Borel subgroup is self-normalizing in general linear group

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Borel subgroup in general linear group (?)) satisfying a particular subgroup property (namely, Self-normalizing subgroup (?)) in a particular group or type of group (namely, General linear group (?)).

Statement

Let $k$ be any field, and $GL_n(k)$ denote the 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 Self-normalizing subgroup (?) inside $GL_n(k)$.

Related facts

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. Further information: Sylow normalizer implies abnormal

Proof

Hands-on proof

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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 $A \in B_n(k)$.