The following question has already appeared on mathstack:
If $B$ is the subgroup of ${\rm GL}(n,F)$ consisting of upper triangular matrices then normalizer of $B$ in ${\rm GL}(n,F)$ is $B$ itself.
I know a proof of this using Bruhat decomposition of ${\rm GL}(n,F)$.
Question: Can we prove above theorem without using Bruhat decomposition?
Why came to this question: Consider the general linear Lie algebra $L=\mathfrak{gl}(n,F)$; in it, let $T=\mathfrak{t}(n,F)$ be the upper triangular sub-algebra. Then normalizer of $T$ in $L$ is $T$ itslef, and this can be proved just by considering a very simple decomposition of ${\mathfrak gl}(n,F)$: write any element as sum of upper triangular matrix and lower triangular matrix whose diagonal is $0$.
But then for problem above, is it necessary to use Bruhat decomposition?
Best Answer
$\DeclareMathOperator{\GL}{GL}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Let $e_0, e_1, \dots, e_{n-1}$ be a basis with respect to which $B$ is upper-triangular, and write $$ V_i = \Span{ e_j : j \ge i}. $$ Allow me to use row vectors, so that the group $G = \GL(n, F)$ acts on the right.
Then $$ B = \Set{b \in G : V_i b \subseteq V_i \text{ for each $i$}}. $$ Moreover,
This is proved by induction on $i$. For $i = 1$ we have that $V_{n-1}$ is the unique common eigenspace for the elements of $B$ (just consider the element of $B$ which is a single Jordan block of size $n \times n$ and eigenvalue $1$, say), then pass to $V / V_{n-1}$ and use induction.
Let $g \in N_{G}(B)$. Then for each $b \in B$ we have $g b g^{-1} \in B$, that is for all $i$ $$ V_{i} g b g^{-1} \subseteq V_i $$ or $$ (V_{i} g) b \subseteq V_i g. $$ It follows from the above that $V_{i} g = V_{i}$, so that $g \in B$.