Irreducibles and Conjugacy Classes

Let $G$ be a finite group. A class function on $G$ is a function $f:G\to \mathbb{C}$ constant on each conjugacy class . The space of class functions is denoted

(See class function .)

Every complex representation $V$ has a character $\chi_V\in \mathrm{Cl}(G)$, and irreducible representations have irreducible characters .

Theorem

Let $G$ be a finite group. Over $\mathbb{C}$, the number of isomorphism classes of irreducible representations of $G$ equals the number of conjugacy classes of $G$.

Equivalent formulation (via class functions)

The irreducible characters $\{\chi_1,\dots,\chi_r\}$ form an orthonormal basis of $\mathrm{Cl}(G)$ with respect to the standard inner product

In particular,

But $\dim_\mathbb{C}\mathrm{Cl}(G)$ is the number of conjugacy classes (a class function is determined by its values on conjugacy classes), so $r$ equals the number of conjugacy classes.

This perspective is tightly linked to character orthonormality (and character orthogonality ).

Examples

Example 1: Abelian groups

If $G$ is abelian, every conjugacy class is a singleton, so the number of conjugacy classes is $|G|$. The theorem implies $G$ has $|G|$ irreducible complex representations. Indeed, all irreducibles are 1-dimensional characters, and there are exactly $|G|$ of them (e.g. for $C_n$, the characters $g\mapsto \zeta^k$, $k=0,\dots,n-1$).

Example 2: $S_3$

The symmetric group $S_3$ has 3 conjugacy classes:

Hence it has exactly 3 irreducible complex representations: the trivial representation, the sign representation, and the 2-dimensional standard representation.

Example 3: Dihedral group $D_8$ (order 8)

Let $D_8=\langle r,s \mid r^4=1,\ s^2=1,\ srs=r^{-1}\rangle$. Its conjugacy classes are

so there are 5 conjugacy classes. Therefore $D_8$ has exactly 5 irreducible complex representations (in fact: four 1-dimensional and one 2-dimensional).