R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
C ARLO C ASOLO
Wielandt series and defects of subnormal subgroups in finite soluble groups
Rendiconti del Seminario Matematico della Università di Padova, tome 87 (1992), p. 93-104
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in Finite Soluble Groups.
CARLO
CASOLO (*)
1. Introduction.
We are interested in this paper in the subnormal structure of finite soluble groups; we show that the Wielandt
length
of a finite solublegroup is bounded in terms of its
Fitting length
and of the maximum de- fect of its subnormalsubgroups.
The Wielandt
subgroup m(G)
of a group G is the intersection of all normalizers of subnormalsubgroups
ofG; clearly
is a characteri- sticsubgroup
of G. The Wielandt series:is defined
by (see [9; § 4.6]):
H. Wielandt
[13] proved
that if G is a finite group thenw(G) 0 1,
forin this case contains all minimal normal
subgroups
of G. It therefollows that for a finite group G there is a smallest
positive integer n
such that Wn
(G)
= G: such n is called the Wielandtlength
ofG;
we deno-te it
by wl(G).
From now on, all groups considered will be finite.We denote
by b(G)
the maximum among the defects of subnormalsubgroups
of the groupG,
andby Bn
the class of all groups G such thatb(G):5
n. H. Wielandt observed thatwl(G) [13];
thus a bound onthe Wielandt
length implies
a bound on the defects of the subnormalsubgroups.
Whendealing
with soluble groups, there is a bound on the derivedlength
of a group in terms of its Wielandtlength (the
exactbound has been determined
by
R.Bryce
and J.Cossey [2],
see[4]
for(*) Indirizzo dell’A.:
Dipartimento
di Matematica eInformatica,
Universitadi
Udine,
Via Zanon 6, 33100Udine, Italy.
Research
partially supported by
MURST funds.an extension to infinite
groups).
On the otherhand,
while solublegroups in
B1
are metabelian[6] [10]
and finite soluble groups inB2
havederived
length
at most 5[5],
no bound exists on the derivedlength
of fi-nite soluble groups in
B3 ,
as first shownby
T. Hawkes[8] (see
also R.Bryce [1]
forexamples
ofB3-groups
with abelianSylow subgroups
andarbitrary
derivedlength). Thus,
afortiori,
no bound on the Wielandtlength
of a group existsdepending only
on the maximum defect of itssubnormal
subgroups. Also,
no bound onb(G) (and
so on the Wielandtlength wl(G))
existsdepending only
on the derivedlength
of the group G: the wreathproduct
of acyclic
group oforder p (a prime number) by
itself is metabelian and has subnormal
subgroups
of defect p.The n-th term
Fn (G)
of theFitting
series of G is definedby Fo (G) =
- 1 and
Fn +
1(G)/Fn (G)
= Fit(G/Fn (G)).
If G issoluble,
theFitting (or nilpotent) length
of G is the smallest n such thatFn (G)
=G,
and it isdenoted
by l(G).
We shall show that there exists a function
f :
N x N - N suchthat
for any finite soluble group G.
In fact this result is a
corollary
of a moregeneral
statement(Theo-
rem
2):
if X is a class of finite soluble groups, closedby homomorphic images
and normalsubgroups,
such that allnilpotent
groupsbelonging
to it have
nilpotency
class at most c, thenwt(G) _ f (L(G), c)
for every G in X.2. A
strong
Wielandtsubgroup.
In order to find bounds on the Wielandt
length
of a finite group, we have found it useful to introduce aspecial subgroup
of the Wielandtsubgroup,
which is in some sense easier to deal with. Then we define acorresponding ascending series,
a bound on whoselength implies
abound on the Wielandt
length.
In our definition and most of our argu-ments,
we have chosen a «local»approach,
as it is done in[2]
for the Wielandtsubgroup.
DEFINITIONS. Let G be a group, p a
prime
number. Weput
~x
EG; [H, x]
for allp’-perfect
subnormal sub- groups H ofG ~ .
