Abelianization of the $\mathbb R \mathbb P^2$#$\mathbb R \mathbb P^2$

Question

I know that the fundamental group of $\mathbb R \mathbb P^2$#$\mathbb R \mathbb P^2$ is $<a,b|a^2b^2>$.

When we abelianize this we have this presentation: $Ab(\pi_1(\mathbb R \mathbb P^2$#$\mathbb R \mathbb P^2)=<a,b|a^2b^2, ab=ba>$.

However, we know that the $H_1(\mathbb R \mathbb P^2$#$\mathbb R \mathbb P^2)=\mathbb Z\times \mathbb Z_2$ (Amstrong's book page 183), i.e.,

$Ab(\pi_1(\mathbb R \mathbb P^2$#$\mathbb R \mathbb P^2)=<a,b|b^2,ab=ba>$.

Both has to be equivalent, but I don't know why.

I need help here.

Thanks a lot.

Answer 1

The first version can be seen as $\langle a,b\mid ab=ba\rangle/(a^2b^2)$, which is $\mathbb Z^2/\mathbb Z(2,2)$, identifying the images of $a$ and $b$ in the abelianization of the free group with $(1,0)$ and $(0,1)$.

Doing the same for the second version, we get $\mathbb Z^2/\mathbb Z(0,2)$.

Now there is an automorphism of $\mathbb Z^2$ which maps $(2,2)$ to $(0,2)$, so from it you can get an iso.