# Mutually Permutable Products of Finite Groups

Authors: Rola A. Hijazi
Journal: ISRN Algebra
 (2011)

---

## Abstract

LetG 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 G=G1G2=G1G3=G2G3 is the product of three supersolvable subgroups G1, G2, and G3, where Gi and Gj are mutually permutable for all i
and j with i≠j 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 Gi(i=1,2,3) whose indices are pairwise relatively
prime, and Gi and Gj are mutually permutable for all i and j with i≠j, then
G is supersolvable.


---

## Body

## 1. Introduction

Throughout this paper,G will denote a finite group. We write σ(G) for the set of prime divisors of the order of G and |σ(G)| 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 G is a minimal non-supersolvable group, then G is solvable and 2≤|σ(G)|≤3. Therefore, if G is a minimal non-supersolvable group with |σ(G)|=3, then G possesses three supersolvable subgroups Gi(i=1,2,3) such that G=G1G2=G1G3=G2G3. In [2], Kegel proved that if G=G1G2=G1G3=G2G3, where G1 and G2 are nilpotent subgroups of G, and G3 is a supersolvable subgroup of G, then G is supersolvable. In [3], Asaad and Shaalan proved the following result: assume that G1 and G2 are supersolvable subgroups of G, G′ is nilpotent and G=G1G2. Assume further that G1 is permutable with every subgroup of G2 and G2 is permutable with every subgroup of G1 (G1 and G2 are mutually permutable). Then G is supersolvable. Further, they proved the following result: Assume that G1 and G2 are supersolvable subgroups of G and G=G1G2 and that every subgroup of G1 is permutable with every subgroup of G2 (G1 and G2 are totally permutable). Then, G is supersolvable.In this paper, we are interested in the following question.Assume thatG=G1G2=G1G3=G2G3, where G1, G2, and G3 are supersolvable subgroups and that Gi and Gj are mutually permutable for all i and j with i≠j. Is G supersolvable? The answer is negative as the following example shows.Example 1 (see [8, pages 8-9]).
LetH=〈x〉×〈y〉, where |x|=|y|=5. The maps α:x→x2, y→y-2, and β:x→y-1, y→x are automorphisms of H and generate a subgroup A≤Aut(H) of order 8 (A is isomorphic with a quaternion group). Take G=[H]A. Then G1=H〈α〉, G2=H〈β〉, and G3=H〈αβ〉 are normal supersolvable subgroups of G and G=G1G2=G1G3=G2G3, but G is not supersolvable.We prove the following result.Theorem 1.1.
IfG=G1G2=G1G3=G2G3  is the product of three supersolvable subgroups  G1,  G2, and  G3 such that  Gi and  Gj are mutually permutable for all  i and  j with  i≠j, and the Sylow subgroups of  G are abelian, then  G is supersolvable.As a corollary of Theorem1.1, we have the following.Corollary 1.2.
IfG possesses three supersolvable subgroups  Gi (i= 1,2,3) whose indices in  G are pairwise relatively prime, and  Gi is permutable with every subgroup of  Gj, for all  i  and  j with  i≠j, then  G is supersolvable.


## 2. Preliminaries

We list here some basic results which are needed in this paper.Lemma 2.1 (see [2]).
LetG=G1G2=G1G3=G2G3  be the product of three subgroups  G1,  G2, and  G3. If  G1,  G2, and  G3 have normal Sylow p-subgroups for a certain prime p, then also  G has a normal Sylow p-subgroup.Lemma 2.2 (see [4]).
LetG=G1G2 be a group such that  G1and  G2 are mutually permutable. (a)
IfG1∩G2=1, then  G1and  G2 are totally permutable.(b)
G1∩G2 is a quasinormal subgroup of  G1 and of  G2.Lemma 2.3 (see [3]).
LetG=G1G2  be a group such that  G1and  G2 are totally permutable subgroups. If  G1 and  G2 are supersolvable subgroups of  G, then  G is supersolvable.Lemma 2.4 (see [5, page 213, Theorem  7.1.2]).
IfH is a quasinormal subgroup of  G, then  H is subnormal in  G.Lemma 2.5 (see [5, page 239, Theorem  7.7.1]).
LetG=G1G2 be a group such that  G1 and  G2 are subgroups of  G. If  H is a subnormal subgroup of  G1 and of  G2, then  H is subnormal in  G.Lemma 2.6 (see [6]).
LetG be a group and  H is a quasinormal subgroup of  G. Then,  H/CorG(H) is nilpotent, where  CorG(H)=∩g∈GHg.Lemma 2.7 (see [7, page 29, Theorem  8.8(a)]).
F(G)=〈H:His subnormal subgroup ofGandHis nilpotent〉.Lemma 2.8 (see [2]).
Let the groupG=G1G2=G1G3=G2G3  be the product of three subgroups  G1,  G2, and  G3. If  G1,  G2, and  G3 are nilpotent subgroups of  G, then  G is nilpotent.Lemma 2.9 (see [8, page 196, Theorem  5.1(15)]).
IfG is a supersolvable group, then  G/Op′,p(G) is abelian of exponent dividing  p-1 for all primes  p.Lemma 2.10 (see [8, page 6, Theorem  1.9]).
LetP be a normal Sylow  p-subgroup of  G. If  G/P is abelian of exponent dividing  p-1, then  G 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 Theorem1.1.
Assume that the result is not true, and letG be a counterexample of minimal order. Since Gi(i=1,2,3) is supersolvable, it follows that Gi has a normal Sylow p-subgroup, where p is the largest prime dividing |G|. Then, by Lemma 2.1, G has a normal Sylow p-subgroup, say P. Certainly, every proper quotient group of G satisfies the hypothesis of the theorem. So every proper quotient group of G is supersolvable by the minimal choice of G. But the class of all supersolvable groups is a saturated formation, so Φ(G)=1, P=F(G), and CG(P)=P. We argue that G1∩G2≠1. If not, G1∩G2=1. Then, G=G1G2 is a totally permutable product of G1 and G2 by Lemma 2.2(a). Then, by Lemma 2.3, G is supersolvable, a contradiction.
Thus,G1∩G2≠1. Analogously, G1∩G3≠1 and G2∩G3≠1. Since G=G1G2 is a mutually permutable product of G1 and G2, it follows that G1∩G2 is a quasinormal subgroup of G1 and of G2 by Lemma 2.2(b). Then, by Lemma 2.4, G1∩G2 is a subnormal subgroup of G1 and of G2. Hence G1∩G2 is a subnormal subgroup of G by Lemma 2.5.
IfCorG1(G1∩G2)=1, then, by Lemma 2.6, G1∩G2 is nilpotent. Hence, G1∩G2 is a subnormal nilpotent subgroup of G. So G1∩G2≤P by Lemma 2.7. Taken into consideration that G1∩G2 is quasinormal in G1, G1∩G2≤P and P is abelian, it follows that G1∩G2 is normal in G1 and so G1∩G2=CorG1(G1∩G2)=1, a contradiction. Thus CorG1(G1∩G2)≠1. Analogously, CorG2(G2∩G3)≠1 and CorG3(G3∩G1)≠1. Set L1=CorG1(G1∩G2), L2=CorG2(G2∩G3) and L3=CorG3(G3∩G1). Then L1G=L1G1G2≤G2, L2G=L2G2G3≤G3, and L3G=L3G3G1≤G1. Now, P≤LiG (i=1,2,3) as P is a unique minimal normal subgroup of G. Hence, P≤Gi (i=1,2,3).
Now, we finish the proof of the theorem. SinceGi is supersolvable, CG(P)=P=F(G) and P≤Gi (i=1,2,3), it follows that CGi(P)=CGi(F(Gi))=P (i=1,2,3) and Gi/P is abelian. Hence, by Lemma 2.8, G/P is nilpotent, and, since the Sylow subgroups of G are abelian, it follows that G/P is abelian. On the other hand, by Lemma 2.9, Gi/P (i=1,2) is of exponent dividing p-1. Hence, G/P is abelian of exponent dividing p-1 and so G is supersolvable, by Lemma 2.10, a final contradiction completing the proof of the theorem.Proof of Corollary1.2.
Assume that the result is not true and letG be a counterexample of minimal order. Let P be a Sylow p-subgroup of G, where p is the largest prime dividing |G|. Then, P is normal in G by Lemma 2.1. Certainly, every proper quotient group of G satisfies the hypothesis of the corollary. So every proper quotient group of G is supersolvable by the minimal choice of G. But the class of all supersolvable groups is a saturated formation, so Φ(G)=1, P=F(G), and CG(P)=P. Since G1, G2, and G3 have coprime indices, we can assume that p does not divide |G:G1| and p does not divide |G:G2|. Then, P≤G1 and P≤G2 and so P=CG1(P)=CG2(P)=F(G2)=F(G1) as P=F(G)=CG(P). Then, G1/P and G2/P are abelian subgroups of G/P by Lemma 2.11. This together with (|G:G1|,|G:G2|)=1 imply that the Sylow subgroups of G/P are abelian. Now Theorem 1.1 implies that G is supersolvable, a contradiction completing the proof of the corollary.



---

*Source: 867082-2011-09-07.xml*