On groups with Frobenius elements. (Q5899618)

Let \(G\) be a group and \(H\) be a proper subgroup of \(G\). An element \(a\in G\) is called \(H\)-\textit{Frobenius} if \(a\in H\) and, for every \(g\in G\setminus H\), the group \(L_g=\langle a,a^g\rangle\) is a Frobenius group with a complement containing \(a\). A nontrivial element \(y\) of a group \(X\) is \textit{finite} if the subgroup \(\langle y,y^x\rangle\) is finite for every \(x\in X\). The main result proved in this paper is the following Theorem. Suppose that \(a\) is an \(H\)-Frobenius element of a group \(G\), \(|a|=3n>3\), \(b\) is an element of order \(3\) in \(\langle a\rangle\), and at least one of the following is fulfilled: (1) \(b^{-1}\cdot b^H\) consists of periodic elements and \(n\) divides by 3; (2) the subgroup \(\langle a\rangle\) contains a \(3'\)-element finite in \(H\). Then the union \(F\) of all kernels of Frobenius subgroups of \(G\) with complement \(\langle a\rangle\) is a nilpotent subgroup normal in \(G\) and \(G=FH\).