Show that if $p\neq 2$, then $\mathbb{Z}_p$ cannot act freely on $\mathbb{C}P^n$

Question

If $p\neq 2$, then the cyclic group $\mathbb{Z}_p$ has no free continuous action on $\mathbb{C}P^n$. My question is how to prove the above fact using Leray-Serre spectral sequence associated to the Borel fibration $ \mathbb{C}P^n\hookrightarrow X_{\mathbb{Z}_p}\rightarrow B_{\mathbb{Z}_p}$.

From the Euler Characteristic argument, $p$ divides $n+1$. Also, $\pi_1(B_{\mathbb{Z}_p})={\mathbb{Z}_p}$ acts trivially on $H^*(\mathbb{C}P^n;\mathbb{Z}_p)$ by using Lefschetz fixed point theorem. $H^*(\mathbb{C}P^n;\mathbb{Z}_p)=\mathbb{Z}_p[b]/\langle b^{n+1}\rangle$ and $H^*(\mathbb{Z}_p;\mathbb{Z}_p)=\bigwedge(s)\otimes\mathbb{Z}_p[t]$. So the only possibility is $d_3(b)=st$. It follows $d_3(sb)=0$ and $d_3{(tb)}=st^2$. After that, I am unable to deduce any contradiction.

Thank you so much in advance.

Answer 1

Consider the cohomology with $\mathbb{Z}$ coefficients (and reduce the 0-th term modulo $p$ to get uniform description of it). Then we have a spectral sequence starting from $\mathbb{F}_p[x,y]/x^{n+1}$ with $deg(x)=deg(y)=2$ and converging to $H^*(\mathbb{C}P^n/\mathbb{Z}_p,\mathbb{Z})$. Since the $E^2$ term is concentrated in even degrees, the spectral sequence degenerates at the $E^2$-term, and so $H^*(\mathbb{C}P^n/\mathbb{Z}_p)$ has arbitrarily high non-zero cohomologies. But is is a finite dimensional manifold (being the quotient of a manifold by a free action), and this is a contradiction.