View source for Borel subgroup is self-normalizing in general linear group

{{subgroup property satisfaction|
group = general linear group|
subgroup = Borel subgroup in general linear group|
property = self-normalizing subgroup}}
==Statement==

Let <math>k</math> be any [[field]], and <math>GL_n(k)</math> denote the [[fact about::general linear group]] of <math>n \times n</math> matrices over <math>k</math>. The [[fact about::Borel subgroup in general linear group|Borel subgroup]] <math>B_n(k)</math> is the subgroup of upper-triangular invertible <math>n \times n</math> matrices. <math>B_n(k)</math> is a [[fact about::self-normalizing subgroup]] inside <math>GL_n(k)</math>.

==Related facts==

===Stronger facts===

* [[Borel subgroup is abnormal in general linear group]]

===Related facts===

* [[Bruhat decomposition theorem]]
* [[Triangulability theorem]]

===For finite fields===

If <math>k</math> is the finite field of order <math>q</math> where <math>q</math> is a power of a prime <math>p</math>, the Borel subgroup <math>B_n(k)</math> equals the normalizer of a <math>p</math>-Sylow subgroup of <math>GL_n(k)</math>: the subgroup of upper-triangular matrices with <math>1</math>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|[[Sylow normalizer implies abnormal]]}}

==Proof==

===Hands-on proof===

{{fillin}}

===Proof in terms of flags===

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

<math>V_0 \le V_1 \le V_2 \le \dots \le V_n</math>

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

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