Sum of sum of elements in conjugacy class is a multiple of them if and only if $G=G'$ I have another question on character/group theory. This one seems to be a bit harder.

Let $Cl(g_1),...Cl(g_r)$ be the conjugacy classes of a finite group, $G$ and let $C_i \in \mathbb{C}(G)$ (the group algebra) be the sum of the elements of $Cl(g_i)$. Then
  $$\sum_{i=1}^rC_i=\alpha \prod_{i=1}^r C_i$$
  for some $\alpha \in \mathbb{C}$ if and only if $G=G'$. 

I am guessing that we somehow want to use the algebra isomorphism $\pi(x)=(\Theta_1(x),...,\Theta_r(x))$ where $\Theta$ are irreps, but honestly, I'm rather clueless.
Any hints, references or other help is much appreciated.
Thanks
 A: The reason is that if G is perfect, then every non-trivial irreducible character of G vanishes somewhere, so if we take any non-trivial homomorphism from the centre of the complex character of G to the complex field, it will vanish on the product of the class sums.
But the homomorphism associated to the trivial character of G does not vanish on that product. Hence the product of the class sums is a nonzero multiple of the sum of the elements of G.the
  On the other hand, if G is not perfect, then it has a non trivial linear character, and the homomorphism
from the centre of the group algebra to the complex field associated to it does not vanish on the product of the class sums, but does vanish on the sum of all elements of G.
In view of the comment by the OP I will expand a bit.
First of all, W. Burnside proved that if $\chi$ is a complex irreducible character of a finite group $G$ and $\chi(1) > 1,$ then $\chi(g) = 0$ for some $g \in G$. On the other hand if $\chi$ is a complex irreducible character of degree one of a finite group $G$, then $|chi(g)|^{2} = 1$ for all $g \in G$, so we never have $\chi(g) = 0$ for such a $\chi$.
Secondly, for each irreducible character $\chi$ of $G$, there is an algebra epimorphism $\omega_{chi} : Z(\mathbb{C}G) \to \mathbb{C}$ defined by 
$\omega_{\chi}(X) = \frac{\chi(X)}{\chi(1)}$ for all $X \in Z(\mathbb{C}G)$.
Since the class sums form a $\mathbb{C}$-basis for $Z(\mathbb{C}G)$, the homomorphism is uniquely specified by the fact that if $C_{y}$ is the class sum of the conjugacy class containing $y \in G$, then $\omega_{\chi}(C_{y}) = \frac{[G:C_{G}(y)]\chi(y)}{\chi(1)}$ whenever $y \in G$. All algebra epimorphisms from $Z(\mathbb{C}G) \to \mathbb{C}$ arise in this way.
Furthermore, since $Z(\mathbb{C}G)$ also has a $\mathbb{C}$-basis 
$\{ e_{\chi} : \chi \in {\rm Irr}(G) \}$, where $e_{\chi}$ is the primitive idempotent $\frac{1}{|G|} \left( \sum_{g \in G} \chi(g^{-1})g \right)$ of $Z(\mathbb{C}G)$, the homomorphisms $\omega_{\chi}$ have the property that 
$\omega_{\chi}(e_{\mu}) = \delta_{\chi,\mu}$ ( and are uniquely determined 
by that property) for irreducible characters $\chi, \mu$ of G.
Now suppose that $G = G^{\prime}.$ Then all irreducible characters of $G$ are non-linear, except the trivial character. Hence all such irreducible characters (except the trivial one) take value $0$ on some element of $G$.
Now let $P$ be the product of the class sums of $G$. Then $\omega_{\chi}(P) = 0$ for each non-trivial irreducible character of $G$, and when $\chi$ is the trivial character, we have $\omega_{\chi}(P) = t $ for some positive integer $t$. But we may write 
$P = \sum_{\mu \in {\rm Irr}(G)} \alpha_{\mu} e_{\mu}$ for unique complex numbers $\alpha_{\mu}$, since the $e_{\mu}'s$ form a $\mathbb{C}$-basis for $Z(\mathbb{C}G)$, and $P$ does lie in $Z(\mathbb{C}G)$.
But then $\omega_{\chi}(P) = \alpha_{\chi}$ for each $\chi \in {\rm Irr}(G)$.
Thus we have $\alpha_{\chi} = 0$ whenever $\chi$ is a non-trivial irreducible character of $G$ and $\alpha_{\chi} = t$ when $\chi$ is the trivial character.
Hence $P = \frac{t}{|G|} \left( \sum_{g \in G} g \right)$.
On the other hand, if $G$ is not perfect, then $G$ has a non-trivial irreducible character $\chi$ of degree one, which vanishes on no element of $G$, soo $\omega_{\chi}(P) \neq 0$ and the idempotent $e_{\chi}$ appears with non-zero coefficient when $P$ is expressed as a $\mathbb{C}$-linear combination of the 
$e_{\mu}'s$. so $P$ is not just a scalar multiple of $e_{1} = \frac{1}{|G|} \left( \sum_{ g \in G} g \right)$.
