Homology group of 3-fold sum of projective planes I want to calculate the homology group of the 3-fold sum of projective planes defined by the labelling scheme $aabbcc$. For this I will use the following corollary from Munkres:
Corollary 75.2: Let $F$ be a free group with free generators $\alpha_1,\dots,\alpha_n$. Let $N$ be the least normal subgroup of $F$ containing the element $x$ of $F$. Let $G=F/N$. Let $p\colon F\rightarrow F/[F,F]$ be projection. Then $G/[G,G]$ is isomorphic to the quotient of $F/[F,F]$, which is free abelian with basis $p(\alpha_1),\dots,p(\alpha_n)$, by the subgroup generated by $p(x)$.
In my case, we have $F$ is the free group with free generators $a,b,c$, N is the least normal subgroup of $F$ containing the element $a^2b^2c^2$, $G$ is the fundamental group and $G/[G,G]$ is the homology group. The corollary says that the homology group is isomorphic to $F/[F,F]$ divided by the least normal subgroup containing the element $q(a^2b^2c^2)=a^2b^2c^2$ (call this normal subgroup M). Since $F/[F,F]$ is free abelian of rank 3, it is isomorphic to $\mathbb{Z}\times \mathbb{Z}\times\mathbb{Z}$. So dividing this by $M$ we have that the homology group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$.
The problem is that the homology group should be $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$. I don't see where I went wrong. Can you help me? Thanks.
 A: You are dividing the groups wrong. $\mathbb{Z}^3 / a^2 b^2 c^2 \cong \mathbb{Z}/2\mathbb{Z} × \mathbb{Z} × \mathbb{Z}$ and not $(\mathbb{Z} / 2 \mathbb{Z})^3$. (Where $a, b, c$ are the generators of $\mathbb{Z}^3$.)
A: You're nearly right up until you quotient $F/[F,F]$ by the normal subgroup generated by $q(a^2b^2c^2)$ which as you rightly point out is just the subgroup of $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}=\{n_1a+n_2b+n_3c\mid n_i\in\mathbb{Z}\}$ generated by $a^2b^2c^2$. It would be a good idea, as we're now working in an abelian group, to instead call this element $2a+2b+2c$.
The trick here is to reparametrise the group so that $2a+2b+2c$ can be written as just a multiple of a single generator. In this case, it looks like taking our generating set to be $\{a,b,a+b+c\}$ will be a good idea. And indeed if we set $a+b+c=d$, we see that $c=d-a-b$ and so this is indeed a generating set.
Now we are instead looking at the group $\langle a\rangle \oplus \langle b\rangle\oplus\langle d\rangle$ and modding out by the subgroup generated by $2d$. It's now much easier to see that this quotient group is isomorphic to $$\langle a\rangle \oplus \langle b\rangle\oplus(\langle d\rangle/\langle 2d\rangle)\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$$ as required.
