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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.
1. W.E.Deskins. On maximal subgroups. Proc. symp. in pure Math., 1959, 1, 100-104.
2. W.E.Deskins. A note on the index complex of a maximal subgroup. Arch. Math., 1990, 54, 3. T.K.Dutta. Some characterisations of π-solvable groups using index-complex. Indian J. Pure Appl. Math., 2002, 33(4), 555-564.
4. Guo Xiu-yun. On Deskins’s conjecture. J. of Pure and Appl. Algebra. 1998, 124, 167-171.
5. Zhao Yao-qing. On the index complex of a maximal subgroup and the supersovability of a finite group. Commun. in Algebra. 1996 24(5), 1785-1791.
6. Xu Ming-yao. An Intruduction of Finite Groups. Beijing: Science Press, 1987. (in Chinese) 7. Wang Pin-chao and Yang Zhao-xing. On the normal index of maximal subgroup in finite group. J. Enginer. Math., 11(1), 1994, 41-48.
8. R.Baer. Classes of finite groups and their properties. Illions J. Math., 1957, 1, 115-167.
9. Wang Yan-ming. C-normality of groups and its properties. J. Algebra. 1996, 180, 954-965.
Beijing Institute of Civil Eng. and Arch. Beijing, 100081, P.R.ChinaE-mail:jiangjl@yahoo.com.cn

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