Presentation of discrete upper triangular group Let $G$ be the nilpotent Lie group consisting of matrices
$$
\begin{pmatrix}
1 & a_{12} & \cdots & a_{1,n}\\
0 & 1 & \ddots & \vdots\\
\vdots & \ddots & \ddots & a_{n-1,n}\\
0 & \cdots & 0 & 1
\end{pmatrix}
$$
where $a_{ij}\in\mathbb R$.
I would like to find a presentation of the group $\Gamma=G\cap\mathrm{GL}_n\mathbb Z$.
The entries on the superdiagonal play a crucial role, in that the $n-1$ matrices with a single $1$ on the superdiagonal ($1$s on the diagonal and $0$s elsewhere) generate all of $\Gamma$.
Trying to write down a presentation, I would start by choosing $n-1$ generators, $a_{12},\dotsc,a_{n-1,n}$ (named after the matrix entries). It is also clear that we need commutation relations, like
$$
[\cdots[[[a_{12},a_{23}],[a_{23},a_{34}]],\ldots]\cdots]=\cdots=e.
$$
(If $n=3$, this would just be $[[a_{12},a_{23}],a_{12}]=[[a_{12},a_{23}],a_{23}]=e$.)
I am just not sure, though, if I explicitly need to require that, say,
$$
[a_{12},a_{34}]=e,
$$
which seems to be directly related to the embedding I am thinking of. I would like a presentation in which any generators can be mapped to any of the matrices with a $1$ in the $(i,i+1)$ position. Is this possible?
 A: There is a paper by D. Biss and S. Dasgupta (A Presentation for the Unipotent Group over Rings
with Identity, J. Algebra, vol. 237, 691-707 (2001)) which gives a presentation of the group of upper triangular matrices over $R$ with diagonal entries equal to 1, where $R$ is any ring with identity. 
(Moreover, in some cases they show that their presentations are minimal both in the sense of having no redundant relations, and in the stronger sense of having the fewest possible number of generators and relations.) 
In the case $R=\mathbb{Z}$, their generators are the same as the ones in the question above, and the relations are $$[a_{i,i+1},a_{j,j+1}] =1\ \ (i<j-1\leq n-2)$$ 
$$[a_{i,i+1},[a_{i,i+1},a_{i+1,i+2}]]=[a_{i+1,i+2},[a_{i,i+1},a_{i+1,i+2}]]=1$$ $$[[a_{i,i+1},a_{i+1,i+2}],[a_{i+1,i+2},a_{i+2,i+3}]]=1.$$
A: I assume that when you are using $a_{ij}$ to denote an element of $\Gamma$, you mean the identity matrix with an extra 1 in the $(i,j)$ position.
It is easy to write down a presentation of $\Gamma$ on the $m := n(n-1)/2$ generators $a_{ij}$ with $i<j$. Just use the $m(m-1)/2$ commutator relations
$[a_{ij},a_{jk}] = a_{ik}$, and $[a_{ij},a_{kl}]=1$ when $j \ne k$, $i \le k$,
and $j<l$ when $i=k$. Of course many of these relations are redundant.
To get a presentation on the generators $a_{i,i+1}$, you can then use the equations $[a_{ij},a_{jk}] = a_{ik}$ to eliminate all of the other generators. Doing that with $n=3$ gives the presentation you have written down.
Although many relations in this presentation are redundant, those like $[a_{12},a_{34}]=1$ between the commuting generators $a_{i,i+1}$ are definitely not redundant, because if you left one of those out, then you could make the group bigger by putting that commutator equal to a new central generator.
