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?

  • $\begingroup$ For generic $a,b,c$? I don't think it would be cyclic in general. $\endgroup$ – Vim Oct 6 '16 at 13:33
  • $\begingroup$ @Vim Is there a way to disprove it? I'm guessing proof by contradiction would work here, right? $\endgroup$ – Andrew Raleigh Oct 6 '16 at 13:56
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Oct 6 '16 at 15:24

Easiest way to prove this group is not generated by single element is to show it is NOT abelian. Take two specific matrices in this format and multiply them in two different ways to see.

  • $\begingroup$ Should I use any real numbers for it, or would it be best to try 0/1? $\endgroup$ – Andrew Raleigh Oct 6 '16 at 14:38
  • $\begingroup$ @AndrewRaleigh just start off with some random numbers. It's highly unlikely you will get a commutative pair. $\endgroup$ – Vim Oct 6 '16 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.