TITLE: The only finite group which can act freely on even dimensional spheres is $C_2$.
QUESTION [7 upvotes]: I don't know how to show this. 
Do I assume $G$ acts on $S^{2n}$ by homeomorphisms? Then, since $S^{2n}$ is Hausdorff I'd know $G$ acts freely and properly discontinuously, and since $\pi_1(S^{2n})={1}$ I'd have $\pi_1(X/G)\cong G$. But I'm not sure whether this is useful.

REPLY [8 votes]: (Of course the trivial group acts freely on all spheres!  Let's assume $G$ is finite and nontrivial.)
Since $G$ is finite, the action is certainly properly discontinuous, and since the action is given to be free, it follows that the quotient $q: X \rightarrow X/G$ is a finite covering map.  Recalling that in any $n$-sheeted covering map $q: X \rightarrow Y$ we have $\chi(X) = n \chi(Y)$ and that the Euler characteristic must be an integer, we are almost done.