(We
remindthat,
if 7r is a set ofprimes,
agroup H
is said to ben-perfect
if0"(~)=~).
And(Here HN
is thenilpotent
residual ofH,
i.e. the smallest normal sub-group K
of H such thatH / K
isnilpotent.)
Then wP
(G)
and15(G)
are characteristicsubgroups
ofG,
and it is im- mediate(G)
is contained inmP(G) (the
intersection of the norma-lizers of all
p’-perfect
subnormalsubgroups
ofG,
as defined in[2]),
and~(G) w(G).
PROPOSITION 1. Let G be a groups,
x(G)
the setof
allprimes
divi-din G ~ , a(G)
the socleof G.
Then:PROOF.
(i)
If H isp’-perfect
then =HN.
Thus(;j(G) ~
for every p E
~(G). Conversely,
let x E U(G)
for every pr=7~G),
andH sn G. For each p,
Hp
= isp’-perfect
and subnormal inG;
so[Hp , HN.
Since H is the normalproduct
of theHp’s,
[H, HN
and x E(ii)
Let M be a minimal normalsubgroup
of G. We have to showthat M S
w(G).
Let H snG;
then MH is subnormal in G and so M:5~
a-(MH).
We may therefore assume MH = G. Then M >_G,
so[M, H]
= 1 or[M, H]
= M. Since the second caseimplies HN ? M,
wehave,
in both cases,[M, m S HN,
as wanted. 0The next lemma states some trivial
properties
of andThe
proof
isstraightforward,
and we omit it.LEMMA 1. Let N be a normal
subgroup of
G. Then:The next
proposition
shows that thesubgroups (G)
anda)(G)
en-joy
aproperty
which is not satisfiedby
thecorresponding
standardWielandt
subgroups.
Since we do not need it in theproof
of our mainresult,
weagain
leave the easyproof
to the reader.PROPOSITION 2. Let G and H be groups, p a
prime.
Then:Let now n be a
positive integer,
we andby put-
ting :
= 1 andfor n b 1.
Observe that if G is
nilpotent,
then theseries - (G). n
eN,
coinci-des with the upper central series of G.
LEMMA 2. Let G be a
p-soluble
group, andK = Op , (G).
Thenand wP(G)/K
=PROOF. We first show
that,
if K =1, wP(G) = w(G).
Let T =wP(G);
then T is a p-group, since
Op - (T)
= 1 and all normalp-sections
of T arecentral. If H sn
G,
then T normalizes OP(H)
because OP(H)
isp-perfect
subnormal and T is a normal
p-subgroup.
Also[T, 0 P (H)l
s T n(H)).
On the otherhand,
T normalizesby
defi-nition and
[T,
In conclusion:Thus T s
~(G).
Since the reverse inclusion istrivial,
weget
T == cJ(G).
,Let now K #
1,
and H ap’-perfect
subnormalsubgroup
of G. ThenH is normalized
by K
and so[K,
n H Hence K~cJ~(G). This, together
with Lemma1, gives:
by
the case discussed above. Tofinish,
we have to show that cc úJP
(G)/K.
LetgK
E H ap’-perfect
subnormalsubgroup
of G.Then H = is normalized
by g
andHence g E
(;jP(G), concluding
theproof
of the lemma.PROOF. This is because all
p’-perfect
subnormalsubgroups
of G liein Op’(G).
By Proposition
1(ii),
for every finite group G there is a minimumpositive integer n
such thatwn (G)
= G. We call such n thestrong
Wie-landt
length
ofG,
and denote itby wl(G).
It is then clear that~
wl(G),
and a bound onwl(G) implies
a bound onwl(G). However,
we donot know the exact relation between and
(and
thus bet-ween the two
corresponding lengths).
Inparticular
we do not knowwhether there exists a
positive integer
m such thatw(G) S wm (G)
forall groups G.
