Are higher dimensional Heisenberg groups free nilpotent? 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.
 A: The higher dimensional Heisenberg groups are the matrix groups G(V,R,f) = { [ 1, x, z ; 0, 1, y ; 0, 0, 1] : x,y in V, z in R } where V is an R-module, R is a ring, and f is an alternating R-bilinear form on V.  Typically V is a free R-module, and R is a field or the integers, and f is a block diagonal matrix with blocks [0,1;-1,0].  Sometimes V is just called a symplectic space over R.
The group has a normal series 1 ≤ ⟨ z: z in R ⟩ ≤ ⟨ y,z : y in V, z in R ⟩ ≤ G.  This gives a very nice presentation of the group as ⟨ x,y,z : [x,y] = f(x,y), [x,z] = [y,z] = 1 ⟩ where the relations from V (for various "x", and also for various "y") and R (for various "z") are implicitly added.
For instance taking V free of rank 2 over R=Z, you get ⟨ x1, x2, y1, y2, z : [ x1, y1 ] = [ x2, y2 ] = z, [ x1, x2 ] = [ y1, y2 ] = [ z, x1 ] = [ z, x2 ] = [ z, y1 ] = [ z, y2 ] = [ x1, y2 ] = [ x2, y1 ] = 1 ⟩.  This is a very pretty presentation, but is not usually "free".  Way too many things commute.
However, these types of groups are very important.  When R=Z/pZ, these are the extra-special groups of exponent p.  Rather than "free products" they are "central products".
These form reasonably nice examples of just-non-abelian groups, since V can be a fairly arbitrary R-module as long as R is Z/pZ.
A: 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). 
