Square of degree of irreducible representation divides order of inner automorphism group in finite nilpotent group

In characteristic zero
Suppose $$G$$ is a fact about::finite nilpotent group and $$K$$ is a splitting field for $$G$$ of characteristic zero. Then, for any irreducible linear representation of $$G$$ over $$K$$, the square of the degree divides the order of the inner automorphism group of $$G$$, which is the same as the index of the center of $$G$$.

General case
Since degrees of irreducible representations are the same for all splitting fields, the truth of the statement for splitting fields in characteristic zero implies its truth for splitting fields in any characteristic not dividing the order of the group.

Similar facts

 * Degree of irreducible representation divides index of abelian subgroup in finite nilpotent group
 * Degree of irreducible representation divides index of center: This holds for all finite groups.
 * Degree of irreducible representation divides index of abelian normal subgroup: This holds for all finite groups.

Opposite facts

 * Square of degree of irreducible representation need not divide group order

Facts used

 * 1) uses::Equivalence of definitions of finite nilpotent group (reduce to the case where $$G$$ has prime power order)
 * 2) uses::Order of inner automorphism group bounds square of degree of irreducible representation
 * 3) uses::Degree of irreducible representation divides order of group

Proof
The proof has two steps:


 * 1) Reduce to the case where $$G$$ is a group of prime power order using Fact (1).
 * 2) 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.