
has infinite index inA;
thus=
oo, that isH 0
Y.(The
group exihibitedby
Heineken and Lennox in[2] shows, furthermore,
that asubgroup
of finite index in aV6-group
need notbelong
to the class of groups with a bound on the defects of their subnormalsubgroups).
This
example shows,
in otherwords,
that G E V does notimply
that every
virtually
subnormalsubgroup
of G isvirtually
normal.On the other
hand,
we have:PROPOSITION 2. Let G be a group; then the
following
areequiva-
lent :
(i)
everyvirtually
subnormalsubgroup of
G isvirtually normal;
(ii)
the relationof
virtualnormality
is transitive in G.PROOF.
(ii)
=>(i)
is obvious.(i)
=>(ii).
Assume that thegroup G
satisfies(i)
and letbe such that and are both finite. Let .L =
==
(NH(K»)H
be the normal core of in,H;
then .L is subnormalin
Na(H),
that is L isvirtually
subnormal in G.By
ourhypothesis, Na(L)
has finite index in G. Hence .M’ =Na(H) r1 NG(l)
has finiteindex in G. But .M’
acts, by conjugation,
on the finite sectionin
particular = 1
is finiteand,
con-sequently,
has finite index in G.Now,
and so .g is
virtually
subnormal in G. Since G satisfies(i),
.K isvirtually
normal inG. 1
In order to prove Theorems 2 and
3,
we need somepreliminary
lemmas.
LEMMA 1..Let A be an abelian group, m a
positive integer
and(~. ),
such thatfor
everyg c .A. (where
HG =a E G».
Then:(a) G
everysubgroup of A/A. [d],
where d = m ! t andA[d]
==~==1}.
(b) If
A isperiodic
andreduced,
there exists a G-invariant sub- group Nof A,
such that N can begenerated by
4melements,
and G every
subgroup of A/N.
PROOF.
(a)
Let then~x~~’: x~ ~ ~ m
andxd~ ~ x~.
Hence is G-invariant and thus G fixes every
subgroup
of Ad =- ~xd; x E A~. Now,
the definedby
forevery x E
.~.,
is aG-homomorphism.
Since 0 issurjective
and Ker(4)) ==
-
A[d],
we conclude that G fixes everysubgroup
of(b) (see [1;
Lemma2.9]). Suppose
that we havealready proved
the assertion when A is
residually
finite. Let B be a basicsubgroup
of A
(see [5; 4.3.4.]).
Thus B is a directproduct
ofcyclic
groups and BG is a finite extension ofB;
hence Ba isresidually
finite.By
our
assumption,
there exists a G-invariantsubgroup
of suchthat N can be
generated by
4melements,
and G acts as a group of powerautomorphisms (that is, fixing
everysubgroup)
onBGIN. Now,
since N is finite and A is
reduced, A/N
is alsoreduced;
moreover, is divisible.By
Lemma 2.2 in[1],
we conclude that G fixesevery
subgroup
ofAIN.
Thus,
it remains to prove(b)
when A isresidually
finite.Let be distinct elements of A such
that,
forany i =
0, 1,
... , k -1:Then,
if X = ... , weget
> k. Infact, proceeding by induction,
we haveI and, if k > 1, using
the inductivehvpothesis :
Thus,
in ourhypotheses,
a subset of Asatisfying
conditions(1)
hasat most m elements.
Let yo == 1,
yi, ... , yr , r m, be such asubset,
with r
maximal,
and let K= yo ,
... ,7 y,)Iy;
then isgenerated by
at most r
+ logs
m m+ logs
melements,
becauseyo , ... , yr~ ~ ~
m.Since A is
residually finite,
and .g isfinite,
there exists asubgroup
Mof finite index in
A,
maximalsubject
to the condition: =1.Then is
generated by
elements and tm + log2 m.
Let Y =
(zi, 7 ... 7 zt)
andput N
= then N isgenerated by
atmost
2(m + + log2 m
= 2m+ 3log2m
4melements,
and it isG-invariant. We now show that G fixes every
subgroup
of itis
enough
to check this forcyclic subgroups.
