Degree of irreducible representation divides index of abelian subgroup in finite nilpotent group
Suppose is a finite nilpotent group, is a splitting field for of characteristic zero, and is an abelian subgroup of . Then, for any irreducible linear representation of over , the degree of the representation divides the index .
This gives a divisibility constraint on the degrees of irreducible representations.
- Square of degree of irreducible representation divides order of inner automorphism group in finite nilpotent group
- Degree of irreducible representation is bounded by index of abelian subgroup: The numerical bounding (without divisibility) holds for non-nilpotent groups as well.
- Degree of irreducible representation divides index of abelian normal subgroup: This fact is true for all finite groups, i.e., if we add the assumption that the subgroup is normal, we can drop the assumption that the whole group is nilpotent.
In the case of a finite nilpotent group, the statement that the degree of any irreducible representation divides the index of any abelian subgroup is prima facie stronger. In practice, for most small orders, the two statements have the same power, due to the Jonah-Konvisser abelian-to-normal replacement theorem which guarantees that, under certain conditions, any abelian subgroup of a given order can be replaced by an abelian normal subgroup of the same order.
- Degree of irreducible representation need not divide index of abelian subgroup
- Square of degree of irreducible representation need not divide group order
- Equivalence of definitions of finite nilpotent group
- Degree of irreducible representation is bounded by index of abelian subgroup
- Degree of irreducible representation divides order of group
The proof has two steps:
- Reduce to the case where is a group of prime power order using Fact (1).
- Now use Facts (2) and (3) and the observation that for two numbers that are powers of the same prime, the smaller one divides the bigger one.