Given a finite group $G = \{a_1 ,a_2,...,a_n\}$ and $\mathbb{C}[G]$ the respective group ring, $\mathbb{C}[G] = \{\sum z_ia_i : z_i \in\mathbb{C} , a_i\in G\}$.  Defining $K_i := \sum_{a_k\in cl_G(a_i)} a_k$ , when $cl_G(a_i)$ the conjugacy class, means $K_i =a_1a_ia_1^{-1} + a_2a_ia_2^{-1} +...+a_na_ia_n^{-1}$.   I wish to show that $(K_1 ,K_2 ,...,K_n)$ spans Z($\mathbb{C}[G])$. I had already shown that for $a_i \in Z(\mathbb{C}[G])$, $K_i = \{a_i\}$ thus $a_i = 1\cdot K_i$, as needed.  But I failed to show the other direction, means that for $w\in span_\mathbb{C}(K_1,K_2,...,K_n)$ then $w$ commutes with any element of $\mathbb{C}[G]$, means $w\in Z(\mathbb{C}[G])$.
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...
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