Let $F$ be a finitely generated free group and let $A$ be a finite index subgroup of $F$.
Does there exist a subgroup $B\subset A$ such that $F/B$ is (elementary) amenable and torsion-free? 


A group $G$ which is amenable and torsion-free has (at least conjecturally) the following nice properties: $\Bbb{Z}[G]$ has conjecturally no zero divisors (this is known if $G$ is say locally indicable or left-orderable) and $\Bbb{Z}[G]$ embeds in its Ore localization. 


I have a certain application for the above question in mind, where using an Ore localization plays a role. But I am also curious since my intuition fails me on that question.
Note that if $F/A$ is solvable, then one could just take $F$ to be an iterated commutator subgroup. But if $F/A$ is non-solvable, then I don't know what to do.



Finite index subgroups of free groups and torsion-free amenable quotients of free groups
If F is a free group and R is a normal subgroup of F, then I thought it was well-known that F/R' is torsion free (cannot find an explicit reference now, though this is stated just after Lemma 5 of [Farkas, Daniel R. Miscellany on Bieberbach group algebras. Pacific J. Math. 59 (1975), no. 2, 427–435]). Of course if F/R is finite, then F/R' is virtually abelian and certainly elementary amenable.


