Degree of irreducible representation is bounded by index of abelian subgroup

In characteristic zero
Suppose $$G$$ is a finite group, $$K$$ is a splitting field for $$G$$ of characteristic zero, and $$H$$ is an abelian subgroup of $$G$$. Then, the fact about::degrees of irreducible representations of $$G$$ over $$K$$ are all at most equal to the index $$[G:H]$$.

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.

Related facts

 * Degree of irreducible representation divides index of abelian normal subgroup
 * Degree of irreducible representation need not divide index of abelian subgroup
 * Degree of irreducible representation divides index of abelian subgroup in finite nilpotent group

Facts used

 * 1) uses::Maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup (this, as stated, requires characteristic zero)
 * 2) uses::Abelian implies every irreducible representation is one-dimensional (this requires us to be over a splitting field)

Proof
The proof follows directly from Facts (1) and (2).