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Let $P$ be a finitely generated pro-$p$ group and let $G$ be a semidirect product $P \rtimes A$, where $A$ is a finite group of order coprime to $p$ that acts faithfully on $P$. Then one can show that $A$ acts faithfully on $P/\Phi(P)$, so the order of $A$ is bounded in terms of $P$. Does it follow that for fixed $P$, there are only finitely many possibilities for the isomorphism type of $G$?

If we restrict to the case where $A$ is a $q$-group for a single prime $q$, the answer is 'yes' by Sylow's theorem applied to $\mathrm{Aut}(P)$. ($\mathrm{Aut}(P)$ can be made into a profinite group because $P$ has a base of neighbourhoods of the identity consisting of open characteristic subgroups.) If $\mathrm{Aut}(P)$ is prosoluble, the answer is 'yes' by Hall's theorem. But I don't know how well Hall's theorem generalises to (prosoluble)-by-bounded groups, so as far as I know, $\mathrm{Aut}(P)$ could have infinitely many conjugacy classes of finite subgroups of order coprime to $p$.

Edit: To answer in the affirmative, it would be enough to prove the following: given a soluble-by-(order $n$) finite group $G$ of order $ab$, where $a$ and $b$ are coprime, then the number of conjugacy classes of subgroups of $G$ of order dividing $a$ is bounded by a function of $a$ and $n$ alone. In fact just the case when $G$ is (prime power order)-by-(order $n$) would suffice, but the more general version would be interesting as a generalisation of Hall's theorem.

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How do you argue when say $A$ is cyclic of order $q$? I agree that $Aut(P)$ is profinite and you need to consider a $q$-Sylow inside. Then you want this $q$-Sylow to have finitely many conjugacy classes of elements of order $q$, why this is so? – YCor Apr 27 '14 at 7:13
@Yves: Aut(P) is virtually pro-p, because you get an open normal pro-p subgroup consisting of the elements that act trivially on $P/\Phi(P)$. Thus the $q$-Sylow is a finite group. – Colin Reid Apr 27 '14 at 7:32

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