What's the significance of using a generator to prove a cyclic subgroup?

Question

I have the main group of a general linear group of order 3, over real numbers. How is it that the below is a cyclic subgroup?

$$H = \begin{bmatrix} 1 & 0 & a\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}, a \in integers$$

The point of the question is to understand the importance of generators, but how do I get the generator of a matrix? From there, I'm assuming I can simply use the two-step subgroup test to prove that it is a subgroup. Any idea on why it would be cyclic though?

Answer 1

This group is generated by $$\begin{pmatrix} 1&0&1 \\ 0&1&0 \\ 0&0&1 \end{pmatrix},$$ which has inverse $$\begin{pmatrix} 1&0&-1 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}.$$ This is the reason it is clearly cyclic --- you can explicitly give the generator.

Answer 2

A cyclic subgroup is the subgroup $\{ x^n \mid n \in \mathbb{Z} \}$ generated by one of its elements $x \in G$, so you would like to find this element.

If you have a matrix $X$ of your form, try calculating its powers $X^n$ for $n\in \mathbb{Z}$. What do you get?