On the index complex of a maximal subgroup and the group-theoretic properties of a finite group

Let G be a finite group, Sp(G), Φ′(G) and Φ1(G) be generalizations of the Frattini subgroup of G. Based on these characteristic subgroups and using Deskins 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.

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 of group 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 partially ordered set by set inclusion relation.If C ∈ I(M ) and C is the maximal element of I(M ), C is said to be a maximal completion for M .If moreover C ¡ G, C then is said to be a normal completion for M .Clearly every normal completion of M is a maximal completion of M .Furthermore, by k(C) we denote the product of all normal subgroups of G which are also proper subgroups of C, k(C) is a proper normal 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 concerning super-solvability of a finite group in the same paper.Deskins's conjecture and other investigations were continued by many successive works [3][4][5].This paper will study the structure of a finite group G. Using the concept of index complex and applying Frattini-Like subgroups such as S p (G), Φ (G) and Φ 1 (G), the paper improves main results of [3][4][5] and obtains some necessary and sufficient conditions for the G to be a p-solvable, π-solvable, solvable, super-solvable and nilpotent group.
Throughout this paper, G denotes a finite group.The terminologies and notations agree with standard usage as in [6].The notation M <• G means M is a maximal 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 |.

Preliminaries
For convenience, we give some notations and definitions firstly.Suppose that p is a prime, put Using subgroups above, one can define Frattini-Like subgroups of G as follows.
Definition 2.1 We begin with a preliminary result which will be used frequently in connection with induction arguments in the next section.
On the other hand, let k(C/N ) = H/N , then H ¡ G and H/N < C/N .Thus, If G is a group with a maximal core-free subgroup, the followings are equivalent: (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 F G 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 ∈ F G with core-free, let p be the unique prime divisor of This leads to a contradiction.Thus P ¡ G and P = N is a nontrivial solvable normal subgroup of G.

Main Results
The following is the main result of the paper which gives a description of psolvable group.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 Proof.It suffices to prove the sufficient condition.Suppose that the result is false and let G be a counterexample of minimal order, now we can claim that: 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 : This contradicts with 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.
That is to say, φ(xk(C)) = φ(yk(C)).Hence the map is well defined.It can be verified 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 minimality of G, G/N is p-solvable.

Similarly, it can be shown that
q be a prime less than p, then |G| divides q!.This leads to another contradiction.Thus |G : M | is 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 completion of 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 maximal subgroups.
Theorem 3.2 Let G be a finite group.G is π-solvable if and only if for every maximal subgroup M of G in F G there exists a normal completion Proof.⇐) Let G be a group satisfying the hypothesis of the theorem.If F G is empty then Φ (G) = G, and G is solvable.Thus assume that F G is not empty.If G is simple, then for every M in F G , 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 can suppose that F G/N is not empty.We will use induction on the order of G.For each M/N in F G/N , by [7,Lemma 3], it follows that M ∈ F G .So by hypothesis there exists a normal completion 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 minimal normal 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 πsolvable, and furthermore G is π-solvable.If N ≤ Φ (G), there exists a maximal subgroup M 0 ∈ F G with N ≤ M 0 .Then Core G M 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 ⇒) The converse is obvious.
The following theorem can be proved similarly as Theorem 3.2, and we omit it here.
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 ) such that G = CM and C/k(C) is cyclic.The following theorem extends this result.
Proof.Let G be a super-solvable group.Then every chief factor of G is a cyclic group of prime order.∀M ∈ F G , 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 Let G be a group satisfying the hypothesis of the Theorem.If F G is empty then G = Φ (G) and G is super-solvable [9].We now assume that F G is not empty and then G is solvable.In the remainder of the proof we will drop the maximality imposed on the completion C in I(M ) in the hypothesis.For each maximal subgroup M in F G , there exists a completion 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 2 2 .First suppose that there exists an M in F G which has a normal completion A such that A/k(A) is elementary abelian of order 2 2 .Let G = G/core G (M ) and C, M , A be the images of C, M and A in G respectively.Then A is an elementary abelian of order 2 2 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 the hypothesis of the theorem, and G is S 4 -free, so G is super-solvable.Now assume that for each maximal subgroup M in F G , M has a normal completion 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, Lemma 3], we can assume that the hypothesis holds for G/N .Using induction, we obtain that G/N is super-solvable.Similar to Theorem 3.1, we can suppose that N is the unique 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 G so that G = N M and core G (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 we have the following result: If G is S 4 -free and solvable, G is super-solvable if and only if for each maximal subgroup M of G in F G , there exists a completion C in I(M ) so that G = CM and C/k(C) is cyclic.
hence G is a cyclic group of prime order.So assume that G is not simple.Let N be a minimal normal subgroup of G. Without loss of generality, suppose that F G/N is not empty.For any maximal subgroup M/N in F G/N , suppose that C/N is an arbitrary normal completion in I(M/N ).From [7, Lemma 3]  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 an elementary abelian p-group.If G is not a p-group, we assume that |G| has a prime factor q different from p.If the subgroup Q = a|a ∈ G and |a| = q ≤ M , this contradicts with the fact that core G M = 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 power of p.This leads to another contradiction.So G must be a p-group and then is a nilpotent group.

Theorem 3 . 5
Let G be a group and M be an arbitrary maximal subgroup of G in F G .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 F G is empty then G we have M in F G .Obviously C is a normal 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 the unique minimal subgroup of G.If N ≤ Φ 1 (G), by [5, Lemma 2.3] G is nilpotent.If N ≤ Φ 1 (G), there exists an M in F G so that G = N M .Clearly, N is a normal completion in I(M ).By hypothesis |N/k(N )| = |N | = |G : M |.For any L in F G with core G (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.