International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 867082 | 4 pages | https://doi.org/10.5402/2011/867082

Mutually Permutable Products of Finite Groups

Academic Editor: L. Vinet
Received19 Jun 2011
Accepted10 Jul 2011
Published07 Sep 2011

Abstract

Let G be a finite group and G1, G2 are two subgroups of G. We say that G1 and G2 are mutually permutable if G1 is permutable with every subgroup of G2 and G2 is permutable with every subgroup of G1. We prove that if ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 is the product of three supersolvable subgroups G1, G2, and G3, where Gi and Gj are mutually permutable for all i and j with ๐‘– โ‰  ๐‘— 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 ๐บ ๐‘– ( ๐‘– = 1 , 2 , 3 ) whose indices are pairwise relatively prime, and Gi and Gj are mutually permutable for all i and j with ๐‘– โ‰  ๐‘— , then G is supersolvable.

1. Introduction

Throughout this paper, ๐บ will denote a finite group. We write ๐œŽ ( ๐บ ) for the set of prime divisors of the order of ๐บ and | ๐œŽ ( ๐บ ) | 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 ๐บ is a minimal non-supersolvable group, then ๐บ is solvable and 2 โ‰ค | ๐œŽ ( ๐บ ) | โ‰ค 3 . Therefore, if ๐บ is a minimal non-supersolvable group with | ๐œŽ ( ๐บ ) | = 3 , then ๐บ possesses three supersolvable subgroups ๐บ ๐‘– ( ๐‘– = 1 , 2 , 3 ) such that ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 . In [2], Kegel proved that if ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 , where ๐บ 1 and ๐บ 2 are nilpotent subgroups of ๐บ , and ๐บ 3 is a supersolvable subgroup of ๐บ , then ๐บ is supersolvable. In [3], Asaad and Shaalan proved the following result: assume that ๐บ 1 and ๐บ 2 are supersolvable subgroups of ๐บ , ๐บ ๎…ž is nilpotent and ๐บ = ๐บ 1 ๐บ 2 . Assume further that ๐บ 1 is permutable with every subgroup of ๐บ 2 and ๐บ 2 is permutable with every subgroup of ๐บ 1 ( ๐บ 1 and ๐บ 2 are mutually permutable). Then ๐บ is supersolvable. Further, they proved the following result: Assume that ๐บ 1 and ๐บ 2 are supersolvable subgroups of ๐บ and ๐บ = ๐บ 1 ๐บ 2 and that every subgroup of ๐บ 1 is permutable with every subgroup of ๐บ 2 ( ๐บ 1 and ๐บ 2 are totally permutable). Then, ๐บ is supersolvable.

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

Assume that ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 , where ๐บ 1 , ๐บ 2 , and ๐บ 3 are supersolvable subgroups and that ๐บ ๐‘– and ๐บ ๐‘— are mutually permutable for all ๐‘– and ๐‘— with ๐‘– โ‰  ๐‘— . Is ๐บ supersolvable? The answer is negative as the following example shows.

Example 1 (see [8, pages 8-9]). Let ๐ป = โŸจ ๐‘ฅ โŸฉ ร— โŸจ ๐‘ฆ โŸฉ , where | ๐‘ฅ | = | ๐‘ฆ | = 5 . The maps ๐›ผ โˆถ ๐‘ฅ โ†’ ๐‘ฅ 2 , ๐‘ฆ โ†’ ๐‘ฆ โˆ’ 2 , and ๐›ฝ โˆถ ๐‘ฅ โ†’ ๐‘ฆ โˆ’ 1 , ๐‘ฆ โ†’ ๐‘ฅ are automorphisms of ๐ป and generate a subgroup ๐ด โ‰ค A u t ( ๐ป ) of order 8 ( ๐ด is isomorphic with a quaternion group). Take ๐บ = [ ๐ป ] ๐ด . Then ๐บ 1 = ๐ป โŸจ ๐›ผ โŸฉ , ๐บ 2 = ๐ป โŸจ ๐›ฝ โŸฉ , and ๐บ 3 = ๐ป โŸจ ๐›ผ ๐›ฝ โŸฉ are normal supersolvable subgroups of ๐บ and ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 , but ๐บ is not supersolvable.

