Borel subgroup is self-normalizing in general linear group

Statement
Let $$k$$ be any field, and $$GL_n(k)$$ denote the fact about::general linear group of $$n \times n$$ matrices over $$k$$. The Borel subgroup $$B_n(k)$$ is the subgroup of upper-triangular invertible $$n \times n$$ matrices. $$B_n(k)$$ is a fact about::self-normalizing subgroup inside $$GL_n(k)$$.

Stronger facts

 * Borel subgroup is abnormal in general linear group

Related facts

 * Bruhat decomposition theorem
 * Triangulability theorem

For finite fields
If $$k$$ is the finite field of order $$q$$ where $$q$$ is a power of a prime $$p$$, the Borel subgroup $$B_n(k)$$ equals the normalizer of a $$p$$-Sylow subgroup of $$GL_n(k)$$: the subgroup of upper-triangular matrices with $$1$$s on the diagonal. The normalizer of any Sylow subgroup in a finite group is self-normalizing; in fact, it is an abnormal subgroup.

Proof in terms of flags
The proof relies on a somewhat different interpretation of $$B_n(k)$$. Let $$e_1, e_2, \dots, e_n$$ denote the standard basis for $$k^n$$. Then, the standard flag is an ascending chain of subspaces:

$$V_0 \le V_1 \le V_2 \le \dots \le V_n$$

where the subspace $$V_j$$ is the span of the vectors $$e_1,e_2, \dots, e_j$$. Then, $$B_n(k)$$ is the subgroup of $$GL_n(k)$$ comprising those linear transformations that preserve this flag. In other words, $$A \in B_n(k)$$ if and only if the flag corresponding to the basis $$Ae_1, Ae_2, \dots, Ae_n$$ is the same as the flag corresponding to the basis $$e_1, e_2, \dots, e_n$$.

Now, given any $$A \in GL_n(k)$$, the subgroup $$AB_n(k)A^{-1}$$ is precisely the subgroup of $$GL_n(k)$$ that stabilizes the flag of $$Ae_1, Ae_2, \dots, Ae_n$$. Thus, for $$A$$ to normalize $$B_n(k)$$ we require that the linear transformations stabilizing the flag of $$e_1,e_2, \dots, e_n$$ are the same as the linear transformations of the flag $$Ae_1, Ae_2, \dots, Ae_n$$. We can now prove, through an inductive argument, that this forces the flags for the two bases to be the same, forcing $$A \in B_n(k)$$.