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 G) we may define a sequence of submodules of G (regarded as a module over ) by Ai equals the linear span of {1,α,α2,,αi}.

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

Given a conjugacy classMathworldPlanetmathPlanetmath C in G, we may set ϕC=gCg. Then ϕC is central in G, as given hG, we have:

ϕCh=hgCh-1gh=hgCg=hϕC

Hence applying ϕC to V induces a G linear map VV. 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 gG 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|=gGχV(g)χV(g)*=CGtr(ϕC)χV(C)*=dCGλCχV(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