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Let $G$ be a finite group and $g$ be a commutator.

It can be shown that if $m$ is a positive integer coprime to order of $g$, then $g^m$ is also a commutator (link).

Q. Is there any example of a finite group such that $g$ is a commutator but for $m$ a divisor of order of $g$, the element $g^m$ is not a commutator?

Beginner
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    Yes, $\mathtt{SmallGroup}(96,3)$. – Derek Holt Jun 28 '19 at 06:47
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    @Derek: I assume that is the smallest example? Maybe it is possible to provide the actual presentation of the group, the element $g$ and divisor $m$. – Nicky Hekster Jun 28 '19 at 16:40
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    I believe that two groups of order $96$ are the smallest groups in which not every element of the commutator subgroup is a commutator. I think only one of these is a counterexample to the question but I would need to check that. And $m=2$. – Derek Holt Jun 28 '19 at 17:16
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    It is $((C4:C2):C4):C3$; one source for presentation is a paper "Commutators: A Survey" by Kappe and Morse. They give the presentation, you can easily find in it. Here $G'$ is Sylow $2$-subgroup (order $32$), but the number of commutators is $29$; they have order $2$ or $4$ (or $1$); among these $29$ members, one of order $4$ is desired $g$. – Beginner Jun 28 '19 at 17:52
  • Paper of Kappe and Morse is here: bit.ly/2xiGWnT – Nicky Hekster Jun 29 '19 at 21:02

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