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(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 forG 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 inM 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 ofM 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. Ifp is a prime, then p denotes the complementary sets of primes and |G : M |p thep-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 moreoverk(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 ofG. 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 everyL ∈ 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 isp-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. LetN be a minimal normal subgroup of G, we will according to cases of N ≤ k(C) orN ≤ 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 Nis 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−1y ∈ k(C), so(xN )−1(yN ) = (x−1y)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/N1 is p-solvable if N is another minimal normal subgroup N1 of G. Thus G = G/N N1, which is isomorphic a subgroup of the p-solvable group G/N × G/N1, 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) withNG(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 completionC 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)isπ-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 thatG/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 M0 ∈ F with N ≤ M 0. Then CoreGM0 = 1 and G = N M0. So N is a normal completion in I(M0). By hypothesis there exists a normal completion C inI(M0) 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 S4-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 S4-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 ) andH/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 Ain 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 S4. AgainS4 and A4 are the only non-super-solvable subgroups of S4, A4 doesn’t satisfy thehypothesis of the theorem, and G is S4-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 S4-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 S4-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 Gin 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) withk(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 seeG/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 elementa 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

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