Number of irreducible representations equals number of conjugacy classes

This article gives a proof/explanation of the equivalence of multiple definitions for the term number of conjugacy classes
View a complete list of pages giving proofs of equivalence of definitions

Statement

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

1. The Number of conjugacy classes (?) in $G$.
2. The number of Irreducible linear representation (?)s (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

Similar arithmetic fact

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

Particular cases

Families

Family	Order of group	Degrees of irreducible representations, indexing set for them	Conjugacy class sizes, indexing set for them	Number of conjugacy classes = number of irreducible representations
dihedral group of even degree $n$	$2n$	1 (4 times), 2 ($(n-2)/2$ times)	1 (2 times), 2 ($(n-2)/2$ times), $n/2$ (2 times)	$(n+6)/2$
dihedral group of odd degree $n$	$2n$	1 (2 times), 2 ($(n-1)/2$ times)	1 (1 time), 2 ($(n-1)/2$ times), $n$ (1 time)	$(n+3)/2$
symmetric group of degree $n$	$n!$	indexed by partitions (see linear representation theory of symmetric groups), described in terms of Young diagram for a partition	indexed by partitions, via cycle type (see cycle type determines conjugacy class)	$p(n)$ the number of unordered integer partitions of $n$

Facts used

1. Splitting implies characters span class functions