A NNALES MATHÉMATIQUES B LAISE P ASCAL
R AFFAELE S CAPELLATO
L IBERO V ERARDI
Bases of certain finite groups
Annales mathématiques Blaise Pascal, tome 1, n
o2 (1994), p. 85-93
<http://www.numdam.org/item?id=AMBP_1994__1_2_85_0>
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BASES OF
CERTAIN FINITE GROUPS
par Raffaele Scapellato and
LiberoVerardi
Resume.
Ce
travail estconsacre
aux groupes dont les ensemblesgenerateurs
minimaux
jouissent
deproprietes proches
a celles des bases d’une matro’ide.1 Introduction
Recently,
considerable attention has beenpaid
toproblems
ongenerators
of groups. This concerns e.g. smallgenerating
sets of finitesimple
groups(see
the survey
[2]). Generating
sets of groups are also relevant from thegraph-
theoretical
point
of view. becausethey give
rise to connectedCayley digraphs (see
e.g.[3] ). Furthermore.
Beneteau[1]
revisited Burnside basistheorem, by restating
it in terms of matroids.Properties
of this kind have also been studiedby
Jones[5]
forsemigroups,
oftenreducing questions
fromsemigroups
to groups.Therefore,
it seems to beinteresting
toinvestigate
groups whose behaviour with respectto generators
isfairly
similar to that of p-groups. Thisstudy
canbe
performed
with thehelp
of an ’abstract* notion ofindependence.
In orderto make this
idea
moreprecise,
we recall thedefinition
of a matroid. Let S bea
set,
and let c :P(S)
--~P(S)
be amapping satisfying,
for all X C S and all’ ’
x. the
following
conditions: ’1..~ C
c(:~-);
:2.
c(c(.~’ )) = e(.~’ ):
3.
c(.V) is
the unionof all c(Y)
for allfinite
K CX;
4. if
y ~ c(.~’ )
and ,r~ c(~’, y)
then y~ c(.K, x),
..1Work
partially supported
by the BIURST.86 Raffaele
Scapellato
and Libero. VerardiThe
pair (S. c)
is then called a matroid. This is one of the many.equivalent
definitions of a matroid
(see
e.g.[6]),
and it seems to be the moreappropriate
for our
purpose...B
subset .~’ of S is said to bec-independent
ifc(~’ j ~ c(.x )
forall Y C X. We say
that
X isa generating
set ifc(X) =6*. Also,
a basis is anindependent generating
set. We will make use of the above definitions even for the moregeneral
case of amapping
csatisfying only (1)-(3).
Forinstance,
letG be a finite group, and let be its Frattini
subgroup (i.e.
the intersection of its maximalsubgroups). Setting c(X)
=X, 03A6(G)
> we obtain such amapping. Following ~7~, [8],
a subset .~’ of a group is said to beindependent
ifit is
c-independent
with respect to this c.Accordingly
with ourprevious
paper[7],
a finite groupG,
is said to be amatroid group if it satisfies the
following
two axioms:(Ml)
the minimalgenerating
sets of G areexactly the
bases of a matroid~~~
(G,):
,(~I2)
eachindependent
subset ~’ of G iscontained
in a minimalgenerating
:setofG:’ :- .. ’ " z.: .
Note that in group
theory
the terms ’basis’ and ’minimalgenerating
set‘ aresynonymous. In this paper we will use the
latter,
in order to avoid confusionwith the ’basis’ of a matroid.
As a consequence of Burnside basis
theorem,
allfinite
p-groups are matroid groups. We studied these groups in( ~j , [8], obtaining
acharacterization
for them((7~.
Lemma 1.1 and Theorem2.5),
that we summarize below in thefollowing
~
way. (Here Op(G)
is themaximum
normalp-subgroup ’of G).
Theorem 1.1 Let G be a
finite
group, and let H =G/03A6(G)
. Then G is amatroid group
if
andonly if
oneof
thefollowing
holds:(1)
G is a p-g~roup: .’ °
(2)
there are twopri-mes
p. q with p ~l. (mod q)
suchthat |H
:Op(H)|
= q.Op(H) is elementary
abelian and everysubgroup of Op(H)
is normal(but
notcentral)
in H .By
the above, all matroid groups are solvable. But a finitesimple
group cannot evensatisfy (M1),
because its minimalgenerating
sets have different numbersof
elements. The situation is less obvious for(VI2).
We are aware ofno non-solvable finite group
satisfying
thiscondition,
and weconjecture
that nosuch group exists.
