I was wondering whether the cyclic group $\mathbb{Z}_2$ acts freely on $S^n \times S^m$ where $n,m\geq 2$? It seems to me that it does not act freely.
I know by Hatcher's book that there are two conditions that a finite group $G$ acting freely on $S^n$ must satisfy:
(a) Every abelian subgroup of $G$ is cyclic. This is equivalent to saying that $G$ contains no subgroup $\mathbb{Z}_p\times \mathbb{Z}_p$ with $p$ prime.
(b) $G$ contains at most one element of order 2.
Thus the cyclic group $\mathbb{Z}_2$ acts freely on $S^n$. But my question is about $S^n\times S^m$ where $n,m\geq 2$.
Surely the answer is that $\mathbb Z_2$ does act freely on $S^n \times S^m$?
Let $e$ and $\sigma$ be the two elements of $\mathbb Z_2$ (with $e$ being the identity element). Suppose that the action of $\mathbb Z_2$ on $S^n \times S^m$ is given by $$ e \ : \ (x, y) \mapsto (x, y), \qquad \qquad \sigma \ : \ (x, y) \mapsto (-x, -y), $$ where $-x \in S^n$ and $-y \in S^m$ are antipodal to $x \in S^n$ and $y \in S^m$ respectively.
Clearly, what I've written down is a free group action.