R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
R OLF B RANDL V ALERIA F EDRI L UIGI S ERENA
On self-centralizing Sylow subgroups of order four
Rendiconti del Seminario Matematico della Università di Padova, tome 95 (1996), p. 189-199
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ROLF
BRANDL (*) -
VALERIAFEDRI (**) -
LUIGISERENA(**)
1. - Introduction.
A well-known result of Gorenstein and Walter
[3],
which confirms aconjecture
of R. Brauer[ 1 ],
states that if G is a finite group of order4g,
g
odd,
with aself-centralizing Sylow 2-subgroup,
then it contains a nor-mal
subgroup
N of odd order such thatG/N
isisomorphic
either to aSylow 2-subgroup
of G or toPSL(2, q) where q
is aprime
power, q= 3,
5 mod 8. Moreover
(see [2,
p. 348 and p.356]),
N must bemetanilpo-
tent.
The
objective
of this paper is toimprove
upon this result in the nonsoluble case
by giving
moreprecise
information on the structure of N.Indeed,
we have:THEOREM Let G be a non soluble group with a
self centrati- zing Sylow 2-subgroup of
order 4 and let N =O(G)
be the maximal nor-mal
subgroup of
Gof
odd order. Then oneof
thefollowing
holds:a) G/N = PSL( 2, q)
with q =pf > 5, q = 3, 5 mod 8,
and N is ap-group;
b) G/N = PSL(2, 5)
and N isnilpotent.
This result is best
possible
in the sensethat,
in Caseb)
of the theo- rem, N need not be a5-group (see Example 2.4).
Moreover the derivedlength
of N ina)
andb)
is not bounded. In fact inExample
3.3. groups Gare constructed with a
self-centralizing Sylow 2-subgroup
of order 4(*) Indirizzo dell’A.: Mathematisches Institut der Universitat, Am Hubland 12, D-97074
Würzburg, Germany.
(**) Indirizzo
degli
AA.:Dipartimento
di Matematica «U. Dini», Universitadegli
Studi di Firenze, VialeMorgagni 67/A,
1-50134Firenze, Italy.
such that N is a p-group of
arbitrary
derivedlength
and.= P,SL(2, p) where p
is aprime
suchthat p = 3,
5 mod 8.In the course of the
proof
of thetheorem,
we shallanalyze
actions ofQ
=PSL(2, pf)
on a group N of odd order such that aSylow 2-subgroup of Q
acts fixedpoint freely
on N. Note that a series of papers deal witha somewhat similar situation in which the orders of
Q
and N arecopri-
me, and
CN(Q)
=1,
see forexample [9], [13]
and the referencesgiven
there.
By
contrast with ourresult,
in that situation N need not benilpotent.
In
[4]
it was shown that ifQ
=P,SL( 2, 2f ), f ; 2,
acts on a2-group N
such that an element of order 3 acts fixed
point freely,
then N is ele-mentary
abelian.If f = 2
and if an element of order 5 acts fixedpoint freely
onN,
then thenilpotent
class of N is ~3(see [5]
and[10]).
Inthese cases, N is of bounded
nilpotent
class(note
that this contrastswith
Example 3.3).
In the
following,
we denoteby
G =[N]H
that G is asplit
extensionof its normal
subgroup
Nby
acomplement
H.Moreover, An
is the alter-nating
group on nletters,
henceA4
=PSL( 2, 3)
andA5 = PSL( 2, 5).
Inaddition,
p will denote aprime
and q isalways
a power of p. The order of theelement g
is denotedby o(g).
All other notation is standard and can be found in
[2], [6]
and[8],
forexample.
All groups in this paper are finite.2. - Actions of
PSL(2, q).
We start
by noting
some well-known facts that will be used several times in thesequel.
LEMMA 2.1. Let G be a group and let N =
O(G).
Assume thatG/N = P,SL(2, q)
where q--- 3, 5
mod 8.If
S ESyl2 (G)
andCG (S) _ S,
then every involutions
PROOF. Since all involutions of G are
conjugate,
the result followsfrom
[2,
p.347].
PROPOSITION 2.2. Let G be a Frobenius group with kernel N and let F be a
finite field of
characteristic r notdividing IN I.
