Krdytkyu

I need to prove that if H is finitely generated, $Qlhd_f H$ (normal of finite index) and $phi in operatorname{Aut}(H)$ then $cap_{i in mathbb Z} phi^i(Q) lhd_f H$.

Given that H is f.g and Q is of finite index, I do know that Q is f.g and thus all $phi^i(Q)$ are f.g . That does not mean however that their intersection is f.g. Other than that I have no clue on how to even approach such a problem and how does H being f.g have anything to do with the intersection being of finite index.

Other than a solution, I would also appreciate any reference to any book which is concerned with these type of topics (the general topic I am trying to study is residual finiteness where these type of questions seem to come up as 'obvious remarks').

Here is a sketch solution. There is no need to assume that $Q$ is normal in $H$. If $|H:Q|=r$, then there is a homomorphism $phi:H to S_r$ with $phi(H)$ equal to the stabilizer of the first point.

Since $H$ is finitely generated, there are only finitely many such homomorphisms, so there are finitely many subgroups of $H$ with index $r$. So the intersection of all of them has finite index in $H$ and, in particular the intersection of $phi(Q)$ for all $phi in {rm Aut}(H)$ has finite index in $H$.

This can be false when $H$ is not finitely generated. For example, if $H$ is a direct prodcut of infinitely many copies of $C_2$, then the intersection of all subgroups of index $2$ is trivial, and all these subgroups can all be mapped to each other by automorphisms of $H$.