Character table of a group determines the set of commutators of the group. We write $[x,y]$ for the commutator $x^{-1}y^{-1}xy$ of $x$ and $y$ in a group $G$.
$(a)$ Let $g \in G$  and fix $x \in G$. Show that $g$ is conjugate to $[x,y]$ for some $y \in G$ iff
$$\sum_{\chi \in \text{Irr}(G)} \frac{|\chi(x)|^{2}\bar{\chi(g)}}{\chi(1)} \neq 0.$$
$(b)$ Show that if $g=[x,y]$ for some $x,y \in G$ iff
$$\sum_{\chi \in \text{Irr}(G)} \frac{\chi(g)}{\chi(1)} \neq 0.$$
I have done the first part. I think that second part is just applying the first part in a correct way but I cannot do it. Any hint or solution will be appreciated. Thanks in advance.
 A: For the sake of completeness, I provide both solutions to Isaacs, CTFG, Problems $(3.10)$$(a)$ and $(b)$.
(3.10)(a)
Let $G$ be a finite group with $k$ conjugacy classes and let  $K_1, \cdots ,K_k$ be the conjugacy class sums in the group ring $\mathbb{C}[G]$. Let $g_1, \cdots , g_k$ be representatives of each of the $k$ $G$-conjugacy classes $Cl(g_i)$ corresponding to the sums of its elements $K_i$. And let the $k^3$ non-negative integers $a_{ij\nu}$ (see Isaacs, CTFG, (2.4) Theorem) with $1 \leq i,j,\nu\leq k$ be defined by
$$K_iK_j=\underset{\nu=1}\sum^k a_{ij\nu}K_{\nu}.\tag{1}$$
Let $g \in G$ and fix $x \in G$. Assume $x \in Cl(g_j)$ and $g \in Cl(g_{\nu})$ for a fixed $\nu$, $1 \leq \nu \leq k$. Note that of course $x^{-1} \in Cl(g_j^{-1}) \iff x \in Cl(g_j) \iff g_j \in Cl(x)$ and that $|Cl(g_j)|=|Cl(g_j^{-1})|$. Let $Cl(g_j^{-1})=Cl(g_i)$. By Isaacs, CTFG, Problem (3.9) we have
$$a_{ij\nu}=\frac{|Cl(g_i)||Cl(g_j)|}{|G|}\underset{\chi \in Irr(G)}\sum \frac{\chi(g_i)\chi(g_j)\overline{\chi(g_{\nu})}}{\chi(1)}=\frac{|Cl(x)|^2}{|G|}\underset{\chi \in Irr(G)}\sum \frac{|\chi(x)|^2\overline{\chi(g)}}{\chi(1)},\tag{2}$$
so with the $i,j$ and $\nu$ chosen as above
$$\underset{\chi \in Irr(G)}\sum \frac{|\chi(x)|^2\overline{\chi(g)}}{\chi(1)}=\frac{a_{ij\nu}|G|}{|Cl(x)|^2}=a_{ij\nu}\frac{|C_G(x)|}{|Cl(x)|}. \tag{3}$$
We see that the left-hand side in equation $(3)$ is positive if and only if $a_{ij\nu} \gt 0$. But $a_{ij\nu}=|\{(u.v): u \in Cl(g_i), v \in Cl(g_j), uv=g\}|$. So, if $a_{ij\nu} \gt 0$, then there must exist a $u \in Cl(x^{-1}), v \in Cl(x)$ with $g=uv$. If $u=s^{-1}x^{-1}s$ and $v=t^{-1}xt$, then $g=s^{-1}[x,ts^{-1}]s$, so $g$ is conjugate to $[x,y]$, with $y=ts^{-1}$. Conversely, if $g=r^{-1}[x,y]r$ for some $r,y \in G$, then $g=r^{-1}x^{-1}y^{-1}xyr=r^{-1}x^{-1}r \cdot r^{-1}y^{-1}xyr=(r^{-1}x^{-1}r)((yr)^{-1}xyr) \in Cl(x^{-1})Cl(x)$. So $a_{ij\nu} \gt 0$.  $\square$
(3.10)(b) From the formula $(2)$ we get
$$\frac{a_{ij\nu}}{|Cl(g_j)|}=\frac{|Cl(g_j)|}{|G|}\underset{\chi \in Irr(G)}\sum \frac{|\chi(g_j)|^2\overline{\chi(g_{\nu})}}{\chi(1)}.$$
Now summing this left and right over $j=1 , \cdots ,k$ gives
$$\underset{j=1}\sum^k\frac{a_{ij\nu}}{|Cl(g_j)|}=\underset{j=1}\sum^k\underset{\chi \in Irr(G)}\sum \frac{|Cl(g_j)|}{|G|}\frac{|\chi(g_j)|^2\overline{\chi(g_{\nu})}}{\chi(1)}=$$
$$\underset{\chi \in Irr(G)}\sum\underset{j=1}\sum^k \frac{|Cl(g_j)|}{|G|}\frac{|\chi(g_j)|^2\overline{\chi(g_{\nu})}}{\chi(1)}=\underset{\chi \in Irr(G)}\sum \frac{\overline{\chi(g_{\nu})}}{\chi(1)}.\frac{1}{|G|}\underset{j=1}\sum^k |Cl(g_j)||\chi(g_j)|^2= \\ \underset{\chi \in Irr(G)}\sum \frac{\overline{\chi(g_{\nu})}}{\chi(1)}.[\chi,\chi]=\underset{\chi \in Irr(G)}\sum \frac{\overline{\chi(g_{\nu})}}{\chi(1)}.$$ So,
$\underset{\chi \in Irr(G)}\sum \frac{\overline{\chi(g_{\nu})}}{\chi(1)}=\underset{j=1}\sum^k\frac{a_{ij\nu}}{|Cl(g_j)|}$ and because the latter is a non-negative rational number, we see that in fact $$\underset{j=1}\sum^k\frac{a_{ij\nu}}{|Cl(g_j)|}=\underset{\chi \in Irr(G)}\sum \frac{\chi(g_{\nu})}{\chi(1)}.\tag{4}$$
The calculations of $(3.10)(a)$ now show with the equation $(4)$ that $a_{ij\nu} \neq 0$ if and only if $g_{\nu}=[u,y]$ for some $u \in Cl(g_j)$ and $ y \in G$.$\square$
