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.

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    $\begingroup$ 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 $\{e\}$. $\endgroup$
    – Corentin B
    Commented Apr 18 at 20:34
  • $\begingroup$ Or rather $G^{p^n}\gamma_n(G)$. $\endgroup$
    – Corentin B
    Commented Apr 18 at 21:56

1 Answer 1


The answer is yes.

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 $$ N= \bigcap_{\substack {I\text{ fully invariant} \\ \text{finite-index}}} I\subseteq \bigcap_{H\text{ finite-index}}I(H) \subseteq \bigcap_{H\text{ finite-index}}H=\{e\}.$$


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