TITLE: Dimension of the center of the group algebra is equal to the number of irreducible representations- Without using character theory
QUESTION [5 upvotes]: In a representation theory course we recently proved that the number of irreducible representations of a finite group $G$ over the complex numbers is equal to the number of conjugacy classes of $G$. The proof we saw used characters, as is standard. But I was interested in whether there exists a proof that does not use characters, and found that (apparently) $\dim_k(Z(k[G]))$ is equal to both the number of conjugacy classes of $G$ and the number of irreducible representations over the field $k$. 
I've proven that it is equal to the number of conjugacy classes, but am struggling in providing a proof that it is equal to the number of irreducible representations that doesn't rely on the original result. This is my progress so far:
First we decompose $k[G]$ into the direct sum of simple unital rings like so: Let $\mathcal{L}$ be the set of isomorphism classes of simple left ideals and for each $l\in\mathcal{L}$ define $R_l=\sum_{L\in l} L$. Then $R_l$ is a unital ring and $R\cong\oplus_{l\in\mathcal{L}}R_l$. I believe that the identities of each $R_l$ give a basis for $Z(k[G])$. It is obvious that they are linearly independent and are in the center, so the issue is how to prove that they actually span the center. 
If we assume that we already know the number of conjugacy classes equal the number of irreducible representations, this becomes a simple matter of counting dimensions, so I'm fairly certain that this is a basis. However I have no clue how to prove the result without using this, and any advice would be hugely appreciated.

REPLY [3 votes]: It is not true for arbitrary fields that the dimension of $Z(k [G])$ is the number of isoclasses of irreducible representations; the characteristic zero assumption is crucial, and it is necessary to assume the field $k$ is large enough relative to $G$ (for instance, algebraically closed; the most precise statement is that $k$ must be a splitting field for $G$).
For instance, let $p$ be a prime number, let $G$ be a cyclic group of order $p$ and let $k$ be a field of characteristic $p$. Evidently the group algebra is abelian, so its center has dimension $p$. But there is one only isoclass of irreducible $k$-linear representations: the trivial representation.
This is easy to see: supposing that $\phi:G \rightarrow \mathrm{GL}(V)$ is an irreducible representation and $g$ is a generator of $G$, the minimal polynomial of $\phi(g)$ divides $X^p-1=(X-1)^p$ and is therefore a power of $X-1$. Irreducibility now implies that the minimal polynomial is just $X-1$.