Every subgroup of infinite index in a surface group is a free group. There are many ways to see this, for instance, using a bit of topology. An infinite index subgroup corresponds to a covering space with infinitely many sheets and this covering space is a non-compact surface. By a result of Whitehead it deformation retracts onto a one-dimensional subcomplex, i.e. a graph, and it follows that the fundamental group is free. But how to find a nice set of free generators?

Today appeared a nice article by Andrew Putman on the arXiv which describes surprisingly simple free generating sets for the commutator subgroups of free and surface groups. Let me briefly state the result for surface groups. Let \(\Sigma_g\) be a closed surface of genus \(g \geq 1\) with fundamental group

$$\pi_1(\Sigma_g) = \langle x_1, x_2,\dots,x_{2g} \mid [x_1,x_2][x_3,x_4]\cdots[x_{2g-1},x_{2g}]\rangle.$$

Then the following is a free generating set for the commutator group of \(\pi_1(\Sigma_g)\):

$$ \{[x_i,x_j]^{x_i^{k_i}x_{i+1}^{k_{i+1}} \cdots x_{2g}^{k_{2g}}} \mid 1\leq i < j \leq 2g, (i,j) \neq (1,2), k_i,\dots,k_{2g} \in \mathbb{Z} \}.$$

The very readable article of Putman provides a nice geometric proof for this result (and a similar result for free groups).