Free nilpotent group on 2 generators of class 2. Classical presentation of Heisenberg group is [x,y,z|[x,y]=z, xz=zx, yz=zy]
https://pdfs.semanticscholar.org/276f/aef6c6b534f6058441a4f96d6260c5f32052.pdf
Here above on page 12 it is written that we can obtain presentation with 2 generators, and then Heisenberg group is free nilpotent group of class 2 on 2 generators. I don't understand how we obtain that 2-generator presentation. In answers, you can use 3x3 matrices instead of x,y,z for better answer. I need it to understand proofs of theorems 5.3 and 5.4.
 A: Your presentation for the Heisenberg group is a polycyclic presentation, but that doesn't stop there from being a presentation of a different sort, say on just 2 generators instead of three.
I suppose the missing remark you might be after is that if a group $G$ is generated by $X$ then the commutator subgroup $[G,G]$ is the normal closure of the set $\{ [x,y] | x,y \in X \}$.  Since presentations $\langle X \mid R\rangle$ only need $R$ to generate up to the normal closure, it is sufficient to write just this set.  In the case |X|=2 then $F_{2,2}=\langle x,y \mid $ class 2 $\rangle$ is the free class-2 group on 2 generators.  The commutator subgroup is generated by $\{ [x,x], [x,y], [y,y] \}$ or rather just by $[x,y]$.  In particular, $[x,y]$ is nontrivial, as you can see by the presentation below where it is nontrivial.  You could add a generator $z$ and a relation $z=[x,y]$ and you obtain what you had.
As matrices
$x=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
and
$y=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
So indeed
$[x,y]=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\neq 1$.
But all triple commutators are 1.  So this models the presentation $\langle x,y\mid $ class 2$\rangle$.
