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
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).
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Statement with symbols
Let be a finite group and an irreducible representation of over an algebraically closed field of characteristic zero (or, more generally, over any splitting field for ). 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 .
We still have the following results:
- Degree of irreducible representation of nontrivial finite group is strictly less than order of group
- Maximum degree of irreducible real representation is at most twice maximum degree of irreducible complex representation
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 .