However,
the solvable case is solvedby
Theorem 1.2 below.(Here F(G )
is theFitting subgroup
ofG,
i.e. its maximum normalnilpotent subgroup).
Theorem 1.2 Let G be a
finite
solvable group, and let H =G/03A6(G).
Then Gsatisfies (1~~2) if
andonly if
oneof
thefollowing
holds:(1)
H isabelian ;
(2) F(H) )
hasprime
index in H and all itssubgroups
arenor:mal
in H.Another characterization
of finite
matroid groups will begiven by using
thefollowing
concept. Let(S. c)
be a matroid and let ~ C S. Then...)[ is said to be c-hierarchical if its elements can be ordered in a sequence x~, x~~, ..., xm such that c(x1, x2, .... xk-1) ~c(x1,
x2, ..., xk) for all k =2, ..., m,
andc(~) ~
For the case of groups(with c(.~’ )
_ .~,’> ),
thisconcept
was introducedby Hamidoune,
Llado andSerra,
who found in[4]
the values ofconnectivity
ofCayley digraphs
associated withhierarchical generating
sets.At a first
glance,
most finite groups have hierarchicalgenerating
sets that arenot minimal. On the other
hand, we
are able to describe the groups all whose hierarchicalgenerating
sets areminimal.
This is the content of thefollowing
theorem, which will be
proved
asa corollary
of aslighty
moregeneral
result(Theorem 3.2 ).
Theorem 1.3 Let G be a
finite
group. Then all hierarchicalgenerating
setsof
G are minimal
generating sets if and only ifC
is a ’matroid group with = I.The rest of the paper is
mostly
devoted to theproof
of Theorems 1.2 and 1.3.Terminology
and notation arestandard,
and can be found in[9] (but
wewrite 03A6(G)
andF(G) for the
Frattini andFitting subgroups).
we will also usesome obvious
shortenings, such
as .~’B
x for ~’B ~x }.
2 Groups satisfying (M2)
First of
all,
we want to proveTheorem
1.2. Note that a group G satisfies(NI2)
if and
only
ifG/ ~ (G’ ) does. Hence,
in Lemmas2.1-2.4,
as w ~ell as in theproof
‘
of the
theorem,
wewill assume that
G’ is afinite solvable
groupsatisfying (M2)
and~(G)=1.. , ’
Lemma 2.1
Euery subgroup
normal in G.PROOF. As 03A6(G) = 1, Op(G)
iselementary
abelian. Let X be a minimalgenerating
set ofOp(G’),
andlet
:~’ U V be a minimalgenerating
set of G, ThenX
U y, is independent
forall y
Er’ . ~ Hence
for each x E.~’ ,
.~’B
> is asubgroup
y >, maximalof
index p. Thus.~’ Bx
> !.~’ Bx: y
>,. so that :’~’
B x
> 4G for all x E‘:~’. Since is the directproduct
of theOp(C).
we are done./
Lemma 2.2 The
group G is supersolvable
and metabelian.Also,
G’F(G).
PROOF. We know
that F(C)
.isabelian
and all itssubgroups
are normal in G.Therefore, by conjugation,
G induces onF(G)
universalpower-automorphisms,
that of course commute each
other.
Then this actioncorresponds
to a homomor-phism
whose kernel isCG (F (G ) )
and whoseimage
isabelian,
so
G/C’c(F(G))
is abelian.By [9], 7.4.7, p.187, G/CG(F(G))
=F(G),
then88 Raffaele
Scapellato
and Libero VerardiG/F(G)
is abelian. HenceG’ ~ F(C).
and G is metabelian. Moreover it isobviously supersolvable
because everysubgroup
of G’ is normal inC,
so thatG has a chief series with
factors
ofprime order. []
Lemma 2.3
Every
elementof G B F(C) belongs
to acomplement of F(G).
PROOF. If X = {x1, x2. .... xr} is a
minimal generating
set ofF(G)
and .Y U Kis a minimal
generating
set ofC,
thenthe subgroup
A/ == ~ > isa com-plement
ofF(G).For otherwise, letting
L’ = .V1,
we can havek =
[
r.Hence, F(G)
==~i~2~..’.Tr,L
> and so G =.ri,j:2 ...~r. ~
> a contradiction. ’If y
% F(G)
then.Y~
= .Y U ~ isindependent,
becauseX~’B
y >== X >==F(G). By
Lemma2.L A~B.r,
>= >=XB~
> ~>~ JT
>for all ~~ e AB Hence there is V such that X’ is a
minimal generating
setof G.
