# Is $\pi_1(\widetilde{X/G})$ always finite if $\pi_1(X)$ is finite?

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.

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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.

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.

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You do not need to assume that X is simply connected. Just lift the action to the universal cover. It will be via another finite group. – Misha Apr 21 '13 at 11:39
Thank you for the nice answer, Francesco. I was a bit surprised to know that my question was not as easy as I had initially expected. – user2013 Apr 22 '13 at 4:42