# Degree of irreducible representation divides order of 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 $G$.