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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?

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  • $\begingroup$ How does $B_k$ act on $F^k$? $\endgroup$ – YCor Aug 22 '14 at 13:15
  • $\begingroup$ By permutation of the factors via the projection $B_k\rightarrow S_k$. $\endgroup$ – Werner Thumann Aug 22 '14 at 13:38
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    $\begingroup$ Then the question does not seem to change when we replace the group by a finite index subgroup, and $B_k\ltimes F^k$ admits the direct product $P_k\times F^k$ as finite index subgroup. $\endgroup$ – YCor Aug 22 '14 at 18:16
  • $\begingroup$ Nice trick! Why don't you post it as a regular answer so that I can upvote and accept it? Question too easy? $\endgroup$ – Werner Thumann Aug 23 '14 at 9:52
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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 $P_k\ltimes F^k=P_k\times F^k$.

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