TITLE: Why is the Borel Subgroup self normalizing?
QUESTION [5 upvotes]: Let $B = B_n(\mathbb{F})$ be the set of upper triangular matrices of order $n$ with entries from $\mathbb{F}$. I am supposed to be showing $N_G(B) = B$
I have been told that what we need to notice here is that the stabilizer of the standard flag (when considered with its natural action due to $GL_n(\mathbb{F})$ is $B$ and we are supposed to be using this fact here, somehow.
My attempt was to try picking an element from the normalizer and showing that it necessarily stabilizes the standard flag, and hence deduce that the elements in the normalizer are in fact in the stabilizer, which is $B$. However, I couldn't succeed with this.
[I have no idea of why flags or Borel groups are considered in general. So, please post an answer which is approachable for a person with no knowledge of Algebraic Groups.]

REPLY [3 votes]: Alternative proof: $G=GL_{\mathbb F}(V)$ acts on the set of (complete) flags $(V_i)_i$ (i.e., the $V_i$ are subspaces of $V$ with $\dim V_i = i$ and $V_i\le V_{i+1}$ where $V_0=0$ and $V_n=V$) and the Borel subgroup $B$ is the stabilizer of the standard flag.
For $v_i\in V_i\setminus V_{i-1}$ the smallest $B$-invariant subspace containing $v_i$ is $V_i$: Picking $v_j\in V_j$ for $j\in\{1,\dots, i-1, i+1, \dots n\}$ gives a base of $V$, so for arbitrary $v\in V_{i-1}$ one can define a linear map $\lambda_v$ sending $v_i$ to $v_i+v$ and fixing all other $v_j$ ($j\ne i$). Clearly $\lambda_v$ stabilizes all $V_i$, i.e., $\lambda_v\in B$. Hence any $B$-invariant subspace of $V$ containing $v_i$ contains also $\lambda_v(v_i)-v_i = v$ for arbitrary $v\in V_{i-1}$, hence it contains also $\langle v_i, V_{i-1}\rangle = V_i$ proving the claim.
Any $B$-invariant subspace $W$ of $V$ of dimension $i$ equals $V_i$ as otherwise it contains some $w\in W\setminus V$. For $j$ minimal with $w\in V_j$ (hence $j>i$) we get by the last paragraph $V_j\le W$ contradicting $\dim V_j = j > i = \dim W$.
This implies that the standard flag is the only (complete) flag stabilized by $B$, i.e., the set of fixed points of $B$ consists of only one element. As the normalizer $N := N_G(B)$ of $B$ acts on the set of fixed points of $B$ (taking $n\in N, b\in B$ and a fixed point $x$ of $B$ one gets $bnx = n(n^{-1}bn)x = nb'x = nx$ as $b':=n^{-1}bn\in B$), it fixes the standard flag, hence $N$ is contained in $B$.