Schur index divides degree of irreducible representation

Statement
Suppose $$G$$ is a finite group, $$\varphi$$ is an fact about::irreducible linear representation over a splitting field for $$G$$, $$\chi$$ is the character of $$\varphi$$, and $$m(\chi)$$ is the fact about::Schur index of $$\chi$$. Then, $$m(\chi)$$ divides the degree of $$\varphi$$ (which is the same as the degree of $$\chi$$). In particular, this bounds the possible values of the Schur index for irreducible representations in terms of the fact about::degrees of irreducible representations.

(Note that instead of requiring $$\varphi$$ to be irreducible over a splitting field, we could instead require $$\varphi$$ to be absolutely irreducible over any field whose characteristic does not divide the order of $$G$$).

In particular:


 * The Schur index of a linear character is $$1$$ (this is obvious even otherwise).
 * The Schur index of an irreducible character of prime degree is either $$1$$ or equal to that prime.

Corollaries
This statement can be combined with results about the degrees of irreducible representations. In particular:


 * Combining with degree of irreducible representation divides index of abelian normal subgroup yields that Schur index of irreducible representation divides index of abelian normal subgroup.
 * Order of inner automorphism group bounds square of degree of irreducible representation yields that order of inner automorphism group bounds square of Schur index of irreducible representation.

Other related facts

 * Schur index of irreducible character in characteristic zero divides exponent
 * Square of Schur index of irreducible character in characteristic zero divides order