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 $G$.