TITLE: Integral homology of $S^{n-1}/\pi$, $H_*(S^{n-1}/\pi; \mathbb{Z}_p)$
QUESTION [5 upvotes]: Let $p$ be an odd prime number. Regard the cyclic group $\pi$ of order $p$ as the group of $p$th roots of unity contained in $S^1$. Regard $S^{2n-1}$ as the unit sphere in $\mathbb{C}^n$, $n \ge 2$. Then $\pi \subset S^1$ acts freely on $S^{2n-1}$ via$$\zeta(z_1, \dots, z_n) = (\zeta z_1, \dots, \zeta z_n).$$Let $L^n = S^{2n-1}/\pi$ be the orbit space; it is called a lens space is an odd primary analogue of $\mathbb{R}P^n$. The obvious quotient map $S^{2n-1} \to L^n$ is a universal covering.
Now, I have two questions.

What is the integral homology of $L^n$, $n \ge2$?
What is $H_*(L^n; \mathbb{Z}_p)$, where $\mathbb{Z}_p = \mathbb{Z}/p\mathbb{Z}$?

Sorrry yet again for two rather remedial questions...

REPLY [3 votes]: That is exactly exercise 5, page 103, of ``A concise course in algebraic topology'',  which gives a hint: mimic the analogous (cellular) computation for $RP^n$.  I won't spoil the fun by giving the answer.