## Spm.uem.br

Bol. Soc. Paran. Mat.

(3s.) v. 23 1-2 (2005):
On the index complex of a maximal subgroup and the group-theoretic
abstract: Let

*G *be a finite group,

*Sp*(

*G*)

*, *Φ (

*G*) and Φ1(

*G*) be generalizationsof the Frattini subgroup of

*G*. Based on these characteristic subgroups and usingDeskins index complex, this paper gets some necessary and sufficient conditions for

*G *to be a

*p*-solvable,

*π*-solvable, solvable, super-solvable and nilpotent group.

Key Words: : index complex; solvable groups; super-solvable groups; nilpo-
The relationship between the properties of maximal subgroups of a finite group
and its structure has been studied extensively. The concept of index complex(see[1]) associated with a maximal subgroup plays an important role in the study ofgroup theory.

Suppose that

*G *is a finite group, and

*M *is a maximal subgroup of

*G*. A
subgroup

*C *of

*G *is said to be a completion for

*M *in

*G *if

*C *is not contained in

*M *while every proper subgroup of

*C *which is normal in

*G *is contained in

*M *.

The set of all completions of

*M *, denote it by

*I*(

*M *), is called the index complex of

*M *in

*G*. Clearly

*I*(

*M *) contains a normal subgroup, and is a nonempty partiallyordered set by set inclusion relation. If

*C ∈ I*(

*M *) and

*C *is the maximal elementof

*I*(

*M *),

*C *is said to be a maximal completion for

*M *. If moreover

*C ✁ G*,

*C *thenis said to be a normal completion for

*M *. Clearly every normal completion of

*M*
*∗ *The project is supported by National Natural Science Foundation of China (10301004) and
Basic Research Foundation of Beijing Institute of Technology

*† *Correspondence should be addressed to Jiang Lining2000

*Mathematics Subject Classification: *20D10, 20F16
is a maximal completion of

*M *. Furthermore, by

*k*(

*C*) we denote the product ofall normal subgroups of

*G *which are also proper subgroups of

*C*,

*k*(

*C*) is a propernormal subgroup of

*C*.

In [2], Deskins studied the group-theoretic properties of the completions and its
influences on the solvability of a finite group. He also raised a conjecture concerningsuper-solvability of a finite group in the same paper. Deskins’s conjecture and otherinvestigations were continued by many successive works [3-5]. This paper will studythe structure of a finite group

*G*. Using the concept of index complex and applyingFrattini-Like subgroups such as

*Sp*(

*G*)

*, *Φ (

*G*) and Φ1(

*G*), the paper improves mainresults of [3-5] and obtains some necessary and sufficient conditions for the

*G *tobe a

*p*-solvable,

*π*-solvable, solvable, super-solvable and nilpotent group.

Throughout this paper,

*G *denotes a finite group. The terminologies and no-
tations agree with standard usage as in [6]. The notation

*M <· G *means

*M *is amaximal subgroup of

*G*, and

*N ✁ G *means that

*N *is a normal subgroup of

*G*. If

*p *is a prime, then

*p *denotes the complementary sets of primes and

*|G *:

*M |p *the

*p*-part of

*|G *:

*M |*.

For convenience, we give some notations and definitions firstly. Suppose that

*p*
*F c *=

*{M *:

*M <· G *and

*|G *:

*M | *is composite

*}*;

*F p *=

*{M *:

*M <· G *and

*M ≥ NG*(

*P *) for a

*P ∈ Sylp*(

*G*)

*}*;
Using subgroups above, one can define Frattini-Like subgroups of

*G *as follows.

*{M *:

*M ∈ F pc} if F pc is nonempty, otherwise Sp*(

*G*) =

*G*;

*{M *:

*M ∈ FG} if FG is nonempty, otherwise *Φ1(

*G*) =

*G*;

*{M *:

*M ∈ FG} if FG is nonempty, otherwise *Φ (

*G*) =

*G.*
We begin with a preliminary result which will be used frequently in connection
with induction arguments in the next section.

