Problem $3.13$ - Character theory of finite groups (M. Isaacs) Let $G$ be a group and $C_{1},C_{2},\dotsc ,C_{s}$ be conjugacy classes of $G$ and $K_{1}, K_{2},\dotsc , K_{s}$ be the class sums in $\mathbb{C}[G]$. Suppose there exists $c \in \mathbb{C}$ such that $\sum K_{i} = c\prod K{i}$. Show that $G=G'$.  Conversely if $G'=G$ then show that there exists $c\in \mathbb{Q}$ such that $\sum K_{i} = c\prod K{i}$.
This is a problem from Isaacs' book of character theory. A hint is given as follows: if $\chi$ is not trivial representation and $\chi$ is irreducible then show that $\frac{h_{i}\chi(g_{i})}{\chi(1)}=0$ for some $i$, where $g_{i}$ is a  class representative of $K_{i}$  and $h_{i}$ is the size of the conjugacy class $C_{i}$.  I need help in this problem. I have neither been able to show the hint nor have any idea what the hint will suggest. 
Thanks in advance!!! 
 A: Theorem 1 Let $G$ be a finite group with $k$ conjugacy classes and let  $K_1, \cdots ,K_k$ be the conjugacy class sums in the group ring $\mathbb{C}[G]$. Assume there exists a $c \in \mathbb{C}$ such that $c\sum_{i=1}^{k}K_i = \prod_{i=1}^{k}K_i$. Then $G$ is perfect, that is $G’=G$.
Proof Since $|G:G’|$ equals the number of linear characters, we need to show that except for the principal character $1_G$, no further linear characters exist. We do this by showing that every non-principal irreducible character must have a zero and hence, cannot be linear. Let $\chi \in Irr(G)$ with $\chi \neq 1_G$. We consider the $\mathbb{C}$-linear algebra homomorphism $\omega_{\chi} : Z(\mathbb{C}[G]) \rightarrow \mathbb{C}$ (for the definition, see Isaacs, CTFG, p. 35) and calculate the image of the above sum and product. Remember that $K_i \in Z(\mathbb{C}[G]) $ for each $i$. Let $g_1, … , g_k$ be representatives of each of the $k$ conjugacy classes $Cl(g_i)$ corresponding to the sums of its elements $K_i$.
Now because of the $\mathbb{C}$-linearity of $\omega_{\chi}$, $$\omega_{\chi}( c\sum_{i=1}^{k}K_i)=c \sum_{i=1}^{k}\omega_{\chi}(K_i)=c\sum_{i=1}^{k}\frac{\chi(g_i)\cdot|Cl(g_i)|}{\chi(1)}=\frac{c}{\chi(1)}\sum_{g \in G}\chi(g)=\frac{c \cdot |G|}{\chi(1)}[\chi,1_G]=0$$ since $\chi$ was chosen to be non-principal. It follows from $\omega_{\chi}$ being also a multiplicative homomorphism that the product
$$\omega_{\chi}(\prod_{i=1}^{k}K_i)=\prod_{i=1}^{k}\omega_{\chi}(K_i)=\prod_{i=1}^{k}\frac{\chi(g_i)\cdot|Cl(g_i)|}{\chi(1)}=0.$$ Hence, that there must be some $i$ such that $\chi(g_i)=0$ and this implies that $\chi$ cannot be linear. $\square$
Remark Put $s=\sum_{i=1}^{k}K_i$. In general the following holds:
$$\omega_{\chi}(s)=\Bigg \{_{0 \text {  if  } \chi \neq 1_G}^{\text {|G| if } \chi=1_G}$$ This follows from the reasoning above, or can also be seen by noting that $s^2=|G|s$, hence $\omega_{\chi}(s)(\omega_{\chi}(s)-|G|)=0.$

Theorem 2 Let $G$ be a finite group with $k$ conjugacy classes and let  $K_1, ...,K_k$ be the conjugacy class sums in the group ring $\mathbb{C}[G]$. Assume that $G$ is perfect. Then there exists a $q \in \mathbb{Q}$ such that $\sum_{i=1}^{k}K_i = q\prod_{i=1}^{k}K_i$. 
For the proof we need a lemma. In the notation above, put $ker(\omega_{\chi})=\{u \in Z(\mathbb{C}[G]): \omega_{\chi}(u)=0 \}$.
