**Notation**. Let be a field and let be a finite group. Let be the conjugacy classes of Let

**Theorem**. The set is a -basis for the center of Thus is equal to the number of conjugacy classes of

*Proof*. are linearly independent over because if then So we only need to show that Let where for all

*Claim*. if and only if for all

*Proof of the claim*. Well, if and only if for all Thus, if and only if

Clearly, fixing we have So if we put then and thus becomes

The claim now follows from

Now, if then, by definition, if and only if for all So for every the coefficients of and in are either both 1 or both 0. Thus, by the claim, for all So

Conversely, if then, by the claim, for all and which are in the same conjugacy class. Thus if for every we choose a then

So