We prove the following result.

Theorem 1.1. If ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 โ€‰โ€‰is the product of three supersolvable subgroups ๐บ 1 , ๐บ 2 , and ๐บ 3 such that ๐บ ๐‘– and ๐บ ๐‘— are mutually permutable for all ๐‘– and ๐‘— with ๐‘– โ‰  ๐‘— , and the Sylow subgroups of ๐บ are abelian, then ๐บ is supersolvable.

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

Corollary 1.2. If ๐บ possesses three supersolvable subgroups ๐บ ๐‘– (i= 1,2,3) whose indices in ๐บ are pairwise relatively prime, and ๐บ ๐‘– is permutable with every subgroup of ๐บ ๐‘— , for all ๐‘– โ€‰โ€‰and ๐‘— with ๐‘– โ‰  ๐‘— , then ๐บ is supersolvable.

2. Preliminaries

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

Lemma 2.1 (see [2]). Let ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 โ€‰โ€‰be the product of three subgroups ๐บ 1 , ๐บ 2 , and ๐บ 3 . If ๐บ 1 , ๐บ 2 , and ๐บ 3 have normal Sylow p-subgroups for a certain prime p, then also ๐บ has a normal Sylow p-subgroup.

Lemma 2.2 (see [4]). Let ๐บ = ๐บ 1 ๐บ 2 be a group such that ๐บ 1 and ๐บ 2 are mutually permutable. (a)If ๐บ 1 โˆฉ ๐บ 2 = 1 , then ๐บ 1 and ๐บ 2 are totally permutable.(b) ๐บ 1 โˆฉ ๐บ 2 is a quasinormal subgroup of ๐บ 1 and of ๐บ 2 .

Lemma 2.3 (see [3]). Let ๐บ = ๐บ 1 ๐บ 2 โ€‰โ€‰be a group such that ๐บ 1 and ๐บ 2 are totally permutable subgroups. If ๐บ 1 and ๐บ 2 are supersolvable subgroups of ๐บ , then ๐บ is supersolvable.

Lemma 2.4 (see [5, page 213, Theoremโ€‰โ€‰7.1.2]). If ๐ป is a quasinormal subgroup of ๐บ , then ๐ป is subnormal in ๐บ .

Lemma 2.5 (see [5, page 239, Theoremโ€‰โ€‰7.7.1]). Let ๐บ = ๐บ 1 ๐บ 2 be a group such that ๐บ 1 and ๐บ 2 are subgroups of ๐บ . If ๐ป is a subnormal subgroup of ๐บ 1 and of ๐บ 2 , then ๐ป is subnormal in ๐บ .

Lemma 2.6 (see [6]). Let ๐บ be a group and ๐ป is a quasinormal subgroup of ๐บ . Then, ๐ป / C o r ๐บ ( ๐ป ) is nilpotent, where C o r ๐บ ( ๐ป ) = โˆฉ ๐‘” โˆˆ ๐บ ๐ป ๐‘” .

Lemma 2.7 (see [7, page 29, Theoremโ€‰โ€‰8.8(a)]). ๐น ( ๐บ ) = โŸจ ๐ป โˆถ ๐ป is subnormal subgroup o f ๐บ and ๐ป is nilpotent โŸฉ .

Lemma 2.8 (see [2]). Let the group ๐บ = ๐บ 1 ๐บ 2 = ๐บ 1 ๐บ 3 = ๐บ 2 ๐บ 3 โ€‰โ€‰be the product of three subgroups ๐บ 1 , ๐บ 2 , and ๐บ 3 . If ๐บ 1 , ๐บ 2 , and ๐บ 3 are nilpotent subgroups of ๐บ , then ๐บ is nilpotent.

Lemma 2.9 (see [8, page 196, Theoremโ€‰โ€‰5.1(15)]). If ๐บ is a supersolvable group, then ๐บ / ๐‘‚ ๐‘ โ€ฒ , ๐‘ ( ๐บ ) is abelian of exponent dividing ๐‘ โˆ’ 1 for all primes ๐‘ .

