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

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(A little formula)
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===A little formula===
===A little formula===
We know that if <math>\rho</math? is irreducible:
We know that if <math>\rho</math> is irreducible:
<math>\sum_c |c| \chi(c) \overline{\chi(c)} = |G|</math>
<math>\sum_c |c| \chi(c) \overline{\chi(c)} = |G|</math>

Revision as of 14:38, 11 May 2007


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.


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.