The quotient group of the Heisenberg group and its center is isomorphic to $(\mathbb{R}^2, +)$. 
Let $$ H = \left\{ \;\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 &
 1 \end{bmatrix} : x,y,z \in \mathbb{R}\right\}, \qquad Z(H) = \left\{
 \begin{bmatrix} 1 & 0 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} :
y \in \mathbb{R}\right\}, $$ then show that $H/Z(H) \cong (\mathbb{R}^2, +)$.

Proof: First I show that $Z(H) \cong (\mathbb{R}, +)$ using the isomorphic map $\phi(h) = y$ for $h \in Z(H)_y$. Then I show that $H \cong (\mathbb{R}^3, +)$ using the isomorphic map $\Phi(h) = (x, y, z)$ for $h \in H_{(x,y,z)}$. So we have
$$
H/Z(H) \cong (\mathbb{R}^3, +)/(\mathbb{R}, +).
$$
We can construct a homomorphic map $\varphi\left((x,y,z)\right) = x + y + z$ with $$\text{ker} \varphi = \{ x,y \in \mathbb{R}: z = -x -y\}$$ which is clearly $(\mathbb{R}^2, +)$ so by the first isomorphism theorem we have
$$
H/Z(H) \cong (\mathbb{R}^3, +)/(\mathbb{R}, +) \cong (\mathbb{R}^2, +)
$$
finishing the proof. Does this sort of iterative isomorphic map proof work? These concepts are new to me so I'm unsure.
 A: Your map $\Phi$ is not an isomorphism. For example let $h_1 \in H$ have coordinates $(x_1,y_1,z_1)=(1,0,0)$, and let $h_2$ have coordinates $(x_2,y_2,z_2) = (0,0,1)$. A short computation shows that $h_3 = h_1 h_2 \in H$ has $z_3=1$ whereas $h_4 = h_2 h_1 \in H$ has $z_4 = 0$. Therefore $h_1 h_2 \ne h_2 h_1$, whereas $\Phi(h_1) + \Phi(h_2) = \Phi(h_2)  + \Phi(h_1)$.
What you should do instead is to construct a map $\Phi : H \to \mathbb R^2$ (not $\mathbb R^3$) which you can prove is a homomorphism, is surjective, and has kernel $Z(H)$.
A: Here is the correct proof:
Proof: For $h \in H$ with
$$
h = \begin{bmatrix}
1 & x & y \\
0 & 1 & z\\
0 & 0 & 1
\end{bmatrix}
$$
we have $\phi : H \to (\mathbb{R}^2, +)$ with $\phi(h) = (x,z)$. Clearly $\phi$ is homomorphic as $\phi(h_1 h_2) = \phi(h_1) + \phi(h_2)$. Furthermore,
$$
\text{ker} \phi = \{ (x,z) : x = z = 0 \} = Z(H).
$$
Lastly we see that $\text{im} \phi = (\mathbb{R}^2, +)$ so by the first isomorphism theorem
$$
H/Z(H) \cong (\mathbb{R}^2, +).
$$
