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 a linear representation|degree]] of <math>\rho</math> divides the order of <math>G</math>.
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===Verbal statement===
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The [[fact about::degrees of irreducible representations]] of a [[group]] over a [[splitting field]] for it all divide the [[fact about::order of a group|order]] of the group.
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===Statement with symbols===
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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 (or, more generally, over any [[splitting field]] for <math>G</math>). Then, the [[degree of a linear representation|degree]] of <math>\rho</math> divides the order of <math>G</math>.
  
 
==Related facts==
 
==Related facts==

Revision as of 22:41, 6 April 2010

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

Statement

Verbal statement

The Degrees of irreducible representations (?) of a group over a splitting field for it all divide the order of the group.

Statement with symbols

Let G be a finite group and \rho an irreducible representation of G over an algebraically closed field of characteristic zero (or, more generally, over any splitting field for G). 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.