Degree of irreducible representation divides order of group

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

This article states a result of the form that one natural number divides another. Specifically, the (degree of a linear representation) of a/an/the (irreducible linear representation) divides the (order) of a/an/the (group).
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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 $G$.

Related facts

Other facts about degrees of irreducible representations

Further information: Degrees of irreducible representations

Breakdown for a field that is not algebraically closed

Let $G$ be the cyclic group of order three and $\R$ be the field. Then, there are two irreducible representations of $G$ over $\R$: the trivial representation, and a two-dimensional representation given by the action by rotation by multiples of $2\pi/3$. The two-dimensional representation has degree $2$, and this does not divide the order of the group, which is $3$.

We still have the following results:

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 $G$.