Degree of irreducible representation divides order of group
This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group
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 .
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 .
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 orthonormality 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 .