Given a discrete group $G$ acting cellularly and cocompactly on a cell-complex $X$ with finite stabilizers, I am struggling to show $X$ is locally finite.
I am trying to show any vertex $x\in X$ is contained in finitely many cells. Since $X/G$ is compact, thus locally finite, which implies there are only finitely many orbits of cells containing $x$. So it suffices to show each orbit of a cell containing $x$ has only finitely many cells (containing $x$).
I am trying to use the orbit-stabilizer theorem to deduce that. However, I think given a cell $C$ containing $x$ with $g\cdot C$ containing $x$ as well, $g$ might not always fix $x$. Thus $g$ might not be in the stabilizer of $x$. So I have no idea why each cell-orbit containing $x$ has only finitely many elements.
Best Answer
Suppose that $X$ is not locally finite.
Then there is some $x$ in the intersection of infinitely many cells. Moreover, since the quotient is compact, there must be some cell $C$, $x\in C$ such that the set $H=\{g\in G|x\in C^g\}$ is infinite.
Now note that $x^{H^{-1}}\subset C$. If this set is infinite, then our group action wasn't discrete, and if this set is finite, then by the pigeonhole principle, there is some infinite $K\subset H$ such that $x^{K^{-1}}$ is a singleton $\{x'\}$, and therefore $KK^{-1}\subset \text{Stab(x')}$ is infinite.