Number of irreducible representations equals number of conjugacy classes

Statement
Consider a finite group $$G$$ and a splitting field $$K$$ for $$G$$. Then, the following two numbers are equal:


 * 1) The fact about::number of conjugacy classes in $$G$$.
 * 2) The number of fact about::irreducible linear representations (up to equivalence) of $$G$$ over $$K$$.

Note that any algebraically closed field whose characteristic does not divide the order of $$G$$ is a splitting field, so in particular, we can always take $$K = \mathbb{C}$$ or $$K = \overline{\mathbb{Q}}$$.

Related facts
For more facts about the degrees of irreducible representations, see degrees of irreducible representations.

Similar facts for irreducible representations of specific types

 * Number of one-dimensional representations equals order of abelianization

Similar facts over non-splitting fields
The key starting fact is this:

Application of Brauer's permutation lemma to Galois automorphism on conjugacy classes and irreducible representations (follows in turn from Brauer's permutation lemma): Suppose $$G$$ is a finite group and $$r$$ is an integer relatively prime to the order of $$G$$. Suppose $$K$$ is a field and $$L$$ is a splitting field of $$G$$ of the form $$K(\zeta)$$ where $$\zeta$$ is a primitive $$d^{th}$$ root of unity, with $$d$$ also relatively prime to $$r$$ (in fact, we can arrange $$d$$ to divide the order of $$G$$ because sufficiently large implies splitting). Suppose there is a Galois automorphism of $$L/K$$ that sends $$\zeta$$ to $$\zeta^r$$. Consider the following two permutations:


 * The permutation on the set of conjugacy classes of $$G$$, denoted $$C(G)$$, induced by the mapping $$g \mapsto g^r$$.
 * The permutation on the set of irreducible representations of $$G$$ over $$L$$, denoted $$I(G)$$, induced by the Galois automorphism of $$L$$ that sends $$\zeta$$ to $$\zeta^r$$.

Then, these two permutations have the same cycle type. In particular, they have the same number of cycles, and the same number of fixed points, as each other.

Using this fact, we can deduce various corollaries:

Opposite facts over non-splitting fields

 * Number of irreducible representations over complex numbers with rational character values need not equal number of conjugacy classes of rational elements

Similar facts under action of automorphism group
The key facts are:


 * Application of Brauer's permutation lemma to group automorphism on conjugacy classes and irreducible representations
 * Cyclic quotient of automorphism group by class-preserving automorphism group implies same orbit sizes of conjugacy classes and irreducible representations under automorphism group: In particular, if the quotient of the automorphism group by the group of class-preserving automorphisms is cyclic, then the orbit sizes both for the set of conjugacy classes and for the set of irreducible representations are equal.
 * Number of orbits of irreducible representations equals number of orbits of conjugacy classes under any subgroup of automorphism group
 * Number of orbits of irreducible representations equals number of orbits under automorphism group

Opposite facts under action of automorphism group

 * Orbit sizes for irreducible representations may differ from orbit sizes for conjugacy classes under action of automorphism group

Similar facts in modular representation theory

 * Number of irreducible Brauer characters equals number of regular conjugacy classes for any fixed prime $$p$$.

Similar arithmetic fact

 * Sum of squares of degrees of irreducible representations equals order of group

Facts used

 * 1) uses::Splitting implies characters span class functions