LEMMA 4. Let P be a normal
p-subgroup o, f
the group G.Suppose
that P has
nilpotency
elass d and thatP / P’ is
Then
(G), where g(r, d) = (2d - 1 ) r - 2d -1
+ 1.PROOF.
By
induction on d.If d =
1,
then P’ =1,
so P Swpr (G) and,
infact, g(r, 1) _ (2 - 1) r -
-1+1=r.Let now d > 1. Write
Wo
=(the
last non trivial term of the lower central series ofP). Then, by
inductivehypothesis:
The
Xn
is normal in G for each n. We show thatXn /Xn - 1
is contained in for every n =1, ...,
2a - 1.Let H be a subnormal
p’-perfect subgroup
of G. Then is nor-malized
by P,
because P is a normalp-subgroup
and is subnor-mal and
p-perfect.
Thus Lett, s E ~ 1,
...,a)
and t + s == n + 1
(n
=1,
...,2a - 1).
Then:(observe that,
if t =1, [Wt -
1, =[Wo ,
=1).
Similarly: [W~ , H,
0 P(H)
1. Since 0 P(H)
1 is normal inPH, by
the threesubgroups
lemma weget:
This is true for all
pairs (t, s)
with t + s = n + 1 and soshowing
that 1 is contained inWe have therefore for every n =
1,
..., 2a -1. Inparticular
Since
P/P’
is contained in we conclude thatand:
REMARKS. The same
argument
of thepreceding proof
can be usedto show that if N is a
nilpotent
normalsubgroup
ofG,
andN / N’
is con-tained
in wr (G / M ’)
then N -5Wk (G),
where kdepends only
on r and onthe
nilpotency
class of N. In thisform,
the result is in the samespirit
ofa well known
nilpotency
criterion of P. Hall[7], stating
that if both N andG / N’
arenilpotent
then G isnilpotent
of class whichdepends
onthe classes of N and of
G/N’.
We have notattempted
to find the bestbounds for k above or
g(r, d)
of Lemma 4.Also,
it is conceivable that asimilar result holds for the standard Wielandt
series,
but we have notbeen able to find a
proof
in this case.3. The main result.
We
begin
this section with three knownresults,
that will be used in theproof
of our main theorem.LEMMA 5. Let A be a normal
p-subgroup of G, compLemented by Q
in
G,
and let S be ap-perfect
subnormalsubgroup of
G. ThenS =
(,S n A)(S
nQ).
PROOF. Since S is subnormal in G:
LEMMA 6. Let
H,
K besubgroups of G,
such thatH,
K are subrcor- mal in J =(H, K),
and let X be the classof
soluble groupsof p-length
at most n
( for
afixed
nEN).
ThenPROOF. This follows from the fact
that,
for every set n ofprimes, H,
K sn J =(H, K ) implies
07r(J) = (0’~ (H), 0’~ (I~~,
and induction onn.
LEMMA 7
(Shult [12];
Carter and Hawkes[3;
Theorem5.15]).
Let F be a saturatedformation
and G a soluble group whose F-residualG F
is abelian. Then
G F
iscomplemented
and any twocomplements
areconjugate.
THEOREM 1. There exists a
function
h: I~T X ~ ~ 1~ such thatif
Gis a soluble group
of p-length
n, and all subnormalp-sections of
G haveclass at most c, then
PROOF. We
proceed by
induction on thep-length n
of G.Observe that if A is a normal
p-subgroup
ofG,
then forevery
prime
q ~ p.Thus,
to prove theTheorem,
it isenough
to showthat
Op(G)
is contained in~~~~,
n)(G).
If n =
1,
thenby
Lemmas 2 and 3:for i =
1, 2,
..., c. HenceOp , , p (G) _ c~ p (G).
Weput h(c, 1)
= c.Let now n > 1 and suppose that we have
proved
the assertion forn - 1 and thus found
h(c, n - 1).
