If $g$ is commutator then so is $g^m$ for $(m,o(g))=1$ 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.)

A character-free 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!
