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? 
 A: 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.
A: Hint: Let $(x,y,z)$ in the center and $(u,v,w)$ in $G$, you have $(x,y,z).(u,v, w)= (x+u,v+xw+y,z+w)$ they commute if and only if $v+xw+y=y+uz+v$ this implies that $xw=uz$,  consider $(0,v,w)$, we have $xw=0$ for every $w$, thus $x=0$, we deduce that $uz=0$ for every $u$ thus $z=0$. So the center is $(0,y,0), y\in R$.
To show that $Z/Z(G)$ is commutative, consider the map from $g:G/Z(G)\rightarrow R^2$ induced by the projection $f(x,y,z)=(x,z)$, $f$ is well defined, and the image of the class of $(x,y,z).(u,v,w)=x+u, v+xw+y,z+w)$ is the $(x+u,z+w)$. So $f$ is a morphism of group. Show that $f$ is bijective.
A: 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.
A: $G/Z(G)$ is nontrivial and abelian if and only if $G$ is $2$-step nilpotent. It is easy to see that $G$, which is the Heisenberg group, is $2$-step nilpotent.
