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

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==Statement== | ==Statement== | ||

− | Let <math>G</math> be a finite group and <math>\rho</math> an irreducible representation of <math>G</math> over an algebraically closed field of characteristic zero. Then, the degree of <math>\rho</math> divides the order of <math>G</math>. | + | Let <math>G</math> be a finite group and <math>\rho</math> an irreducible representation of <math>G</math> over an algebraically closed field of characteristic zero. Then, the [[degree of a linear representation|degree]] of <math>\rho</math> divides the order of <math>G</math>. |

+ | ==Related facts== | ||

+ | |||

+ | ===Other facts about degrees of irreducible representations=== | ||

+ | |||

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

==Proof== | ==Proof== | ||

## Revision as of 13:10, 9 October 2008

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

## 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 .

## 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 .