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? 

The multiplication is
$$
\begin{pmatrix}
1 & a & b \\ 
0 & 1 & c \\ 
0 & 0 & 1 
\end{pmatrix}
\begin{pmatrix}
1 & x & y \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix}
=
\begin{pmatrix}
1 & x+a & y+az+b \\ 
0 & 1 & z+c \\ 
0 & 0 & 1 
\end{pmatrix}
$$
Similarly
$$
\begin{pmatrix}
1 & x & y \\ 
0 & 1 & z \\ 
0 & 0 & 1 
\end{pmatrix}
\begin{pmatrix}
1 & a & b \\ 
0 & 1 & c \\ 
0 & 0 & 1 
\end{pmatrix}
=
\begin{pmatrix}
1 & a+x & b+xc+y \\ 
0 & 1 & c+z \\ 
0 & 0 & 1 
\end{pmatrix}
$$
Denote by $\mu(a,b,c)$ the matrix
$$
\begin{pmatrix}
1 & a & b \\ 
0 & 1 & c \\ 
0 & 0 & 1 
\end{pmatrix}
$$
Saying that $\mu(a,b,c)$ is in the center is the same as saying that, for every $x,y,z$,
$$
y+az+b=b+xc+y
$$
that is, $az=xc$. Taking $z=1$ and $x=0$ we have $a=0$; with $z=0$ and $x=1$ we get $c=0$. It follows easily that the center consists of the matrices of the form $\mu(0,b,0)$.
Now consider the map $f\colon G\to M^2$ (considering $M^2$ a group with respect to componentwise addition) defined by
$$
f(\mu(x,y,z))=(x,z)
$$
The computation above shows that
$$
f(\mu(a,b,c)\mu(x,y,z))=(a+x,c+z)=f(\mu(a,b,c))+f(\mu(x,y,z))
$$
so $f$ is a homomorphism, obviously surjective. It's also clear that $\ker f=Z(G)$. By the isomorphism theorem,
$$
G/Z(G)=G/\ker f\cong M^2
$$
is abelian.
