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 $G$.

  • $\begingroup$ 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
    Aug 21 '21 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 $\mathcal{F}$. $\endgroup$ Aug 21 '21 at 19:53
  • 1
    $\begingroup$ 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
    Aug 21 '21 at 20:30

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 $N\rtimes C_2$.

  • $\begingroup$ 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? $\endgroup$ Aug 22 '21 at 12:35
  • $\begingroup$ Try replacing $\mathbb Z$ by a residually finite hyperbolic linear group with property (T), say $SO(n,1)$. $\endgroup$
    – markvs
    Aug 22 '21 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.