Let $F$ be a free group of finite rank, and $K\subset F$ a finite index characteristic subgroup.

Let $\hat{F}$ be the profinite completion of $F$ (i.e. a free profinite group of same rank), and $\bar{K}$ the closure of $K$ in $\hat{F}$.

What are necessary and/or sufficient conditions for $\bar{K}$ to be characteristic in $\hat{F}$?

Is there an example of such $K$ for which $\bar{K}$ is not characteristic?

Motivation: I'm trying to answer the question whether there exists a finite (non-abelian) simple characteristic quotient of $F$. This will be possible only with a characteristic subgroup whose profinite closure is not characteristic.

This question was asked here in a more general setting. A sufficient condition was given in the answer, namely that $K$ be "hyper-characteristic" (or "isomorph-free"). However this doesn't help me, as the quotient by such a subgroup cannot be a non-abeian simple group, so I'm looking for other conditions.

  • $\begingroup$ "(i.e., a free profinite group of the same rank)" The profinite completion is not just a profinite group, it's a profinite group along with a homomorphism from the original group. $\endgroup$ – YCor Mar 15 '18 at 10:45
  • $\begingroup$ The first question is precisely a duplicate of the linked question (mathoverflow.net/questions/250809) by @rtz. It's a pity that rtz accepted his/her own answer, because it would be better gather new answers there, and because it would be worth isolating the second question (about existence of an example). $\endgroup$ – YCor Mar 15 '18 at 11:10
  • $\begingroup$ @YCor The linked question considered general $F$, I was hoping that restricting to free $F$ might help to gather new answers. $\endgroup$ – ChanaG Mar 15 '18 at 12:10
  • $\begingroup$ The other point is that finding an example can be quite energy-consuming (unless there's a pointer to the literature, or a simple example I would fail to see), so it would be more motivating if it's part of a single question. $\endgroup$ – YCor Mar 15 '18 at 12:24
  • $\begingroup$ Here is something in answe to myself. I'm not 100% sure it's correct, and I don't know yet if it can be used to find concrete examples. Let $K_{n}$ be the intersection of all normal subgroups of $F$ of index $n$. Then $\hat{F}$ is the inverse limit of $F/K_{n}$, and the automorphism group of $\hat{F}$ is the inverse limit of automorphism groups of these quotients. Hence thw closure of $H\subset F$ is characteristic in $\hat{F}$ iff the image of $H$ is characteristic in $F/K_{n}$ for every $n$, $\endgroup$ – ChanaG May 23 '18 at 9:00

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.