Why is the group of unit upper triangular matrices solvable?

Question

Let $GL_n(k)$ be the $n$ by $n$ general linear group over $k$, $B_n(k)$ be the subgroup of $GL_n(k)$ consisting of all upper triangular matrices, and $U_n(k)$ be the subgroup of $B_n(k)$ whose diagonal elements are all $1$.

To show $B_n(k)$ is solvable, I'm proving it now by following steps:

1. $U_n(k)$ is a subgroup of $B_n(k)$. (done)
2. $U_n(k)$ is normal in $B_n(k)$. (done)
3. $U_n(k)$ is solvable. (question)
4. $B_n(k) / U_n(k)$ is also solvable. (not yet)
5. $B_n(k)$ is solvable. (by the below thm)

I'll use a theorem to verify $B_n(k)$ is solvable.

$G$ is solvable if and only if $H$ and $G/H$ are solvable for some normal subgroup $H$ of $G$.

So, I have to prove both step 3 and step 4. But I have no idea about them. How to prove them? Since my knowledge is not enough, I don't want to show them using Lie theory.

Thanks in advance.

Answer 1

If $A$ is an $n\times n$ matrix, I call the entries $a_{i,i+\ell}$ of $A$ the $\ell$-th diagonal of $A$. In particular, the $0$-th diagonal of $A$ is its actual diagonal.

You can prove that if $A,B \in U_n(k)$ and $C = [A,B]$ then the first diagonal of $C$ is $0$. More in general, if the first $\ell$ diagonals (except the $0$-th) of $A$ are $0$ and the first $m$ diagonals (except the $0$-th) of $B$ are $0$, then the first $\ell+m+1$ diagonals of $[A,B]$ are $0$.

In order to prove the statement about the quotient, you can observe that $B_n/U_n$ is indeed isomorphic to the group of diagonal matrices.

Answer 2

You can prove $U_n$ solvable by induction on $n$. Consider $H_n$, the group of matrices of the form $$ \begin{pmatrix} 1 & & & a_1 \\ & \ddots & & \vdots \\ & & 1 & a_{n-1} \\ & & & 1 \end{pmatrix} $$ It is then easy to see that $H_n \triangleleft U_n$, that $H_n$ is abelian (in fact isomorphic to $k^{n-1}$), and that $U_n/H_n\cong U_{n-1}$.