Let M be anFG-module and assume that
CM (N)
= 0. Let A be a Frobeniuscomplex-
ment
for
G. Then M has a basis which ispermuted by
A with orbitsof
size . In
particular, if I M [ = rt ,
we have[ =rt/IAI.
PROOF. See
[8,
p.270].
We now deal with extensions of a group N of odd order
by PSL(2, q),
which have aself-centralizing Sylow 2-subgroup.
The follo-wing
result can be read off from the(modular)
character table ofPSL(2, q)
and mayalready
be known. Wepresent
here anelementary proof.
PROPOSITION 2.3. Let G =
PSL(2, q)
with q =pf ,
q= 3,
5 mod 8and assume q > 5 . Let S E and let M be a nontrivial and irre- ducible module
for
G over afinite field
Fof
characteristic r, whereThen
CM (S) ~
0.PROOF. We
proceed by
way of contradiction.Suppose
thatCM(S)
= 0.First,
Lemma 2.1implies I M I
= for somepositive
inte-ger
h,
and for each involution Q ES,
we haveI
=rh .
Take u E Gwith
o(u) = (q - 1)/2. Since q # 3,
we have thatNG ((u))
is a dihedralgroup and we can write U = y 1 y 2 where y 1, y 2 are suitable involutions in
NG ((u)).
We have[M, y 1]
n[M, y2] ~ CM (u).
Set c[M, y 1]
nn
[M, y 2].
ThenC * ~ ~ rh , since [M, I = r2h
for i =1,
2. Now let P be aSylow p-subgroup
ofG,
normalizedby
u. ThenMi : = [M, P]
andM2 : = CM (P)
are invariant under the action of u. Since M =Mi E9 M 2,
we have
CM ( u )
=CM2(U).
Inparticular,
every element of C*can be written in a
unique
way as sum of an element ofCMl (u)
and anelement of
CM (u).
We have C * fl = o. Infact,
if x E C * flfl
CM 2 (u)
then~x~
is invariantby
P andNG (u).
So(x)
is invariant for G.Since M is irreducible for
G,
weget x
= 0. Hence we have~
rh .
On the otherhand,
sinceM1
is a faithful module forNG (P)
which sa-tisfies the conditions of
Proposition 2.2,
it follows thatI
=final contradiction.
It may be observed that there are modules for
A4
andA5
such thatthe
previous proposition
does not hold:EXAMPLE 2.4.
a)
Let F beafield of
characteristicdifferent from
2and let
with entries in F. Then G =
A4 .
Let M be the natural vector space on which G acts and let S ESyl2 (G).
Then M is an irreducible andfaithful
FG-module and we have
CM (,S)
= 0.b)
Now let G =A5,
choose P ESyl5 (G)
and set H =NG (P).
Let Fbe
a fzeld of
characteristicdifferent from
2.Suppose
thatM1
is the non-trivial FH-module
of
dimension 1. Let M =MG
be the induced module.It is easy to prove that
Syl2 (G),
thenCM (S)
= 0. SoCM (S)
=0 for
every
composition factor
Mof
M.If we
take two such modulesM1, M2 over fields of different
odd cha-racteristic,
then the naturalsplit
extensionof M1 EÐ M2 by
G shows thatin
part b) of the theorem,
the group N need not a p-groupfor
noprime
whatsoever.
The
following
resultprovides
a criterion when every proper sub- group ofPSL(2, q)
has trivial intersection with at least oneSylow
2-subgroup.
It will be seen in theproof
of 2.6 that thestrange-looking hypothesis
of thefollowing
lemma is satisfied in our case.LEMMA 2.5. Let G =
PSL(2, q),
where q =3, 5
mod 8 and assumethat q > 5.
Suppose
that the propersubgroups of G
are either soluble orisomorphic
toA5. If H G,
then there existsS e Syl2(G)
such thatHfIS=1.
PROOF. Of course it suffices to consider the case when
H ~ I
is even.By
Dickson’s theorem(see [6,
p. 213f.])
and ourhypothesis,
the sub-groups of G are the
following:
1)
Dihedral groupsDz
of order 2z with ±1)/2.
2) Groups isomorphic
toA4.
3) Groups isomorphic
toA5.
4)
Asubgroup Q
ofNG (P)
where P eSylp (G)
if q = 5 mod 8.If Q is an involution of G and
S,
S areSylow 2-subgroups
of G sati-sfying a
e S nS,
then(S, S) CG (a).
