Finite groups can be generated by a \(\pi\)-subgroup and a \(\pi^\prime\)-subgroup (Q6584651)

A subgroup \(H\) of a group \(G\) is intravariant if for any \(\alpha \in \mathrm{Aut}(G)\), \(H\) and \(\alpha(H)\) are conjugate in \(G\) (e.g. Sylow subgroups of finite groups).\N\NThe first result in this paper is Theorem 1.1: Let \(\pi \not\) be a set of primes and \(G\) a finite group. Then there exist intravariant subgroups \(P\) and \(R\) of \(G\) such that \(G =\langle P, R \rangle\) with \(P\) a \(\pi\)-subgroup and \(R\) a \(\pi'\)-subgroup.\N\NUsing the proof of Theorem 1.1 and Iwasawa's criterion for freeness, the authors (thanks to a remark by D. Haran and A. Lubotzky), prove Theorem 1.3: Let \(\pi\) be a set of primes. Let \(A\) be the free pro-\(\pi\) group on countably many generators and \(B\) be the free pro-\(\pi'\) group on countably many generators. Then \(A \ast B\) is a free profinite group.\N\NAnother result in this paper is Theorem 1.5: Let \(S\) be the free prosolvable group on countably many generators, then \(S\ast S\) is a free profinite group. If \(\pi\) is a set of primes and either \(2 \in \pi\) or \(\{3, 5\} \subseteq \pi\) and \(A\) is the free pro-\(\pi\)-group on countably many generators then \(A \ast S\) is a free profinite group.