There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for $Z(\mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post I don't see that...
My approach to this is:
Be $x \in Z(\mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g \in G$. Writing $x = \sum\limits_{h \in G} \lambda_h h$, we get $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b \in K$ (same conjugacy class) you get $\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s$, since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar.
But from here I don't really know how I can generate the $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$ with the $e_K$ ...
I would be really thankful, if someone could give me quite a basic proof that even I understand...
Notice that $\sum \lambda_h h = \sum \lambda_h ghg^{-1}$ implies that $\lambda_h = \lambda_{ghg^{-1}}$ for all $g$, i.e $\lambda_h$ is constant along conjugacy classes. It follows that an element is the center can be written $ \sum \lambda_r c_r$ where $r$ runs along the conjugacy classes and $c_r = \sum_{h \in r}h$. Since the $c_r$ are obviously linearly independant the claim follows.