TITLE: Can $S^2$ be turned into a topological group?
QUESTION [45 upvotes]: I know that $S^1$ and $S^3$ can be turned into topological groups by considering complex multiplication and quaternion multiplication respectively, but I don't know how to prove or disprove that $S^2$ can.
This is just a recreational problem for me. Here's the work I have done:
We know that $S^2$ is homogenous and doesn't have the fixed-point property, so those don't help. I know that for any subgroup $H$ of $S^2$ that $S^2/H$ (the coset space with quotient topology) is homogenous. So maybe I can find a subgroup that would have to create a non-homogenous space. The only subgroups that I  know would have to exist are cyclic subgroups. But I have no clue what $S^2/\langle x\rangle$ would even look like for any $x\in S^2$.
Any help is appreciated.
EDIT
Mike Miller's answer below certainly makes short work of my problem. But if someone comes up with a more elementary approach, I'd be appreciative.

REPLY [7 votes]: The only non-trivial group which can acts freely on $S^n$, $n$ even, is the cyclic group of order $2$.  
Indeed, if $f$ is a homeomorphism from $S^n$ to $S^n$ without a fixed point, then $f$ and the antipodal map are homotopic via $H(t,x)=f(t,x)/|f(t,x)|$, where $f(t,x)=tf(x)-(1-t)x$.  Thus $f$ and the antipodal map induces the same automorphism on the group $H_n(S^n) \cong \mathbb{Z}$, but the group automorphism induced by the antipodal map is $m \mapsto (-1)^{n+1}m=-m$.  Now if $G$ is a group which acts freely on $S^n$, we can define a homomorphism $\phi$ from $G$ to $\operatorname{Aut}(\mathbb{Z}) \cong \mathbb{Z}_2$, which map each $g \in G$ to the automorphism induced by $g$ on $H_n(S^n)$.  We have seen that if $g$ is not trivial then $\phi(g)$ is not trivial.  This shows that $\phi$ is injective and we are done.
Now if $S^n$, $n$ even, has the structure of a topological group, then it acts on itself freely by left translation (as noticed by Qiaochu Yuan), which implies that $S^n$ and $\mathbb{Z}_2$ are isomorphic.