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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27
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Complete flags are G/B

Statement

Let n≥1 and let q be a prime power. Put G=GL⁡n(Fq) and let V=Fqn. Then:

  1. G acts on the set X of complete flags 0=F0<F1<⋯<Fn=V with dim⁡FqFi=i by g⋅F:=( g(F0),…,g(Fn) );
  2. the standard flag V∙ with Vi=⟨e1,…,ei⟩, where e1,…,en is the standard basis of V, is a complete flag whose stabiliser is the standard Borel subgroup B of Standard subgroups of finite general linear groups;
  3. consequently gB↦g⋅V∙ is a G-equivariant bijection G/B→X, so the complete flags are in bijection with the left cosets of B.

Facts & Assumptions

Given: An integer n≥1, a prime power q, the group G=GL⁡n(Fq) acting by matrix-vector multiplication on V=Fqn, the standard basis e1,…,en of V, and the set X of complete flags of V.

[L1]

B={ b∈G:b is upper triangular }, T≤B, U≤B, T∩U={In} and B=T⋉U; T, U and B are the standard torus, the standard maximal unipotent subgroup and the standard Borel subgroup of G, with U⊴B (Standard subgroups of finite general linear groups).

[L2]

The standard flag of V=Fqn is the chain V0:={0}⊊V1:=⟨e1⟩⊊⋯⊊Vn:=⟨e1,…,en⟩=V of standard coordinate subspaces, and each Vi has dim⁡FqVi=i with Vi−1⊊Vi for 1≤i≤n (Standard subgroups of finite general linear groups).

[L3]

A matrix A=(aij)∈Mn(Fq) is upper triangular when aij=0 for i>j (Upper triangular, lower triangular and diagonal square matrices over a commutative ring).

[L4]

A left action of G on a set X is a map G×X→X, (g,x)↦g⋅x, with e⋅x=x and (gh)⋅x=g⋅(h⋅x) for all g,h∈G, x∈X; the action is transitive when every x,y∈X satisfy g⋅x=y for some g∈G (Left group actions, transitive actions, and faithful actions).

[L5]

If X is a transitive G-set and x∈X, the orbit map G/Gx→X, gGx↦g⋅x, is an equivariant isomorphism from the left-coset action to the given action (Every transitive G-set is equivariantly isomorphic to G/Gx for any chosen point x).

[L6]

For A∈Mn(F) the following are equivalent: A is invertible; and x↦Ax is a linear isomorphism (Invertible matrix theorem: invertibility, full pivot rank, RREF I, trivial nullspace and unique solvability are equivalent).

[L7]

For an ordered basis b1,…,bm of a vector space and a linear map T, the j-th column of the matrix of T is the coordinate column of T(bj); conversely a matrix with prescribed columns defines the linear map sending bj to the corresponding vector (Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

[L8]

⟨e1,…,ei⟩ has e1,…,ei as a basis, so every g∈G satisfies g(⟨e1,…,ei⟩)=⟨g(e1),…,g(ei)⟩ and dim⁡Fqg(⟨e1,…,ei⟩)=i (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis, Linear subspace of a vector space, Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T).

Proof

technique · direct
1.1

The map G×X→X, (g,F∙)↦g⋅F∙ with (g⋅F)i:=g(Fi), is a well-defined left action: for g∈G and F∙∈X each g(Fi) is a linear subspace with g(F0)={0} and g(Fn)=V, and by [L6] the map x↦gx is injective, so restricting it to Fi and applying [L8] gives dim⁡Fqg(Fi)=dim⁡FqFi=i; dimensions therefore increase by one at each step, so g(Fi−1)⊊g(Fi) and g⋅F∙∈X. Moreover Inx=x and (gh)x=g(hx) for all x∈V, so In⋅F∙=F∙ and (gh)⋅F∙=g⋅(h⋅F∙); thus X is a G-set.

L4L6L8given
1.2

The standard flag V∙ lies in X: by [L2] each Vi=⟨e1,…,ei⟩ is a linear subspace of V with V0={0}, Vn=V, dim⁡FqVi=i and Vi−1⊊Vi.

L2given
1.3

The stabiliser of V∙ is B. Let g∈G have matrix A=(aij) in the standard basis, so that g(ei)=∑kakiek and g(ei)∈Vi if and only if aki=0 for every k>i, that is, if and only if A is upper triangular by [L3]. If g⋅V∙=V∙ then g(ei)∈g(Vi)=Vi for every i; conversely, if g(ei)∈Vi for every i, then g(Vi)=⟨g(e1),…,g(ei)⟩⊆Vi by [L8], and both spaces have dimension i by [L8] and [L2], so g(Vi)=Vi by [L10]. Hence g stabilises V∙ if and only if A is upper triangular, that is, if and only if g∈B by [L1], so GV∙=B.

L1L2L3L7L8L10given
1.4

Let F∙∈X be an arbitrary complete flag. Since Fi−1⊊Fi, for each 1≤i≤n there is a vector vi∈Fi∖Fi−1.

constructgiven
2.1

The list v1,…,vi is a basis of Fi for every i, by induction on i: v1≠0 spans the line F1; and if v1,…,vi−1 is a basis of Fi−1, then Fi−1+⟨vi⟩⊆Fi has dimension (i−1)+1−dim⁡(Fi−1∩⟨vi⟩)=i by [L9], because vi∉Fi−1 forces Fi−1∩⟨vi⟩={0}, while dim⁡FqFi=i; hence Fi−1+⟨vi⟩=Fi by [L10], so v1,…,vi spans Fi, and it is linearly independent because a relation ∑j≤icjvj=0 with ci≠0 expresses vi as an element of span⁡(v1,…,vi−1)=Fi−1, contrary to the choice of vi, while ci=0 reduces the relation to the independent list v1,…,vi−1. In particular v1,…,vn is a basis of Fn=V.

step 1.4L9L10
3.1

Let g:V→V be the linear map with g(ei)=vi for all i; by [L7] it has a unique matrix in the standard basis. It carries the basis e1,…,en of V onto the basis v1,…,vn of V from step 2.1, hence is bijective, so its matrix is invertible by [L6] and g∈G. Moreover, for every i, the members Vi of the standard flag of [L2] satisfy g(Vi)=⟨g(e1),…,g(ei)⟩=⟨v1,…,vi⟩=Fi by [L8], the last equality by step 2.1, so g⋅V∙=F∙.

step 2.1L2L6L7L8
4.1

By step 3.1 every F∙∈X has the form g⋅V∙ for some g∈G, so the action of step 1.1 is transitive in the sense of [L4]; with GV∙=B by step 1.3, the orbit map of [L5], applied to the transitive G-set X and the point V∙, is a G-equivariant bijection G/B→X, gB↦g⋅V∙. This is the asserted bijection between complete flags and left cosets of B. ∎

step 1.1step 1.3step 3.1L4L5

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