Letting V
= V U ~/. also .Y U F is a minimalgenerating
set of (7. Thus.V = F > is a
complement
of Gcontaining
K.N
Lemma 2.4
IfG is
nonabelian. then the indexof F(G)
in G isprime.
. PROOF.
By contradiction,
suppose thatC/F(C)
has notprime
order. Let - Y ={~i...~r}
be .a minimalgenerating
setof F(C),
andC C B F(C).
Inview of Lemma
2.3,
we can assumethat y
hasprime
order. Let H ==.Y,
y >.°
By
Lemma2.2.
we have G’F(G) H,
so JFf ] G.Letting
X’ ={xy|x
6.Y},
we
get
H =X’, y
>. Let us show that X’ U y isindependent. Put X1
=6 .Y
B x1},
where j’i .Y is fixed. In view of Lemmas 2.1 and2.3,
wehave A’’ >== >=
AB
> x1y> ~
H.Similarly.
for j- .V we getYB.r.~>==~B.r>~>~.
.
By contradiction,
suppose that can be embedded in a minimal genera-.
ting
set.Then,
inparticular,
there is z%
jFf such thatX’~{y, z}
isindependent.
In view of Lemma
2.3.
there is acomplement
M ofF (G) containing
:.Suppose
that
Cc(~).
From Lemma 2.2 it follows that M n G’ =1,
so M is abelian.Furthermore
H M =C,
and so = G. Therefore z =Z(C).
Butby [9],
7.4.7, p. 167, F (G)
isself-centralizing,
so inparticular Z(C) ~ F(C) ~ ~,
henceH,
a contradiction. Then there is w 6 X’~ y
such that[w, z] ~
1.’
By
Lemma 2.2 we haveG’ ~ F(G), so [aB z] F(C).
If[y, z] ~
1 we canchoose w = y and then there is
xk
~ X that may bereplaced by
a power of[y, z]
having prime
order. Thisgives
rise to another minimalgenerating
set ofF(G).
Therefore .r~ ....~~.~/. ~ >= ~ ~ >= .ri~...,.r~-~.r~-i~..~r~~ >=
X’
B
xky, y. z >, a contradiction. Thus[y, z]
= 1. Since n’ X’ U y, andby
the
above w ~
y, we have w = xiy for a suitable i.Now = = = e xi >, and so we get
.
- ~ ~ >, which
implies y
e >. Therefore.Y~,~
>= >= >=A~,~
>,a
contradiction..
PROOF OF
THEOREM 1.2.
Let G be a groupsatisfying (1~I2).
We mayassume that
~(G’)
= 1.~f ’G
isnonabelian,
we must prove(2).
In view ofLemmas
2.1-2.~,
itremains to show
thatthe Sylow p-subgroups
P of G areelementary
abelian. IfP F(G)
this is clear.Otherwise,
in view of Lemma2.~:.
P nF(G)
has index p inP,
then~(P) F(G). A
non-trivialcyclic subgroup
C ofF(P)
isnormal
inG and abelian,
hence it has acomplement
in G
([9], i.~.14, p. 169).
Thus C hasalso
acomplement
inP,
a contradictionbecause C _ ~(P). Hence ~(P)
= 1 and we are done. Therefore(2)
follows.Conversely,
let G be a groupsatisfying
either(1)
or(2).
The formerclearly
imply (1~I2 ).
Assume now that
(2) holds.
Let X beindependent
and suchthat
H = .k‘ >is different from
G,
and prove that there isy ~ X
such thatX ~ y
isindependent,
too.
If
F(C)
is not contained inH,
choose any y EF(G) B
H. Now AU y
is.
independent. In fact,
for all x E .’~’ w’e have.~’ ~ x, y
>==.~C ~ x
> y>~
~ ~ >. as AB.r >~ ~.
,If
F ( G )
is contained inH
then H =F (G ).
becauseF (G is maximal
in G.Let y
E GB F(G). Clearly,
the set .~’ U Y generates G.Moreover,
for all j? E .~’,K = .~’
B
:~> F (G’ ), and
so K j G’. It follows that~’ ~ x, ;y ~=
K y >,which is a proper
subgroup of
G. This proves that .x U Y isindependent,
andso G satisfies
(1~~I2). 1
.By
Theorem1.2,
anelementary example
of a nonabelian groupsatisfying (’~i2)
isgiven by
any dihedral group.It is w orth to
mention
some consequences of Theorem 1.2.Corollary 2.5
Let Gbe a solvable
groupsatisfying (M2),
and let .4 be anabelian>
group with(|A|.|F(G)| ) =
I . Then G x .4satisfies (M2).
