#### Abstract

Let *G* be a finite group and *G*_{1}, *G*_{2} are two subgroups of *G*. We
say that *G*_{1} and *G*_{2} are mutually permutable if *G*_{1} is permutable with every
subgroup of *G*_{2} and *G*_{2} is permutable with every subgroup of *G*_{1}. We prove
that if is the product of three supersolvable subgroups *G*_{1}, *G*_{2}, and *G*_{3}, where *G*_{i} and *G*_{j} are mutually permutable for all *i*
and *j* with and the Sylow subgroups of *G* are abelian, then *G* is supersolvable. As a corollary of this result, we also prove that if *G* possesses three
supersolvable subgroups whose indices are pairwise relatively
prime, and *G*_{i} and *G*_{j} are mutually permutable for all *i* and *j* with , then
*G* is supersolvable.

#### 1. Introduction

Throughout this paper, will denote a finite group. We write for the set of prime divisors of the order of and for their number. In [1], Doerk determined the structure of minimal non-supersolvable groups (nonsupersolvable groups and all of whose proper subgroups are supersolvable). He proved that if is a minimal non-supersolvable group, then is solvable and . Therefore, if is a minimal non-supersolvable group with , then possesses three supersolvable subgroups such that . In [2], Kegel proved that if , where and are nilpotent subgroups of , and is a supersolvable subgroup of , then is supersolvable. In [3], Asaad and Shaalan proved the following result: assume that and are supersolvable subgroups of , is nilpotent and . Assume further that is permutable with every subgroup of and is permutable with every subgroup of ( and are mutually permutable). Then is supersolvable. Further, they proved the following result: Assume that and are supersolvable subgroups of and and that every subgroup of is permutable with every subgroup of ( and are totally permutable). Then, is supersolvable.

In this paper, we are interested in the following question.

Assume that , where , , and are supersolvable subgroups and that and are mutually permutable for all and with . Is supersolvable? The answer is negative as the following example shows.

*Example 1 (see [8, pages 8-9]). *Let , where . The maps , , and , are automorphisms of and generate a subgroup of order 8 ( is isomorphic with a quaternion group). Take . Then , , and are normal supersolvable subgroups of and , but is not supersolvable.

We prove the following result.

Theorem 1.1. *If is the product of three supersolvable subgroups , , and such that and are mutually permutable for all and with , and the Sylow subgroups of are abelian, then is supersolvable.*

As a corollary of Theorem 1.1, we have the following.

Corollary 1.2. *If possesses three supersolvable subgroups (i= 1,2,3) whose indices in are pairwise relatively prime, and is permutable with every subgroup of , for all and with , then is supersolvable.*

#### 2. Preliminaries

We list here some basic results which are needed in this paper.

Lemma 2.1 (see [2]). *Let be the product of three subgroups , , and . If , , and have normal Sylow p-subgroups for a certain prime p, then also has a normal Sylow p-subgroup.*

Lemma 2.2 (see [4]). *Let be a group such that and are mutually permutable. *(a)*If , then and are totally permutable.*(b)* is a quasinormal subgroup of and of .*

Lemma 2.3 (see [3]). *Let be a group such that and are totally permutable subgroups. If and are supersolvable subgroups of , then is supersolvable.*

Lemma 2.4 (see [5, page 213, Theorem 7.1.2]). *If is a quasinormal subgroup of , then is subnormal in .*

Lemma 2.5 (see [5, page 239, Theorem 7.7.1]). *Let be a group such that and are subgroups of . If is a subnormal subgroup of and of , then is subnormal in .*

Lemma 2.6 (see [6]). *Let be a group and is a quasinormal subgroup of . Then, is nilpotent, where .*

Lemma 2.7 (see [7, page 29, Theorem 8.8(a)]). *is subnormal subgroup andis nilpotent.*

Lemma 2.8 (see [2]). *Let the group be the product of three subgroups , , and . If , , and are nilpotent subgroups of , then is nilpotent.*

Lemma 2.9 (see [8, page 196, Theorem 5.1(15)]). *If is a supersolvable group, then is abelian of exponent dividing for all primes .*

Lemma 2.10 (see [8, page 6, Theorem 1.9]). *Let be a normal Sylow -subgroup of . If is abelian of exponent dividing , then is supersolvable.*

Lemma 2.11 (see [8, page 5, Theorem 1.6]). *The commutator subgroup of a supersolvable group is nilpotent.*

#### 3. Proofs

*Proof of Theorem 1.1. *Assume that the result is not true, and let be a counterexample of minimal order. Since is supersolvable, it follows that has a normal Sylow -subgroup, where is the largest prime dividing . Then, by Lemma 2.1, has a normal Sylow -subgroup, say . Certainly, every proper quotient group of satisfies the hypothesis of the theorem. So every proper quotient group of is supersolvable by the minimal choice of . But the class of all supersolvable groups is a saturated formation, so , , and . We argue that . If not, . Then, is a totally permutable product of and by Lemma 2.2(a). Then, by Lemma 2.3, is supersolvable, a contradiction.

Thus, . Analogously, and . Since is a mutually permutable product of and , it follows that is a quasinormal subgroup of and of by Lemma 2.2(b). Then, by Lemma 2.4, is a subnormal subgroup of and of . Hence is a subnormal subgroup of by Lemma 2.5.

If , then, by Lemma 2.6, is nilpotent. Hence, is a subnormal nilpotent subgroup of . So by Lemma 2.7. Taken into consideration that is quasinormal in , and is abelian, it follows that is normal in and so , a contradiction. Thus . Analogously, and . Set , and . Then , , and . Now, () as is a unique minimal normal subgroup of . Hence, ().

Now, we finish the proof of the theorem. Since is supersolvable, and (), it follows that () and is abelian. Hence, by Lemma 2.8, is nilpotent, and, since the Sylow subgroups of are abelian, it follows that is abelian. On the other hand, by Lemma 2.9, () is of exponent dividing . Hence, is abelian of exponent dividing and so is supersolvable, by Lemma 2.10, a final contradiction completing the proof of the theorem.

*Proof of Corollary 1.2. *Assume that the result is not true and let be a counterexample of minimal order. Let be a Sylow -subgroup of , where is the largest prime dividing . Then, is normal in by Lemma 2.1. Certainly, every proper quotient group of satisfies the hypothesis of the corollary. So every proper quotient group of is supersolvable by the minimal choice of . But the class of all supersolvable groups is a saturated formation, so , , and . Since , , and have coprime indices, we can assume that does not divide and does not divide . Then, and and so as . Then, and are abelian subgroups of by Lemma 2.11. This together with imply that the Sylow subgroups of are abelian. Now Theorem 1.1 implies that is supersolvable, a contradiction completing the proof of the corollary.