A minor point in Serre's Representation Theory Let $R$ be a commutative ring of characteristic zero, $G$ a finite group. For each conjugacy class $[x]$ in $G$, define $e_{[x]} = \sum_{s\in[x]}s$. It is easily checked that these $\{e_{[x]}\}_{x\in G}$ form a basis for $Z(\mathbb{C}[G])$. Serre makes the claim that

"... each product $e_{[x]}e_{[y]}$ is a linear combination with integer coefficients of the $e_{[z]}$."

While I can see how these $e_{[x]}$'s form a basis for $Z(\mathbb{C}[G])$, for some reason I can't see this fact as easily. In fact, I can't even intuit why $e_{[x]}e_{[y]}$ should also be an element of $Z(\mathbb{C}[G])$ when written out as
$$
e_{[x]}e_{[y]} = \left(\sum_{g\in G}gxg^{-1}\right)\left(\sum_{h\in G}hyh^{-1}\right) = \sum_{g,h\in G}gxg^{-1}hyh^{-1}
$$
To me it's not clear at all that the above double sum is "a linear combination with integer coefficients of the $e_{[z]}$." Can someone provide some insight?
 A: Well, since $e_{[x]}$ and $e_{[y]}$ are both in $Z(\mathbb{C}[G])$, so is their product.  Moreover, it is clear that $e_{[x]}e_{[y]}$ is an integer linear combination of elements of $G$ (since it is just some sum of elements of $G$).  But since it is in $Z(\mathbb{C}[G])$, the coefficients must be constant on each conjugacy class, and so it is in fact an integer linear combination of $e_{[z]}$'s.
To see this more explicitly from the expression $$e_{[x]}e_{[y]} = \sum_{g,h\in G}gxg^{-1}hyh^{-1},$$ note that $$k(gxg^{-1}hyh^{-1})k^{-1}=(kg)x(kg)^{-1}(kh)y(kh)^{-1}.$$ So, this sum is unchanged if you conjugate by $k$: the terms just get permuted, with the term indexed by $(g,h)$ becoming the term indexed by $(kg,kh)$.
A: If $z=gxg^{-1}hyh^{-1}\in[x][y]$ and $\gamma\in G$ we have
$$
\gamma z\gamma^{-1}=(\gamma g)x(g^{-1}\gamma^{-1})(\gamma h)y(h^{-1}\gamma^{-1})\in[x][y]
$$
so that $[x][y]$ is a union of conjugacy classes.
The coefficient of $z\in G$ is the cardinality of the set
$$
P_z=\{(g,h)\in G\times G\,|\, z=gxg^{-1}hyh^{-1}\}.
$$
The same formula above says that multiplication by $\gamma$ embedded diagonally in $G\times G$ gives a bijection between $P_z$ ad $P_{\gamma z\gamma^{-1}}$ so that the coefficients are the same for all the elements in the same conjugacy class.
