Cocompact action with finite stabilizer implies locally finite

Question

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.

Answer 1

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.

Answer 2

There is a standard lemma in the theory of CW complexes, namely Proposition A.1 in the Appendix of Hatcher's Topology:

Every compact subset of a CW complex $X$ is contained in a finite subcomplex of $X$.

For the action of $G$ on $X$ to be cocompact means, by definition, that there is a compact $K \subset X$ such that $G \cdot K = X$. Let $L \subset X$ be a finite subcomplex containing $K$, and so $G \cdot L = X$ (one sees from this argument that cocompactness of the action is equivalent to the existence of a finite subcomplex $L$ such that $G \cdot L = X$).

Arguing by contradiction, suppose there exists a $0$-cell $x \in X$ such that $X$ is not locally finite at $x$. By definition of local finiteness, this means that there are infinitely many (open) cells $e \subset X$ of the given CW structure such that $x \in \overline e$. Referring to $\overline e$ as a closed cell, this is equivalent to saying that there are infinitely many closed cells $\overline e$ containing $x$.

Note also that the number of closed cells contained in $L$ is finite (because the number of (open) cells contained in $L$ is finite). It follows that for each $a \in G$ the number of closed cells contained in $a \cdot L$ is finite. But there are infinitely many closed cells that contain $x$, and so by applying the pigeonhole principle it follows that the set $$A = \{a \in G \mid x \in a \cdot L\} = \{a \in G \mid a^{-1} \cdot x \in L\} $$ is infinite.

Knowing that $x$ is a $0$-cell, it follows that each $a^{-1} \cdot x$ is a $0$-cell. But $L$ has only finitely many $0$-cells. By another application of the pigeonhole principle, there exists an infinite subset $B \subset A$ such that $b^{-1} \cdot x$ is constant independent of $b$. Picking one $b' \in B$, it follows that $b \cdot (b')^{-1}\cdot x = x$ for all $b \in B$. This proves that $x$ has infinite stabilizer, contradicting the hypothesis.