Corollary
2.6 Let G be a solvable groupsatisfying (M2),
let p be the indexof F(G ) in G.
Then theSylow p-subgroup of F(G’) )
is a directfactor of G. Also, if
G has no abelian direct
factors.
thenF(G)
is a Hallsubgroup of
G..
PROOF.
Let P be ap-Sylow subgroup
of G.Since Q
= and P isabelian,
we have[Q. G]
=[Q,
=1,
henceQ Z (G ).
NowQ
isnormal, .abelian.and ~(G)
=1,
soQ has a complement
C in G. FromQ _ Z(G’)
weget
Yet another easy consequence of Theorem 1.2 is the
following.
Corollary
2.7 .4finite
solvablegroup
Gsatisfying (M2)
is a matroid groupif
and
only if F{G)/~{G’)
is a p-group.’
The
following proposition provides
a criterion to prove that a group does notsatisfy (M2).
Note that a group does notsatisfy (M2)
if andonly
if it containsa maximal
independent
set that is not a minimalgenerating
set.90 Raffaele
Scapellato
and Libero VerardiProposition
2.8 Let G be a group u;ith =1,
~ JB C and let AT*be the intersection
of all subgroups of
Gproperly containing
K. Assume thatneither K nor
G, and
that K* has a -minimalgenerating
setX,
withK >= AT. .
Then
G does notsatisfy (1~I2}.
PROOF.
Suppose
.~C’ U Y >= G for some ~r CG,
and prove that the set isdependent. By
thehypothesis
on .~’ we have h’ _.~C’ >,
where .~.’’
== .~’ > UY’. Thisinclusion
is proper. Infact, .~’
is aminimal
generating
set of ~i and~’ ~ 0,
otherwise we would have K* = G.The definition of h’ *
gives
us h’ * C.~.’’ >,
so ~’C .~’’
>. Since V C!~’’,
itfollows that G = .~C’ >. But .~‘
B
K is non-empty, or else K * = K. Hence a’is a proper subset of .~ U
Y’,
and.’~’
U r’ isdependent. ’
’We can
apply Proposition 2.8 in
order to provethat..46
does notsatisfy (NI2).
Let x =(123456)
and ~i = x >. Then K* is adihedral
gr.oup of order12,
and h‘* -:VG(k }.
Now h * = x, y >, where y has order2,
and~
~
~ h =~x ~
is a minimalgenerating
set ofK.
3 Hierarchical generating sets
.Unlike those of vector spaces,
independent
sets in finitegroups
cannot ingeneral
be embedded in bases.
Namely,
thishappens only
for groupssatisfying (NI2).
On the other
hand,
each c-hierarchical subset .~’ of a group G is contained in ac-hierarchical
generating
set of G.Let G be a
(finite)
group and let F be asubgroup
of G. C’onsider themapping
cF = c :P(G)
-~P(G )
definedby c(.~’ }
_ F U .~’ >..~ subset .~’ of G will be said to beF-independent (F-hierarchical)
if it iscF-independent (resp.
cF-hierarchical). A
set X isF-independent
if andonly
if its is F-hierarchical for all of its total orders...
C’learly,
c satisfies the conditions(1)-(3)
of thedefinition
of a matroid. We will prove that(4)
holdsexactly
when G is a matroid group.Note that
(M1)
does notimply (4)
for the map c. Asimple counterexample
is
provided by
the group G =.4.~. Letting Q
be itsSylow 2-subgroup,
defineci
by }
= X > if X~ Q
C 1 andci(.Y) =
XU Q
> othenvise. Thus(G. c1) is
a matroid whose bases are the minimalgenerating
sets ofC,
so(Ml)
holds. However. G does not
satisfy (NI2),
because itscl-independent
subset~(12}{3~). {1~~(2;3}}
cannot be embedded in a minimalgenerating
set of G.Lemma
3.1 .4 group G is a ’matroid groupif
andonly if
G is a03A6(G)-matroid
group.
PROOF. A
subset of G isindependent if
andonly
if it is~(G}-independent.
We will write c =
c03A6(G).
Let G be a matroid
group. Let C
G and c G. In order to show(4),
wemay assume that .~’ is
c-independent.
