I was wondering whether the cyclic group $\mathbb Z/4\Bbb Z$ acts freely on $S^{2k} \times \Bbb CP^n$ where $n>1$? It seems to me that it does not act freely. In case it acts freely then the induced action on cohomology must be non-trivial as the Euler characteristic is non-zero. I was trying to prove using Lefschetz fixed point theorem. But I could not able to derive any contradiction.
Thank you so much for your help.
Nothing really changes from Will Sawin's answer here.
Given a self-homeomorphism $\sigma: S^{2k} \times \Bbb{CP}^n$, you want to know what the induced map is on cohomology, written via the Kunneth decomposition as $\Bbb Z[c, S]/(c^{n+1}, S^2)$, where $|c| =2$ and $|S| = 2k$.
Then for any $k$, the classes $c$ and $-c$ are the only primitive degree 2 classes $x$ which do not square to zero and for which $x^{n+1} = 0$, as $(ac + bS)^{n+1} = a^n b c^n S$ (which is only relevant if $k = 1$), so we have $\sigma^*(c) = \pm c$. For any $k$, because $\pm S$ are the unique primitive classes in $H^{2k}$ which have non-trivial products with $c^n$ (which is preserved by $\sigma$ up to a sign) but trivial cup-square, we may write $\sigma^*(S) = \pm S$.
Therefore $\sigma^2$ fixes both $c$ and $S$, and hence acts by the identity on cohomology. The Lefschetz number is then $\chi(S^{2k} \times \Bbb{CP}^n) = 2n+2 \neq 0$. So for any self-homeomorphism $\sigma$, the map $\sigma^2$ has fixed points, and thus there is no free $\Bbb Z/2j$ action on these spaces for any $j > 1$.