Let X be the class of all soluble groups ofp-length
n -1;
then X is a saturatedformation,
and it is closedby
ta-king
normalproducts.
Given a group K we denoteby K
its X-residual.We prove the
following
fact:(*)
let G be a soluble group ofp-length
n. Assume that all subnormalp-sections
of G have class at most c and that G is an abelian p- group. Then G Sc.~ ~~c, n -
1)(G).
Let G be a minimal
counterexample.
Write A = G and let N =Gx
bethe X-radical of
G;
alsoput
d =h(c, n -1 ).
1) Op - (G)
= 1 andG / N is
a p-group.Let K =
Op’ (G);
thenAK/K
=( G/K )
is abelian and a p-group; ifK;e 1 then
AK/K S
and so,by
Lemma2,
A S~g(G).
HenceK = 1. Furthermore
(G). Since, by
Lemma3,
_S
15$/ (G),
our choice of Gimplies
G =0 p ~ (G).
NowG / N
hasp-length 1,
and
Op, (GIN)
=1,
soG / N
is a p-group.2)
A is notdecomposable
as the directproduct of
two proper G-inva- riantsubgroups.
Suppose
thisfalse,
and let R x ,S = A be a proper direct G-decom-position
of A. Now(G/R) = A/R
so,by
our choice ofG,
S
~ d (G/R).
For i =1,
...,d let Ri /R (G/R) f1 A/R
andXi
=Ri
flS;
thus
XiR
=Ri . Now, by
Lemma7,
A iscomplemented
inG,
sayby ~o .
It there follows that R is
complemented
in Gby Q = SQO .
Let H be ap’- perfect
subnormalsubgroup
of G.Then,
fori = 1, ..., d (and
withRo
=R):
In
particular [Ri , H]
Now0 P (H)
isp-perfect, whence, by
Lemma5,
nQ)(OP(H)
nR).
We have therefore:showing
thatXi/Xi-I
1 is contained for all i =1,
... , d.Thus ,S =
Xd s Applying
the sameargument,
we obtain also~ d (G)
and socontradicting
the choice of G. Thus2)
isproved.
Let now M = {H; N H G, H A and T = A for all T > H}. Ob-
serve
that,
since N =1,
0. Let H E m.By
Lemma7,
H iscomple-
mented in H
by Q ,
say, and all suchcomplements
areconjugate.
LetCH= An QH ;
then 1 ~ and A = H xCH .
Observe inparticular
that H and
CH
are normalsubgroups
of N.3)
For every H E ~,Since
H/H
hasp-length n - 1, A/H
is containedby
our inductivehypothesis.
Then followsby
the sameargument
used inpart 2).
Thus if we show that[CH , S] :5
S
OP(S)
for anyp’-perfect
subnormalsubgroup
,S of G not containedin H. Let therefore S be such a
subgroup,
andput Si
= N,S.Then,
in
particular, S,
> N and(S, H)
> H.By
Lemma 6:and,
since _H E 3l1: S ~ H_ _(S, H~
= A.Let
X/S
=Op, (SI /9).
I claim that S =[A, X].
Infact,
if V =[A, X],
then V
S
becauseA/S
andXlg
are normalsubgroups
ofS1 /S
ofcopri-
me order. 0_n the other
hand, AX / Y is nilpotent and,
ifP/S
==
°P’, P (Si IS),
P/AX
is a p-group. Itfollows
that P/ V
hasp-length 1,
and so
belongs
toX;
thisyields
V andconsequently
S = V.Now,
X:5N,
so X normalizes both H andCH .
Thus:Hence we have: A = S · H =
H[CH , X],
whichgives [CH , X]
=CH .
The-refore S >
CH .
Inparticular CH ,
and so[CH ,
Thisshows that
4) Proof of (*)
concluded.c 3l1.
