Let $H_3(\mathbf{Z})$ be the set of all matrices of the form
\begin{equation}
\begin{pmatrix}
1&a&c\\
0&1&b\\
0&0&1\\
\end{pmatrix}
\end{equation}with $a,b,c\in\mathbf{Z}$ and matrix multiplication as the binary operation.Prove that $H_3(\mathbf{Z})$ is a nonabelian group.This group is called the Heisenberg group.
Proof:
\begin{equation}
\begin{pmatrix}
1&a_1&c_1\\
0&1&b_1\\
0&0&1\\
\end{pmatrix}\begin{pmatrix}
1&a_2&c_2\\
0&1&b_2\\
0&0&1\\
\end{pmatrix}=\begin{pmatrix}
1&a_2+a_1&c_2+a_1b_2+c_1\\
0&1&b_2+b_1\\
0&0&1\\\end{pmatrix}\end{equation}
So the matrix multiplication is a binary operation.The identity element is
\begin{equation}
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
\end{equation}
The inverse element of
\begin{equation}\begin{pmatrix}
1&a_1&c_1\\
0&1&b_1\\
0&0&1\\
\end{pmatrix}
\end{equation}is
\begin{equation}
\begin{pmatrix}
1&-a_1&b_1a_1-c_1\\
0&1&-b_1\\
0&0&1\\
\end{pmatrix}
\end{equation}