Warning: This is a very shoddily organized post and can be vastly improved. The method of proof is still nice however.
Let be a finite group of size and an algebraically closed field such that in it. Let be the irreducible representations of over and the corresponding characters.
A class function is a set function such that is constant on conjugacy classes. That is, for all . It is easy to see that any character is a class function by cyclicity of the trace function.
Let us denote by the vector space of class functions and the vector space of functions spanned by the within . It is natural to wonder about the size of relative to . In fact, the two vector spaces are equal!
To prove this, let us consider the algebra . By our hypothesis and the usual argument of Maschke’s theorem, this is a semisimple ring. That is, is semisimple as a representation. Let be the character attached to this so called regular representation. A simple computation shows us that and otherwise.
Moreover, for any irreducible representation , we see from last time that:
Therefore, by semisimplicity
as representations. Moreover, the orthogonality relations from last time also show that the are linearly independent. That is, irreducible characters.
Taking the endomorphism ring on both sides (as modules over ), we obtain:
as algebras. Considering dimensions, we have incidentally shown that . Moreover, let us consider the central elements in . They are easily seen to exactly be the elements of the form for $f:G \to k$ a class function.
On the other hand, the central elements in are always just . Therefore, the dimension of the center of (= vector space of class functions) is equal to the number of irreducible representations. Therefore, and we are done.
The above map can be thought of as the representation of a class function in the basis defined by the functions for some appropriate constants . This is because the element:
maps to a a diagonal element in that can be calculated by taking the trace. That is to say, the image of in $latex is given by:
by the orthogonality relations. We see that .