Recall the fundamental group of a genus-2 surface:
$$ \pi_1(\Sigma_2) = \langle a_1, b_1, a_2, b_2 \mid [a_1, b_1][a_2, b_2] = 1 \rangle $$
By which I mean a free group of four variables, divided out by the group generated by the single element $[a_1, b_1][a_2, b_2]$.
Now, I want to know what the commutator subgroup is. In particular, I want to draw its generators as loops on a torus with two holes.
I know that the commutator subgroup, which I will denote by $G'$, is generated by all homologically trivial loops, because the first homology group $H_1$ is isomorphic to the abelianization of $\pi_1$.
I read in here that since $G'$ is infinite, it must be free. I do not get why it should be free. If $G'$ is indeed free, can I conclude that it is the free group generated by $[a_1, a_2]$, $[a_1, b_2]$, $[a_2, b_2]$, i.e. all commutators of $\pi_1$? Because the fundamental group already has the identity $[a_1, b_1][a_2, b_2] = 1$, wouldn't this imply that I can drop a generator in $G'$ ? How do I use the fact that $G'$ is free here?
If I take the torus (genus 1), I see that the commutator subgroup is indeed trivial, because the only possible commutator $[a_1, b_1]$ is already trivial in the fundamental group.
How do I extend this reasoning to a genus-2 surface?
Note that this is a homework assignment, and that I'm a physicist following a mathematics course. All feedback appreciated.