If we denoteby
na the number ofSylow 2-subgroups containing
o, we have:a)
na =(q
+1)/4 if q
--- 3mod 8, b)
na =(q - 1)/4 if q
= 5 mod 8.Now let !1 H be the number of involutions which are contained in the
subgroup H
of G. Then we have:when q
= 5 mod 8 .So
if p
* is the maximum number of involutions which are contained in aproper
subgroup
ofG,
we have:if q = 5 mod 8 and
if q
--- 3 mod 8 .If m denotes the number of
Sylow 2-subgroups
which intersect non tri-vially
with a propersubgroup H
ofG,
we have:On the other
hand,
the number y ofSylow 2-subgroups
of G isequal
toq(q
+1)(q - 1)/24. By
an easy calculation weget
m y forq ~
11.Thus if q > 5 and H is a proper
subgroup
ofG,
there is aSylow
2-sub-group S such that 1.
The next two results deal with modules for the groups
occurring
inthe theorem.
They
will be used to excludenonnilpotent
normal sub- groups0(G).
PROPOSITION 2.6. Let a be the class
of all
groups Gfor
which there exists a normalsubgroup.
Nof
G such that:a)
N is an abelian p-group(possibly
theidentity) 2;
b) G/N = PSL( 2, q)
where q =pf , q ~ 3, 5
andq = 3,
5 mod 8.If G
E rz and M is an FG-module where F is afinite field of
charac-teristic # 2, #
p, thenfor
all S ESyl2 (G)
we have 0.PROOF.
By
way ofcontradiction,
assume that there exists a counte-rexample (G, M)
where G E a and M is an FG-modulesatisfying
thehypothesis
of theproposition.
Choose thispair
such thatG ~ I
+I M I
isminimal. Then
(G, M)
has thefollowing properties:
1)
If p #3, 5,
thenG/N
=PSL(2, p).
In
fact, by
Dickson’s theorem there exists asubgroup
H ~ G suchthat
HIN =- PSL(2, p).
If p ~3, 5,
we have H E a and H contains someSylow 2-subgroup S
of G. If H G then 0by minimality
of(G, M).
But this is a contradiction and so we have G = H.2)
If p = 3 or p =5,
thenG/N = PSL( 2, 3f)
orG/N
=P,SL( 2, 5m)
where
f
and m areprimes.
In
fact,
if G E ci wehave f > 1,
and so there is aprime t
with tf (in
asimilar way there is a
prime
t such thatt I m). By
Dickson’s theoremthere exists H ~ G such that
H/N
=P,SL(2, 3t)
orH/N = P,SL(2, 5t).
As in
1),
it follows that G = H.We observe that
by 1)
and2),
theonly subgroups
ofG/N
are eithersoluble or
isomorphic
toA5 ,
so that thehypothesis
of Lemma 2.5 holds.3)
M is an irreducible and faithful FG-module.In
fact,
letM1
be an irreducible FG-module withM1
M.By
mini-mality
of(G, M),
we haveCMl (S) ~
0 and so0,
a contradiction.Thus
M1
= M and M is irreducible. Let K be the kernel of the action of G on M. Of course, we have K ~ N and(GIN, M)
satisfies thehypothe-
ses of the
proposition.
If K #1,
then 0 ~CM (SK/K)
=CM (S) by
mini-mality
of(G, M),
a contradiction. So we have K = 1.4)
This follows from
Proposition
2.3.5)
N is not contained inZ(G).
In
fact, suppose N Z( G ). Then, by properties
of the Schur multi-plicator
ofPSL(2, q) (see [6, p. 646]
and[12, p. 257]),
we have G = NLfor a suitable
subgroup
L ~ G. Also(L, M)
is acounterexample,
butthis is
against
theminimality
of(G, M).
6)
N is notcyclic.
Otherwise N would be
central,
but this contradicts5).
7)
M is an induced module.Let M be a
homogeneous component
ofM,
considered as FN-modu- le.Suppose
that M = M. Since N is notcyclic,
the kernel of the action of _G on M is nontrivialagainst
the faithfulness of the action of G on M. So M # M and[6;
p.565] implies
that M is induced.8)
Final contradiction.Let I be the stabilizer of M in G. So
7) implies
M =(M)G .
