Let $S$ be a compact orientable surface, and let $G$ be a nontrivial finite group acting freely on $S$ and preserving orientation (note the the action being free is a strong condition, since automorphisms usually have fixed points). Then $H^1(S)$ also has an action of $G$. I know how to prove, using Riemann-Hurwitz and Artin's induction theorem, that $$ H^1(S,\mathbb{Q}) \cong \mathbb{Q}^2 \oplus \mathbb{Q}[G]^{2k} $$ for some integer $k\ge 0$, where $\mathbb{Q}$ denotes the trivial $\mathbb{Q}[G]$-module.

What is the $\mathbb{Z}[G]$-module $H^1(S,\mathbb{Z})$ ?

I could not figure this out, although it feels like it should be well-known. In particular, when do we have $H^1(S,\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ ?

EDIT: In case it helps, here is a proof over $\mathbb{Q}$:

Let $k = g(M/G)-1$ and $V = \mathbb{Q}^2\oplus\mathbb{Q}[G]^{2k}$, where $g(\cdot)$ denotes the genus. For every subgroup $H\le G$, the covering $M\to M/H$ is unramified, so we have $g(M)-1 = |H|\cdot(g(M/H)-1)$, so that $$ \dim H^1(M,\mathbb{Q})^H = \dim H^1(M/H,\mathbb{Q}) = 2+2\frac{\dim H^1(M,\mathbb{Q})-2}{|H|}\cdot $$ On the other hand, we have $$ \dim V^H = 2+2k[G:H] = 2+2\frac{\dim V-2}{|H|}\cdot $$ By Artin's induction theorem, $\mathbb{Q}[G]$-modules are characterized by the dimensions of the spaces fixed by the subgroups of $G$, so $H^1(M,\mathbb{Q})\cong V$.