Is $\begin{bmatrix} a &b \\0 &1 \end{bmatrix}$ a cyclic group

Question

I appreciate all the help I can get with this task.
$$
G=\left \{ \begin{bmatrix}
a &b \\0
&1
\end{bmatrix},a,b\in \mathbb{Z}_3, a\neq 0 \right \}
$$

1. Is G a cyclic group?

2. Does a subgroup with 3 elements exists?

All elements in G:
$$\begin{bmatrix}
1 &0 \\0
&1
\end{bmatrix},\begin{bmatrix}
1 &1 \\0
&1
\end{bmatrix},\begin{bmatrix}
1 &2 \\0
&1
\end{bmatrix},\begin{bmatrix}
2 &0 \\0
&1
\end{bmatrix},\begin{bmatrix}
2 &1 \\0
&1
\end{bmatrix},\begin{bmatrix}
2 &2 \\0
&1
\end{bmatrix}$$

What do I do now?

EDIT: Thanks for your comments.

3. Is this one subgroup of order 3? $a=1$ generates 3 matrices.
   $$\begin{bmatrix}
   1 &0 \\0
   &1
   \end{bmatrix},\begin{bmatrix}
   1 &1 \\0
   &1
   \end{bmatrix},\begin{bmatrix}
   1 &2 \\0
   &1
   \end{bmatrix}$$

From my book

A group is said to be cyclic if it contains an element x such that every member of G is a power of x.

4. Can $x$ be one of the six elements above?

EDIT: Thanks for all the help! I appreciate it very much.

Answer 1

A quicker way to see that $G$ is not cyclic: note that $$ \pmatrix{2 & 0\\0 & 1} \pmatrix{1 & 1\\0&1} = \pmatrix{2 & 2\\0 & 1}, \\ \pmatrix{1 & 1\\0 & 1} \pmatrix{2 & 0\\0 & 1} = \pmatrix{2 & 1\\0 & 1}. $$ That is, we have found elements $g,h \in G$ with $hg \neq gh$. Because $G$ is not abelian, it cannot be cyclic.