Degrees of irreducible representations need not cover all prime factors

Statement

It may happen for a finite group that the set of prime factors of the Degrees of irreducible representations (?) of the group is a proper subset of the set of all prime factors of the group. Here, degrees of irreducible representations stands for the degrees of all the irreducible linear representations of the finite group over $\mathbb {C}$.

Related facts

Partial opposite fact

• Ito-Michler theorem states that if a particular prime $p$ does not divide the degree of any irreducible representation, then the $p$-Sylow subgroup must be an abelian normal subgroup.

Other related facts

• Degree of irreducible representation divides order of group
• Degree of irreducible representation divides index of center
• Degree of irreducible representation divides index of Abelian normal subgroup
• Degree of irreducible representation need not divide exponent

Proof

Example of an abelian group

For a finite Abelian group, the degrees of irreducible representations are all $1$, which certainly does not cover all prime factors of the group.

Example of the symmetric group

Further information: symmetric group:S3

Consider the symmetric group on three letters. This is a group of order six, and the degrees of its irreducible representations are $1,1,2$. The prime factor $3$ of the order of the group does not occur as a prime factor of the degree of any irreducible representation.

Example of dihedral group, or any group with a very big Abelian normal subgroup

Further information: Degree of irreducible representation divides index of Abelian normal subgroup

Any dihedral group has an Abelian normal subgroup of index two. Thus, all its degrees of irreducible representations are either $1$ or $2$, and so if the group is not a $2$-group, there are certainly prime factors of the order of the group that are not covered in the degrees of irreducible representations.