4
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ "(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
    Commented Mar 15, 2018 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
    Commented Mar 15, 2018 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
    Commented Mar 15, 2018 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
    Commented Mar 15, 2018 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
    Commented May 23, 2018 at 9:00

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .