Let $F$ be a free group on a finite set $X$. Let $A \subseteq X$ be a subset of $X$ contained in some $H \leq F$, a subgroup of finite index in $F$. Must there be a basis (free generating set) for $H$ containing $A$?


The existence of some basis is guaranteed since $H$ is a free group (Nielsen-Schreier Theorem). Even the case when $A$ is a singleton is not clear to me. On the other hand, It may well be that the there is no need to assume that the index is finite, and just take $H$ to be finitely generated. Furthermore, I find the following profinite analogue interesting too:


Let $F$ be a free profinite group on a finite set $X$. Let $A \subseteq X$ be a subset of $X$ contained in some $H \leq\_o F$. Must there be a basis (free topological generating set) for $H$ containing $A$?



Bases of free groups
As Derek Holt says in comments, the answer to your first question is 'yes'. You can argue topologically.


There is a rose $R$ corresponding to $X$ with $\pi\_1R\cong F$. The subset $A$ defines a connected subrose $R'\subseteq R$. The subgroup $H$ corresponds to a based, finite-sheeted covering space $S\to R$. The assertion that $A\subseteq H$ implies that $R'$ lifts homeomorphically to a rose $\widehat{R}$ in $S$ at the basepoint. Collapsing a maximal tree $T$ in $S$ we obtain a basis for $H$. Since $\widehat{R}$ has no edges in $T$, it survives as a subrose in $S/T$.


Translating back into group theory, this means precisely that $A$ is a subset of the corresponding basis of $H$.


