Subgroup of a free group that is characteristic but not totally characteristic

Looking for a counter example (if it exists) and a reference for further reading. Can there be a subgroup of finite index in a finitely generated free group that is characteristic but not totally characteristic.

Definition: A subgroup $H\le G$ is said to be totally characteristic if for all endomorphisms, $\psi\in\mathrm{End}(G)$ we have that $\psi(H)\subseteq H$.

This was previously asked as math.stackexchange.com/questions/2260574

• The stackexcahnge post is math.stackexchange.com/questions/2260574 Note that an equivalent question is whether all characteristic subgroups of finite index in a finitely generated free group are verbal. – Derek Holt May 10 '17 at 12:21
• usual terminology is "fully characteristic" (in arbitrary groups) or "verbal" (in free groups) – YCor May 10 '17 at 13:04
• This question is answered in Thurston's answer to mathoverflow.net/questions/51053/… – Benjamin Steinberg May 10 '17 at 13:08
• @YCor: I've never seen "fully characteristic" or "totally characteristic". For arbitrary groups, the terminology I'm familiar with is "fully invariant". Verbal subgroups are always fully invariant, and in free groups the two coincide. – Arturo Magidin May 16 '17 at 5:47
• @ArturoMagidin fine to me, you can't say any longer that you've never seen "fully characteristic" :) – YCor May 16 '17 at 17:37

A source for such groups: take a finite $2$-generator group $G$ for which the formation $for(G)$ generated by $G$ is different from the finite variety (pseudovariety) $var(G)$ generated by $G$. Choose an integer $n\ge 2$ for which there is an $n$-generator member of $var(G)$ which is not in $for(G)$. [I think in every such case one can choose any $n\ge 2$.] Let $F$ be the free group of rank $n$ and $N$ be the intersection of all normal subgroups $K$ of $F$ for which $F/K\in for(G)$. $N$ is characteristic: for every automorphism $\phi$ of $F$, $\phi(K)$ belongs to the collection in question since $F/\phi(K)$ also belongs to $for(G)$. $N$ is not fully invariant (=verbal): if it were then $F/N$ would be the $n$-generator free object in some variety and since $G$ is a quotient of $F/N$ it would be the $n$-generator free object in $var(G)$. But since all $n$-generator members of $var(G)$ are quotients of $F/N$ they would in turn belong to $for(G)$, a contradiction.
(In some sense) minimal examples for $G$: every simple non-abelian group $G$: $for(G)$ consists of all (finite) direct products of $G$ while $var(G)$ contains all subgroups of $G$, in particular abelian groups; every dihedral group $D_p$ of order $2p$ with $p$ an odd prime: $for(D_p)$ does not contain the cyclic group $C_p$ of order $p$ while $var(D_p)$ does contain $C_p$.