On some product of SM-groups (Q6570135)

In the article under review, the authors prove that the class of finite SM-groups is closed under products of tcc-subgroups. To be more precise, let us recall some concepts as follows.\N\NLet \(A\) and \(B\) be subgroups of a group \(G\). We say that\N\begin{itemize}\N\item[1.] \(A\) and \(B\) are \textit{totally permutable} if \(UV=VU\) for all subgroups \(U\) of \(A\) and all subgroups \(V\) of \(B\).\N\item[2.] \(A\) and \(B\) are \textit{cc-permutable} if there exists an element \(x\in \langle A, B\rangle\) such that \(A\) and \(B^x\) are permutable.\N\item[3.] \(A\) and \(B\) are \textit{totally cc-permutable} if every subgroup of \(A\) is cc-permutable with every subgroup of \(B\).\N\end{itemize}\N\N\textit{Definition}. A subgroup \(A\) of a group \(G\) is called a \textit{tcc-subgroup} if the following two conditions hold:\N\begin{itemize}\N\item[1.] There exists a subgroup \(Y\) of \(G\) such that \(G=AY\);\N\item[2.] \(A\) is totally cc-permutable with \(Y\).\N\end{itemize}\N\N\textit{Definition}. A group in which every subnormal subgroup is permutable with every maximal subgroup is called a SM-group.\N\NThe main result of the article is the following theorem.\N\N\textit{Theorem}. Let \(G\) be a finite group. Assume that \(G=AB\), where \(A\) and \(B\) are tcc-subgroups of \(G\). If \(A\) and \(B\) are SM-groups, then \(G\) is a SM-group.