ON CERTAIN PRODUCTS OF PERMUTABLE SUBGROUPS

Abstract In this paper, we study the structure of finite groups 
$G=AB$
 which are a weakly mutually 
$sn$
 -permutable product of the subgroups A and B, that is, A permutes with every subnormal subgroup of B containing 
$A \cap B$
 and B permutes with every subnormal subgroup of A containing 
$A \cap B$
 . We obtain generalisations of known results on mutually 
$sn$
 -permutable products.


Introduction
All groups considered here will be finite.
Mutually permutable products, that is, products G = AB such that A permutes with every subgroup of B and B permutes with every subgroup of A, have been extensively studied by many authors [3]. In recent years, some other permutability connections between the factors have also been considered. In particular, the rich normal structure of a mutually permutable product of two nilpotent groups [3,Ch. 5] has motivated interest in the study of mutually sn-permutable products. DEFINITION 1.1. We say that a group G = AB is the mutually sn-permutable product of the subgroups A and B if A permutes with every subnormal subgroup of B and B permutes with every subnormal subgroup of A. [2] On certain products of permutable subgroups 279 THEOREM 1.

[1, Theorem B]. Let G = AB be the mutually sn-permutable product of the subgroups A and B, where A is supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is supersoluble.
Following [8], we say that a subgroup H of a group G is P-subnormal in G whenever either H = G or there exists a chain of subgroups H = H 0 ≤ H 1 ≤ · · · ≤ H n−1 ≤ H n = G such that |H i : H i−1 | is a prime for every i = 1, . . . , n. It turns out that supersoluble groups are exactly those groups in which every subgroup is P-subnormal. Having in mind this result and the influence of the embedding of Sylow subgroups on the structure of a group, the following extension of the class of supersoluble groups introduced in [8] seems to be natural.
The class of all finite w-supersoluble groups, denoted by wU, is a saturated formation of soluble groups containing U, the class of all supersoluble groups, which is locally defined by a formation function f, such that for every prime p, f (p) is composed of all soluble groups G whose Sylow subgroups are abelian of exponent dividing p − 1 [8, Theorems 2.3 and 2.7]. Not every group in wU is supersoluble [8,Example 1]. However, every group in wU has an ordered Sylow tower of supersoluble type [8,Proposition 2.8].
In [4], mutually sn-permutable products in which the factors are w-supersoluble are analysed. The following extension of Theorem Obviously, mutually sn-permutable products are weakly mutually sn-permutable, but the converse is not true in general, as the following example shows. EXAMPLE 1.6. Let G = Σ 4 be the symmetric group of degree 4. Consider a maximal subgroup A of G which is isomorphic to Σ 3 , and B = A 4 , the alternating group of degree 4. Then G = AB is the weakly mutually sn-permutable product of the subgroups A and B. However, the product is not mutually sn-permutable because A does not permute with a subnormal subgroup of order 2 of B.
The first goal of this paper to prove weakly mutually sn-permutable versions of the aforesaid theorems. We show that Theorem 1.4 holds for weakly mutually sn-permutable products.
THEOREM A. Let G = AB be the weakly mutually sn-permutable product of the subgroups A and B, where A is w-supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is w-supersoluble.
The next corollary follows from the proof of Theorem A and generalises Theorem 1.2.

COROLLARY B. Let G = AB be the weakly mutually sn-permutable product of the subgroups A and B, where A is supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is supersoluble.
The second part of the paper is concerned with weakly mutually sn-permutable products with nilpotent derived subgroups. Our starting point is the following extension of a classical result of Asaad and Shaalan [2].

THEOREM 1.7 [1, Theorem C]. Let G = AB be the mutually sn-permutable product of the supersoluble subgroups A and B. If the derived subgroup G of G is nilpotent, then G is supersoluble.
A natural question is whether this result is true for weakly mutually sn-permutable products under the same conditions. The following example answers this question negatively.
Note that B is a normal subgroup of G; therefore, it permutes with every subgroup of A. Moreover, A ∩ B = c 3 and the unique subnormal subgroup of B containing A ∩ B is the whole of B. It is not difficult to see that B is supersoluble. Therefore, A and B are supersoluble and G is nilpotent. Moreover, A is nilpotent and B is a normal subgroup of G. Thus, in particular, it permutes with every Sylow subgroup of A.
However, an additional assumption allows us to get supersolubility.

Preliminary results
In this section we will prove some results needed for the proofs of our main results. We start by showing that factor groups of weakly mutually sn-permutable products are also weakly mutually sn-permutable products.

