Calculate the first homology group of $P^2\#T$, that is $H_1(P^2\#T)$

I've already found that the fundamental group of the connected sum $P^2\#T$, by the labelling scheme $aabcb^{-1}c^{-1}$, to be $F_3/<aabcb^{-1}c^{-1}>$. How would I find the first homology group? The first homology group is defined to be: $H_1(X) = \pi_1(X,x_0)/[\pi_1(X,x_0),\pi_1(X,x_0)]$. I believe I have to use the following theorem: Let $F$ be a group; let $N$ be a normal subgroup of $F$; let $q$: $F \to F/N$ be the projection. The projection homomorphism $P: F \to F/[F,F]$ induces an isomorphism $\phi: q(F)/[q(F),q(F)] \to p(F)/p(N)$. But I can't seem to figure out a nicer representation for the first homology group with this theorem.

• Your notation $\mathbb{Z}^3 / \langle aabcb^{-1}c^{-1} \rangle$ is flawed. $\mathbb{Z}^3$ refers to the free abelian group of rank 3. What you want there instead is $F_3 = \mathbb{Z} * \mathbb{Z} * \mathbb{Z}$, the free product of three copies of $\mathbb{Z}$, which is the free non-abelian group of rank 3. – Lee Mosher Jun 20 '15 at 13:56
• Thanks, I completely agree. Sorry for that. – Endosred Jun 20 '15 at 14:51

As Lee Mosher has pointed out in the comments, the correct representation for the fundamental group should be $F_3/\langle aabcb^{-1}c^{-1} =1\rangle$, or in other words, $\langle a, b, c|aabcb^{-1}c^{-1} = 1\rangle$.
$H_1$ is precisely abelianization of this group, and the abelianization just attaches the relators $[a, b] = [b, c] = [a, c] = 1$ to the presentation, i.e., $\langle aabcb^{-1}c^{-1} = [a, b] = [b, c] = [a, c] = 1\rangle$. As $[b, c] = 1$, $aabcb^{-1}c^{-1}$ reduces to $aa$. Thus, your group is $\langle a, b, c | a^2 = [a, b] = [b, c] = [a, c] = 1\rangle$ which is isomorphic to $\Bbb Z/2\Bbb Z \oplus \Bbb Z^2$
Hence, $H_1(P^2 \# T) \cong \Bbb Z^2 \oplus \Bbb Z/2\Bbb Z$.
• Hey thanks, I understand. The crucial step I missed is that its an abelization and thus attaches the relators $[a,b] = [b,c] = [a,c] = 1$. – Endosred Jun 20 '15 at 14:59
• Maybe the same question, suppose I have the group $F_4/<acadbcb^{-1}d>$, what is the first homology group? I tried your method but it doesn't nicely result in any relationship. Other than $a^2c^2d^2=1$. That is $<a,b,c,d|a^2c^2d^2=[a,b]=[b,c]=[c,d]=[a,c]=[a,d]=1$> – Endosred Jun 20 '15 at 18:54
• @Endosred You can do a trick and avoid getting into the mess of word-manipulations. Note that $\Bbb RP^2 \# T$ is really homeomorphic to $\Bbb RP^2 \# \Bbb RP^2 \# \Bbb RP^2$. So $\pi_1(\Bbb RP^2 \# T)$ is equivalently $G = \langle a, b, c | a^2b^2c^2 = 1\rangle$ by an appropriate cell decomposition. Abelianization of this, as we proved, is $\Bbb Z^2 \oplus \Bbb Z/2\Bbb Z$. Now all you have to do is to note that the group you have wrote down is just $G^{ab} \times \Bbb Z$, which is in turn isomorphic to $\Bbb Z^3 \oplus \Bbb Z/2\Bbb Z$. – Balarka Sen Jun 20 '15 at 19:43
• Also : the relator $[b, d] = 1$ is missing from your presentation. – Balarka Sen Jun 20 '15 at 19:44
• @Endosred $G^{ab}$ is the abelianization of $G$. Note that your presentation $\langle a, b, c, d | a^2 c^2 d^2 = 1 = [a, b] = [a, c] = [a, d] = [b, c] = [b, d] = [c, d]\rangle$ is $\langle a, c, d, | a^2 c^2 d^2 = 1 = [a, c] = [a, d] = [c, d]\rangle \oplus \langle b \rangle$, which is precisely isomorphic to $(\Bbb Z^2 \oplus \Bbb Z/2\Bbb Z) \oplus \Bbb Z \cong \Bbb Z^3 \oplus \Bbb Z/2\Bbb Z$ by the logic I gave in the previous comment. – Balarka Sen Jun 21 '15 at 11:22