I have a simple algebraic topology question. Let $M$ and $N$ be 2-dimensional oriented manifolds (say $H^{2}(M,\mathbb{Z})\cong \mathbb{Z}\alpha_{M}$ and $H^{2}(M,\mathbb{Z})\cong \mathbb{Z}\alpha_{N}$). Assume that a finite group $G$ acts on $M$ and $N$ in such a way that $G$ preserves the orientation of $M$ and $N$ and the induced action of $G$ on $M\times N$ has no fixed point. Then $$ X=(M\times N)/G $$ is a 4-dimensional oriented manifold.
I would like to understand the intersection form on the middle cohomology $H^{2}(X,\mathbb{Z})$. There is a ono-to-one correspondence between $$ H^{2}(X,\mathbb{Z}) \ \longleftrightarrow H^{2}(M\times N,\mathbb{Z})^{G}, $$ i.e. any $G$-invariant element of $H^{2}(M\times N,\mathbb{Z})$ descends to $H^{2}(X,\mathbb{Z})$ and any element of $H^{2}(X,\mathbb{Z})$ can be pulled back to $H^{2}(M\times N,\mathbb{Z})^{G}$ by the quotient map. Since the $G$-action is free, the intersection form on $H^{2}(X,\mathbb{Z})$ is given by the intersection form on $H^{2}(M\times N,\mathbb{Z})^{G}$ divided by $|G|$. So, any intersection number on $H^{2}(M\times N,\mathbb{Z})^{G}$ must be a multiple of $|G|$.
On the other hand we have $$ p_{1}^{*}(\alpha_{M}), \ p_{2}^{*}(\alpha_{N})\in H^{2}(M\times N,\mathbb{Z})^{G} $$ (because $G$ preserves both $M$ and $N$) and $$ p_{1}^{*}(\alpha_{M})\cup \ p_{2}^{*}(\alpha_{N})=\alpha_{M\times N} $$ where $p_{i}$ is the $i$-th projection of $M\times N$ and $H^{2}(M\times N,\mathbb{Z})\cong \mathbb{Z}\alpha_{M\times N}$. This means that the intersection number $p_{1}^{*}(\alpha_{M})\cup \ p_{2}^{*}(\alpha_{N})$ is 1, not divisible by $|G|$ (unless $G$ is trivial).
Could anyone point out what is wrong with my argument?
Thank you in advance.
Why not try out both arguments on a specific example?
Take $M = N = S^1\times S^1$ and $G = \mathbb{Z}/2\mathbb{Z}$ acting by $-1 *(\theta,\phi) = (\theta +\pi,\phi +\pi)$. Then this action is free on each of $M$ and $N$, so the induced action on $M\times N$ is certainly free.
In this case, $(M\times N)/ G \cong (S^1)^4$, since, for example, $(M\times N)/ G$ is a compact abelian Lie group. (I'm sure there's an easy direct way to see it).
So, what do both computations give you? Which one must be wrong?