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>.
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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. Then, the [[degree of a linear representation|degree]] of <math>\rho</math> divides the order of <math>G</math>.
  
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==Related facts==
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===Other facts about degrees of irreducible representations===
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{{further|[[Degrees of irreducible representations]]}}
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* [[Degree of irreducible representation divides index of center]]
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* [[Degree of irreducible representation divides index of Abelian normal subgroup]]
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* [[Order of inner automorphism group bounds square of degree of irreducible representation]]
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* [[Sum of squares of degrees of irreducible representations equals order of group]]
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===Breakdown for a field that is not algebraically closed===
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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

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.

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.