Borel subgroup is self-normalizing in general linear group

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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 (?)).


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

Stronger facts

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 1s 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


Hands-on proof


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).