Lemma 2.10 (see [8, page 6, Theoremโ€‰โ€‰1.9]). Let ๐‘ƒ be a normal Sylow ๐‘ -subgroup of ๐บ . If ๐บ / ๐‘ƒ is abelian of exponent dividing ๐‘ โˆ’ 1 , then ๐บ is supersolvable.

Lemma 2.11 (see [8, page 5, Theoremโ€‰โ€‰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 ๐บ be a counterexample of minimal order. Since ๐บ ๐‘– ( ๐‘– = 1 , 2 , 3 ) is supersolvable, it follows that ๐บ ๐‘– has a normal Sylow ๐‘ -subgroup, where ๐‘ is the largest prime dividing | ๐บ | . Then, by Lemma 2.1, ๐บ has a normal Sylow ๐‘ -subgroup, say ๐‘ƒ . Certainly, every proper quotient group of ๐บ satisfies the hypothesis of the theorem. So every proper quotient group of ๐บ is supersolvable by the minimal choice of ๐บ . But the class of all supersolvable groups is a saturated formation, so ฮฆ ( ๐บ ) = 1 , ๐‘ƒ = ๐น ( ๐บ ) , and ๐ถ ๐บ ( ๐‘ƒ ) = ๐‘ƒ . We argue that ๐บ 1 โˆฉ ๐บ 2 โ‰  1 . If not, ๐บ 1 โˆฉ ๐บ 2 = 1 . Then, ๐บ = ๐บ 1 ๐บ 2 is a totally permutable product of ๐บ 1 and ๐บ 2 by Lemma 2.2(a). Then, by Lemma 2.3, ๐บ is supersolvable, a contradiction.
Thus, ๐บ 1 โˆฉ ๐บ 2 โ‰  1 . Analogously, ๐บ 1 โˆฉ ๐บ 3 โ‰  1 and ๐บ 2 โˆฉ ๐บ 3 โ‰  1 . Since ๐บ = ๐บ 1 ๐บ 2 is a mutually permutable product of ๐บ 1 and ๐บ 2 , it follows that ๐บ 1 โˆฉ ๐บ 2 is a quasinormal subgroup of ๐บ 1 and of ๐บ 2 by Lemma 2.2(b). Then, by Lemma 2.4, ๐บ 1 โˆฉ ๐บ 2 is a subnormal subgroup of ๐บ 1 and of ๐บ 2 . Hence ๐บ 1 โˆฉ ๐บ 2 is a subnormal subgroup of ๐บ by Lemma 2.5.
If C o r ๐บ 1 ( ๐บ 1 โˆฉ ๐บ 2 ) = 1 , then, by Lemma 2.6, ๐บ 1 โˆฉ ๐บ 2 is nilpotent. Hence, ๐บ 1 โˆฉ ๐บ 2 is a subnormal nilpotent subgroup of ๐บ . So ๐บ 1 โˆฉ ๐บ 2 โ‰ค ๐‘ƒ by Lemma 2.7. Taken into consideration that ๐บ 1 โˆฉ ๐บ 2 is quasinormal in ๐บ 1 , ๐บ 1 โˆฉ ๐บ 2 โ‰ค ๐‘ƒ and ๐‘ƒ is abelian, it follows that ๐บ 1 โˆฉ ๐บ 2 is normal in ๐บ 1 and so ๐บ 1 โˆฉ ๐บ 2 = C o r ๐บ 1 ( ๐บ 1 โˆฉ ๐บ 2 ) = 1 , a contradiction. Thus C o r ๐บ 1 ( ๐บ 1 โˆฉ ๐บ 2 ) โ‰  1 . Analogously, C o r ๐บ 2 ( ๐บ 2 โˆฉ ๐บ 3 ) โ‰  1 and C o r ๐บ 3 ( ๐บ 3 โˆฉ ๐บ 1 ) โ‰  1 . Set ๐ฟ 1 = C o r ๐บ 1 ( ๐บ 1 โˆฉ ๐บ 2 ) , ๐ฟ 2 = C o r ๐บ 2 ( ๐บ 2 โˆฉ ๐บ 3 ) and ๐ฟ 3 = C o r ๐บ 3 ( ๐บ 3 โˆฉ ๐บ 1 ) . Then ๐ฟ ๐บ 1 = ๐ฟ ๐บ 1 ๐บ 2 1 โ‰ค ๐บ 2 , ๐ฟ ๐บ 2 = ๐ฟ ๐บ 2 ๐บ 3 2 โ‰ค ๐บ 3 , and ๐ฟ ๐บ 3 = ๐ฟ ๐บ 3 ๐บ 1 3 โ‰ค ๐บ 1 . Now, ๐‘ƒ โ‰ค ๐ฟ ๐บ ๐‘– ( ๐‘– = 1 , 2 , 3 ) as ๐‘ƒ is a unique minimal normal subgroup of ๐บ . Hence, ๐‘ƒ โ‰ค ๐บ ๐‘– ( ๐‘– = 1 , 2 , 3 ).
