Ito-Michler theorem

Statement

Suppose $G$ is a finite group and $p$ is a prime number. The following are equivalent:

1. $p$ does not divide any of the degrees of irreducible representations of $G$ over $\mathbb{C}$ (or more generally, over some splitting field).
2. The $p$-Sylow subgroup of $G$ is a Normal Sylow subgroup (?) and is also abelian.

Note that in the case that $p$ does not divide the order of $G$ at all, (2) is satisfied, so we do not need to assume that $p$ divides the order of $G$. However, making that assumption does not weaken our theorem.

Related notions

• Character degree graph of a finite group is an undirected graph associated with any finite group. Its vertex set is given precisely by the Ito-Michler theorem.

Related facts

• Degrees of irreducible representations need not cover all prime factors

Facts used

1. Degree of irreducible representation divides index of abelian normal subgroup

Proof

(2) implies (1)

This follows directly from fact (1), since the index of an abelian normal $p$-Sylow subgroup is relatively prime to $p$, hence all irreducible representations have degree dividing a number relatively prime to $p$, forcing the degrees to be relatively prime to $p$.

(1) implies (2)

This is the hard part!