Cohomology of Heisenberg Group

Question

If we let $G :=$ the Heisenberg group over the integers, then I can show that the center of this group is isomorphic to the integers $\mathbb{Z}$ and that the abelianization of G is isomorphic to $\mathbb{Z} \times \mathbb{Z} = \mathbb{Z}^2$. Can I use these two facts to more easily compute $H^2(\mathbb{Z} \times \mathbb{Z}, \mathbb{Z})$? I am struggling to compute this second cohomology group and found hints to consider recasting the problem using the upper triangular matrices, but this doesn't seem to make the problem any easier.

Answer 1

The Heisenberg group over ${\mathbb Z}$ consists of the $3 \times 3$ upper unitriangular matrices over ${\mathbb Z}$. This group has the presentation $$G = \langle x,y,z \mid [x,y] = z, [x,z]=[y,z]=1 \rangle.$$ Note that the subgroup $Z = \langle z \rangle$ satisfies $Z \le [G,G] \cap Z(G)$, and $G/Z \cong {\mathbb Z} \times {\mathbb Z}$.

So ${\mathbb Z}$ is a quotient group of $H^2({\mathbb Z} \times {\mathbb Z})$.

On the other hand, suppose that $H$ is a group with subgroup $Y \le H$ with $Y \le [H,H] \cap Z(H)$ and $H/Y \cong {\mathbb Z} \times {\mathbb Z}$, and choose $u,v \in H$ such that $H = \langle u,v,Y \rangle$.

Then $[H,H]$ is generated by commutators $[u^iv^j,u^kv^l]$ with $i,j,k,l \in {\mathbb Z}$. But since these commutators are central, we have $[u^iv^j,u^kv^l] = [u,v]^{il-jk}$ and so $Y = \langle [u,v] \rangle$ is cyclic.

So $H^2({\mathbb Z} \times {\mathbb Z})$ is cyclic, and it follows that it is isomorphic to $\mathbb{Z}$.