$G$ is an extension of $\mathbb{Z}^2$ by $\mathbb{Z}$

Question

Show that $G=\mathbb{Z}^3$ is an extension of $\mathbb{Z}^2$ by $\mathbb{Z}$, where the binary operation of $G$ is

$$(x_1,y_1,z_1).(x_2,y_2,z_3)=(x_1+x_2,y_1+y_2,z_1+z_2+y_1x_2)$$

Is this extension split?

Here is what I have so far

$G$ is a non-abelian group with identity $e:=(0,0,0)$ and for all $a:=(x,y,z)\in G$, $a^{-1}=(-x,-y,-z+xy)$.

Now let $H’=\langle (0,0,1) \rangle$. For $(x,y,z)\in G$, we have

$g(0,0,1)^kg^{-1}=(x,y,z)(0,0,k)(-x,-y,-z+xy)=(x,y,z+k)(-x,-y,-z+xy)=(0,0,k)=(0,0,1)^k$

Hence $H’$ is a normal subgroup of $G$, and $H’$ is isomorphic to $\mathbb{Z}$. Furthermore, $K=\mathbb{Z}^2$ is isomorphic to $G/H’$.

However, I am a little bit stuck at determining wether it is split or not. I think that it is split using $K’=\langle (0,1,1) \rangle$, and $K’$ is isomorphic to $\mathbb{Z}^2$, but I am still not sure about it.

Definition of extension: Let $H,K,$ and $G$ be groups. We say that $G$ is an extension of $K$ by $H$ if there exists a normal subgroup $H’$ of $G$ and isomorphisms $H\cong H’$, $K\cong G/H’$. A split extension is such an extension, together with a subgroup $K’\le G$ such that $K’ \to G/H’, \ x\to xH’$ is an isomorphism.

Answer 1

This is a central extension, because the subgroup $\{(0,0,z) \mid z \in {\mathbb Z}\}$ is in the centre of $G$.

So if the extension was split, then the group would be isomorphic to $({\mathbb Z}^3,+)$, which is abelian. But the operation defined is not commutative, so the extension is not split.

This group is known as the Heisenberg group over ${\mathbb Z}$.

Answer 2

As noted in my comment, we can determine if the extension is split by determining if $G$ is a semidirect product of $\mathbb{Z}$ and $\mathbb{Z}^2.$ We can define the semidirect product $\mathbb{Z}^2\ltimes \mathbb{Z}$ as follows. The underlying set is $\mathbb{Z}^2\times \mathbb{Z},$ and the group operation is $$((x_1,y_1),z_1)\cdot ((x_2,y_2),z_2)=((x_1+x_2,y_1+y_2),z_1+\varphi(x_1,y_1)(z_2)),$$ where $\varphi: \mathbb{Z}^2\to \text{Aut}(\mathbb{Z})\cong \mathbb{Z}_2$ is a homomorphism. However, note that this forces $\varphi(x_1,y_1)(z_2)=\pm z_2$ which both result in groups isomorphic to the direct product, $(\mathbb{Z}^3,+).$ Thus, $G$ is not a semidirect product and therefore the extension is not split.