Now, we finish the proof of the theorem. Since ๐บ ๐‘– is supersolvable, ๐ถ ๐บ ( ๐‘ƒ ) = ๐‘ƒ = ๐น ( ๐บ ) and ๐‘ƒ โ‰ค ๐บ ๐‘– ( ๐‘– = 1 , 2 , 3 ), it follows that ๐ถ ๐บ ๐‘– ( ๐‘ƒ ) = ๐ถ ๐บ ๐‘– ( ๐น ( ๐บ ๐‘– ) ) = ๐‘ƒ ( ๐‘– = 1 , 2 , 3 ) and ๐บ ๐‘– / ๐‘ƒ is abelian. Hence, by Lemma 2.8, ๐บ / ๐‘ƒ is nilpotent, and, since the Sylow subgroups of ๐บ are abelian, it follows that ๐บ / ๐‘ƒ is abelian. On the other hand, by Lemma 2.9, ๐บ ๐‘– / ๐‘ƒ ( ๐‘– = 1 , 2 ) is of exponent dividing ๐‘ โˆ’ 1 . Hence, ๐บ / ๐‘ƒ is abelian of exponent dividing ๐‘ โˆ’ 1 and so ๐บ 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 ๐บ be a counterexample of minimal order. Let ๐‘ƒ be a Sylow ๐‘ -subgroup of ๐บ , where ๐‘ is the largest prime dividing | ๐บ | . Then, ๐‘ƒ is normal in ๐บ by Lemma 2.1. Certainly, every proper quotient group of ๐บ satisfies the hypothesis of the corollary. So every proper quotient group of ๐บ is supersolvable by the minimal choice of ๐บ . But the class of all supersolvable groups is a saturated formation, so ฮฆ ( ๐บ ) = 1 , ๐‘ƒ = ๐น ( ๐บ ) , and ๐ถ ๐บ ( ๐‘ƒ ) = ๐‘ƒ . Since ๐บ 1 , ๐บ 2 , and ๐บ 3 have coprime indices, we can assume that ๐‘ does not divide | ๐บ โˆถ ๐บ 1 | and ๐‘ does not divide | ๐บ โˆถ ๐บ 2 | . Then, ๐‘ƒ โ‰ค ๐บ 1 and ๐‘ƒ โ‰ค ๐บ 2 and so ๐‘ƒ = ๐ถ ๐บ 1 ( ๐‘ƒ ) = ๐ถ ๐บ 2 ( ๐‘ƒ ) = ๐น ( ๐บ 2 ) = ๐น ( ๐บ 1 ) as ๐‘ƒ = ๐น ( ๐บ ) = ๐ถ ๐บ ( ๐‘ƒ ) . Then, ๐บ 1 / ๐‘ƒ and ๐บ 2 / ๐‘ƒ are abelian subgroups of ๐บ / ๐‘ƒ by Lemma 2.11. This together with ( | ๐บ โˆถ ๐บ 1 | , | ๐บ โˆถ ๐บ 2 | ) = 1 imply that the Sylow subgroups of ๐บ / ๐‘ƒ are abelian. Now Theorem 1.1 implies that ๐บ is supersolvable, a contradiction completing the proof of the corollary.

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Copyright © 2011 Rola A. Hijazi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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