# Free partners in semi-direct products

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 $$G$$.

• 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).
– YCor
Aug 21 at 15:53
• 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 $\mathcal{F}$. Aug 21 at 19:53
• 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 $K$.
– YCor
Aug 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 $$N\rtimes C_2$$.
• 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? Aug 22 at 12:35
• Try replacing $\mathbb Z$ by a residually finite hyperbolic linear group with property (T), say $SO(n,1)$. Aug 22 at 13:08