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It seems there is a result by Frobenius that states that the number of ways an element $g$ of a finite group can be written as a commutator ($\phi(g) = | \{(x,y) \in G \times G: g = [x,y]\}|$) is given by $\phi(g) = \sum_{\chi} \frac{|G| \chi(g)}{\chi(1)}$, where the sum is taken over all the irreducible characters of $G$.

I can't find the original paper and am having trouble on proving this. I'm trying to make use of the class algebra constants, but it's of no use so far. Would anybody kindly provide some advice?

Thank you!

José Siqueira
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    The result is Proposition 4.1 in Aner Shalev, Commutators, words, conjugacy classes and character methods, available at http://journals.tubitak.gov.tr/math/issues/mat-07-31-suppl/mat-31-suppl-10-0709-19.pdf but the paper may not be too helpful (and maybe it's where you found the result in the first place). – Gerry Myerson Jul 03 '13 at 11:01
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    I believe this is also proved in Isaacs, I will take a look when I get home. – Alexander Gruber Jul 03 '13 at 18:53
  • Could someone check the following argument? We look at $\sum_{x,y} xyx^{-1}y^{-1} \in \mathbb{C}G$. Notice that we may bracket the sum as $\sum_{x,y} (xyx^{-1})y^{-1}=\sum_y |C_G(y)| \hat{y} y^{-1}$. If we have $m$ conjugacy classes, this would turn out to be $\sum_{i=1}^m |C_G(y_i)| \hat{y_i} \hat{y_i^{-1}}$ – José Siqueira Jul 11 '13 at 09:29
  • So using the class algebra constant $\alpha_{\hat{y_i}\hat{y_i^{-1}}\hat{z}}$, we may write this sum as $\sum_{i=1}^m|C_G(y_i)| \alpha_{\hat{y_i}\hat{y_i^{-1}}\hat{z}} \hat{z}$. From the calculations that would follow, we'd get that the coefficient of $\hat{z}$ is exactly $\sum_{\chi} \frac{|G| \chi(z)}{\chi(1)}$, and this coefficient should count in how many ways an element in $\hat{z}$ could be written as $[x.y]$, right? – José Siqueira Jul 11 '13 at 09:40
  • I can't read those chicken scratches in the comments, they're way too small. Why not post it as part of your answer, and use displays? Also, I don't know what $y$ with something over it (tilde? caret?) means. – Gerry Myerson Jul 20 '13 at 13:12
  • Please, disregard those comments. I've found a better way to calculate this, and I've posted here an answer to my own question today. – José Siqueira Jul 20 '13 at 14:20

4 Answers4

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More general results are proved in Alon Amit and Uzi Vishne, Characters and solutions to equations in finite groups, J Alg Appl 10 (2011) 675-686. For the result of Frobenius, they refer to pages 1 to 37 of his Gesammelte Abhandlungen, Band III. They also cite A M A Alghandi and F G Russo, A generalization of the probability that the commutator of two group elements is equal to a given element, arXiv:1004.0934 (now also Bull. Iranian Math. Soc. 38 (2012), 973-986), and T Tambour, The number of solutions of some equations in finite groups and a new proof of Ito's theorem, Commun Alg 28 (2000) 5353-5362. I haven't looked at any of these.

Another paper that might interest you is M R Pournaki and R Sobhani, Probability that the commutator of two group elements is equal to a given element, J Pure Appl Alg 212 (2008) 727-734. But this paper just cites the result of Frobenius, and refers to the collected works, without page numbers.

Victor Wang
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Gerry Myerson
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  • Thank you for this. At the moment I'm trying to prove it before looking at the original proof. So far, I've got a way of calculating in how many ways a given group element can be expressed as a word on two letters. Any tips on how to get the number of ways it can be expressed as a commutator? – José Siqueira Jul 04 '13 at 10:54
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    The funny thing is, the question, how many ways can a given element be expressed as a commutator, occurred to me about a month ago, as I was teaching an intro Group Theory class. I couldn't figure it out, and I thought about posting it as a question here, but I never got around to it. So I was very happy to see you not only ask the question, but give the answer. But as to how to get that answer, I'm sorry, that's something I still don't know. – Gerry Myerson Jul 04 '13 at 12:39
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Actually, the solution follows easily if one combines the result of Problem 3.23 in Isaacs' book on characters with the orthogonality relation (once one replaces $h$ by $g^-1$ in that problem).

To my mind, a different question related to the above is interesting: to get a character-free formula for that number of ordered pairs $(x,y)$ giving $g=[x,y]$ for a fixed $g$ in $G$.

Hope this helps,

Nea Marin

nea marin
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Uber Gruppencharaktere. Sitzungsber. der Berl. Ak., 1896, Seite 985-1021. (See Sect.3 in it).

user1729
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Boris Novikov
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  • What is the full name of the journal? (Also, I have corrected the dates - Frobenius was publishing in 1896 not 1986...) – user1729 Jul 03 '13 at 12:17
  • I have only Russian translation of this paper, and there is not the full name of the journal in it. \ Zentralblatt MATH gives the name: Sitzungsberichte der Königlich Preussischen Akademie der Wissenschaften. – Boris Novikov Jul 03 '13 at 13:53
  • Hmm - there is something not right about that though, as I was presuming the Berl. would be Berlin. I suppose "Preussischen" is Prussia, and Berlin was in Prussia. So maybe. – user1729 Jul 03 '13 at 14:10
  • Yes, Berlin was in Prussia. The publisher of the journal was: Königlich Preussische Akademie der Wissenschaften, Berlin – Boris Novikov Jul 03 '13 at 14:18
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Thank you for all your help, but I've managed to find out in how many ways an arbitrary group element can be written as a given word (and of course this proves this topic's statement). I had opened this Number of ways a group element of a finite group can be written as a given word for this more general question.

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José Siqueira
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