LetC/N
be acyclic
subgroup
ofA/N.
Since there exists suchthat C =
~x, N~ .
But then K r1x~ c .g’
r1 if = 1and,
because... , is a maximal subset of A
satisfying
conditions(1),
weget
x~ a : (K, 0153) =1;
inparticular, N,
= C°’ =~N, 0153)
= C. ThusG fixes every
cyclic subgroup
ofA/N,
and the Lemma isproved.
We remind that the group of
automorphisms fixing
every sub- group of a group(called
the group of powerautomorphisms)
is abelian.We will make use of this fact in the
sequel.
Inparticular,
under thehypotheses
of theprevious Lemma,
G’ centralizes bothA/A[d]
andA/N.
We observe also that the order of
N,
inpoint (b)
of theLemma,
cannot be bounded
by
a function of m. Let n be anypositive
inte-ger, C =
x~
acyclic
group of order 2n+2 and B the directproduct
of n
+ 1 cyclic
groups of order4;
let a be anautomorphism
of A === B X C
centralizing
C andacting
as the inversion map on B. If G =a> Aut (A),
then = 2 and it is easy to check that(and
indeed for every But if N is a G-invar- iantsubgroup
of A. andthen G
does not act as a group of powerautomorphisms
onLEMMA 2. There exists a
function a : N - N,
suchthat, if
G ETm
and G is
solubte,
then G(2) has order at mosta(m).
Observe
that, by
D. Robinson’s result onT-groups (see [5; 13.4.2]),
we may
put a(1)
= 1.PROOF. Let G be a soluble
Tm-group (m
If m =1, then, by
thequoted
result ofRobinson,
G is metabelian. Hence assume m > 1. Let F be theFitting
radical ofG;
then F isnilpotent by
Lem-ma 3.1
in [1].
Thus everysubgroup
of .F is subnormal inG ; and,
in
particular,
for everyBy
a Theorem of I. D.Macdonald [4;
Theorem5.14],
cm900(logs
m)3 - IfFl/F’
isthe
Fitting
radical ofG/F’,
then.F’1/I"
isnilpotent
and so,by
a wellknown
nilpotency
criterion of P. Hall(see [5; 5.2.10]), ~’1
isnilpo-
tent. Thus
.F’1
= F andF/I"
is theFitting
radical ofG/I".
We may therefore assume, from now on, F’ - 1. Observethat, G being
sol-uble,
thisimplies OG(F)
= .F’.Now, /HG:H/
c m for every.H c F;
weare therefore in a
position
toapply
Lemma1 (ac).
If d =m !,
weget
that every
subgroup
ofF/I’[d]
is normalizedby G,
and thus G’ cen-tralizes
FfF[d]. Now, F[d]
is reduced andperiodic; by part (b)
ofLemma
1,
we obtain asubgroup
N ofF[d],
normal inG,
such that G normalizes everysubgroup
ofF[d]/N
and N can begenerated by
4melements. Since
exp (N) d,
we haveMoreover,
G’ central- izes whence[F, G’, G’] c N. By
the threesubgroups lemma, [14’, G~2~] c N. Now,
if K = then the indexI
isbounded,
sayG : .g ~ c a2(m). Futhermore,
stabilizes a finite series of F. Because (? issoluble,
thisimplies K n and,
con-sequently :
In~
particular, G~2~ : G~2~ n PI c a2(m).
But[G(2)
G~2~r1.Z’] c N;
thus thecentre of G(2) has index at most in
G(2),
modulo N. It followsthat
IG(3)NfNI
isbounded; indeed, Wiegold [6]
has obtainedThus,
we may assume G(3) =1;
soG(2) F
andG’.F’/I’
is thereforeabelian.
By repeating
thearguments
usedabove,
we find asuch that and
[F, G’,
In this case is nil-potent. Applying again
lVlacdonald’sTheorem,
weget
Nl being finite,
thiscompletes
theproof.
