Let $X$ be an irreducible variety, $G$ a finite subgroup of automorphisms of $X$. Then there is an open subset $U \subset X$ where $G$ acts freely.

Question

I'm looking for a reference (or proof), of the following well known fact:

Let $X$ be an irreducible variety, $G$ a finite subgroup of automorphisms of $X$. Then there is an open subset $U \subset X$ where $G$ acts freely.

Answer 1

For $1\neq g\in G$, the set $$Z_g=\{x\in X\mid gx=x\}$$ is Zariski closed (It is the inverse of the diagonal by the map $(\mathrm{id},g)\colon X\mapsto X\times X$). Then, $\cup_g Z_g$ is Zariski closed and $G$ acts freely over its complement.