Let $\Gamma$ be a group acting properly and cocompactly by isometries on a metric space $X$. Then, $\Gamma$ has finitely many conjugacy classes of point stabilizers.
The proof of this well-known result goes like this:
Let $x\in X$, and let $\Gamma_x$ denote the stabilizer of $x$ under the $\Gamma$-action. Since the action is cocompact, there exists a compact subset $K \subset X$ such that the $\Gamma$-translates of $K$ cover $X$. Hence there exists $\gamma \in \Gamma$ such that $x \in \gamma K$.
Since $\gamma^{-1} \Gamma_x \gamma = \Gamma_{\gamma^{-1}x}$, and $\gamma^{-1}x \in K$, for each $\alpha \in \gamma^{-1} \Gamma_x \gamma$, $\alpha K \cap K \neq \emptyset$.
Usually the proof concludes by saying that the result follows by properness of the action. I'm not sure how this follows, since we have only shown that for a specific element $\gamma \in \Gamma$, $\gamma^{-1} \Gamma_x \gamma$ is contained in a finite set. Wouldn't we have to show that this holds for all $\gamma \in \Gamma$?
Best Answer
From what is shown so far, it follows that for each $x \in X$ there exists $y \in K$ such that $\Gamma_x$ is conjugate to $\Gamma_y$.
So, what's left to check is that as $y$ varies over $K$, there are only finitely many different subgroups of the form $\Gamma_y$.
This will follow if one can prove that the set of elements $$\bigcup_{y \in K} \Gamma_y = \bigcup_{y \in K} \{\delta \in \Gamma \mid \delta(y)=y\} $$ is finite.
And that will follow if one can prove that the set of elements $$\{\delta \in \Gamma \mid \delta(K) \cap K \ne \emptyset\} $$ is finite. Well, that's the definition of properness of the $\Gamma$ action on $X$, applied to the compact set $K$.