Let2/ ~ c(~’). Then ~’’ =
E.k’}
is
independent
inG / ~ (G )
andy ~ (G ) ~
~’’ >.By ~7~ . p. 192,
allsubgroups
of are matroid groups,
hence
also .~’ > is a matroid group.It contains the
independent
set~’’,
soby (~I2)
each of its minimalgenerating
subsets must have at least
1 elements.
It follows that.~’’
U{ y~ (G ) }
is
independent
in hence X U y isindependent
in G’.By
the sameargument,
if ,r~ c(.~’, y)
then{x. y}
isindependent,
hencec(:~’, x).
Thus
(G, c)
is a matroid whose bases are the minimalgenerating
sets ofG,
andso it coincides with ~~t
(G ).
Conversely,
let G a03A6(G)-matroid
group, and let X be ac-independent
subset of G. Then.B is contained in a basis B of the matroid
(G, c).
FromG =
c(B)
= >, weget
G = B >. Furthermorec(BBy) ~ c(B)
= Gfor
all y
E B. hence B is a minimalgenerating
set of G’. Thus from X C B weget(M2).
Finally,
a minimalgenerating
set B of Gsatisfies
B >= B,~(G)
>= Gand for
all
proper subsets K of B we haveY. 03A6(G ) > ~ B, 03A6(G)
>, so°
c(B)
= G butc(Y) ~
G. This proves(M1),
withM(G)
=(G. c). []
Theorem 3.2 Let G be a group and let F G . . Then G is av
F-matroid
groupif
andonly if
allof
its F-hierarchical sets areindependent.
PROOF. Assume that G is an F-matroid group. Let X =
{xl, ..., xm}
be an F-hierarchical subset of G.
By contradiction,
supposethat
X is notF"-independent.
Let i such that xi Ec(x1, ...,
xi-1, xi+1, ..., xm). There is k suchthat
Ec(.r~...
but x~~ c(x~... .... x~_1).
Let
.~ ’’
={.r~....,
x t_ i.. r I~i... Thenb
thehypothesis
we have~ c(X’, xi),
hence(4) implies
xt~ c(X’, xk),
a contradiction. ThusX
must beF-independent.
Conversely,
assume that allF-hierarchical
subsets of G areF-independent.
V’e will prove
(4)
in thefollowing
form:y E
c(.~’. x ) ,
=~ x Ey).
Once
again,
we may assume that X isF-independent.
IfX
isF-hierarchical,
both X U x and turn: out to be
F-hierarchical,
so alsoF-independent.
On. the other
hand, from
y Ec(:~’..~ )
itfollows
that ~,’ U x U y isnot F-independent,
~
hence not even
F-hierarchical.
Thusc(.~’, y)
=c(.~’.
y,x),
and(4)
follows.1
PROOF oF THEOREM 1.3.
By
Lemma3.1,
G is amatroid
group if andonly
if G is a03A6(G)-matroid
group.The
assertion follows now fromTheorem
3.2.[]
Proposition
3.3If
G is~zn
F’ -matroid group, then~ (G )
F .PROOF. If y E G
B F,
there is an F-basis X U y. Thenc(.~, y)
= G~ c(.~ ).
that is
F, X, y
>= G~
F.X >. Hencey ~
92 Raffaele
Scapellato
and Libero VerardiProposition
3.4If
G is an F-matroid group and F H G, then G is alsoan H-matroid
group.
PROOF. In view of Theorem
3.2,
wehave to
show that everyH-hierarchical
subset{y1. ..., yn}
of G isH-independent.
£Since G is an F-matroid group, it hasan F-independent
subset~xl, ..., xm~,
such thatF: x ~, ..., x,~ >= ~H.
Then the set
~x l, ..., xm. yl...., yn ~
isF-hierarchical,
so alsoF-independent by
Theorem :3:2.
Suppose, by contradiction,
that{y1, ..., yn}
is notH-independent.
Then.. there is yi such that
’ ’
.
but then
>_
F, xI , .... xm, yl, ..., yn
>; ;a
contradiction,
because ..., y1, ...,yn}
isF-independent. []
The converse of
Proposition
3.4 is not true. Obviouscounter-examples
arisefrom the fact that if H is a maximal
subgroup
of a groupG,
then G isalways
an H-matroid group.
Acknowledgement
.We are
grateful
to the referee for his constructiveremarks,
that led to animprovement
of the paper.References
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L.Bénéteau,
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