PROOF. Let us consider
Interchanging A and B and arguing in the same manner proves the result. Observe that Lemma 2.2 implies that if G = AB is the weakly mutually sn-permutable product of A and B, H is a subnormal subgroup of A such that A ∩ B H and K is a subnormal subgroup of B such that A ∩ B K, then HK is a weakly mutually sn-permutable product of H and K. In the next result we analyse the behaviour of minimal normal subgroups of weakly mutually sn-permutable products containing the intersection of the factors. LEMMA 2.3. Let G = AB be the weakly mutually sn-permutable product of A and B. If N is a minimal normal subgroup of G such that A ∩ B N, then either A ∩ N =  B ∩ N = 1 or N = (N ∩ A)(N ∩ B).

PROOF. Observe that
By the same argument, N) is a normal subgroup of G. By the minimality of N,

LEMMA 2.5. Let G = AB be the weakly mutually sn-permutable product of the subgroups A and B, where A is soluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is soluble.
PROOF. Suppose that the theorem is false, and let G be a minimal counterexample. If N is a minimal normal subgroup of G, then G/N = (AN/N)(BN/N) is the weakly mutually sn-permutable product of the subgroups AN/N and BN/N by Lemma 2.1. Since BN/N permutes with each Sylow subgroup of AN/N, it follows that G/N is soluble by the minimality of G. Let N 1 and N 2 be two minimal subgroups of G. Then G G/ (N 1 ∩ N 2 ) is soluble, a contradiction. Hence G has a unique minimal normal subgroup N of G and we may assume that N is nonabelian. This means that F(G) = 1.
On the other hand, A ∩ B F(G) using Lemma 2.4. Therefore A ∩ B = 1 and then G = AB is the totally sn-permutable product of A and B. The result then follows by applying [5,Theorem 6]. LEMMA 2.6 [1, Lemma 3]. Let G be a primitive group and let N be its unique minimal normal subgroup. Assume that G/N is supersoluble. If N is a p-group, where p is the largest prime dividing |G|, then N = F(G) = O p (G) is a Sylow p-subgroup of G.

Main results
We are ready to prove our main results. PROOF OF THEOREM A. Suppose the theorem is not true and let G be a minimal counterexample. We shall prove our theorem in five steps. (c) N is the Sylow p-subgroup of G and p is the largest prime dividing |G|. Let q be the largest prime dividing |G| and suppose that q p. Suppose first that q divides |BN|. Since BN has a Sylow tower of supersoluble type, BN has a unique Sylow q-subgroup, say (BN) q . This means that (BN) q centralises N. Thus (BN) q = 1, since C G (N) = N, a contradiction.
We may assume that q divides |A| but does not divide |BN|. Since A has a Sylow tower of supersoluble type, A has a unique Sylow q-subgroup, A q say. This means that A q is normalised by (N ∩ B) is w-supersoluble by the choice of G. It follows that A q (N ∩ B) has a unique Sylow q-subgroup since it has a Sylow tower of supersoluble type. In other words, A q is normalised by N ∩ B. Hence A q is normalised by (N ∩ A)(N ∩ B) = N. This means that A q centralises N, a contradiction. We may assume that A q (N ∩ B) = G. Then N ∩ B = B and so B is an elementary abelian p-group. Moreover, contradiction. Therefore p is the largest prime dividing |G|.
We now prove that N is the Sylow p-subgroup of G. Since G is a primitive soluble group, G = NM, where M is a maximal subgroup of G and N ∩ M = 1. Then M G/N is w-supersoluble. By [6,Theorem A.15.6], O p (M) = 1. If p divides |M|, then since M has a Sylow tower of supersoluble type, O p (M) 1, a contradiction. Hence p does not divide |M| and therefore N is the unique Sylow p-subgroup of G.
(d) N is contained in A and N is not contained in B. Suppose that B is a p-group.
A and so G = AN = A, a contradiction. So we may assume that B is not a p-group. If N is contained in B, then since B is nilpotent and N = C G (N), it follows that B is a p-group, a contradiction. Therefore N is not contained in B. Hence B has a nontrivial Hall p -subgroup, B p , which is normal in B. Consequently, AB p = A (A ∩ B)  PROOF OF THEOREM C. Assume the result is not true and let G be a minimal counterexample. It is clear that G 1, A and B are proper subgroups of G, and G is a primitive soluble group. Hence there exists a unique minimal normal subgroup N of G, such that N = F(G) = C G (N). Moreover, G = N. We may assume that A 1 and B 1, otherwise A or B is nilpotent and the result follows from Corollary B. If A ∩ B = 1, then G is the mutually sn-permutable product of A and B. By [1, Theorem C], the group is supersoluble, a contradiction. Thus we may assume A ∩ B 1. Since A permutes with every Sylow subgroup of B and B permutes with every Sylow subgroup of A, it follows that A ∩ B permutes with every Sylow subgroup of A and every Sylow subgroup of B. Hence A ∩ B is subnormal in A and it is a subnormal subgroup of B. Let N 1 denote a minimal normal subgroup of A such that N 1 ≤ A . Since A is supersoluble, it is clear that |N 1 | = p. Note that N 1 (A ∩ B) is a subnormal subgroup of A. Therefore