Lemma 2.1 Let

*M *be a maximal subgroup of a group

*G *and

*N *a normal subgroupof

*G*. If

*C ∈ I*(

*M *) and

*N ≤ k*(

*C*), then

*C/N ∈ I*(

*M/N *) and

*k*(

*C/N *) =

*k*(

*C*)

*/N *.

Proof. Since

*C ∈ I*(

*M *),

*C ≤ M *. Also

*C/N ≤ M/N *. And if

*A/N < C/N *,

*A/N ✁ G/N *, then

*A < C *and

*A ✁ G*. Since

*A ≤ M *,

*A/N ≤ M/N *, and

*C/N ∈*
On the index complex of a maximal subgroup

*I*(

*M/N *). Also

*C ≤ M *means

*k*(

*C*) =

*C*. Then

*k*(

*C/N *)

*≤ C/N *and moreover

*k*(

*C*)

*/N ≤ M/N *. So

*k*(

*C/N *)

*≤ k*(

*C*)

*/N *.

On the other hand, let

*k*(

*C/N *) =

*H/N *, then

*H ✁ G *and

*H/N < C/N *. Thus,

*H < C *and

*k*(

*C/N *) =

*H/N ≤ k*(

*C*)

*/N *. Therefore,

*k*(

*C/N *) =

*k*(

*C*)

*/N *.

Lemma 2.2[2] Let

*C *and

*D *be normal completions of a maximal subgroup

*M *of

*G*. Then

*C/k*(

*C*)

*∼*
The order of

*C/k*(

*C*), where

*C *is a normal completion of

*M *, is called the normal
index of

*M *in

*G*, denoted by

*η*(

*G *:

*M *).

Lemma 2.3[7] Φ1(

*G*) is a nilpotent group; Φ (

*G*) is a Sylow tower group.

Lemma 2.4 If

*G *is a group with a maximal core-free subgroup, the followings areequivalent:
(1) There exists a nontrivial solvable normal subgroup of

*G*.

(2) There exists a unique minimal normal subgroup

*N *of

*G *and the index of
all maximal subgroups of

*G *in

*FG *with core-free are powers of a unique prime.

Proof. Using Ref.[7], it suffices to prove that (2) implies (1). Indeed for every

*L ∈ FG *with core-free, let

*p *be the unique prime divisor of

*|G *:

*L|*. Since

*N ≤ L*,

*G *=

*LN *. Moreover

*|G *:

*L| |N |*, thus

*p |N |*. Let

*P ∈ Sylp*(

*N*). If

*P*
the Frattini argument we have

*G *=

*N · NG*(

*P *). Suppose that

*NG*(

*P *)

*≤ M <· G*,there exists

*Gp ∈ Sylp*(

*G*) satisfying

*NG*(

*P *)

*≥ NG*(

*Gp*). This means

*M ≥ NG*(

*Gp*)and therefore

*M ∈ FG*. But

*N ≤ M*, by the uniqueness of

*N *we get that

*M *iscore-free. By the hypothesis,

*p |G *:

*M |*. Since

*M ≥ NG*(

*Gp*),

*p*
leads to a contradiction. Thus

*P ✁ G *and

*P *=

*N *is a nontrivial solvable normalsubgroup of

*G*.

The following is the main result of the paper which gives a description of

*p*-
Theorem 3.1 Let

*p *be the largest prime divisor of the order of

*G*. The

*G *is

*p*-solvable if and only if for each non-nilpotent maximal subgroup

*M *of

*G *in

*F pc*,there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*) is a

*p *-group.

