Let $F$ be a group which is strongly type $F\_\infty$ in the sense that every subgroup is of type $F\_\infty$. Here, type $F\_\infty$ means that the group admits a classifying space with compact skeleta.


Consider the semidirect product $B\_k\ltimes F^k$ where $B\_k$ is the braid group on $k$ strands. Think of its elements as braids with strands labelled with elements of $F$.
It is of type $F\_\infty$ because $B\_k$ and $F$ are of type $F\_\infty$ and (semi)direct products of type $F\_\infty$ groups are of type $F\_\infty$.


Let $H,G<F$ be subgroups. Then the subgroups $B\_k\ltimes H^k$ and $B\_k\ltimes G^k$ of $B\_k\ltimes F^k$ are of type $F\_\infty$ because $H$ and $G$ are. Also, the intersection $B\_k\ltimes H^k\cap B\_k\ltimes G^k$ is of type $F\_\infty$ because it is $B\_k\ltimes(H\cap G)^k$.


Now let $\alpha\in F^k$. I guess that the intersection
$$\alpha(B\_k\ltimes H^k)\alpha^{-1}\cap(B\_k\ltimes G^k)$$
is of type $F\_\infty$. Is it true? Is there an easy argument?



Semidirect products with braid groups and type $F_\infty$
In order to close this question, I cite YCor's answer in his comment above: One can replace $B\_k$ with the finite index subgroup $P\_k$ and in this case $P\_k\ltimes F^k=P\_k\times F^k$.


