Let $G$ be the nilpotent Lie group consisting of matrices
$$
\begin{pmatrix}
1 &amp; a_{12} &amp; \cdots &amp; a_{1,n}\\
0 &amp; 1 &amp; \ddots &amp; \vdots\\
\vdots &amp; \ddots &amp; \ddots &amp; a_{n-1,n}\\
0 &amp; \cdots &amp; 0 &amp; 1
\end{pmatrix}
$$
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?


Question: Is it possible to have a convergent subsequence of a divergent sequence?
Answer: Sure. Consider $0, 1, 0, 1, 0, 1, \dots$

Furthermore, the Bolzano-Weierstrass Theorem says that every bounded sequence has a convergent subsequence.