Degree of irreducible representation divides order of group

Statement
Let $$G$$ be a finite group and $$\varphi$$ an irreducible representation of $$G$$ over an algebraically closed field of characteristic zero (or, more generally, over any splitting field of characteristic zero for $$G$$). Then, the degree of $$\varphi$$ divides the order of $$G$$.

Other facts about degrees of irreducible representations

 * Degree of irreducible representation divides index of center
 * Degree of irreducible representation divides index of abelian normal subgroup
 * Order of inner automorphism group bounds square of degree of irreducible representation
 * Number of irreducible representations equals number of conjugacy classes
 * Sum of squares of degrees of irreducible representations equals order of group

Similar fact about irreducible projective representations
See degree of irreducible projective representation divides order of group

Breakdown for a field that is not algebraically closed
Let $$G$$ be the cyclic group of order three and $$\R$$ be the field. Then, there are two irreducible representations of $$G$$ over $$\R$$: the trivial representation, and a two-dimensional representation given by the action by rotation by multiples of $$2\pi/3$$. The two-dimensional representation has degree $$2$$, and this does not divide the order of the group, which is $$3$$.

We still have the following results:


 * Degree of irreducible representation over reals divides twice the group order
 * Degree of irreducible representation over any field divides product of order and Euler totient function of exponent
 * Degree of irreducible representation of nontrivial finite group is strictly less than order of group
 * Maximum degree of irreducible real representation is at most twice maximum degree of irreducible complex representation

Proof
Given: A finite group $$G$$, an irreducible linear representation $$\varphi$$ of $$G$$ over a splitting field of characteristic zero for $$G$$, with character $$\chi$$ and degree $$d$$. Note that $$d$$ equals $$\chi(1)$$, i.e., the value of $$\chi$$ at the identity element of $$G$$.

To prove: $$d$$ divides the order of $$G$$.

Proof: