Let $G$ be a (non-abelian) group, and let $G\_2$ denote its commutator subgroup. Then the abelianization $G\_2^{ab} = H\_1(G\_2,\mathbb{Z})$ is a module over the group ring $\mathbb{Z}[G^{ab}]$. The action of $G^{ab}$ on $G\_2^{ab}$ is induced by the conjugation action of $G$ on $G\_2$. I am interested in understanding the structure of this module when $G$ is a finitely generated free group, $F$, or the fundamental group of a compact Riemann surface $S\_g$ of genus $g\geq 2$. In either case, I find it convenient to think of $\mathbb{Z}[G^{ab}]$ as a ring of Laurent polynomials with integer coefficients.


$\textbf{What I know}$:


1) The group structure. In case $G = F\_g$ or $G = \pi\_1(S\_g,\*)$ with $g\geq 2$, the commutator subgroup $G\_2$ is a free group of infinite rank. Consequently, $H\_1(G\_2,\mathbb{Z})$ is a free abelian group of infinite rank. 


2) Commutators of pairs of generators of $G$ form a generating set for $H\_1(G\_2,\mathbb{Z})$ as a module over $\mathbb{Z}[G^{ab}]$. For example, if $G = \langle x\_1,..., x\_n \rangle$ then $H\_1(G\_2,\mathbb{Z})$ is generated over $\mathbb{Z}[x\_1^{\pm 1},..., x\_n^{\pm 1}]$ by the classes of the elements $[x\_i,x\_j]$ for $i < j$. I am fairly certain that $H\_1((F\_2)\_2,\mathbb{Z})$ is a free module of rank $1$ over $\mathbb{Z}[x\_1^{\pm1}, x\_2^{\pm 1}]$ with the single generator $[x\_1,x\_2]$. 


3) If $G = \pi\_1(S\_g,\*) = \langle x\_1,...,x\_{2g} | [x\_1,x\_2]\cdots [x\_{2g-1},x\_{2g}] = 1\rangle$, the $[x\_i,x\_j]$ generate $H\_1(G\_2,\mathbb{Z})$ over $\mathbb{Z}[x\_1^{\pm 1},..., x\_{2g}^{\pm 1}]$, but there is definitely a relation of the form $\sum\_{i=1}^g [x\_{2i-1},x\_{2i}] = 0$. 


$\textbf{My Question}$:


1) For what values of $n\geq 2$ is $H\_1((F\_n)\_2,\mathbb{Z})$ freely generated over $\mathbb{Z}[x\_1^{\pm 1},...,x\_n^{\pm 1}]$ by the commutators $[x\_i,x\_j]$ with ${i<j}$?


2) In case $H\_1((F\_{2g})\_2,\mathbb{Z})$ is a free $\mathbb{Z}[x\_1^{\pm 1},...,x\_{2g}^{\pm 1}]$ module, let $\langle \Theta \rangle$ be the submodule spanned by $\Theta = \sum\_{i=1}^g [x\_{2i-1},x\_{2i}]$. Is there a module isomorphism $$ H\_1((F\_{2g})\_2,\mathbb{Z})/\langle \Theta \rangle\simeq H\_1((\pi\_1(S\_g,\*))\_2, \mathbb{Z})?$$ 


My apologies if the level of this question is too low. I feel that the answer must be well-known, but I have been unable to find it despite an extensive search. 



Module structure of the abelianization of the commutator subgroup
To answer your **Q1**, the group in question is $H\_1$ of the complex
$$0\to\Lambda a\_1\oplus\ldots\oplus\Lambda a\_n\overset\partial\to\Lambda\to0,$$ 
where $\Lambda$ is your ring of Laurent polynomials and $\partial\colon a\_i\mapsto(x\_i-1)$. (Clearly, the $a\_i$ are in a correspondence with the generators of $F\_n$.) So, it's very unlikely that $\ker\partial$ is a free module; in any case, you suggest too many generators!


For the other group, you replace the leftmost $0$ with another copy of $\Lambda$, mapping its generator to 
$$(1-x\_2)a\_1+(x\_1-1)a\_2+(1-x\_4)a\_3+(x\_3-1)a\_4+\ldots,$$
and still compute $H\_1$ of the resulting complex. 


The elements $(1-x\_j)a\_i+(x\_i-1)a\_j$ are the commutators (up to sign, depending on the definitions), so, to an extent, the answer to your **Q2** is "yes". As to **Q1**, I would say that the module **is** generated by the commutators (probably, a bit of algebra is needed), but definitely not freely; e.g., there are obvious relations for each triple of indices $(i,j,k)$.


All statements are immediate from the explicit CW-complex realizing the group: a graph in the former case and a graph with a single $2$-cell in the latter.