The
following
Lemma isprobably
well known.LEMMA 3. There exists a
function do : N - N,
suchthat, if
G is agroup and
for
somepositive integer
r,then, denoting by
Wthe
nilpotent
residualof G, G/W’
isnilpotent and
PROOF. Let L =
Zr(G),
r apositive integer. Then, by
a resultof Baer
(see [5; 14.5.1.]),
is finite. Thisimplies
at once thatW is finite and is
nilpotent.
We have to show that the order of W is bounded.Now,
it is well known that[lY, L]
=1,
thusand so
By
analready quoted
result ofWiegold [6],
9 it is Withoutloss of
generality,
we may assume, from now on, that TVP is abelian(observe
thatLet
C = Ca(W) ;
thenC>LW.
Inparticular GIC
isnilpotent,
and
I
divides For eachprime
pdividing I
letbe the
Sylow p-subgroup
ofGIC,
andWp,
thep’-component
of W.Since
W1J’
isabelian,
we have:Moreover,
and soL)
is ap’-group.
Thus
W1JI
r1L Owp,(G1J) and, consequently :
Put I~ _ p divides
BG:OI);
then E« G and the order of 1~ is at mostNow,
for anyprime
pdividing C] , [yV’,
=== for some
s(p)
E ~.Therefore,
if s = pdividing
we have:Since
G/w’
isnilpotent.,
it follows that isnilpotent,
and so.1~ = w’. In conclusion we
get:
PROOF OF THEOREM 2. Let and
H ~ G, with ~
== n(m, n
eN).
Let S be a subnormalsubgroup
ofH;
we will prove that is less orequal
tod(m, n),
where d is a function from NxN toN, whence,
inparticular, n).
Let K =
Hg,
then the order ofGIK
dividesn !,
and K is sub-normal in .g. Thus .g’ r1 S is subnormal in G and so, if U =
(g
rl~S’)°‘,
we have
U K
andNow,
U8 sn H andhence,
without loss ofgenerality,
we may assume U= 1and,
conse-quently,
K =1.Let T = SG,m be the m-th term of the normal closure series of S in G.
Now, G
ETm clearly implies
that every subnormalsubgroup
of G has defect at most m in G. Thus T is the minimal subnormal
subgroup
of Gcontaining ~S,
and 8’ = T.Furthermore, by
Theorem 1((i)
~(iii)), IT:81 I
is finite.Because 181
_8)
=I
di-vides
n !,
we have that T is a finiteTm-group.
Let L = K n
T,
then and L =1;
since 8 is subnormalin
L~ c .g,
it follows that 8 stabilizes a finite series[L, ;8]
of L.Thus, 8/Cs(L)
isnilpotent; being finite,
thisyields [L, 8] CS(L)/CS(L) nilpotent,
and so[L, S]
isnilpotent.
Let 1~ be theFitting
radicalof
L ;
then F« T and[L, ~’] c .F’. Tm
and all of itssubgroups
are
subnormal; by
Macdonald’s Theorem[4;
Theorem5.14(i)], m900(log2
m)8 ==Furthermore,
T actsby conjugation
- in such a waythat for every
BF. By
Lemma1 (b),
there exists anN =
generated by
at most 4melements,
such that Na T and Tfixes, by conjugation,
everysubgroup
of.F/N.
Since[L, S] c I’,
weconclude that T normalizes every term of the series
[L, iS]
mod. Nof
E7 i
=1, 2, ...
But every factor of this series is centralizedby S
(because
and so, from 8" =T,
we infer that T itself centralizes every such factor. Thus where s is thedefect of S in and
ZS(T/N)
is the s-th term of the upper central series ofTIN.
Similarly, by point (a)
in Lemma1,
T acts as a group of powerautomorphisms
on.F’/ Y,
where(P/F’)[m!]. Arguing
asbefore,
we obtain
Thus,
if weget:
Now,
is an abelian group whoseexponent
dividesm !,
and it isgenerated by
4m elements orless,
whence hasorder at most
(m !)4m.
