Square of degree of irreducible representation need not divide group order

Statement
It is possible to have a finite group $$G$$ such that the square of one of its fact about::degrees of irreducible representations (i.e., the degree of an irreducible linear representation over an algebraically closed field of characteristic zero) does not divide the order of the group.

Similar facts

 * Degree of irreducible representation need not divide exponent
 * Degree of irreducible representation need not be less than exponent

Opposite facts
Additive, rather than divisibility, bounds:


 * Sum of squares of degrees of irreducible representations equals order of group: In particular, all the squares of degrees of irreducible representations are bounded by the order of the group.
 * Order of inner automorphism group bounds square of degree of irreducible representation

Divisibility facts:


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

Proof
The simplest example is that of particular example::symmetric group:S3, a group of order 6 that has an irreducible linear representation of degree $$2$$, even though $$2^2$$ does not divide $$6$$.