# Module structure of the abelianization of the commutator subgroup

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.

• Look up Alexander module/polynomial and Fox calculus. – Alex Degtyarev Jan 18 '14 at 19:44

## 1 Answer

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.

• Thanks, Alex. I guess I just didn't know where to look for this stuff. – Kevin Jan 19 '14 at 0:17