Since we have wheredl(m)
=Moreover Z/if
is contained in thehypercentre
of and:
If is the
nilpotent
residual ofT/M,
it follows from Lemma 3:and, consequently,
SinceT / W
isnilpotent,
~S W is subnormal in T =ST ;
thus 8W =T, yielding:
Finally,
T = 8",m is subnormal inG,
whenceITG:TI
==Reminding
that we assumed at thebeginning
we find:PROOF OF COROLLARY 3. Let G E
Um ,
H asubgroup
of G of indexn, and 8 a subnormal
subgroup
of H. SinceU. C Tm ,
theproof
ofTheorem 2
gives
whence But= is subnormal in
G,
and soSince, clearly,
I we
get
whered(m, n)
=d(m, n) ! m,
andthus
Observe
that,
in Theorem 2 andCorollary 3,
since the classesfi~
andUm
are closed under normalsubgroups, nothing changes
ifwe assume instead of
PROOF OF THEOREM 3. Let G be a
Tm-group,
and let H be a sub-group of finite index in
G, containing
some term of the derived se-ries of G. Let ~S be a subnormal
subgroup
of H. If .g =H,
I then is a finite soluble group, and S is subnormal in G. As in theproof
of Theorem2,
since we may assume Sn ~=1.Let T = be the m-th term of the normal closure series of S in G. Like in the
proof
of Theorem2,
S’ = T and T is finite.Let L = .g r1
T;
then and Now 8 stabilizes the series[L, ;S] (i
EN)
of .L. Since SL isfinite,
we have that[L, ~S]
is
nilpotent ;
whence ~S centralizes the factor where F is the Fit-ting
radical of L. Because ST =T,
we obtain thatLjF
is a centralfactor of T . In
particular,
.L is soluble. SinceT /L = TIT
r1 K ~is also
soluble,
we conclude that T is a(finite)
so-luble
Tm-group. By
Lemma2, T~2~ ( c a(m).
_We now use a bar to denote
subgroups
of T modulo T(2) . Let Wb_e_the _nilpotent
residual of T =TIT(2).
Since ST =T,
we have~S W = T . Moreover,
y W isabelian,
and so:and,
because isnilpotent.
Thisfact,
to-gether
with(1)
and the finiten~ess ofT, implies:
and, consequently:
~ andBut 8 sn H m T
implies
~ST~2> sn(.g
r1T)T(2)
and so,by (2),
weget
ST(2) =T) T(2),
whichyields
H f1 T ST(2).Con~sequently:
Now,
that is Therefore we findbecause T is subnormal in G.
Together
with(2),
thisgives:
Finally,,
yrecalling
that we assumed we obtain:This holds for every subnormal
subgroup
ofH,
and so HET6(.),
where
b(m)
=m2 a(m).
It is at once evident thatb(1)
=1,
and thiscompletes
theproof
of the Theorem and itsCorollary.
Needless to say,
b(m)
mayactually
bestrictly greater
than m.For
instance,
the groupGL(2, 3)
iseasily
seen to be a solubleT2-group,
while its
Sylow 2-subgroups belong
toT4",T2.
REFERENCES
[1] C. CASOLO,
Groups
withfinite conjugacy
classesof
subnormalsubgroups,
to appear in this
journal.
[2] H. HEINEKEN - J. LENNOX,
Subgroups of finite
index inT-groups,
Boll.Un. Mat. It., (6) 4-B (1985), pp. 829-841.
[3] J. C. LENNOX - S. E. STONEHEWER, Subnormal
Subgroups of Groups,
Oxford Mathematical
Monographs,
Clarendon Press (1987).[4] I. D. MACDONALD, Some
explicit
bounds in groups withfinite
derived groups, Proc. London Math. Soc., (3) 11 (1961), pp. 23-56.[5] D. J. S. ROBINSON, A Course in the
Theory of Groups, Springer,
New York-Heidelberg-Berlin
(1982).[6] J. WIEGOLD,
Groups
withboundedly finite
classesof conjugate
elements,Proc.
Roy.
Soc. London, Ser. A, 238 (1957), pp. 389-401.Manoscritto