Generating finite groups with conjugates of a subgroup. II (Q5925814)

Let \(G\) be a finite group, \(H\) be a subgroup of \(G\), and \(G\) is generated by the conjugates of \(H\). The author considers the following question: when can \(G\) be generated by two conjugates of \(H\)? The chain length \(\text{cl}_G(H)\) of \(H\) in \(G\) is the maximal positive integer \(n\) such that there is a chain \(H=K_0<K_1<\cdots<K_n=G\). If \(\text{cl}_G(H)=1\), then \(H\) is maximal in \(G\) and the answer to the question is yes. In part I [Commun. Algebra 23, No. 11, 3947-3959 (1995; Zbl 0840.20028)], the author proved that if \(\text{cl}_G(H)=2\), \(H_G=1\), and \(G\) cannot be generated by two conjugates of \(H\), then \(G\) is a Frobenius group with cyclic complement \(H\) and elementary Abelian kernel. In the present paper, he describes the structure of \(G\) in the case when \(\text{cl}_G(H)=3\), \(H_G=1\), \(N_G(H)=H\), and \(G\) cannot be generated by two conjugates of \(H\). In particular, in this case \(G\) is not simple. Some problems for further study are also suggested.