This time I want to show that $H(R)$ ( the heisenberg group attached to a conmutative ring $R$ is nilpotent). But I dont know how to proceed, I was thinking in first serch the normal subgroups if the Heisenberg group and then try to figure out the quotient group of that, from there get the centre, etc, but I dont think it would be a good idea (In fact I thought to use Gap, but I dont know how to ask this questions in Sage), Is there a better way to do this? Thanks a lot.
For me
$H(R) = \left \{ \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}: \ \ a,b,c \in R \right \}$
For a conmutative ring $R$
Hint: The Heisenberg group $H(R)$ is the subgroup $$\begin{bmatrix} 1 & \ast & \ast \\ 0 & 1 & \ast \\ 0 & 0 & 1 \end{bmatrix}$$ of $\mathrm{GL}_3(R)$.
You should start by just taking the commutator of two such elements and seeing what it looks like. It should be pretty easy to identify $[H(R), H(R)]$ and then take a guess at what $[H(R), [H(R), H(R)]]$ is.