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.


1 Answer 1


I don't have an answer for you, but I have some relevant comments. Firstly, you don't expect $G'$ to be finitely generated in general. If $G=\langle g_i|r_i\rangle$ then $G'=\langle [x,y]:x,y\in G\rangle$, and there's no general reason to expect every such $[x,y]$ to be in $\langle [g_i,g_j]\rangle$.

Secondly, you really don't want to draw the generators of $G'$ on the surface of $\Sigma_2$. What you want to do instead is to find a covering space $E$ with a covering map $E\to\Sigma_2$ such that $\pi_1(E)=G'$. The standard example is the figure-eight $B$, which has fundamental group $\pi_1(B)=\langle x,y\rangle$ the free group on 2 generators. Now $E$ is the grid $(\mathbb Z\times\mathbb R)\cup(\mathbb R\times\mathbb Z)$. Note how every square in the grid is a generator of $G'$, so you can see $G'$ very clearly.


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