Using that all groups that act freely on some sphere $S^n$ have periodic cohomology, one can see that $\mathbb Z/2 \times \mathbb Z/2$ can not act freely on any $S^n$. But can it act freely on $S^{\infty}$?

(By $S^{\infty}$ I mean, as topologists usually do, the colimit of $S^i$, with the weak topology. It is contractible.)