Some Conditions for P-Solubility of Finite Groups

A subgroup H of a group G is c-subnormal in G if G has a subnormal subgroup T such that HT=G and T 3 H ⊆ HG. Using this concept, in Jaraden obtain some new conditions for solubility of a finite group are given. Here we obtain local versions of these results.


INTRODUCTION
All groups that we consider are finite. Let M be a maximal subgroup of a group G. Then normal index |G: M|n of M in G is equal to |H/K| where H/K is a chief factor of G such that K ⊆ M and H ⊄ M (we note that every two chief factors with such property are isomorphic). This concept was introduced by Deskins [2] where the following nice result was proved: A group G is soluble if and only if for every its maximal subgroup M it is true that |G: M| = |G: M|n. Local versions of this result were obtained by many researchers [3][4][5][6] . In Wang [7] , analyzing the concept of normal index, introduced the following important concept: A subgroup H of a group G is said to be c-normal if there exists a normal subgroup T of G such that HT = G and T 3 H ⊆ HG (where HG is the intersection of all G-conjugates of H, i.e., the unique largest normal subgroup of G contained in H). Using this concept Wang obtained [7] several new interesting results on soluble and supersoluble groups. The concept of c-normal subgroup was used and analyzed. In particular, by Jaraden [1] the following its generalization was considered.

Definition:
A subgroup H of a group G is said to be c-subnormal in G if there exists a subnormal subgroup T such that HT = G and T 3H ⊆ HG.
Using this concept, by Jaraden [1] obtained some new conditions for solubility of a group were obtained. Here we prove the following theorems.

Theorem 1:
A group G is p-soluble if and only if every maximal subgroup M with p | |G: M|n is c-subnormal in G.

Theorem 2:
A group G is p-soluble if and only if it has a p-soluble maximal subgroup M such that either p| |G: M|n or M is c-subnormal in G.
Lemma 1: Let G be a group, H be a subgroup of G. Then the following statements hold: • If H is subnormal in G and M ≤ G, then H 3 M is subnormal in M.
• If K G and H is subnormal in G, then HK/K is subnormal in G/K.

Lemma 2:
Let L be a minimal normal subgroup of a group G and T be a subnormal subgroup of G. Then L ⊆ NG(T).
The following useful lemma was proved by Beildleman and Spencser [4] . Recall that a primitive group is a group G such that for some maximal subgroup U of G, UG =1.
A primitive group is of one of the following types (see [8; A,(15.2) ] ): • Soc(G), the socle of G is an abelian minimal normal subgroup of G, complemented by U.
• Soc(G) is a non-abelian minimal normal subgroup of G.
• Soc(G) is the direct product of the two minimal normal subgroups of G which are both non-abelian and complemented by U.

Lemma 5:
Let M be a maximal subgroup of G with MG = 1, where G is a primitive group of type 2 [11] . Let R = Soc(G) be the socle of G. If R \ M = 1, then M is a primitive group of type 2 and the simple component of R is isomorphic to a section of a simple component of Soc(M).
We shall also need the following observations on c-subnormal subgroups.
Lemma 6: Let G be a group and H a subgroup of G [1] . Then the following statements are true:

PROOFS OF THEOREM 1 AND 2
Proof of Theorem 1: First assume that G is a p-soluble group. Let M be a maximal subgroup of G. Assume that p | |G: M|n. Let H/MG be a chief factor of G. Then p | |H/MG| and so H/MG is an abelian p-group. Hence H 3 M = MG. Thus M is c-subnormal in G. Now assume that every maximal subgroup M of G with p | |G: M|n is c-subnormal in G. We shall show that G is p-soluble. Assume that it is false and let G be a counterexample with minimal order. Then Since by (4) R ⊄ M, we have MG = 1 and so p| |R| = |G: M|n. Hence by hypothesis M is c-subnormal in G. Therefore G has a subnormal subgroup T such that TM = G and T 3 M ⊆ MG = 1.

• Final contradiction.
Let L be a minimal subnormal subgroup of G contained in T. Let L G be the normal closure of L in G. Then L G ≠ 1 and so R ⊆ L G . Assume that L ⊄ R. Then by Lemma 1, L 3 R is a subnormal subgroup of G and 1⊆ L 3 R ⊆ L. Hence L 3 R = 1, since L is a minimal subnormal subgroup of G. By Lemma 2, R ⊆ NG(L). Hence < L, R >= LR = L × R. But then L ⊆ CG(R). Since CG(R) G and R ⊆ CG(R). Then R is an abelian group. This contradiction shows that L ⊆ R. Since R is a minimal normal subgroup of G, R = A1 ×... × At, where A1 ≈ A2 ≈.. ≈ At ≈ A and A is a anon-abelian simple group. Hence L ≈ A. Clearly p divides the order |A| of the group A. Hence p divides the order |L| of the group L. By Lagrange's theorem the order |L| of the group L divides the order T of the group T. Hence the prime p divides |T|. We have known that G = TM and T 3 M = 1. Hence |G| = |T||M| = |G: M||M| and so |T| = |G: M|. But the prime p does not divide the index |G: M| of M in G. Hence p does not divide |T|. This contradiction shows that G is a p-soluble group [4] . The theorem is proved.
Proof of Theorem 2: n view of Theorem 1 we have only to prove the sufficiency. Assume that it is false and let G be a counterexample with minimal order. Then Final contradiction.
Let A be a composition factor of H. In view of (2), The group G is primitive of type 2 and so by (5) and Lemma 5, A is isomorphic to some section D/L where D ≤ Soc(M). But by hypothesis M is p-soluble and so A is p-soluble. Then H is a p-soluble group and therefore H is a p-group, contrary to (2). The theorem is proved.

SOME APPLICATIONS
Theorems 1: and 2 have many corollaries. The most important of them we consider in this section.

Corollary 1:
A group G is soluble if every its maximal subgroup M is c-subnormal in G [1] .

Corollary 2:
A group G is soluble if it has a soluble maximal subgroup M which is c-subnormal in G [12] .

Corollary 3:
A group G is soluble if every its maximal subgroup M is c-normal in G [7] .

Corollary 4:
A group G is soluble if it has a soluble maximal subgroup M which is c-normal in G [7] .
It was proved that for a maximal subgroup M of a group G the following conditions are equivalent [7] : • M is c-normal in G; • |G: M| = |G: M|n.
Thus one can obtain from Theorem 1,2 the following known results. Corollary 5: (W.E. Deskins [2] ). A group G is soluble if for every its maximal subgroup M we have |G: M| = |G: M|n.
Corollary 6: (A.Ballester-Bolinches [5] ). A group G is p-soluble if for every its maximal subgroup M we have either p | |G: M|n or |G: M| = |G: M|n.