Square of degree of irreducible representation need not divide group order

This fact is related to: linear representation theory
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Statement

It is possible to have a finite group $G$ such that the square of one of its 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.

Related facts

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

Further information: linear representation theory of symmetric group:S3

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