Lemma $\bigcap_{\chi \in Irr(G)}ker(\omega_{\chi})=\{0\}$
This Lemma has an obvious consequence.
Corollary Let $u, v \in Z(\mathbb{C}[G])$. Then $u=v$ if and only if $\omega_{\chi}(u)=\omega_{\chi}(v)$ for all $\chi \in Irr(G)$.
Proof of the Lemma Let $u=\sum_{j=1}^{k}u_jK_j \in Z(\mathbb{C}[G])$ with $u_j \in \mathbb{Z}.$ Asssume $u \in ker(\omega_{\chi_i})$ for all $\chi_i \in Irr(G), i=1, ..., k.$ This yields $$\omega_{\chi_i}(u)=\omega_{\chi_i}(\sum_{j=1}^{k}u_jK_j)=\sum_{j=1}^{k}u_j\omega_{\chi_i}(K_j)=\sum_{j=1}^{k}u_j\frac{\chi_i(g_j)|Cl(g_j)|}{\chi_i(1)}=\frac{1}{\chi_i(1)}\sum_{j=1}^{k}u_j\chi_i(g_j)|Cl(g_j)|=0$$ And this implies that for $i=1, ...k$ we get $k$ equations:
$$\sum_{j=1}^{k}u_j\chi_i(g_j)|Cl(g_j)|=0$$
Now we are going to apply a bit of linear algebra. Abbreviate $|Cl(g_j)|:=h_j$. Let us define the $k \times k$-matrix over $\mathbb{C}$, also known as the group determinant, by $X=(\chi_i(g_j))_{i,j=1}^k$, so 
$$
  X=
\left[ {\begin{array}{ccccc}
\chi_1(g_1) & \chi_1(g_2) &  \chi_1(g_3) & \cdots &  \chi_1(g_k)\\
\chi_2(g_1) & \chi_2(g_2) &  \chi_2(g_3) & \cdots &  \chi_2(g_k)\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
\chi_k(g_1) & \chi_k(g_2) &  \chi_k(g_3)& \cdots &  \chi_k(g_k)\\
  \end{array} } \right]
$$
We are going to show that $X$ has an inverse, or equivalently, $det(X) \neq 0$ and this will guarantee that $u_i=0$ for all $i=1,...k$, since $X \cdot$u=0, where the vector is u$=(u_1h_1,...,u_kh_k)^T$. Now if we look at the Hermitian transpose of $X$, $X^H:=\overline{X}^T$, then by the Second Orthogonality Relation, it follows that $$X^H\cdot X=
\left[ {\begin{array}{ccccc}
|C_G(g_1)| &0 & 0 & \cdots &  0\\
0 & |C_G(g_2)| &  0 & \cdots & 0\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
0 & 0 &  0 & \cdots & |C_G(g_k)|\\
  \end{array} } \right]
$$ 
It follows that $det(X^H \cdot X)=|det(X)|^2=\prod_{i=i}^k|C_G(g_i)|$, from which we conclude that $det(X) \neq 0$, which proves the Lemma.$\square$
We are now ready to prove Theorem 2.
Proof of Theorem 2 Put $s=\sum_{i=1}^kK_i$ and $t=\prod_{i=1}^kK_i$. For the principal character we have $\omega_{1_G}(K_i)=|Cl(g_i)|$, so $\omega_{1_G}(s)=\sum_{i=1}^k|Cl_G(g_i)|=|G|$. Likewise, $\omega_{1_G}(t)=\prod_{i=1}^k|Cl_G(g_i)|$. Put $q=\frac{|G|}{\prod_{i=1}^k|Cl_G(g_i)|}$. Then $q$ is a rational number. If $\chi \in Irr(G)$, and $\chi \neq 1_G$, then, since $G$ is perfect, $\chi$ in non-linear and by a theorem of Burnside (CTFG Theorem (3.15)) $\chi$ must have a zero. It follows that $\omega_{\chi}(t)=0$. From the Remark above we also know that $\omega_{\chi}(s)=0$. In summary, for all $\chi \in Irr(G)$ we have $\omega_{\chi}(s)=\omega_{\chi}(qt)$. The previously proved Corollary now yields $s=qt$ as wanted.$\square$
