I think it is true that there is no free-action of $\mathbb{Z}\_p(p\neq 2$) on product of $\mathbb{CP}^n(n$ odd) and $\mathbb{S}^{2m}$. But I don't know how to prove it. Any solution will be helpful.


Thanks in advance.



What about the action on this space?
I think the following will work for $p > 3$. I am not sure if it can be made to work for $p =3$, but maybe it can. I will the appeal to the representation theory of $\mathbb{Z}/p\mathbb{Z}$, but probably this is overkill. 


Given a continuous action of $\mathbb{Z}/p\mathbb{Z}$ on your space $X = \mathbb{CP}^n\times S^{2m}$, the rational cohomology groups are all representations of $\mathbb{Z}/p\mathbb{Z}$. There are two irreducible representations of $\mathbb{Z}/p\mathbb{Z}$ over $\mathbb{Q}$, the trivial representation and a non-trivial representation of dimension $p-1$. Each cohomology group can be expressed as a direct sum of copies of these representations. 


On the other hand, the Künneth theorem shows that the only non-zero cohomology groups of $X$ have dimension 1 or 2, and that these occur in even degree. Since $p-1 > 2$, the cohomology groups must be direct sums of copies of the trivial representation. In other words, each element $g$ of $\mathbb{Z}/p\mathbb{Z}$ acts trivially on the cohomology of $X$. It follows easily now that the Lefschetz number of any such element is positive, and therefore that $g$ has a fixed point. 


