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 step3 and step4. But I have no idea about them. How to prove? Since my knowledge is not enough, I don't wanna show using Lie theory.
Thanks in advance.
Best Answer
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 $U_n/H_n\cong U_{n-1}$.