# 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).

View other divisor relations |View congruence conditions

## Contents

## Statement

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

## Related facts

### Other facts about degrees of irreducible representations

`Further information: Degrees of irreducible representations`

- Degree of irreducible representation divides index of center
- Degree of irreducible representation divides index of Abelian normal subgroup
- Order of inner automorphism group bounds square of degree of irreducible representation
- Sum of squares of degrees of irreducible representations equals order of group

### Breakdown for a field that is not algebraically closed

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

We still have the following results:

- degree of irreducible representation of nontrivial finite group is strictly less than order of group

## 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 with character , and any conjugacy class , the number:

are algebraic integers. Note that is the degree of .

### A little formula

We know that if is irreducible:

by the ortho*normality* of the irreducible characters.

Dividing by , we get:

Note that since the characters are algebraic integers and so are the values taken by , the overall left-hand side is an algebraic integer. Thus is an algebraic integer.

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