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Related to this post of my coauthor Sebastien, I should also mention that one can also prove the following dual result:

For any finite group G, the following identity holds: $$ \left(\prod_{j=0}^m \frac{C_j}{|\mathcal C^j|}\right)^2=\frac{1}{|G'|}\sum_{\mathcal C^j\subseteq G'} C_j $$ where $G'$ is the commutator subgroup of $G$. Also $\{\mathcal C^j\}_{j=0}^m$ are all the conjugacy classes of $G$ and $|\mathcal C^j|$ is the size of the conjugacy class. The element $C_j:=\sum_{g \in \mathcal C^j}g$ is the class sum associated to $\mathcal C^j$.

Question: Probably this identity is known for finite groups but I couldn't locate it exactly in the literature. Could you please give me an exact reference for this identity?

In particular this identity proves that $|G'|\big{|} \prod_{j=0}^m |\mathcal C^j|$ which should also probably be known.

I should also mention that this identity can also be obtained as a particular case of Theorem 1.8 of this paper.

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    $\begingroup$ I think this is in one of Feit's books. It may have been noted by Harada somewhere. it may also be in one .of Gabriel Navarro's books, I have certainly seen it in more than one place in the literature. As you probably realised if you proved it yourself, the key is that every non-linear complex irreducible character vanishes somewhere, so all central characters of (the centre of) the complex group algebra of a finite group $G$ vanishes on the product of the class sums. $\endgroup$
    – Geoff Robinson
    Commented Jun 17, 2023 at 20:03
  • $\begingroup$ @Geoff Yes, this is also the proof we have in the more general settings of fusion categories. For the other identity, $$\left(\prod_{\chi \in \mathrm{Irr}(G)} \frac{\chi}{\chi(1)} \right)^2 =\frac{|Z(G)|}{|G|}\left(\sum_{\chi \in \mathrm{Irr}(G/Z(G))}\chi(1) \chi\right)$$ we use the dual situation. We have to know that all the columns of the character table, except those coming from the central conjugacy classes, contain at least one zero entry. $\endgroup$
    – Sebastian Burciu
    Commented Jun 17, 2023 at 20:14
  • $\begingroup$ That is the reason why we called such groups "dual-Burnside groups". $\endgroup$
    – Sebastian Burciu
    Commented Jun 17, 2023 at 20:21
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    $\begingroup$ @Geoff: Thanks! I have found it as Corollary 4.15 in Gabriel Navarro's book "Character Theory and the McKay Conjecture". I will write as an answer to this question. $\endgroup$
    – Sebastian Burciu
    Commented Jun 17, 2023 at 20:48

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At Geoff's suggestion I have found that the identity for the product of all conjugacy classes is attributed to Harada and it can be found for example Corollary 4.15 in the book "Character Theory and the McKay Conjecture" by Gabriel Navarro.

It is written even in a little more general context, without a square, as

$$ \prod_{j=0}^m \frac{C_j}{|\mathcal C^j|}=\frac{1}{|G'|}\sum_{g\in G'} gx $$

where $x:=x_1x_2\dots x_m$ is the product of $m$-group elements arbitrarily chosen with $x_j\in \mathcal C_j$.

If one squares this product then one obtains the identity from the question since it is easy to see that $x^2G'=G'$.

Indeed if $x_i$ is not conjugate to $x_i^{-1}$ then one can choose $x_i$ and $x_i^{-1}$ as representatives of the two distinct conjugacy classes. Then getting them one next to the other does not change the coset modulo $G'$.

On the other hand if $x_i$ is conjugate to $x_i^{-1}$ then I can choose in the square of the product an element $x_i$ in the first factor and and then an element $x_i^{-1}$ in the second factor, since the coset of the product is independent of the representatives $x_i$.

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    $\begingroup$ For further information: The paper of K. Harada which contains this result has the title "On a Theorem of Brauer and Wielandt". The special case of the result when $G = G^{\prime}$ was proved by Brauer and Wielandt: If $G$ is a finite group with $G = G^{\prime}$, then the product of all class sums of the conjugacy classes of $G$ contains every element of $G$ with the same multiplicity. I think this version is stated in one of Feit's books. $\endgroup$
    – Geoff Robinson
    Commented Jun 17, 2023 at 22:16

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