Degree of irreducible representation divides order of group

From Groupprops
Revision as of 14:38, 11 May 2007 by Vipul (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search


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</math? is irreducible:

<math>\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.