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I had previously asked about the number of ways a group element in a finite group could be written as a commutator (the question is still open for a proof, by the way) In how many ways can a group element in a finite group be written as a commutator?

Let $G$ be a finite group. Suppose we have a word on $k$ letters and we want to write a given $g \in G$ using that word (for example, if $k=2$, we'd be looking at the number of ways $g$ can be written as a product, which should be $|G|$).

For the case where the word is a product of $m$ commutators (so that $k=2m$), the following procedure should give the number of ways we can write the identity element as that word:

Let $\delta$ be the regular character of $G$ and $w$ be the word (which is a product of $m$ commutators). So $\delta$ takes value $|G|$ at the identity and $0$ at every other element. So $\sum_{\vec{g} \in G^{2m}} \delta(w(\vec{g}))=|G| \gamma(w)$, where $\gamma(w)$ is the number of ways we can write the identity with $w$. So we are able to calculate $\gamma$(w).

If $w$ is a product of $m$ commutators, one may also prove that $\gamma(w)=|G|^{2m-1}\zeta(2m-2)$, where $\zeta$ is Witten's zeta function ($\zeta(k)=\sum_{\chi} \chi(1)^{-k}$, where the sum is taken over all irreducible characters).

Is it possible to get the number of ways to write a general element $g$, from this result for the identity (or an analogous method)? At least for the case where $w$ is a product of commutators?

Thank you very much!

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José Siqueira
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  • This question is unclear, and I don't see what commutators have to do with it. I would say the answer is $|G|^{k-1}$ independently of the target element$~g$, since one can choose $k-1$ letters freely in $G$, and then the final letter is determined by the requirement that the total product must be$~g$. If this is not correct, the question should be reformulated to make it clear why not. – Marc van Leeuwen Jul 26 '13 at 12:31
  • @MarcvanLeeuwen I just mentioned the case of the commutators as it's simpler to calculate (and was my original question). Maybe the answer below makes it clear why it wouldn't work as suggested? Sorry if I was unclear, though. – José Siqueira Jul 27 '13 at 11:57

1 Answers1

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Nevermind, I've sorted it out. In the end you get that the number of ways you can write a group element $g$ with a word $w$ is given by $\sum_{\chi} \frac{\overline{\chi(g)}}{|G|} \sum_{\vec{g} \in G^n} \chi(w(\vec{g}))$. For that, just look at the class function $f$ that takes value $|G|$ at elements in the conjugacy class of $g$ and $0$ everywhere else (then do the analogous procedure). This is nice, since it also proves the statement I wanted to prove in the commutator question (by taking $w=[a,b]$).

Thanks anyways!

José Siqueira
  • 933
  • Nice quanswer. $+1$ – Alexander Gruber Jul 20 '13 at 19:00
  • @AlexanderGruber Thank you. Do you have any nice suggestions for calculating $\sum_{G^n} \chi(w(\vec{g}))$ for a general word? Useful tricks, methods? It is surely simpler for words such as products of commutators, or sums over $x \in G$ of $\chi(xgx^{-1})$, but what about other sorts of words? Say, $a^2b^2$? – José Siqueira Jul 21 '13 at 12:56
  • What does $w(\vec{g})$ mean? – Ben Nov 29 '24 at 01:49