# Process behind proving a matrix subgroup is a cyclic subgroup?

I have this matrix below, which is a subgroup of the general linear group with order 3 and over real numbers. I know it is a subgroup, but how can I tell that it is cyclic?

$$K = \begin{bmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{bmatrix} | \ a, b, c \in real \ numbers$$

I know that a subgroup is the subgroup $\{ x^n \mid n \in \mathbb{Z} \}$ generated by one of its elements $x \in G$. Would proving that it is cyclic involve multiplying it by itself?

• For generic $a,b,c$? I don't think it would be cyclic in general. – Vim Oct 6 '16 at 13:33
• @Vim Is there a way to disprove it? I'm guessing proof by contradiction would work here, right? – Andrew Raleigh Oct 6 '16 at 13:56
• Take $(a,b,c)=(1,0,0)$ for $K_1$ and $(a,b,c)=(0,1,0)$ for $K_2$. The Heisenberg relation says that the group is non-abelian, and $2$-step nilpotent. In particular, it is not cyclic. – Dietrich Burde Oct 6 '16 at 15:24