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. This problem is important in the sense that it determines all the commutators of a group from the character table. The set of commutator generally does not form a subgroup. Any hint or solution will be appreciated. Thanks in advance.

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    $\begingroup$ Is this homework? It certainly looks like it. $\endgroup$ Apr 14, 2017 at 13:06
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    $\begingroup$ If you have correctly proved the first part, you will know that the first sum given is non-negative for any $x,g,$ and that for a fixed choice of $x,g$ it is strictly positive if and only if $g$ is a commutator of the form asked for in the first part of the question. Clearly a sum of non-negative real quantities is non-negative, and is strictly positive if and only if one of the terms in the sum is strictly positive. $\endgroup$ Apr 14, 2017 at 13:18
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    $\begingroup$ There are many paper on such questions, eg by Isaacs and his students, and by G. Navarro. There is a generalizaton of the above commutator result givenn as an exercise in Burnside's book: an element $g \in G$ is expressible as a product of $n$ commutators if and only if $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi(g)}{\chi(1)^{2n-1}} \neq 0.$ $\endgroup$ Apr 14, 2017 at 15:10
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    $\begingroup$ For future reference: if you are quoting exercises from a book, it is good manners to give the actual citation (e.g. Isaacs's book). The way you phrased your questions suggests very strongly that they are someone else's words, so you should acknowledge that, to help people appreciate the context and motivation for the questions $\endgroup$
    – Yemon Choi
    Apr 14, 2017 at 17:42
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    $\begingroup$ (This is indeed Exercise 3.10 from Isaacs's Character Theory book.) $\endgroup$ May 4, 2017 at 11:45


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