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Let $G = N \rtimes K$ be a semi-direct product of groups and suppose that $K$ is a finite group. Call the set $\mathcal{F} = \{ \alpha \in G \mid \langle \alpha, K \rangle = \langle \alpha \rangle \ast K \}$ the set of free partners of $K$ in $G$. I was wondering the following:

  • Can $\mathcal{F}$ be profinitely-dense in $G$?

Recall that profinitely-dense means dense in the profinite topology i.e. $\mathcal{F}$ has non-trivial intersection with every coset of every finite-index normal subgroup. In other words $\mathcal{F}H=G$ for any finite index subgroup $H$ of $G$.

Edit: My case of interest would mainly be for a finitely generated group $G$.

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  • $\begingroup$ If there exists a free partner (in the above sense: I would rather said a cyclic free factor), obviously $K$ can't be profinitely dense (unless $K=G$). In general (=if $K$ is an arbitrary retract, there are easy examples for which $K$ is profinitely dense (e.g., take $N$ to be infinite simple). $\endgroup$
    – YCor
    Commented Aug 21, 2021 at 15:53
  • $\begingroup$ Thanks! I agree that, in general, one can take $N$ to be a group such that K is profinitely dense. However, what is not clear to me is the first part of your remark: that the existence of an element of $\mathcal{F}$ implies that $K$ is not profinitely-dense. Besides, I am really wondering about the profinite-density of the set $\mathcal{F}$. $\endgroup$
    – Doryan Temmerman
    Commented Aug 21, 2021 at 19:53
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    $\begingroup$ If $G=K\ast F$ with $F$ residually finite, then the kernel of $G\to F$ is profinitely closed and contains $K$, so contains the profinite closure of $K$. $\endgroup$
    – YCor
    Commented Aug 21, 2021 at 20:30

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Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the trivial group, $N=G$. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense. If you do not want $K$ to be trivial, then take $G={\mathbb Z}\wr C_2$, $K=C_2$, you get a trivial $\mathcal F$. On the other hand take $G={\mathbb Z} * C_2$ and represent it as $N\rtimes C_2$.

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  • $\begingroup$ I agree that for the example $\mathbb{Z} \ast C_2$ the set $\mathcal{F}$ is profinitely dense, and so that it can happen, thanks! The reason why I did not think of this example is that actually the groups $G$ I am considering also satisfy a stronger property (Kazhdan's Property (T)) which would imply that it can not be a free product itself. Note that for such groups the set $\mathcal{F}$ can be non-empty. Do you think that there is still an example among such groups? $\endgroup$
    – Doryan Temmerman
    Commented Aug 22, 2021 at 12:35
  • $\begingroup$ Try replacing $\mathbb Z$ by a residually finite hyperbolic linear group with property (T), say $SO(n,1)$. $\endgroup$
    – markvs
    Commented Aug 22, 2021 at 13:08

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