Let $G$ be a finite group and suppose that $P$ is a soluble $\{2, 3\}^\prime$-subgroup of $G$. The reader will lose only a little by assuming that $P$ is a subgroup of prime order $p > 3$.Define\[\Sigma_G(P) = \{A \le G \mid A \hbox{ is soluble and } A = \langle P, P^a\rangle\hbox{ for some }a \in A\}.\] This set is partially ordered by inclusion and we let\[\Sigma^*_G(P)\] denote the set of maximal members of $\Sigma_G(P)$.