Central extension of the Discrete Heisenberg group $H_3(\Bbb Z)$ I want to use the  Discrete Heisenberg group $(H_3(\Bbb Z),\times)$ as an example for a presentation on central extensions.
$H_3(\Bbb Z) = \begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix}x,y,z\in \Bbb Z$
Now I have shown that the center of $H_3(\Bbb Z)$ is $Z = \begin{bmatrix}1&0&z\\0&1&0\\0&0&1 \end{bmatrix}$

I want a short exact sequence:
$$I\hookrightarrow Z \hookrightarrow H_3(\Bbb Z) \twoheadrightarrow X\twoheadrightarrow I$$
Now for this to work, I need $X\cong H_3(\Bbb Z) / Z$ I believe, but I don't fully understand quotient groups apparently. Now I know that I can take this quotient group, since $Z$ is normal of course and I believe that we will get $$X= \begin{bmatrix}1&x&0\\0&1&y\\0&0&1 \end{bmatrix}$$
I can also see that $$Z^{-1}\begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix}=\begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix}Z^{-1}=X$$
Where $Z^{-1} = \begin{bmatrix}1&0&-z\\0&1&0\\0&0&1 \end{bmatrix}$

Does my guessed $X$ have $X\cong  H_3(\Bbb Z) / Z$

Can I let my maps be explicitly:
$$I\overset{\phi_1}{\hookrightarrow} Z \overset{\phi_2}{\hookrightarrow} H_3(\Bbb Z) \overset{\phi_3}{\twoheadrightarrow} X\overset{\phi_4}{\twoheadrightarrow} I$$
$$\phi_1:A\mapsto A$$$$\phi_2:A\mapsto A$$$$\phi_3:A\mapsto Z^{-1}A$$$$\phi_4:A\mapsto I$$
 A: $\renewcommand{\phi}{\varphi}$$\newcommand{\Z}[0]{\mathbb{Z}}$$X$ will be (isomorphic to) the group $\Z \times \Z$. Your maps will be
$$
\phi_{2} : z \mapsto \begin{bmatrix}1&0&z\\0&1&0\\0&0&1 \end{bmatrix}
$$
and
$$
\phi_{3} : \begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix} \mapsto (x, y).
$$
The facts that $\phi_2$ is a homomorphism, and that $\phi_{3}$ is a (surjective) homomorphism follow from
$$
\begin{bmatrix}1&x_1&z_1\\0&1&y_1\\0&0&1 \end{bmatrix}
\cdot
\begin{bmatrix}1&x_2&z_2\\0&1&y_2\\0&0&1 \end{bmatrix}
=
\begin{bmatrix}1&x_1 + x_2&z_2 + x_1 y_2 + z_1\\0&1&y_1+y_2\\0&0&1 \end{bmatrix}.
$$
Moreover it is clear that 
$$
\ker(\phi_3)
=
\left\{ \begin{bmatrix}1&0&z\\0&1&0\\0&0&1 \end{bmatrix} : z \in \Z \right\}
=
\text{image of $\phi_2$},
$$
which gives you exactness.

Note that I'm taking the sequence as
$$I\overset{\phi_1}{\hookrightarrow} \mathbb{Z} \overset{\phi_2}{\hookrightarrow} H_3(\Bbb Z) \overset{\phi_3}{\twoheadrightarrow} \mathbb{Z} \times \mathbb{Z} \overset{\phi_4}{\twoheadrightarrow} I,$$
and that there's only one choice for $\phi_1, \phi_4$.
