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Let $F$ be a free finitely generated group, $L \leq H \leq F$ subgroups of finite index.

  1. Given bases $B$ of $F$ and $C$ of $H$, must there be a Schreier transversal $T$ for $H$ in $F$ such that the basis constructed from $T$ and $B$ is $C$ (using the standart basis construction procedure, as in the proof of the Nielsen-Schreier theorem)?

  2. Given a basis $C$ of $H$, must there be a Schreier transversal $T$ for $H$ in $F$ and a basis $B$ of $F$ such that the basis constructed from $T$ and $B$ is $C$?

  3. If a basis $B$ for $H$ is constructed from a basis $A$ for $F$ and some transversal, and a basis $C$ for $L$ is constructed from the basis $B$ for $H$ and some transversal, must there be a transversal giving rise to $C$ from $A$?

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  • $\begingroup$ It's easy to find counterexamples to 1 with $F$ of rank $2$ and $|F:H|=2$. There are only four Schreier transversals to consider. $\endgroup$
    – Derek Holt
    Commented Jul 22, 2014 at 8:07
  • $\begingroup$ I mainly suspect that the transitivity condition in 3 is true... $\endgroup$
    – Pablo
    Commented Jul 22, 2014 at 12:48

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