Orthogonal group
label: 正交群 @zh
label: 직교군 @ko
label: 直交群 @ja
label: Ортогональна група @uk
label: Orthogonale groep @nl
label: زمرة متعامدة @ar
label: Grupo ortogonal @es
label: Orthogonale Gruppe @de
label: Groupe orthogonal @fr
label: Ortogonální grupa @cs
label: Ортогональная группа @ru
label: Gruppo ortogonale @it
label: Grup ortogonal @ca
label: Ortogonalgrupp @sv
description: groupe des automorphismes laissant invariant un produit scalaire et sa norme associée @fr
description: mathematical group consisting of transformations of a Euclidean space which preserve distance and a fixed point
description: grupo matemático formado por las transformaciones de un espacio euclidiano que conservan la distancia y un punto fijo @es
description: Тип группы математики @ru
description: mathematische Gruppe bestehend aus Transformationen eines euklidischen Raumes, die Distanz und einen festen Punkt bewahren @de
subject: Category:Lie groups
subject: Category:Quadratic forms
subject: Category:Euclidean symmetries
subject: Category:Linear algebraic groups
drop: hidden
date: November 2019
date: May 2020
id: p/o070300
proof: For studying the orthogonal group of , one can suppose that the matrix of the quadratic form is because, given a quadratic form, there is a basis where its matrix is diagonalizable. A matrix belongs to the orthogonal group if , that is, , , and . As and cannot be both zero , the second equation implies the existence of in , such that and . Reporting these values in the third equation, and using the first equation, one gets that , and thus the orthogonal group consists of the matrices : where and . Moreover, the determinant of the matrix is . For further studying the orthogonal group, it is convenient to introduce a square root of . This square root belongs to if the orthogonal group is , and to otherwise. Setting , and , one has : If and are two matrices of determinant one in the orthogonal group then : This is an orthogonal matrix with , and . Thus : It follows that the map is a homomorphism of the group of orthogonal matrices of determinant one into the multiplicative group of . In the case of , the image is the multiplicative group of , which is a cyclic group of order . In the case of , the above and are conjugate, and are therefore the image of each other by the Frobenius automorphism. This meant that and thus . For every such one can reconstruct a corresponding orthogonal matrix. It follows that the map is a group isomorphism from the orthogonal matrices of determinant 1 to the group of the -roots of unity. This group is a cyclic group of order which consists of the powers of , where is a primitive element of , For finishing the proof, it suffices to verify that the group all orthogonal matrices is not abelian, and is the semidirect product of the group and the group of orthogonal matrices of determinant one. The comparison of this proof with the real case may be illuminating. Here two group isomorphisms are involved: : where is a primitive element of and is the multiplicative group of the element of norm one in ; : with and In the real case, the corresponding…
reason: most notations are undefined; no context for explaining why these consideration belong to the article. Moreover, the section consists essentially in a list of advanced results without providing the information that is needed for a non-specialist for verifying them
reason: This seems like it is the associated polar form B′=Q−Q−Q . When expressed in the same terms , the expression for a reflection is the same for all cases.
title: Orthogonal group
title: Proof: