Irreducible character degree divides the group order

For every irreducible complex character $\chi$ of a finite group $G$,Indeed, row orthogonality expresses $|G|/\chi (1)$ as a sum of products of algebraic integers:It is a rational algebraic integer and therefore an integer.

A degree-two irreducible representation of a nonabelian finite simple group would be faithful. Its determinant is a linear character and hence trivial, so its image lies in ${\mathrm{SL}}_2(\mathbb{C})$. Degree divisibility makes the group order even; an involution must map to the unique nonidentity involution $-I$ in ${\mathrm{SL}}_2(\mathbb{C})$ and would therefore be central, a contradiction.

For a prime $p$, either a group of order $2p$ is abelian and all irreducible character degrees are one, or $p$ is odd and its degrees are

If $p>q$ are prime and $G$ is nonabelian of order $pq$, then it has $q$ linear characters and $(p-1)/q$ irreducible characters of degree $q$. It consequently hasconjugacy classes.