Let $F$ be the free group of rank $\aleph_0$, let $L \leq F$ be a finite index subgroup, and let $M$ be a proper characteristic subgroup of $L$. Is it possible that there exists a finite subset $S \subseteq F$ for which we have $\langle M,S \rangle = F$?

$\begingroup$ @GeoffRobinson you are right. It is just a special case of the one I am interested in, where $M$ is some characteristic subgroup of $L$ and not only the commutator. I will edit... $\endgroup$– PabloMay 11, 2015 at 12:16

$\begingroup$ A subgroup of $L$ has finite corank in $F$ iff it has finite corank in $L$. And "is it possible" makes unclear your use of quantifiers. So the question should be stated as: Let $F$ be the free group of rank $\aleph_0$. Does there exist a proper characteristic subgroup of in $F$ of finite corank, in the sense that $F$ generated by the union of $M$ and a finite subset? $\endgroup$– YCorMay 11, 2015 at 13:14

$\begingroup$ @YCor I am sorry for the ambiguity. The question asks whether there exists a finite index subgroup $L \leq F$ and a proper characteristic subgroup $M \ \mathrm{char} \ L$ such that the corank of $M$ in $F$ is finite, which means that there exists a finite subset $S \subseteq F$ such that $\langle M \cup S \rangle = F$. You claimed that: "A subgroup of L has finite corank in F iff it has finite corank in L". This is not correct. $\endgroup$– PabloMay 11, 2015 at 13:42

$\begingroup$ Yes I've realized it's unclear. What's a counterexample? and what if you assume it to be normal in $L$? $\endgroup$– YCorMay 11, 2015 at 14:02

$\begingroup$ @YCor Even if $M$ is a normal subgroup of $L$, there exists a counterexample. It is essentially constructed here mathoverflow.net/questions/203285/… and the example is very simple. $\endgroup$– PabloMay 11, 2015 at 14:29
1 Answer
The answer is no:
for every free group of infinite rank $F$, every finite index subgroup $L$ and every proper characteristic subgroup $M$ of $L$, the subgroup $M$ has infinite corank in $F$.
First in the case $L=F$, the question can be rewritten as: let $F=F_I$ be a free group of infinite rank over the generators $(e_i)_{i\in I}$. Does there exist a characteristic subgroup $M$ of $F$ such that $F/M$ is a nontrivial finitely generated group?
The answer is no. Indeed, let $G$ be the symmetric group on the set of generators. Then the action of $G$ on the generators stabilizes $M$ and hence induces an action of $G$ on the finitely generated group $F/M$, giving rise to a homomorphism $G\to\mathrm{Aut}(F/M)$, and $F/M$ is countable because $F/M$ is finitely generated. But it is known that $G$ has no proper countable quotient (e.g., by the SchreierUlam theorem, see this MS post). Hence the action of $G$ on $F/M$ is trivial. This means that $e_ie_j^{1}\in M$ for all $i,j$. Now pick two distinct elements $u,v$ in $I$, and let $\psi$ be the automorphism mapping $e_u$ to $e_ue_v$ and mapping $e_i$ to itself for any $i\neq u$. Then $\psi(e_ue_v^{1})=e_u$. Since $e_ue_v^{1}\in M$ and $M$ is characteristic, we deduce that $e_u\in M$; hence $e_i\in M$ for all $i$ and hence $M=F$.
One can be too optimistic and believe that the general case ($L$ not necessarily equal to $F$) reduces to this, but it's quite unclear (as mentioned by Pablo, a subgroup of $L$ of finite corank in $F$ need not have finite corank in $L$). Instead, an elaboration of the previous argument works:
Let $L'$ be a normal subgroup of finite index of $F$ contained in $L$; write $S=F/L'$. The projection $F\to F/L'$ yields a finite partition $(I_s)_{s\in S}$ of $I$, where $I_s$ is the inverse image of $s$ in $I$. We can change the basis to modify this partition as follows: if $I_s$ is nonempty, pick an element $i\in I_s$, and for any other $j\in I_s$, replace $e_j$ with $e_je_i^{1}$. After this modification, we have ensured that all but finitely many generators $e_i$ are in the kernel of $F\to S$, say for $i\in J$ with $IJ$ finite.
The infinite symmetric group $\mathrm{Sym}(J)$ acts on the generators of $F$ (permuting $J$ and fixing the remaining finite subset); this action preserves the fibers of the canonical projection $F\to F/L'$ and hence preserves every subgroup containing $L'$; in particular it preserves $L$, and hence acts on $L$, thus preserving $M$. By the previous argument on quotients of the infinite symmetric group, we obtain that the action of $\mathrm{Sym}(J)$ on $L/M$ is trivial.
Thus the image $\gamma$ in $L/M$ of $e_i$ for any $i\in J$ does not depend on $i\in I$. If $u,v\in I$ are distinct, the automorphism of $F_I$ mapping $e_u$ to $e_ue_v$ and all other $e_i$ to itself preserves $L'$ and acts trivially on the quotient, and hence preserves $L$. Thus it maps $M$ onto itself and we deduce, as in the special case $L=F$, that $\gamma=1$. So we have proved that $F_J\subset M$.
Actually, we have proved that for any basis $B$ of $F$ such that $B\cap L'$ is infinite, we have $B\cap L'$ contained in $M$. We can thus apply this to any conjugate of the previous basis; as a result, any conjugate of $F_J$ is contained in $M$ and thus the normal subgroup $N$ generated by $F_J$ is contained in $M$. Since $F/N$ being finitely generated, so is its finite index subgroup $L/N$, and it follows that its quotient $L/M$ is finitely generated as well. (Here it sounds easy but the issue was to find a large subgroup of $M$ normal in all of $F$!) Applying to $L$ the special case of the beginning, we deduce that $M=L$.

$\begingroup$ There is an inaccuracy: "By the previous argument on quotients of the infinite symmetric group, we obtain that the action of $\mathrm{Sym}(J)$ on $L/M$ is trivial. " this requires $L/M$ to be finitely generated, which is not a priori clear. $\endgroup$– PabloMay 12, 2015 at 7:05

$\begingroup$ That's correct, I'll have to think twice. For the moment it's fine only when $L=F$. $\endgroup$– YCorMay 12, 2015 at 8:28