Proof. It suffices to prove the sufficient condition. Suppose that the result is falseand let

*G *be a counterexample of minimal order, now we can claim that:
i)

*F pc *is not empty. Indeed if

*F pc *is empty, then

*Sp*(

*G*) =

*G*. Using [9, Lemma
2.2],

*Sp*(

*G*) is

*p*-closed. So

*P ∈ Sylp*(

*G*)

*✁ G *and

*G *is

*p*-solvable. This leads to acontradiction.

ii) Every maximal subgroup

*M *of

*G *in

*F pc *must be non-nilpotent. Indeed if
there exists a maximal subgroup

*M *in

*F pc *which is also nilpotent, then

*|G *:

*M |p *=
1 and

*G *is

*p*-solvable. It is a contradiction.

iii)

*G *has a unique minimal normal subgroup

*N *such that

*G/N *is

*p*-solvable.

Indeed if

*G *is simple, then for every

*M *of

*G *in

*F pc*,

*G *is the only normal completionin

*I*(

*M *) with

*k*(

*G*) = 1. By hypothesis,

*G *=

*G/k*(

*G*) is a

*p *-group. This contradictswith the fact that

*p *is the largest prime dividing

*|G|*, hence

*G *is not simple. Let

*N *be a minimal normal subgroup of

*G*, we will according to cases of

*N ≤ k*(

*C*) or

*N ≤ k*(

*C*) prove that

*G/N *satisfies the hypothesis of the theorem.

If

*N ≤ k*(

*C*), then

*N ≤ C *and

*C/N *is a normal completion for

*M/N *in

*G/N *.

By Lemma 2.1,

*C/N k*(

*C/N *) =

*C/N k*(

*C*)

*/N ∼*
=

*C/k*(

*C*). Again

*C/k*(

*C*) is a

*p *-group, so

*C/N k*(

*C/N *) is a

*p *-group.

If

*N ≤ k*(

*C*), then

*N ≤ C*. For otherwise, either

*N *=

*C *or

*N < C*, so either

*G *=

*M C *=

*M N *=

*M *or

*N < k*(

*C*). Each of which is a contradiction. Since

*N*is a minimal normal subgroup of

*G*, we have either

*C*
*N *=

*N *, then

*N ≤ C*. It is also a contradiction. So

*C*
is a normal completion for

*M/N *in

*G/N *. We are to show that

*C/N k*(

*C/N *) is a

*p *-group. Since

*k*(

*C*)

*< C *and

*C*
*N *= 1, it follows that

*k*(

*C*)

*N < CN *, and hence

*k*(

*C*)

*N/N < CN/N *. Also

*k*(

*C*)

*N/N ✁ G/N *, so we have

*k*(

*C*)

*N/N ≤ k*(

*CN/N *).

We define a map

*φ*:

*C/k*(

*C*)

*→ CN/N k*(

*CN/N *), by

*φ*(

*xk*(

*C*)) =

*xN k*(

*CN/N *)
for all

*xk*(

*C*)

*∈ C/k*(

*C*). Now

*xk*(

*C*) =

*yk*(

*C*) implies that

*x−*1

*y ∈ k*(

*C*), so(

*xN *)

*−*1(

*yN *) = (

*x−*1

*y*)

*N ∈ k*(

*C*)

*N/N ≤ k*(

*CN/N *) and
(

*xN *)

*k*(

*CN/N *) = (

*yN *)

*k*(

*CN/N *)

*.*
That is to say,

*φ*(

*xk*(

*C*)) =

*φ*(

*yk*(

*C*)). Hence the map is well defined. It can beverified that

*φ *is an epimorphism and

*CN/N k*(

*CN/N *) is an epimorphic image of
a

*p *-group. Thus

*G/N *satisfies the hypothesis of the theorem. By the minimalityof

*G*,

*G/N *is

*p*-solvable.

Similarly, it can be shown that

*G/N*1 is

*p*-solvable if

*N *is another minimal
normal subgroup

*N*1 of

*G*. Thus

*G *=

*G/N*
*N*1, which is isomorphic a subgroup
of the

*p*-solvable group

*G/N × G/N*1, is

*p*-solvable. So in the following supposethat

*N *is the unique minimal normal subgroup of

*G*.

