Let $F(S)$ be the free group on a (possibly infinite) set $S$. Let $T$ be a subset of $F(S)$ with the following two properties.

  1. $T$ generates $F(S)$.

  2. $T$ injectively projects to a basis for the free abelian group $H_1(F(S);\mathbb{Z})$.

Question : Must $T$ be a free basis for $F(S)$?

If $S$ is finite, then this follows from the standard fact that any generating set for $F(S)$ of cardinality $|S|$ is a free basis for $F(S)$ (for example, this is Proposition 2.7 in Lyndon and Schupp's book "Combinatorial Group Theory"). However, I don't see how to adapt this to the case where $S$ is infinite.

I really only care about the case where $S$ is countable, but I can't imagine that this is true for countable $S$ but false for uncountable $S$.

  • $\begingroup$ Why do you write the abelianization in form of a cohomology group? $\endgroup$ – Martin Brandenburg May 24 '11 at 17:22
  • $\begingroup$ @Martin : Habit. I do a lot of group cohomology stuff... $\endgroup$ – Andy Putman May 24 '11 at 18:10

If I understand the question correctly, you are concerned that there might be hidden relations between the elements of $T$. Moreover, since the images of $T$ are linearly independent in the abelianization, these would have to be commutator relations.

This kind of thing can't happen: if there were a nontrivial relation involving elements of $T$, then said relation would involve only finitely many elements of $T$, so it would be a relation on the free subgroup generated by that finite set of elements. Essentially, condition (2) guarantees that the elements of any finite subset of $T$ are algebraically independent, and therefore all of the elements of $T$ are algebraically independent.

  • 2
    $\begingroup$ Duh! I somehow convinced myself this morning that an argument like that wouldn't work, but I was mistaken... $\endgroup$ – Andy Putman May 24 '11 at 18:09

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