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There are certain theorems in finite group theory whose proofs involve character theory and for which there are still no character-free proofs. Among such is Frobenius theorem on transitive permutation groups. (Another was Burnside's $pq$ theorem; but there is now group theoretic proof.

I am considering one such theorem, whose proof is based on following theorem.

Theorem 1. An element $g\in G$ is a commutator if and only if $\sum_{\chi\in{\rm Irr}(G)} \frac{\chi(g)}{\chi(1)}\neq 0.$

So if $m$ is an integer relatively prime to $o(g)=n$, then consider the Galois automorphism $\sigma$, which, on $n$-th roots of unity acts by $\zeta_n\mapsto \zeta_n^m$. Then $(\chi(g))^{\sigma}=\chi(g^m)$. Thus, if $g$ is a commutator, then applying $\sigma$ to inequality in theorem we get that $g^m$ is also a commutator. Thus,

Theorem 2. If $g\in G$ is a commutator then so is $g^m$ for $(m,o(g))=1$.

The Theorem 2 is purely group theoretic; but proof involves character theory arguments. Question is now simple:

Q. Is there character-free proof for Theorem 2?

(I do not know whether this question has been considered before by anyone.)

Nicky Hekster
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    With HNN extension (which needs no character theory) you can find a group $X$ containing $G$ such that $g$ and $g^m$ are conjugate in $X$. So then at least in $X$, the element $g^m$ is a commutator. Just an idea, I don't know if this could be used to show that $g^m$ is a commutator in $G$. – spin May 01 '17 at 14:17
  • Also, the inverse of a commutator $[x,y]$ is equal to $[y,x]$, so the claim is clear if $m = o(g) - 1$. Now the smallest $o(g)$ where we need more than this is $o(g) = 5$: We should show that if $[x,y]^5 = 1$, then $[x,y]^2$ is a commutator. Not sure how to do even this case. – spin May 01 '17 at 14:31
  • In the few examples I have considered (abelian, symmetric, alternating, dihedral groups), for each $g\in G$ and $m$ coprime to $g$, there is an automorphism of $G$ taking $g$ to $g^m$ (whether or not $g$ is a commutator). Is it possible this is always the case? – Julian Rosen May 02 '17 at 13:30
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    @JulianRosen: Definitely not. Now $g$ is conjugate in $G$ to $g^m$ for all $m$ coprime to $o(g)$ if and only if $g$ takes rational value on all irreducible characters. So it would be enough to find $G$ without outer automorphisms such that the character table contains irrational entries. Probably there are much easier examples, but looking at some tables from books it seems the Mathieu group $M_{11}$ is one example like this. – spin May 02 '17 at 21:22

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A character-free and elegant proof of this result of K. Honda (1953) has recently been found by Hendrik Lenstra. It has been published in Operations Research Letters, Powers of Commutators, Volume 51, Issue 1, January 2023, Pages 17-20. Here is the link. The proof involves calculations in (several) group rings. It is fun and a joy to read!

Nicky Hekster
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