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I don't know much about (algebraic/etale) fundamental groups, so sorry if this question sounds stupid. I am interested in quotient singularities (quotients $X$ of $\mathbb{C}^n$ by a finite subgroup $G$ of $GL(n,\mathbb{C})$).

The divisor class group $Cl(X)$ of $X$ is the abelianization $G/[G,G]$ of $G$. Is there also a connection with the fundamental group (defined in any way)? Maybe even $G=\pi_1(X)$, as stated in a similar way here: https://mathoverflow.net/a/50638. I am not sure if my case fits in this setting. I would be very happy with references, since I did not find any convenient papers myself.

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  • $\begingroup$ The MO-answer you cite implies $\pi_1(X)$ is trivial because every element of the group fixes the origin. In your setting of varieties over $\mathbb{C}$, the étale fundamental group is the profinite completion of the topological fundamental group (Riemann existence theorem). This reduces questions about the étale fundamental group to topological computations. One of the standard references for the comparison étale vs topological is SGA1, Corollaire XII.5.2. $\endgroup$ – Matthias Wendt May 10 '16 at 12:48
  • $\begingroup$ I think that you should specify that $G$ acts without pseudoreflections. If $G$ acts with pseudoreflections, then $X/G$ may well be $\mathbb{A}^n$. This has trivial class group, even when $G$ is Abelian. $\endgroup$ – Jason Starr May 10 '16 at 12:48
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If $S$ is a contractible $G$-variety over $\mathbb{C}$ for $G$ reductive, then $S/\!\!/G$ is contractible.

This follows from Kempf-Ness Theory and Oliver's proof of the Conner Conjecture. See Schwarz's nice exposition of this in The Topology of Algebraic Quotients.

Therefore, $\mathbb{C}^n/\!\!/G$ is contractible and hence simply connected.

As stated in the comments, if the topological fundamental group of a variety is trivial, then the algebraic fundamental group is trivial too.

Slightly more general is this: If $G$ is further assumed connected, and $S$ is simply connected, then $S/\!\!/G$ is still simply connected (topologically or algebraically). This follows, again from Kempf-Ness Theory, since the GIT quotient map has the path-lifting property.

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    $\begingroup$ Is the contractibility of $S//G$ "easy" to prove when $G$ is finite? $\endgroup$ – Ariyan Javanpeykar May 10 '16 at 13:52
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    $\begingroup$ Contractability of $\mathbb{C}^n/G$ is certainly easy ($G$ is linear, and hence commutes with the scaling action, which indues a $\mathbb{C}^*$ action on $\mathbb{C}^n/G$ which contracts it to a point). $\endgroup$ – Geordie Williamson May 10 '16 at 21:30

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