Take any two elements \(\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H\). Then,
\[\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1+a_2 & b_1+b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\]
Since, \(a_1+a_2\) and \(b_1+b_2\) should be congruent to \(0, 1\) or \(2\) under modulo \(3\) we have, \(a_1+a_2,\,b_1+b_2\in\mathbb{Z}_3\). Therefore,
\[\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H\]
Hence \(H\) is closed under matrix multiplication.
Since matrix multiplication is associative \(H\) under matrix multiplication also satisfies associativity.
Note that for each \(\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H\) we get,
\[\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\]
Therefore the identity element of \(H\) is, \(\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\).
Also for each, \(\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H\) we have,
\[\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & -a & -b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\]
Therefore an inverse exists for each element of \(H\).
Finally,
\[\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1+a_2 & b_1+b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\]
Hence \(H\) is Abelian.
Since we have three choices for both \(a\) and \(b\) (0, 1 and 2) there are \(3\times 3=9\) elements in the group \(H\). By the Fundamental Theorem of Finite Abelian Groups we have \(H\cong \mathbb{Z}_3\oplus\mathbb{Z}_3\).
