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

Question

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.

Answer 1

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, $H_1(P^2 \# T) \cong \Bbb Z^2 \oplus \Bbb Z/2\Bbb Z$.