If a finite group acts freely, then it acts properly discontinuously.

136 Views Asked by At

In my differential geometry course, we saw that a group of diffeomorphism $G$ of a smooth manifold $M$ acts properly discontiuously on $M$ if

  • each $x \in M$ admits an open neighborhood $U$ such that $\varphi(U) \cap U = \varnothing$ for all $\varphi \in G\backslash \{\text{id}\}$,
  • for any pair $x, y \in M$ such that $x$ is not in the orbit of $y$ under $G$, there exists two neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $\varphi(U) \cap \psi(V) = \varnothing$ for all $\varphi, \psi \in G$.

And then the professor add a remark saying that if a finite group acts freely on $M$, then it also acts properly discontinuously. I really have no idea where to start, does anyone have a hint? Thank you!