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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.

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Why does $S^{2n}$ being Hausdorff imply that $G$ acts freely? Why does $\pi_1(X/G) \cong G$? I don't see how either of your conclusions follow. Anyway, this can be solved by considering the degree of the maps in $G$. – Carl Apr 24 '12 at 6:11
Euler characteristic is multiplicative for covers. – user641 Apr 24 '12 at 6:20
Use Lefschetz fixed point theorem, you'll get all fixed-point-free homomorphism from $S^{2n}$ to itself must be orientation reversing. – Yuchen Liu Apr 24 '12 at 7:58
Any map $S^{2n}\to S^{2n}$ without fixed points is homotopic to the antipodal map $x\mapsto -x$, and this map has degree $-1$. Since degrees are multiplicative, the only way how all non-$1$ elements of $G$ have degree $-1$ is that there is at most one such element. – user8268 Apr 24 '12 at 10:02

(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.

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Note: this expands out Steve D's answer from almost a year ago. The homework assignment was probably due before now! – Pete L. Clark Mar 20 '13 at 16:11

Do you know the definition of degree? You can arguing as follows. Suppose a group G act on 2n-sphere. Then there is a homomorphism from G to {1,-1} which sends an element g to the degree of g-action. But if g is not identity, then g(x) is not x, thus the g-action is homotopic to the antipodal map. And in sphere of even dimension, the antipodal map is of degree -1(because it reverse the orientation). Let's see what do we have:(1) a homomorphism p:G->$C_2$; (2)p(g)=-1 if g is not identity. So now you can conclude that G actually has only two element.(So I don't understand why you assume that G is a finite group.)

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This is already covered in the comments above. – user641 Jan 15 '13 at 15:31
I didn't see the comments at first. You can take it as a expand. – lee Jan 16 '13 at 1:12

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