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
 A: This question is answered in Thurston's answer to Counting characteristic subgroups.
Putting this here as CW so the question won't pop up as unanswered.
A: 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 $C_p$.
