Proving that there doesn't exist a free $\mathbb{Z}_2\times \mathbb{Z}_2$ action on $S^n$. 
Prove that there doesn't exist a free $\mathbb{Z}_2\times \mathbb{Z}_2$ action on $S^n$.

I know  that  an action of a group $G$ on a space $X$ is a homomorphism from $G$
to the group $Homeo(X)$ of homeomorphisms $X\to X$, and the action is free if the
homeomorphism corresponding to each nontrivial element of $G$ has no fixed points.
And $\mathbb{Z}_2=\{-1,1\}$ can act freely on $S^n$ via
$(1,x)\mapsto x$ and $(-1,x)\mapsto -x$. I also know that $S^n/\mathbb{Z}_2\cong \mathbb{R}P^n$
But how to analyze the action of $\mathbb{Z}_2\times \mathbb{Z}_2$? I got a little confused and don't know where to start. Any helps will be highly appreciated!
Added:
I have learned some homology theory and I got some results based on that. But I haven't learned the cohomology theory. I would like to show you some of the results I've got, and I hope to find a method that doesn't go beyond homology. Any enlightening ideas will be highly appreciated!

  
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*The only nontrivial group that can act freely on $S^{2n}$ is $\mathbb{Z}_2$

It's known that if $f:S^n\to S^n$ has no fixed point, then $f$ is homotopic to the antipodal map whose degree is $(-1)^{n+1}$. Define $d:G\to \{-1,1\}$ by taking $g\in G$ to the degree of the homeomorphism derived by $g$. Then we know
$$G\cong G/\ker d \cong \text{Im } d \le \mathbb{Z}_2$$


  
*$\mathbb{Z}_2\times \mathbb{Z}_2$ can't act freely on $S^1$.
  

It's known that if $G$ is a finite group, $Y$ is path connected and locally path connected, then the quotient map $p:Y\to Y/G$ is a normal covering map, and G is the group of deck transformations of $Y\to Y/G$ and
$$G\cong \pi_1(Y/G)/p_*(\pi_1(Y))$$
For $S^1$ we know that $S^1/G$ is homeomorphic to $S^1$. Hence G must be cyclic.
 A: I'm not sure where I first encountered (a  version of) this proof, but it's definitely not original.
For ease of writing, I will write $G = \mathbb{Z}_k\oplus\mathbb{Z}_k$.  Note that if $G$ acts freely, then any subgroup of $G$ acts freely.  Hence, we may assume $k$ is prime.
Assume for a contradiction that $G$ acts freely on $S^n$ with $n$ odd.  You have already handled the case $n = 1$, so we will assume $n\geq 3$.  Because $n$ is odd, a simple application of Lefschetz shows that $M:=S^n/G$ is a closed orientable manifold.  Further, because $n > 1$, $G\cong \pi_1(M)\cong H_1(M)\cong H_1(M;\mathbb{Z}_k)$.  Poincare duality now forces $H^{n-1}(M;\mathbb{Z}_k)\cong G$.
Now, $\pi_k(M) = 0$ for $k = 2,..., n-1$.  So, we can turn $M$ into an Eilenburg-Maclane space $K(G,1)$ by attaching cells of dimension $n+1$ or higher.  In particular, by uniqueness (up to homotopy) of a $K(G,1)$, we have $H^\ast(M)\cong H^\ast(K(G,1))$ for $\ast\leq n-1$.
Thus, $H^{n-1}(K(G,1);\mathbb{Z}_k)\cong G$.
On the other hand, a model for $K(G,1)$ is a product of 2 infinite dimensional lens spaces $L_k$.  Each of these has $H^\ast(L_k;\mathbb{Z}_k)\cong Z_k$ for all $\ast$.
Now, because $k$ is prime, $\mathbb{Z}_k$ is a field, so the Kunneth theorem is particularly nice:  $H^{n-1}(K(G,1); \mathbb{Z}_k)\cong \bigoplus_{s=0}^{n-1} H^s(L_k;\mathbb{Z}_k)\otimes H^{n-1-s}(L_k;\mathbb{Z}_k)$.  Since each cohomology group is $\mathbb{Z}_k$ and $\mathbb{Z}_k\otimes \mathbb{Z}_k\cong \mathbb{Z}_k$, so $H^{n-1}(K(G,1);\mathbb{Z}_k)\cong \bigoplus_{s=0}^{n-1} \mathbb{Z}_k \cong \mathbb{Z}_k^n$.
Thus, we conclude $$\mathbb{Z}_k^2 \cong G\cong H^{n-1}(M;\mathbb{Z}_k)\cong H^{n-1}(K(G,1);\mathbb{Z}_k)\cong \mathbb{Z}_k^n$$
In other words, $n=2$.  Since $n$ was the (odd!) dimensions of our sphere, we have a contradiction.
