Commutator subgroup of a free group Let $F_k$ be the free group of rank $k$.
If $k=2$ it is not hard to see that the set $\{[s_1^{n_1},s_2^{n_2}] \mid n_i\neq 0\}$ is a basis for  $F_2'$. (Prime denotes the  commutator subgroup).
What is a basis for $F_k'$ if $k\geq3$?
 A: It is straightforward to write down the Schreier generators of a subgroups of finite index of a group given by a finite presentation. When the group is free, this will give you a free basis, by the proof of Schreier that subgroups of free groups are free. The free basis depends on a choice of well-ordering of words in the generators of $G$ and on a transversal of the subgroup in $G$. It will not necessarily give the "nicest" free basis, and it gives a slightly more complicated basis than yours in the 2-generator case.
Let $G$ be free on $x_1,\ldots,x_k$. Then the obvious right transversal for $G'$ in $G$ is $\{x_1^{n_1}\cdots x_k^{n_k} \mid n_i \in \mathbb{Z} \}$ and (if I have got this right), this gives rise to the free basis
$\{ x_1^{n_1}\cdots x_m^{n_m} x_l (x_1^{n_1}\cdots x_l^{n_l+1} \cdots x_m^{n_m})^{-1} \mid n_i \in \mathbb{Z}, 1 \le l < m \le k, n_m \ne 0\}$
of $G'$.
A: The same works in the case where you have a free group $F$ on infinitely many generators, as long as those generators are ordered. In general, in the setting of Schreier transversals you would require a well-ordering but in the case of the commutator subgroup the Schreier transversal is so easy to write that just a total ordering is enough.
