 Solution

Every nonidentity biholomorphism in the finite group $G$ has only finitely many fixed points, by the identity theorem on a Riemann surface and compactness. Because the group action is a faithful group action, the union of these finite sets does not exhaust $X$. A point outside this union has trivial stabilizer subgroup, so its group orbit has $|G|$ points by the orbit-stabilizer theorem. The generic fiber of a function of the projection is such an group orbit, givingMore precisely, put $H=G_x$ and $m=|H|$. Differentiation at $x$ gives a group homomorphism $H\to {\mathbb{C}}^{\times }$. It is injective: if a nonidentity finite-order germ had the form $h(z)=z+a_k{z}^k+O(z^{k+1})$ with $k\ge 2$ and $a_k\ne 0$, then its $q$th iterate would begin $z+q{a}_k{z}^k$, so no positive iterate could be the identity. A germ equal to the identity gives the identity on $X$ by the identity theorem on a Riemann surface. Thus $H$ is a cyclic group, with derivative image the $m$th roots of unity.

For completeness, an explicit local coordinate linearizing $H$ isIt has $w^{\prime }(x)=1$, and reindexing the sum gives $w(hp)=h^{\prime }(x)w(p)$. Therefore a coordinate on the local quotient space is $w^m$. Other translates of a sufficiently small neighbourhood of $x$ are disjoint unless their translating element lies in $H$, so this local quotient space is also a chart for $X/G$. The local cyclic quotient of a Riemann surface consequently gives