I am interested in calculating of $H_2(G,\mathbb{Z})$, where $G$ is a group given by presentation with two relators $\langle a,b| r_1 = r_2 = 1\rangle$.
Moreover, I am interested in such presentations, that $H_2(G,\mathbb{Z}) = \mathbb{Z}^2$, it would be great to learn how to construct a lot of such examples.
Best Answer
If $G$ is a perfect group (i.e. $G=[G,G]$, i.e. $G_{ab}=0$) which admits a presentation with two generators and two relations, then $H_2G=0$.
More generally, if $G$ is a group which admits a presentation with $n$ generators and $m$ relations then $H_2G$ can be generated by $m-n+\text{rk}_\mathbb{Z}(G_{ab})$ elements. This appears as Exercise II.5.5(b) in Ken Brown's group cohomology book. Here is my old solution:
With the presentation $G=\langle s_1,\cdots,s_n\,|\,r_1,\cdots,r_m\rangle$ we associate the $2$-complex $\;Y=(\bigvee_s S^1)\cup_{r_1}e^2\cdots\cup_{r_m}e^2\;$ so that $\pi_1Y\cong G$. By computing the Euler characteristic $\chi(Y)$ two different ways we obtain the equation $\sum(-1)^i\text{rk}_\mathbb{Z}(H_iY)=\sum(-1)^ic_i$, where $c_i$ is the number of $i$-cells. Then $1-\text{rk}_\mathbb{Z}(G_{ab})+\text{rk}_\mathbb{Z}(H_2Y)=1-n+m$ and so $\text{rk}_\mathbb{Z}(H_2Y)=m-n+r$, where $r=\text{rk}_\mathbb{Z}(G_{ab})=\dim_\mathbb{Q}(\mathbb{Q}\otimes G_{ab})$. Now $H_2Y=\text{ker}(\partial_2)$ is a free abelian group (subgroup of cellular 2-chain group), and by applying Theorem II.5.2[Brown] we get a surjection $H_2Y\rightarrow H_2G$ (from the exact sequence in that theorem -- this theorem is a play on the Hurewicz theorem). Thus $H_2G$ $\textit{can}$ be generated by $m-n+r$ elements. Now in my special scenario, $m-n+r=2-2+0=0$.