On the exponent of mutually permutable products of two abelian groups. (Q308095)

Two subgroups \(A\) and \(B\) are called mutually permutable (totally permutable), if for all \(X\subseteq A\) and \(Y\subseteq B\) we have \(XB=BX\) and \(AY=YA\) (\(XY=YX\)).NEWLINENEWLINE The authors prove: If \(G=AB\) is the product of the totally permutable abelian subgroups \(A\) and \(B\), then \(\exp(G)=\mathrm{lcm}(\exp(A),\exp(B))\). -- If \(G=AB\) and \(A,B\) are abelian and mutually permutable, the \(\exp(G)\) divides \(k(\mathrm{lcm}(\exp(A),\exp(B))\), where \(k\) is the product of all prime divisors of \(|G|\). In addition \(\exp(G')\) divides \(\mathrm{lcm}(\exp(A),\exp(B))\) (Corollary 1, Corollary 2, Theorem 3). This leads to the following more structural facts: If \(G=AB\) is the product of the mutually permutable \(p\)-supersoluble subgroups \(A,B\), then \(G'\) is \(p\)-supersoluble. If instead \(A\) and \(B\) are \(p\)-soluble with \(p\)-lengths \(l_p(A),l_p(B)\) not exceeding \(1\) and \(G'\) is \(p\)-supersoluble, then \(l_p(G)\leq 1\).