Let $S$ be a compact connected 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 $H^1(S,\mathbb{Z})$ ?
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> 
> 


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




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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 $H^1(M,\mathbb{Q})\cong V$.



$G$-action on the integral homology of a compact surface
I don't see how to answer this is general, but the following partial result might be of interest to you.


Theorem: If $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then $G$ does not contain $(\mathbb{Z}/p)^3$ for any prime number $p$.


Proof: Suppose $P = (\mathbb{Z}/p)^3 \leq G$. Then considered as a $\mathbb{Z}[P]$-module we still have $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[P]^{2\ell}$ for some $\ell$. We consider the Serre spectral sequence for the fibration sequence
$$S \to S/P \to BP$$
in $\mathbb{F}\_p$-cohomology, which has the form $E\_2^{p,q} = H^p(P ; H^q(S;\mathbb{F}\_p)) \Rightarrow H^{p+q}(S/G;\mathbb{F}\_p)$.


The surface $S$ has genus $2+2\ell\vert P\vert$, and by Euler characteristic the surface $S/P$ therefore has genus $2+2\ell$. The spectral sequence gives an exact sequence
$$0 \to H^1(P;\mathbb{F}\_p) \to H^1(S/P;\mathbb{F}\_p) \to H^0(P; H^1(S;\mathbb{F}\_p)) \overset{d\_2}\to H^2(P;\mathbb{F}\_p) \to X \to 0$$
and $X$ injects into $H^2(S/G;\mathbb{F}\_p)$. This exact sequence is
$$0 \to \mathbb{F}\_p^3 \to \mathbb{F}\_p^{2+2\ell} \to \mathbb{F}\_p^{2+2\ell} \overset{d\_2}\to \mathbb{F}\_p^{6} \to X \to 0$$
so $X \cong \mathbb{F}\_p^3$, but this cannot inject into $H^2(S/G;\mathbb{F}\_p) = \mathbb{F}\_p$. QED


The point of this argument is that if $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then (i) the same is true of any subgroup, and (ii) it follows from the spectral sequence argument that there are classes $a,b,c,d \in H^2(G;\mathbb{Z})$ such that
$$0 \to H^i(G;\mathbb{Z}) \overset{(a,b)}\to H^{i+2}(G;\mathbb{Z}) \oplus H^{i+2}(G;\mathbb{Z}) \overset{c+d}\to H^{i+4}(G;\mathbb{Z}) \to 0$$
is exact (in fact, with any ring coefficients not just $\mathbb{Z}$). This ought to give many non-examples.


EDIT: In fact, the argument obstructs being $\text{projective} \oplus \mathbb{Z}^2$, not just $\text{free} \oplus \mathbb{Z}^2$.


