Matrix representation of Heisenberg group Equiped with the law  $(a,b,c)\circ (a',b',c')  = (a + a', b + b', c + c' + ab'),$ the matrix representation of the Heisenberg group $H^3$ is given by 
$$
\begin{pmatrix}
 1 & a & c\\
 0 & 1 & b\\
 0 & 0 & 1\\
\end{pmatrix}; \quad 
a,b,c\in \mathbb{R}.
$$
My question, what  the matrix representation of $H^3$, if the law in $H^3$ is given by 
$$(x,y,z) . (x',y',z')  = (x + x', y + y', z + z' + xy'-yx').$$
thank you in advance
 A: I think the answer is as follows, we have:


*

*$$ \begin{pmatrix}
 1 & a & c\\
 0 & 1 & b\\
 0 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
 1 & a' & c'\\
 0 & 1 & b'\\
 0 & 0 & 1\\
\end{pmatrix}=
\begin{pmatrix}
 1 & a+a' & c+c'+ab'\\
 0 & 1 & b+b'\\
 0 & 0 & 1\\
\end{pmatrix}  $$
is corresponds to $(a,b,c)\circ (a',b',c') = (a + a', b + b', c + c' + ab')$.


. And $$ \begin{pmatrix}
 1 & 0 & 0 & y \\
 x & 1 & b & z \\
 0 & 0 & 1 & - x\\
0 & 0 & 0 & 1 \\ 
\end{pmatrix} 
\begin{pmatrix}
 1 & 0 & 0 & y' \\
 x' & 1 & y' & z' \\
 0 & 0 & 1 & - x'\\
0 & 0 & 0 & 1 \\ 
\end{pmatrix}
=\begin{pmatrix}
 1 & 0 & 0 & y+y' \\
 x+x' & 1 & y+y' & z + z' + xy'-yx' \\
 0 & 0 & 1 & - x-x'\\
0 & 0 & 0 & 1 \\ 
\end{pmatrix}$$
is corresponds to $(x,y,z).(x',y',z')=(x + x', y + y', z + z' + xy'-yx')$.
It follows that, the matrix representation of $H^3$, if the law in $H^3$ is $(x,y,z) . (x',y',z')  = (x + x', y + y', z + z' + xy'-yx')$, is given by 
$$\begin{pmatrix}
 1 & 0 & 0 & y \\
 x & 1 & y & z  \\
 0 & 0 & 1 & - x\\
0 & 0 & 0 & 1 \\ 
\end{pmatrix}, \quad x,y,z \in \mathbb R$$
thank you for any remark or comment.
