Let $\Sigma$ be a compact oriented surface of genus $g\ge 1$.
Its fundamental group is given by
$\Gamma:= \langle x_1,y_1,\dots,x_g,y_g\ \lvert\ \prod_{i=1}^{i=g}[x_i,y_i]\rangle$.
It is known that $b_1(\Gamma):=\operatorname{dim}_{\Bbb Z}H^1(\Gamma,\Bbb Z) = 2g$.
Let $A$ be a one-dimensional representation of $\Gamma$ over $F_q$, the field of $q=p^r$ elements for a prime $p$.
I have been told that $b^1_A:=\operatorname{dim}_{\Bbb Z}H^1(\Gamma,A) = 2g-2$. Trying to prove it, I see a contradiction but I don't know where exactly.
My attempt
I don't know if there are other tools but the first thing that comes to mind is the universal coefficient theorem. We have a short exact sequence: $0\to \operatorname{Ext}^1_{\mathbb Z}(H_0(\Gamma,\Bbb Z),A)\to H^1(\Gamma,A)\to Hom_{\Bbb Z}(H_1(\Gamma,\Bbb Z),A)\to 0$ with a non-canonical splitting.
Looking at the right hand side: \begin{equation} Hom_{\Bbb Z}(H_1(\Gamma,\Bbb Z),A)\cong Hom_{\Bbb Z}(\Bbb Z^{2g},A)\cong (\Bbb Z^{\oplus 2g})^{\vee}\otimes_{\Bbb Z}A \cong A^{\oplus 2g} \end{equation}
And here it is already problematic since it is already higher than the dimension of the middle term of the short exact sequence. But let me continue for the sake of completeness. On the left hand side we have:
\begin{equation} \operatorname{Ext}^1_{\mathbb Z}(H_0(\Gamma,\Bbb Z),A)\cong \operatorname{Ext}^1_{\mathbb Z}(\Bbb Z^s,A)=0 \end{equation}
for some non-negative integer $s$, since $\mathbb Z^{\oplus s}$ is free, so a projective $\Bbb Z$-module.
So all in all, adding the dimensions we get: $b_1^A = 2g+0=2g$.
I don't know where is my mistake, or if my assumptions are wrong somewhere.
Thank you for your help.