*|N | *or

*N *is a

*p*-group, then

*N *is

*p*-solvable and so

*G *is

*p*-solvable. It is
a contradiction. Hence,

*|N |p *= 1 and

*N *=

*Np ∈ Sylp*(

*N*). Let

*M *be a maximalsubgroup of

*G *such that

*NG*(

*Np*)

*≤ M*. By the Frattini argument, we obtainthat

*G *=

*N · NG*(

*Np*). Using [7, lemma 5], there exists a

*Gp ∈ Sylp*(

*G*) with

*NG*(

*Np*)

*≥ NG*(

*Gp*), so

*M ∈ F p *and

*|G *:

*M|p *= 1. If

*|G *:

*M| *=

*q *be a prime lessthan

*p*, then

*|G| *divides

*q*!. This leads to another contradiction. Thus

*|G *:

*M | *is
On the index complex of a maximal subgroup
composite and

*M ∈ F pc*. By ii) and hypothesis, there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*) is a

*p *-group. Obviously

*N *is a normal completionof

*M *. Combining with Lemma 2.2, we have

*C/k*(

*C*)

*∼*
=

*N/k*(

*N *) =

*N *. Thus

*N *is
a

*p *-group, which leads to the final contradiction. This completes the proof.

As we have known in [3], a group

*G *is

*π*-solvable if and only if for every maximal
subgroup

*M *of

*G *there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*)is

*π*-solvable. We now extend this result by considering a smaller class of maximalsubgroups.

Theorem 3.2 Let

*G *be a finite group.

*G *is

*π*-solvable if and only if for everymaximal subgroup

*M *of

*G *in

*F *there exists a normal completion

*C *in

*I*(

*M *) such
that

*C/k*(

*C*)is

*π*-solvable.

Proof.

*⇐*) Let

*G *be a group satisfying the hypothesis of the theorem. If

*F*
empty then Φ (

*G*) =

*G*, and

*G *is solvable. Thus assume that

*F *is not empty. If

*G*
is simple, then for every

*M *in

*F *,

*G *is the only normal completion in

*I*(

*M *) with

*k*(

*G*) = 1 and thus

*G *=

*G/k*(

*G*) is

*π*-solvable. So suppose that

*G *is not simple.

Let

*N *be a minimal normal subgroup of

*G*. Without loss of generality, one cansuppose that

*F*
is not empty. We will use induction on the order of

*G*. For
, by [7, Lemma 3], it follows that

*M ∈ F *. So by hypothesis
there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*) is

*π*-solvable.

Similar to the proof in Theorem 3.1,

*CN/N k*(

*CN/N *) is

*π*-solvable. Thus

*G/N *satisfies the hypothesis of the theorem. Using the induction we obtain that

*G/N *is

*π*-solvable. Furthermore, we can assume that

*N *is the unique minimalnormal subgroup of

*G*. By the same way,

*G/N *is still a

*π*-solvable group.

Now if

*N ≤ *Φ (

*G*), then from Lemma 2.3 Φ (

*G*) is solvable. Thus,

*N *is

*π*-
solvable, and furthermore

*G *is

*π*-solvable. If

*N ≤ *Φ (

*G*), there exists a maximalsubgroup

*M*0

*∈ F *with

*N ≤ M*
0. Then

*CoreGM*0 = 1 and

*G *=

*N M*0. So

*N *is a
normal completion in

*I*(

*M*0). By hypothesis there exists a normal completion

*C *in

*I*(

*M*0) such that

*C/k*(

*C*) is

*π*-solvable. By Lemma 2.2,

*N/k*(

*N*) =

*N ∼*
Again

*C/k*(

*C*) is

*π*-solvable, therefore

*N *is

*π*-solvable and moreover,

*G *is

*π*-solvable.

*⇒*) The converse is obvious.

The following theorem can be proved similarly as Theorem 3.2, and we omit it
Theorem 3.3 Let

*G *be a finite group.

*G *is solvable if and only if for everymaximal subgroup

*M *of

*G *in

*F *there exists a normal completion

*C *in

*I*(

*M *) such
that

*C/k*(

*C*)is solvable.

