Show that the group is abelian Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix}$ with $x,y,z\in M$. 
I have shown that all the elements of the center $Z(G)$ are the matrices of the form $\begin{pmatrix}
1 & 0 & \tilde{y} \\ 
0 & 1 & 0 \\ 
0 & 0 & 1 
\end{pmatrix}$. 
How could I show that $G/Z(G)$ is abelian? 
We have that $$G/Z(G)=\{gZ(G)\mid g\in G\}$$ 
Do we have to take $A=g_1Z(G)$ and $B=g_2Z(G)$ and show that $AB=BA$ ? 
Or do we have to take the same $g$ just an other element of the center? 

Pick an arbitrary
$$g = \begin{pmatrix}
1 & x & y \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix} \in G$$
In the quotient $G/Z(G)$, $g$ is a representative of the coset
$$gZ(G) = \left\{ \begin{pmatrix}
1 & x & y+ \tilde{y} \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix}: \tilde{y} \in M \right\}$$
so you can choose another (more clever) representative for $gZ(G)$, namely
$$g' = \begin{pmatrix}
1 & x & 0 \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix}$$
(i.e. $gZ(G)=g'Z(G)$). This will make computations easier.
Now, pick two arbitrary matrices
$$a=\begin{pmatrix}
1 & x & 0 \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix} , b=\begin{pmatrix}
1 & x' & 0 \\ 
0 & 1 & z' \\ 
0 & 0 & 1 
\end{pmatrix}$$
and compute
$$ab= \begin{pmatrix}
1 & x+x' & xz' \\ 
0 & 1 & z+z' \\ 
0 & 0 & 1 
\end{pmatrix} , ba= \begin{pmatrix}
1 & x+x' & x'z \\ 
0 & 1 & z+z' \\ 
0 & 0 & 1 
\end{pmatrix}$$
it is easily verified that $abZ(G) = baZ(G)$. This is enough to conclude that $G/Z(G)$ is abelian.
