I've been playing around with this proof for a while and I can't seem to figure out where to go from here:
I have that $M$ is a manifold, $G$ is a finite group (say, of order $n$), and the action of $G$ on $M$ is $\Phi:G\times M\rightarrow M$.
I want to show that if $\Phi(g,p)\neq p,$ $\forall p\in M, \forall g\in G$ then $\Phi$ is properly discontinuous.
So far have:
Suppose $\Phi$ is not properly discontinuous. Then $\forall p\in M, \forall g\in G-\{1_{g}\}, \forall U_{p}\in M$ where $U_{p}$ are neighborhoods of $p$ in $M$, we have: $$\Phi(g, U_{p})\cap U_{p}\neq\emptyset.$$ So pick some $q\in\Phi(g, U_{p})\cap U_{p}$ then $q\in U_{p}$ and $\Phi(g, s) = q$ for some $s\in U_{p}$. My plan originally was to show (by applying $\Phi(g, s)$ $m$ times, where $m$ is the order of the element $g\in G$) that $G$ being finite $\implies s=q$, thus contradicting the fact that $\Phi$ has no fixed points. But I can't seem to figure out how to show it and I'm starting to think my original strategy was wrong. Any ideas?
Let $p$ be a point in $M$. Since $M$ is Hausdorff and $G$ is finite, there is an open set $U$ such that $p\in U$ and $gp\not\in U$ for all $g\in G\setminus\{1\}$. Since $M$ is regular, there is an open set $V$ such that $p\in V\subseteq\overline V\subseteq U$. Now the set $W=V\setminus\bigcup_{g\in G\setminus\{1\}}g(\overline V)$ is open, contains $p$. What is $W\cap g(W)$ when $g\in G\setminus\{1\}$?