Is there a profinite group $G$, a continuous automorphism $\alpha$ of $G$ and a topologically finitely generated closed subgroup $H \leq G$ such that $\alpha(H) \lneq H$ ?

Note that if an example exists, then $G$ is not topologically finite generated, and $\alpha$ is not given by conjugation by an element of $G$.

If we allow $H$ not to be topologically finitely generated, then the example given by Yves de-Cornullier in Wild automorphisms of a free group gives rise to an example here.

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  • $\begingroup$ Since I answered the other question in a comment (because it was really a trivial example), I would rather suggest you edit that other question and erase this one. $\endgroup$ – YCor Nov 24 '15 at 13:59
  • $\begingroup$ @YCor maybe this is a good idea but I do not know how to erase a question. $\endgroup$ – Pablo Nov 24 '15 at 14:02
  • $\begingroup$ An artefact would be to first edit the other one, then to edit this one as "Now a duplicate of... " and the question will be quickly closed as duplicate (and even erased by a moderator). $\endgroup$ – YCor Nov 24 '15 at 14:06


Let $K_n = K_n(G)$ be the intersection of all open normal subgroups of $G$ of index at most $n$. Then $\alpha(K_n) = K_n$. If we replace $G$ with $G/K_n$, we still have $\alpha$ acting as an automorphism on $G/K_n$ and for $n$ large enough, $\alpha(H)K_n/K_n$ is a proper subgroup of $HK_n/K_n$ (since $H$ and $\alpha(H)$ are both closed, we can distinguish them using finite images of $G$). So we may assume $K_n(G)$ is trivial for some $n$. It then follows that $K_n(H)$ is trivial. But if a topologically finitely generated profinite group $H$ satisfies $K_n(H) = 1$ for some $n$, then $H$ must be finite, since $H$ only has finitely many finite quotients of any given order. This gives a contradiction.

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