Possible finite groups that can act properly discontinuously on $S^2$ Let $G$ act on $S^2$ such that for every $x\in X$ and every $g\in G, g\neq e$, there is a neighborhood $U$ of $x$ such that $g(U)\cap U\neq \varnothing$.
If $S^2/G$ is a surface and $G$ is a finite group, then what are the possibilities for $G$? $1$ acts on $S^2$ trivially, with $S^2/1\cong S^2$ and $\mathbb{Z}_2$ acts on $S^2$ to give $S^2/\mathbb{Z}_2\cong RP^2$. Are there other possibilities?
 A: Surprisingly, $\Bbb Z/2\Bbb Z$ is the only nontrivial group which acts freely on $S^2$ (and even on $S^{2n}$). Hatcher proves this as proposition 2.29.
The idea is that an action of $G$ on $S^{2n}$ is equivalent to a map $G\to \operatorname{Homeo}(S^{2n})$, where $\operatorname{Homeo}(S^{2n})$ is the group of homeomorphisms of $S^{2n}$. If we denote this map $g\mapsto f_g$, then we can use this to define a group homomorphism $G\to\{\pm1\}$ where $g$ maps to the degree of $f_g$. If this action is properly discontinuous, then any nonidentity element has no fixed point. Such a map is homotopic to the antipodal map, so for even dimensions it has degree $-1$ . But in a free action, every element other than the identity must have no fixed point, so the only possibility is that $G\to\{\pm1\}$ is injective, so $G=\Bbb Z/2\Bbb Z$.

I also wanted to mention that it is super easy to see why any map of $S^n$ with no fixed points is homotopic to the antipodal map: just visualize the sphere centered at the origin, and note since $f(x)\neq x$, the straight line from $f(x)$ to $-x$ does not pass through the origin. So simply define
$$h_t(x)=\frac{(1-t)f(x)-tx}{|(1-t)f(x)-tx|}.$$
