Degree of irreducible representation need not divide index of abelian subgroup

Statement
It is possible to have a finite group $$G$$, an abelian subgroup $$H$$ of $$G$$, and an irreducible representation $$\rho$$ of $$G$$ over $$\mathbb{C}$$ such that the degre of $$\rho$$ does not divide the index $$[G:H]$$.

Opposite divisibility facts
These are all for irreducible representations over $$\mathbb{C}$$, or more generally, over splitting fields of characteristic zero (and more generally, characteristic coprime to the order of the group).


 * Degree of irreducible representation divides group order
 * Degree of irreducible representation divides index of center
 * Degree of irreducible representation divides index of abelian normal subgroup

Opposite bounding facts

 * Degree of irreducible representation is bounded by index of abelian subgroup
 * Sum of squares of degrees of irreducible representations equals group order
 * Order of inner automorphism group bounds square of degree of irreducible representation

Similar non-divisibility/non-bounding facts

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

Proof
Let $$G$$ be particular example::symmetric group:S3 and $$H$$ be a subgroup of order 2 in $$G$$ (i.e., particular example::S2 in S3). Then, $$[G:H] = 6/2 = 3$$, whereas $$G$$ has an irreducible linear representation of degree $$2$$.