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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.)

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    $\begingroup$ Isn't $S^\infty$ homeomorphic to $S^\infty \times S^\infty$? That would answer your question. I think you can construct the map fairly explicitly, $S^n \times S^n \to S^{2n}$. You think of this map as crushing the two factors $S^n \times \{1\} \cup \{1\} \times S^n$ to a common point, with the map a homeomorphism otherwise. If you do this fairly naturally, this induces a map of the colimits $S^\infty \times S^\infty \to S^\infty$. This map is not a homeomorphism itself but it looks like it can be fixed as it fails to be a homeomorphism only by crushing two contractible subspaces. $\endgroup$
    – Ryan Budney
    Commented Jun 1, 2015 at 6:59
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    $\begingroup$ Alternatively, $S^\infty$ is homeomorphic to $\mathbb{R}^\infty$ (see this answer, for instance), and clearly $\mathbb{R}^\infty$ is homeomorphic to its square. $\endgroup$
    – Eric Wofsey
    Commented Jun 1, 2015 at 12:53
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    $\begingroup$ Another point of view is that $S^\infty$ carries a structure of topological abelian group of exponent 2 (i.e., a topological vector space over $\mathbb{Z}/(2)$) as in this answer: mathoverflow.net/a/43047/2926. Translation by any two linearly independent elements would also answer your question affirmatively. $\endgroup$
    – Todd Trimble
    Commented Jun 1, 2015 at 17:30

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