# On intersection of finite index fully invariant subgroup

Let $$G$$ be a group. A subgroup $$H$$ of $$G$$ is said to be fully invariant if for every endomorphism $$\phi$$ of $$G$$, we have $$\phi(H) \subseteq H$$. For a finitely generated residually finite group $$G$$, let $$N$$ be the intersection of all finite index fully invariant subgroup of $$G$$. It seems in all trivial cases (finite and finitely generated abelian) that $$N$$ must be trivial. I want to know whether or not $$N$$ is trivial in every finitely generated residually finite group and in particular free group. What can be said about the intersection of finite index verbal subgroup in these groups? I do not know how to go ahead.

• For residually $p$-groups, as free groups, one can take $G^p\gamma_n(G)$ and get a sequence of finite-index verbal subgroups going to $\{e\}$. Commented Apr 18 at 20:34
• Or rather $G^{p^n}\gamma_n(G)$. Commented Apr 18 at 21:56

For each subgroup $$H$$ of $$G$$, we may define its fully invariant core as $$I(H)=\bigcap_\phi \phi^{-1}(H)$$ where $$\phi$$ ranges in the set of all endomorphisms of $$G$$. Then $$I(H)$$ is a fully invariant subgroup of $$G$$ and $$I(H)\subseteq H$$.
If $$G$$ is finitely generated and $$H$$ is a finite-index subgroup of $$G$$, we can add that $$I(H)$$ has finite index in $$G$$. Indeed, each subgroup $$\phi^{-1}(H)$$ has index at most $$[G:H]$$ in $$G$$. Using a lemma of Marshall Hall (here $$G$$ finitely generated is key), there are only finitely many such subgroups. Hence $$I(H)$$ has finite index in $$G$$.
If in addition $$G$$ is residually finite, we deduce that $$N= \bigcap_{\substack {I\text{ fully invariant} \\ \text{finite-index}}} I\subseteq \bigcap_{H\text{ finite-index}}I(H) \subseteq \bigcap_{H\text{ finite-index}}H=\{e\}.$$