As we have known [4], if

*G *is

*S*4-free, then

*G *is super-solvable if and only if for
each maximal subgroup

*M *of

*G*, there exists a maximal completion

*C *in

*I*(

*M *) suchthat

*G *=

*CM *and

*C/k*(

*C*) is cyclic. The following theorem extends this result.

Theorem 3.4 Suppose that

*G *is

*S*4-free.

*G *is super-solvable if and only if for each
maximal subgroup

*M *of

*G *in

*F *, there exists a maximal completion

*C *in

*I*(

*M *)
such that

*G *=

*CM *and

*C/k*(

*C*) is cyclic.

Proof. Let

*G *be a super-solvable group. Then every chief factor of

*G *is a cyclicgroup of prime order.

*∀M ∈ F *, it is clear that the set

*S *=

*{T ✁ G|T ≤ M } *is
not empty. Choose an

*H *to be the minimal element in

*S*. Clearly,

*H ∈ I*(

*M *) and

*H/k*(

*H*)is a chief factor of

*G*, hence

*H/k*(

*H*) is cyclic.

Let

*G *be a group satisfying the hypothesis of the Theorem. If

*F*
then

*G *= Φ (

*G*) and

*G *is super-solvable [9]. We now assume that

*F*
empty and then

*G *is solvable. In the remainder of the proof we will drop themaximality imposed on the completion

*C *in

*I*(

*M *) in the hypothesis. For eachmaximal subgroup

*M *in

*F *, there exists a completion

*C *in

*I*(

*M *) such that

*G *=

*CM *and

*C/k*(

*C*) is cyclic. From [5, Lemma 2], we can get a normal completion

*A*in

*I*(

*M *) such that

*A/k*(

*A*) is either cyclic or elementary abelian of order 22.

First suppose that there exists an

*M *in

*F *which has a normal completion

*A*
such that

*A/k*(

*A*) is elementary abelian of order 22. Let

*G *=

*G/coreG*(

*M*) and

*C*,

*M *,

*A *be the images of

*C*,

*M *and

*A *in

*G *respectively. Then

*G *=

*C · M *=

*A · M *. Itis easy to verify that

*k*(

*A*) =

*A*
*coreGM*, so

*A/k*(

*A*)

*∼*
=

*A coreGM/coreGM *=

*A*.

Since

*core M *= 1,

*k*(

*A*) = 1,

*A *is a minimal normal subgroup of

*G*.

*A *is an
elementary abelian of order 22 and

*M*
*A *= 1. Considering the permutation
representation of

*G *on 4 cosets of

*M *,

*G *is isomorphic to a subgroup of

*S*4. Again

*S*4 and

*A*4 are the only non-super-solvable subgroups of

*S*4,

*A*4 doesn’t satisfy thehypothesis of the theorem, and

*G *is

*S*4-free, so

*G *is super-solvable.

Now assume that for each maximal subgroup

*M *in

*F *,

*M *has a normal com-
pletion

*A *so that

*A/k*(

*A*) is cyclic. Let

*N *be a minimal normal subgroup of

*G*.

Obviously, that

*G *is

*S*4-free is quotient-closed. By [4, Lemma 3] and [7, Lemma3], we can assume that the hypothesis holds for

*G/N *. Using induction, we obtainthat

*G/N *is super-solvable. Similar to Theorem 3.1, we can suppose that

*N *is theunique minimal normal subgroup of

*G*. If

*N ≤ *Φ (

*G*), then

*G *is super-solvable.

If

*N ≤ *Φ (

*G*), there exists a maximal subgroup

*M *in

*F*
*coreG*(

*M*) = 1. Obviously

*N *is a normal completion in

*I*(

*M*). By hypothesis,there exists a normal completion

*A *so that

*A/k*(

*A*) is cyclic. By Lemma 2.2,

*A/k*(

*A*)

*∼*
=

*N/k*(

*N *) =

*N *. Thus

*N *is cyclic and

*G *is super-solvable.

