Difference between revisions of "Degree of irreducible representation divides order of group"

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Revision as of 23:25, 7 May 2008

This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group

Statement

Let G be a finite group and \rho an irreducible representation of G over an algebraically closed field of characteristic zero. Then, the degree of \rho divides the order of G.

Proof

Introduction of some algebraic integers

Further information: Convolution algebra on conjugacy classes

Using the convolution algebra on conjugacy classes, we can show that for any representation \rho with character \chi, and any conjugacy class c, the number:

\omega(c,\rho) = \frac{|c|\chi(c)}{\chi(1)}

are algebraic integers. Note that \chi(1) is the degree of \rho.

A little formula

We know that if \rho is irreducible:

\sum_c |c| \chi(c) \overline{\chi(c)} = |G|

by the orthonormality of the irreducible characters.

Dividing by \chi(1), we get:

\sum_c \omega(c,\rho) \overline{\chi(c)} = |G|/\chi(1)

Note that since the characters are algebraic integers and so are the values taken by \omega, the overall left-hand side is an algebraic integer. Thus |G|/\chi(1) is an algebraic integer.

But since it is a rational number, it must be a rational integer, or in other words, the degree of \rho divides the order of G.