I know that the Heisenberg group { x,y | [x,[x,y]]=[y,[x,y]]=1 } is free nilpotent; what 
about the higher dimensional ones? Do the higer dimensional Heisenberg groups have nice presentations? By higher dimensional Heisenberg groups I mean nxn upper triangular matrices with integral entries. Thanks.



Are higher dimensional Heisenberg groups free nilpotent?
Let $U\_n$ be the group of upper triangular integer matrices of size $n$ by $n$ with ones on the diagonal. Then $U\_n$ is generated by the elements $x\_i$, $i=1,...,n-1$, with Serre relations 
$$
[x\_i,[x\_i,x\_{i+1}]]=[x\_{i+1},[x\_i,x\_{i+1}]]=1,
$$
and $[x\_i,x\_j]=1$ if $|i-j|\ge 2$ (EDIT: one needs additional relations, see below). Indeed, the group $G\_n$ generated in this way maps surjectively to $U\_n$ by $x\_i\mapsto 1+E\_{i,i+1}$, and one can check that this map is injective (EDIT: with additional relations) by writing every element of $G\_n$ as an ordered product of powers of $x\_{ij}:=[x\_i[x\_{i+1}...x\_j]]$, $i\le j$. The corresponding groups over $\Bbb Z/m\Bbb Z$ are then obtained by adding the relations $x\_{ij}^m=1$. 


EDIT: As Jack kindly pointed out, I erroneously ignored the treacherous 2-torsion. 
According to Theorem 1 of the paper 


D. Biss, S. Dasgupta, "A presentation for the unipotent groups over rings with $1$",
J. of Algebra, v. 237, pp.691-707, 2001,


the above statements are correct over $\Bbb Z/m\Bbb Z$ when $m$ is an odd integer. 
In general (i.e. for any integer $m$, including $m=2$ and $m=0$), 
it suffices to add the relation 
$$
[[x\_i,x\_{i+1}],[x\_{i+1},x\_{i+2}]]=1
$$
for $i=1,...,n-3$. 
Moreover, over $\Bbb Z/m\Bbb Z$ with $m\ne 0$ it is enough to impose the relations $x\_i^m=1$ (the relations $x\_{ij}^m=1$ will then follow). 