We haveI/N G/N =- PSL(2, q). By
Lemma2.5,
there existsSN/NE SyI2(G/N)
with i.e. SN n I = N. Then = N. Since
= 1 it follows that S n 1= 1. Let T be a set of double coset
representatives
withrespect
to S and I in G. We may assume 1 e T.By Mackey’s
theorem[6,
p.557],
we have:Since
1,
we have that is direct sum ofregular
FS-modules. Therefore the above
implies
thatCM (S)
0 and soCM(S) # 0.
The
following
deals with groupshaving A5
as nonsoluble chief fac- tor :PROPOSITION 2.7. Let ~3 be the class
of all
groups Gfor
which there exists a nontrivial normalsubgroup
N such that:a) G/N = A5 ;
b)
N is abelianof
oddorder;
c) If
S is aSylow 2-subgroup of G,
thenCN (S)
= 1.If
G e 1B and M is afaith, fut
and irreducible FG-module where F isof
oddcharacteristic,
thenCM (S) ~
0.PROOF. Let G E 83 and let S E
SY12 ( G ).
Assume that the FG-module M is acounterexample,
that isCM (S)
= 0. We then have:1)
M is an induced module.Since
CN (S)
=1,
we see that N isnot cyclic.
As inpart 7)
of theproof
ofProposition 2.6,
it follows that if M is ahomogeneous
compo- nent ofM,
restricted toN,
then M # M. So the stabilizer I of M is pro-perly
contained in G and M =MG .
Let L =
G/N = A5
and let H be the normalizer in L of aSylow
5-subgroup
P of L.2)
We haveI/N =
H.An
inspection
of the propersubgroups
ofA5
shows thatI/N
=H,
because
otherwise,
there would exist aSylow 2-subgroup S
of G suchthat
I/N
fl,SN/N
=N/N.
As inpart 8)
of theproof
ofProposition 2.6,
we
get CM (S) #
0. But this is a contradiction.3)
For all involutionsI,
we have = 0. Inparticular
a is notcontained in the kernel K of the
representation
of I on M.In
fact,
suppose 0. Let T be aright
transversal of I in G.Then
1) implies
that We may assume that1,
r ET,
where(Q, r)
= S. We then havea contradiction.
4) [N, a] c K.
Otherwise,
the group[[N, a]
nK)]~Q~
is a Frobenius group.But
here, [7, p. 411] implies 0, against 3).
5) N ~ I
is not divisibleby
5.In
fact,
otherwise there would exist a minimal normal5-subgroup
Rof G with R ~ N. Now R is a faithful and irreducible module for
L,
and,S acts fixed
point freely
on R.So,
from Lemma 2.1 and[7,
p. 38ff]
it fol-lows that
R ~ - 53 . Moreover, considering
the action of L on R it can beseen that
[R, I
=52 . By 4)
we have[R, or]
c K. Since M is a faithfulmodule,
weget [R, or]
But this is a contradiction because[R, Q]
is not normalizedby
P.6) CN (P)
n 1.Since
IHI)
=1,
it follows that N =(N fl K)
fl3No
whereNo
isH-invariant. Moreover
No
must becyclic.
Also we haveNo
=[No , P] fl3
?
CNo (P).
Since[No , P]
is invariant for the nonabelian groupH,
weget [No , P] = 1,
soNo c CN (P).
On the otherhand, by 4),
we haveNo c CN (a)-
7)
Final contradiction.Let N =
N1
>N2
> ... >Nh
= 1 bepart
of a chief series of G. SinceCN (H) ~
1 and( I N I ,
1 it follows that there exists a chief factorNe /Ne
+ 1 of G such thatCNe/Ne+l 1 (H) ~
1. Without loss ofgenerality,
wemay assume that N is a minimal normal
subgroup
of G. So N can be vie-wed as an irreducible and faithful
G/N-module
and we will use the ad-ditive notation. Let
No
be the trivial H-module. Thenby 6), No
is a sub-module of
NIH,
so that Hom(No,
0. Thereforeby Nakayama’s reciprocity
law[7, p. 50]
weget
Hom(Nt, N) #
0 and so there exists anon-trivial
homomorphism
fromNo
toN,
which is anepimorphism
be-cause N is irreducible.
