# 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.

## 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!