I am trying to understand the following example, which I came across in a research article. I am posting it as a question below. 


$\bf{Question}$. Let $\Sigma$ be a curve of genus two with the automorphism group $G$, and $p\_1$, $p\_2$, and $p\_3$ are three points on $\Sigma$. Let $a\_{1}$, $b\_{1}$, $a\_{2}$, $b\_{2}$ denote the generators of $\pi\_{1}(\Sigma)$ and $\gamma\_{1}$, $\gamma\_{2}$, $\gamma\_{3}$ are simple loops about the points $p\_{1}$, $p\_{2}$, and $p\_{3}$. Consider the following map $f$ from $\pi\_{1} (\Sigma \setminus p\_{1}, p\_{2}, p\_{3})$ to $\mathbb{Z}\_{3}$ given by $\gamma\_i \rightarrow 1$ and $a\_{i}, b\_{i} \rightarrow 0$. What it means the covering (unramified) $\Pi\_{f} :\Sigma' \rightarrow \Sigma$ associated to the stabilizer of $f$? How one can think of such covering map in terms of the automorphism group $G$ of $\Sigma$? What one can say about the degree of $\Pi\_{f}$? 



finite covering 
By Riemann Existence Theorem, the surjection $f \colon \pi\_1(\Sigma \setminus p\_1, p\_2, p\_3) \to \mathbb{Z}\_3$ gives a Galois cover with Galois group $\mathbb{Z}\_3$ branched only at the points $p\_1, p\_2, p\_3$. 


This cover is precisely your covering $\Pi\_f \colon \Sigma' \to \Sigma$. Then $\deg \Pi\_f = |\mathbb{Z}\_3|=3$. (By the way, here I do not understand why you wrote that the cover is "unramified", which is false. Maybe you meant "ramified"?)


The Galois group of $\Pi\_f$ acts naturally on $\Sigma'$, and the action is free outside the preimages of $p\_1$, $p\_2$, $p\_3$. At these preimages the stabilizer is obviously $\mathbb{Z}\_3$ itself, so the cover $\Pi\_f$ is totally ramified over $p\_1$, $p\_2$, $p\_3$.


Then the genus of $\Sigma'$ can be computed by Riemann-Hurwitz formula, obtaining
$$2g(\Sigma')-2= |\mathbb{Z}\_3| \bigg(2g(\Sigma)-2 + 3 \bigg(1-\frac{1}{3} \bigg) \bigg),$$ 
that is $g(\Sigma')=7$.


