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 fact about::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) uses::Degree of irreducible representation divides index of abelian normal subgroup

(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!