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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 $A$?

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  • $\begingroup$ It is possible that one can somehow use Nielsen transformations to tackle this. $\endgroup$
    – Pablo
    Jun 25, 2014 at 7:06
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    $\begingroup$ A reduced word of length $n$ in a Nielsen reduced free generating $Y$ of $H$ has length at least $n$ as a word over $X$, so if its $X$-length is 1, then it must be in $Y$ or $Y^{-1}$. $\endgroup$
    – Derek Holt
    Jun 25, 2014 at 7:57
  • $\begingroup$ If I understand your comment correctly, this means that the answer to the question is positive. Right? Could you please give a reference for your claim? I will gladly accept this as an answer. $\endgroup$
    – Pablo
    Jun 25, 2014 at 8:15
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    $\begingroup$ Corollary 2.4 of Lyndon & Schupp. $\endgroup$
    – Derek Holt
    Jun 25, 2014 at 8:20

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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 $H$.

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    $\begingroup$ A for the profinite case, doesn't that follow? Let $G$ be the abstract free group generated by $X$. Then $H\cap G$ is of finite index in $G$, so has a free basis extending $A$, which is also a free basis for $H$. $\endgroup$
    – HJRW
    Jun 27, 2014 at 11:59
  • $\begingroup$ 'A' -> 'As'${}$ $\endgroup$
    – HJRW
    Jun 28, 2014 at 23:31

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