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 $\rho </math?isirreducible:<math>\sum _{c}|c|\chi (c){\overline {\chi (c)}}=|G|$

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 $G$.