How can we derive the generators of the Discrete Heisnberg Group?

Everyone seems to just state this as a given and never actually derive it from scratch.

I'm looking for a (somewhat) elementary derivation

  • 2
    $\begingroup$ The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation. $\endgroup$ – Lorenzo Dec 18 '18 at 15:52
  • $\begingroup$ I have no idea how i missed that, but still a derivation from scratch would be nice $\endgroup$ – user371732 Dec 18 '18 at 16:13

Since the discrete Heisenberg group is defined to be the subgroup of $GL_3(\Bbb{Z})$ consisiting of upper-unitriangular matrices, it is clear that the generators are given by $x,y,z$, where $$ \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}=y^bz^cx^a\, $$ see Wikipedia. Here it is enough to consider $x$ and $y$ since $z=[x,y]$ by matrix multiplication.


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