Let $\pi_1(\Sigma_g) = \langle\text{$x_1,\ldots,x_{2g}$ $$ $[x_1,x_2]\cdots[x_{2g1},x_{2g}]$}\rangle$ be a surface group. Can anyone tell me an explicit free basis for the commutator subgroup of $\pi_1(\Sigma_g)$? I would prefer one consisting of conjugates of the elementary commutators $[x_i,x_j]$.
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2$\begingroup$ See the comments here. $\endgroup$ – Dietrich Burde Mar 23 '18 at 19:48

3$\begingroup$ @DietrichBurde: I hadn't seen that question, but the comments don't look like they contain any nontrivial information. $\endgroup$ – Linda Mar 23 '18 at 20:26

2$\begingroup$ @HJRW: I think you must have a typo or something in your 1st sentence: the conjugates of commutators of basis elements generate $F'$, but they are not a free basis  if they were, then they would project to a $\mathbb{Z}$basis for the abelianization $[F,F]^{\text{ab}}$ and it would follow that this abelianization is a free $\mathbb{Z}[F_n]$module (while in fact the $F_n$action descends to an action of $\mathbb{Z}^n$; what is more, it is not even a free $\mathbb{Z}[\mathbb{Z}^n]$module). Can you give me the precise reference in Serre's book so I can figure out what you are trying to say? $\endgroup$ – Linda Mar 24 '18 at 15:41

4$\begingroup$ @HJRW: I see! It is true that for $n=2$ these form a basis, but I don't think this is true for higher $n$. The issue is that $[F,F]^{\text{ab}}$ is not a free $\mathbb{Z}[\mathbb{Z}^n]$module for $n \geq 3$. There are a lot of ways to see this; one easy one is to use the fact that $H_k(\mathbb{Z}^n;[F_n,F_n]^{\text{ab}}) \cong \wedge^{k+2} \mathbb{Z}^n$ together with the fact that if $M$ if a free $\mathbb{Z}[G]$module, then $H_k(G;M)=0$ for $k \geq 1$. $\endgroup$ – Linda Mar 24 '18 at 16:44

4$\begingroup$ nb: to see that $H_k(\mathbb{Z}^n;[F_n,F_n]^{\text{ab}}) \cong \wedge^{k+2} \mathbb{Z}^n$, simply use the fact that the spectral sequence for the extension $1 \rightarrow [F_n,F_n] \rightarrow F_n \rightarrow \mathbb{Z}^n \rightarrow 1$ has to degenerate to give the homology of $F_n$, so all of the differentials have to be isomorphisms. $\endgroup$ – Linda Mar 24 '18 at 16:46