By
induction or r= I oz 1,
we show thatIf r = 1: A = H x
CH (where K = IHI) by
definition ofCH .
Let r > 1 and let:no By
inductivehypothesis
A= Co
xHo ,
where
Co
=(CK ;
K E:no)
andHo
=f l K.
LetX/H
=Op’ (H/H).
As wesaw in
point 3),
H =[A, X].
Also[CH,
=1,
and thisyields CH
=CA (X),
as X actsby coprime
action on the abelian group A.Now, Co
andHo
are normal inN, whence,
inparticular, they
are normalizedby
X. Thus we have:as wanted.
Let now C =
~CH ; H
E3ll). Clearly
C~G and C iscomplemented by
T = n H in
A, by
what weproved
above. But T is also normal inG,
H E M
so,
by point 2),
T =1,
becausecertainly
C # 1. Hence C = A.By point
3), CH ~ d (G)
for all H E 3l1; thus a contradiction which concludes theproof
of claim(*).
We now finish the
proof
of the Theorem. We have d =h(c, n -1 ) by
inductive
hypothesis.
Let G be a soluble group ofp-length n,
all of who-se subnormal
p-sections
havenilpotency
class at most c.By
Lemma2,
we may assumeOp’ (G)
= 1. Let P =Op (G).
ThenG / P
has Hence A =
P,
where X is the class of soluble groups ofp-length n -
1.By (*)
By applying
Lemma4,
seget
A _~,9~d, ~> (G).
SinceP/A
is contained in~(G/A) by
inductivehypothesis,
wefinally
havewhere
h(c, n)
=g(d, c)
+ d =g(h(c, n -1), c)
+(c, n -1 ), which, by
what we observed at the
beginning
of theproof,
it isenough
to conclu-de that P is contained
Now observe that the function h of Theorem 1
(as
well as the fun-ction g
of Lemma4)
does notdepend
on theprime
p. Thus if weput:
we obtain the
following
COROLLARY 1. Let G be a soluble group.
(i) If
G hasp-Length n,
an all subnormalp-sections of
G haveclass at most c:
(ii) If
G hasFitting length l(G)
and allnilpotent
subnormal sec-tions
of
G have class at most c:PROOF. It follows
easily
from Lemma2,
Theorem1,
and induction on n. mWe restate
part
of thecorollary
to make moretransparent
ourappli-
cation to the classes
Bm.
THEOREM 2. Let X be a class
of
soluble groups, closedby
homo-morphic images
and normalsubgroups,
and suppose that everynilpo-
tent group in X has
nilpotency
class at most c. Then:for
all G in X.This in
particular applies
to the classeBm (m E N); they
are closedby homomorphic images
and normalsubgroups and, by
a theorem of J.Roseblade
[11], nilpotent
groups inBm
havenilpotency
class boundedby
a function of m. Thus Theorem 2yields:
COROLLARY 2. There exists a
function ,f:
N x ~T -~ I~ suchthat, for
every soluble group G:An
interesting particular
case is the classB2.
In fact if G is a solublegroup in
B2 (i.e.
ifb(G) 5 2),
thenI(G) 5
4(see [5]).
Hence we haveCOROLLARY 3. Soluble groups in
B2
have bounded(strong)
Wie-landt
length.
By
combined results of H. Heineken and S. K.Mahdavianary,
nil-potent
groups inB2
have class at most three. Thusf(3, 4)
of Theorem 2provides
such a bound.However,
directcomputation
of itgives
a ra-ther unrealistic
value;
we leave it open theproblem
ofdetermining
theexact bound for the Wielandt
length
of a finite solubleB2-group (I
donot know of any
example
of a finite solubleB2-group
whose Wielandtlength
exceeds3).
Acknowledgement.
I amgrateful
to R.Bryce
for discussions on thesubject
of this paper, whichparticularly helped
to state the results in thepresent generality.
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Manoscritto pervenuto in redazione il 15 novembre 1990.