Let $X$ be a smooth complex manifold with finite fundamental group. Suppose that a finite group $G$ acts on $X$ and let $\widetilde{X/G}$ be a resolution of singularities. Is $\pi\_1(\widetilde{X/G})$ always finite?


I think this is true but don't know a way to prove or a reference.



Is $\pi_1(\widetilde{X/G})$ always finite if $\pi_1(X)$ is finite?
The answer is **yes** when $X$ is simply connected. This can be proven as follows.


$\underline{\textrm{Step 1.}}$ The fundamental group $\pi\_1(X/G)$ is finite. More precisely, $\pi\_1(X/G)= G/N$, where $N$ is the smallest normal subgroup generated by those elements in $G$ which have fixed points on $X$. 


For a proof, see [M. A. Armstrong, *Calculating the fundamental group of an orbit space*, Proceedings of the American Mathematical Society **84** (1982)]. 


$\underline{\textrm{Step 2.}}$ Since $X/G$ has only quotient singularities, one has $\pi\_1(X/G)=\pi\_1(\widetilde{X/G})$. This is proven in Theorem 7.8 of [J. Kollar, *Shafarevich maps and plurigenera of algebraic varieties*, Inventiones Mathematicae **113** (1993)]. See also this MathOverflow question: 


[Comparing fundamental groups of a complex orbifolds and their resolutions.](https://mathoverflow.net/questions/50580/comparing-fundamental-groups-of-a-complex-orbifolds-and-their-resolutions)


The fact that $X$ is simply connected is only used in Step 1. Maybe this step can be refined in order to make it working also in the more general case where $X$ has finite fundamental group.