Remark Let

*G *be a solvable group. To obtain the conclusion in Theorem 3.4,the condition of maximality imposed on the completion

*C *is nonsignificant. So wehave the following result: If

*G *is

*S*4-free and solvable,

*G *is super-solvable if andonly if for each maximal subgroup

*M *of

*G *in

*F *, there exists a completion

*C *in

*I*(

*M *) so that

*G *=

*CM *and

*C/k*(

*C*) is cyclic.

Theorem 3.5 Let

*G *be a group and

*M *be an arbitrary maximal subgroup of

*G*in

*FG*. Then

*G *is nilpotent if and only if for each normal completion

*C *of

*M*,

*|C/k*(

*C*)

*| *=

*|G *:

*M |.*
Proof.

*⇐*) Let

*G *be a group satisfying the hypothesis of the theorem. If

*FG *is
On the index complex of a maximal subgroup
empty then

*G/N *= Φ1(

*G/N*). Using [9, Lemma 2.3],

*G/N *is nilpotent. If

*G *issimple, then for every

*M *in

*FG*,

*G *is the only normal completion in

*I*(

*M*) with

*k*(

*G*) = 1. By hypothesis

*|G/k*(

*G*)

*| *=

*G *=

*|G *:

*M |*,

*M *= 1, hence

*G *is a cyclicgroup of prime order. So assume that

*G *is not simple. Let

*N *be a minimal normalsubgroup of

*G*. Without loss of generality, suppose that

*FG/N *is not empty. Forany maximal subgroup

*M/N *in

*FG/N *, suppose that

*C/N *is an arbitrary normalcompletion in

*I*(

*M/N *). From [7, Lemma 3] we have

*M *in

*FG*. Obviously

*C *is anormal completion in

*I*(

*M *) and

*|C/k*(

*C*)

*| *=

*|G *:

*M |*. Using Lemma 2.1,

*|C/N k*(

*C/N *)

*| *=

*|C/N k*(

*C*)

*/N | *=

*|C/k*(

*C*)

*| *=

*|G *:

*M | *=

*|G/N M/N |.*
Thus

*G/N *satisfies the hypothesis of the theorem. Applying induction one can see

*G/N *is nilpotent. Similar to the proof in Theorem 3.1, we may assume

*N *is theunique minimal subgroup of

*G*.

If

*N ≤ *Φ1(

*G*), by [5, Lemma 2.3]

*G *is nilpotent. If

*N ≤ *Φ1(

*G*), there exists
an

*M *in

*FG *so that

*G *=

*NM*. Clearly,

*N *is a normal completion in

*I*(

*M*).Byhypothesis

*|N/k*(

*N *)

*| *=

*|N | *=

*|G *:

*M |*. For any

*L *in

*FG *with

*coreG*(

*L*) = 1,obviously

*N ≤ L *and

*G *=

*N L*.

*N *is also a normal completion in

*I*(

*M *), so

*|N/k*(

*N *)

*| *=

*|N | *=

*|G *:

*L|*. By Lemma 2.4

*G *has a nontrivial solvable subgroup

*K*,so

*N ≤ K *and

*N *is solvable. Since

*G/N *is nilpotent,

*G *is solvable. Thus

*N *is anelementary abelian

*p*-group. If

*G *is not a

*p*-group, we assume that

*|G| *has a primefactor

*q *different from

*p*. If the subgroup

*Q *=

*a|a ∈ G *and

*|a| *=

*q ≤ M *, thiscontradicts with the fact that

*coreGM *= 1. So there exists an of order

*q *element

*a *in

*G − M *. This implies that

*G *=

*M, a *. However,

*|N | *=

*|G *:

*M | *is a powerof

*p*. This leads to another contradiction. So

*G *must be a

*p*-group and then is anilpotent group.

*⇒*) The converse holds obviously.

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*Beijing Institute of Civil Eng. and Arch.*
*Beijing, 100081, P.R.ChinaE-mail:jiangjl@yahoo.com.cn*
Source: http://www.spm.uem.br/bspm/pdf/vol23-1-2/art5.pdf

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