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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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 of characteristic zero 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
The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
|Fact no.||Statement||Steps in the proof where it is used||Qualitative description of how it is used||What does it rely on?||Other applications|
|1||Character orthogonality theorem: The part relevant for us is: for an irreducible representation over a splitting field of characteristic zero,||Step (1)||click here|
|2||Size-degree-weighted characters are algebraic integers: This states that for an irreducible linear representation of a finite group over an algebraically closed field of characteristic zero (or more generally, over any splitting field), a conjugacy class in and an element , the number (with denoting the identity element of the group) is an algebraic integer.||Step (3)||click here||3||Characters are algebraic integers||Step (4)||click here|
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Given: A finite group , an irreducible linear representation of over a splitting field of characteristic zero for , with character and degree . Note that equals , i.e., the value of at the identity element of .
To prove: divides the order of .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||The following holds: where the sum is over all conjugacy classes of , and where denotes the value of at any element of .||Fact (1)||is irreducible over a splitting field of characteristic zero, with character .||Follows from fact (1). The comes because for each conjugacy class , elements of the class appear in the full statement of the column orthogonality theorem.|
|2||Step (1)||Divide both sides of step (1) by .|
|3||Each is an algebraic integer for each conjugacy class .||Fact (2)||is irreducible over a splitting field of characteristic zero, with character .|
|4||Each is an algebraic integer for each conjugacy class .||Fact (3)||is a character.||The complex conjugate of an algebraic integer is also an algebraic integer.|
|5||is an algebraic integer.||Steps (3), (4)||The set of algebraic integers forms a ring, so a finite sum of products of algebraic integers is an algebraic integer.|
|6||is an algebraic integer.||Steps (2), (5)||By Step (5), the left side of Step (2) is an algebraic integer, hence so is the right side.|
|7||is a positive integer, so divides .||Step (6)||Both and are positive integers, hence their quotient is a positive rational number. The only way a rational number can be an algebraic integer is if it is an integer, hence the conclusion.|