proof that dimension of complex irreducible representation divides order of group


Theorem Let G be a finite groupMathworldPlanetmath and V an irreduciblePlanetmathPlanetmath complex representation of finite dimensionMathworldPlanetmathPlanetmath d. Then d divides |G|.

Proof: Given any α in the group ring of G (denoted ℤ⁢G) we may define a sequence of submodules of ℤ⁢G (regarded as a module over ℤ) by Ai equals the ℤ linear span of {1,α,α2,⋯,αi}.

ℤ⁢G is Noetherian as a module over ℤ so we must have Ai=Ai-1 for some i. Hence αi may be expressed as a ℤ linear combinationMathworldPlanetmath of lower powers of α. In other α solves a monic polynomial of degree i with coefficients in ℤ.

Given a conjugacy classMathworldPlanetmathPlanetmath C in G, we may set ϕC=∑g∈Cg. Then ϕC is central in ℤ⁢G, as given h∈G, we have:

ϕC⁢h=h⁢∑g∈Ch-1⁢g⁢h=h⁢∑g∈Cg=h⁢ϕC

Hence applying ϕC to V induces a ℂ⁢G linear map V→V. By Schur’s lemma this must be multiplication by some complex numberMathworldPlanetmathPlanetmath λC. Then λC is an algebraic integerMathworldPlanetmath as it solves the same monic polynomial as ϕC.

Also any g∈G has finite order so the map it induces on V must have eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath which are roots of unityMathworldPlanetmath and hence algebraic integers. Hence the sum of the eigenvalues, χV⁢(g), must also be an algebraic integer.

Now V is irreducible so,

|G|=∑g∈GχV⁢(g)⁢χV⁢(g)*=∑C⊂Gtr⁢(ϕC)⁢χV⁢(C)*=d⁢∑C⊂GλC⁢χV⁢(C)*

Therefore |G|/d is both rational and an algebraic integer. Hence it is an integer and d divides |G|.

Title proof that dimension of complex irreducible representation divides order of group
Canonical name ProofThatDimensionOfComplexIrreducibleRepresentationDividesOrderOfGroup
Date of creation 2013-03-22 17:09:04
Last modified on 2013-03-22 17:09:04
Owner whm22 (2009)
Last modified by whm22 (2009)
Numerical id 8
Author whm22 (2009)
Entry type Proof
Classification msc 20C99