Mutually permutable products of finite groups. (Q643172)

Summary: 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 \(G=G_1G_2=G_1G_3=G_2G_3\) 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 \(i\neq j\) 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 \(G_i\) (\(i=1,2,3\)) whose indices are pairwise relatively prime, and \(G_i\) and \(G_j\) are mutually permutable for all \(i\) and \(j\) with \(i\neq j\), then \(G\) is supersolvable.