It follows that dim N ~
6,
and Lemma 2.1implies
dim NE ~ 3, 6}.
Ifdim N = 6 then N =
Nt,
soCN (S) ~
0 and we have a contradiction. If dim N =3,
then[N, P] decomposes
into a direct sum ofregular
modu-les for
~o~ ~ by Proposition
2.2. Hence we have dim[N, P]
= 2. SinceCN(P)
fl0,
it followsby 6)
that dimCN(a)
= 2. But this isagainst
Lemma 2.1. m3. - Conclusion.
3.1 PROOF OF THE THEOREM. Let S be a
Sylow 2-subgroup
of G. Sin-ce G is non
soluble,
the result of Gorenstein and Walter[3] implies
thatG/O(G) = PSL(2, q) with q = 3, 5
mod 8. Let N =O(G). By [2, p. 348],
we know that N’ is
nilpotent.
Wesplit
theproof
into two cases:q > 5: Let t be a
prime dividing I NIN’ I
and let t # p.By Proposi-
tion 2.3 we have
CN~N- (S) ~
1. SoCN (S) # 1,
a contradiction.Hence
N/N’
is a p-group.By
way of contradiction suppose that N’ is not a p-group. As N’ isnilpotent,
we can choose achief factor
N’ /K
of G which is ap’-group.
But thenProposi-
tion
2.6, applied
to M =N’ /K, yields CN, (,S) ~ 1,
a contradic- tion.q = 5:
Suppose by
way of contradiction that N is notnilpotent.
Thenthere exists a chief factor N of G below
N,
which is central inN’ ,
but not in N. So N is a faithful and irreducible module for which satisfies the conditions ofProposition
2.7.Then 1 and so but this is the final contradiction.
If G is assumed to be soluble in the statement of the
theorem,
then0(G)
need not benilpotent.
Infact,
it is easy to constructexamples
inwhich G is
2-nilpotent
with aself-centralizing Sylow 2-subgroup
of or-der 4 such that the normal
2-complement
is ofFitting length
two. Forthe convenience of the reader we
give
anexample
of a group G such thatG/N = A4
where N =0(G)
is ofFitting length
two and theSylow 2-subgroups
of G areself-centralizing.
EXAMPLE 3.2. The group H =
A4
can actfaithfully
and irreduci-bly
on a vector space Vof
dimension 3 overGF(3) (see Example 2.4).
Let
VI
be asubspace of dimension 1,
invariantfor
,S ESyl2 (H).
LetV2
bean S-invariant
complement of VI
in V Set T =[V] ,S and
G =[V]
H.There exists an irreducible T-module
MI of
dimension 2 overGF(5)
with kerneL
V2 .
Let be the induced module. Then it is easy toverify
thatCM (S) _ ~ 0 ~.
Set N =MV,
then we haveCN (S) _ ~ 0 ~
and Nis not
nilpotent.
Finally,
for everyprime p > 5,
we construct a finite group G with aself-centralizing Sylow 2-subgroup
such that =PSL(2, p)
andOp ( G )
is ofprescribed
derivedlength.
We are indebted to the refereefor
greatly improving
upon ouroriginal example.
EXAMPLE 3.3. Let p be a
prime, p
=3,
5 mod 8. LetZp
be thering ofp-adic integers
and consider the groupSL(2, Zp)
and its normal sub-group N
consisting of
all matricesof
theform I
Notethat N is the group
ml1,
i,1of[6,
p.387].
Let F =PSL(2, Zp).
Weidentify
N with a
subgroup of F,
so that we haveFIN
=PSL(2, p).
Moreover(see [6,
p. 387ff.])
it is known thatfor
everypositive integer d,
thefac-
tor group
N /NCd)
is a p-groupof
derivedlength precisely
d. It is easy tocheck that G =
F/NCd)
isa finite
group with aself-centralizing Sylow
2-subgroup of
order 4 in whichOp (G)
isof
derivedlength
d.Acknowledgment.
The first named author wishes to thank his friends(including
Isabella andAlessandro)
andcolleagues
in Firenzefor their warm
hospitality during
the time ofwriting
this paper and the G.N.S.A.G.A. of the C.N.R. for its financialsupport.
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Manoscritto pervenuto in redazione il 28 ottobre 1994
e, in forma revisionata, il 9 febbraio 1995.