Throughout is a finite group**.**

**Lemma.** There exist a finite set of mutually non-equivalent irreducible representations of such that every irreducible representation of is equivalent to for some

*Proof*. By the corollary in this post, is a semisimple ring. Thus there exist simple -modules which are mutually non-isomorphic and any simple -module is isomorphic to for some For every let be the representation of corresponding to i.e. is defined by for all and Since is simple, is irreducible by definition. Also and are equivalent iff because iff If is any irreducible representation of then is a simple -module and hence for some That means is equivalent to

**Definition**. The set in the lemma is called a **complete set** of irreducible representations of

**Theorem**. Let be a complete set of irreducible representations of Then

1) The number is equal to the number of conjugacy classes of

2)

*Proof*. 1) As we mentioned at the beginning of the proof of the above lemma, is semisimple. Thus by the Wedderburn – Artin theorem

for some integers and some -division algebras Since is finite dimensional over each has to be finite dimensional, and hence algebraic, over Thus because is algebraically closed. So becomes

Now, since taking the center of both sides in gives us

Thus We also proved in here that is equal to the number of conjugacy classes of So is equal to the number of conjugacy classes of

2) Let By we have and also