A question about matrices I was working on the McLain example about a characteristically simple group. 
McLain 
Unfortunaly I have no access to the original paper, so I was trying to make it out by myself. I found (on Google) that McLain use a division ring to create a group of automorphisms of a vector space which (I hope) has to be the characteristically simple one. Now I proved that this group has to be locally nilpotent and that every its element can be embedded in an unitriangular group (over the same division ring). 
So my question is the following one:
Are unitriangular matrices, over a division ring, nilpotent groups? How can I show that?
(Nilpotency will help me to make a further step towards the proof)
(I know that this works for unitriangular matrix over a field.)
(Obviously if you can tell to me some information on the McLain example it would be great too!)  
Thanks for the attention!
 A: The question might be a little misleading asMcLain's group is not nilpotent, and yet it is formed from unitriangular matrices.
McLain's group is not nilpotent. No non-abelian nilpotent group is characteristically simple since the derived subgroup is a characteristic subgroup. McLain's group is perfect, $G=[G,G]$, and centerless, $Z(G)=1$.
However, McLain's group is a group of unitriangular matrices over a field (or division ring if you'd like). The trick is that the matrices are not of finite size, so the explanation in Jim's answer does not apply to the whole of $G$, only to its finitely generated subgroups.
Let $R$ be a ring, and $I$ a totally ordered set. Define $\newcommand{\UT}{\operatorname{UT}}\UT(I,R)$ to be the subgroup of $\operatorname{Aut}(R^{(I)})$ generated by all $E_{i,j}(r)$ for $i<j$, $i,j \in I$, and $r \in R$, where $E_{i,j}(r)$ takes the basis vector $e_i$ to $e_i + r \cdot e_j$, and takes all other basis vectors $e_k$ to $e_k$.
Each $E_{i,j}(r)$ is a unipotent matrix, and any finitely generated subgroup lies inside a unitriangular matrix group of the sort described in Jim's answer, so is nilpotent. Hence $\UT(I,R)$ is always locally nilpotent.
If $I$ has the property that for any $i<j \in I$ there is some $k \in I$ with $i<k<j$, then $\UT(I,R)$ is perfect.
If $I$ has the property that for any $i<j, k < l \in I$ there is some permutation order-preserving $\pi$ of $I$ such that $\pi(i) = k$ and $\pi(j)=l$ and $R$ is a field, then $\UT(I,R)$ is characteristically simple.
I forget which hypotheses guarantee it is centerless, but clearly if $|I| > 2$ then $\UT(I,R)$ is not abelian, so if $I$ also satisfies the order-preserving permutation condition, then $Z(\UT(I,R))=1$ as the center is characteristic.
If $R$ has characteristic $p$, then $\UT(I,R)$ is a $p$-group. If $R$ is a domain of characteristic $0$, then $\UT(I,R)$ is torsion-free.
In particular, $\UT(\mathbb{Q},\mathbb{Q})$ is a characteristically simple, locally nilpotent, perfect, centerless, torsion-free group, and $\UT(\mathbb{Q},\mathbb{Z}/p\mathbb{Z})$ is a characteristically simple, locally finite, locally nilpotent, perfect, centerless, $p$-group. Neither is nilpotent.
A: The group of unitriangular matrices over any ring is nilpotent.  See here for example.
