TITLE: Which groups act freely on $S^n$?
QUESTION [20 upvotes]: When $n$ is even, it is easy to classify groups which act freely on $S^n$ using degree theory: if $G$ acts on $S^n$, then associating to each element $g \in G$ the degree of the map obtained from multiplication by $g$, one gets a map $d : G \to \{\pm 1\}$. It is easy to verify this is a homomorphism. If $G$ acts freely, multiplication by any element $g$ is a fixed point free map, thus $d(g) = (-1)^{n+1} = -1$, making $d$ injective. The only nontrivial group which injects into $\Bbb Z/2$ is itself, so we're done.
However, a lot of groups act freely on $S^n$ for odd $n$. For example, $\Bbb Z/p$ acts on $S^3$ freely for all prime $p$ (so called lens space action). What do we know about such groups? Is it possible to classify them? 
If $G$ is a finite group acting freely on $S^3$, then as free action of finite groups on Hausdorff spaces is properly discontinuous, quotient map $S^3 \to S^3/G$ is a covering projection. Thus $S^3/G$ is a closed 3-manifold with fundamental group $G$, hence $G$ is a closed 3-manifold group. On the other hand, if $G$ is a closed 3-manifold group, let $M$ be that manifold and $G$ must act on $\tilde{M}$ freely. But $\tilde{M}$ is a simply connected closed 3-manifold hence homeomorphic to $S^3$ by Poincare conjecture, so $G$ must act on $S^3$ freely. 
Thus, finite groups acting freely on $S^3$ are precisely the finite closed 3-manifold groups. But what about infinite groups? Given an infinite group, how can we tell if it acts on $S^3$ or not? More generally, what about groups acting freely on $S^n$ for some fixed odd $n > 1$?

[edit] I am only interested in actions of discrete groups. Also, any sort of general remark (long enough to not fit as a comment) or partial answers (like the answers below) are welcome to me, you can post them as answers.

REPLY [5 votes]: This has almost but not quite been stated a few times, so to clear the air: the answer is known for finite groups, it is due to Madsen, Thomas, and Wall, and it says that a finite group $G$ acts freely on some sphere if and only if

all of the abelian subgroups of $G$ are cyclic; equivalently, the cohomology is periodic; equivalently, $\mathbb{Z}_p \times \mathbb{Z}_p$ does not occur as a subgroup for any prime $p$; and
every element of order $2$ is central.

The necessity of the first condition is due to Smith and the necessity of the second condition is due to Milnor. This is taken from the introduction to Alejandro Adem's Constructing and Deconstructing Group Actions.