I get stuck with this question :

Let $g$ be an element in a finite group $G$, and let $k$ be an integer coprime to $|g|$. Prove that $g$ is a commutator in $G$ if and only if $g^k$ is a commutator.

It is obvious that if $g$ is a commutator, i.e., $g\in G'$ then $g^k\in G'$ since $G'$ is a subgroup of $G$. I dont know how to proceed the converse. Do you have any hint?

Thanks in advance.

  • $\begingroup$ "$g$ is a commutator" is not the same thing as "$g$ is in $G'$". The subgroup $G'$ is generated by the commutators, but it may contain elements that are not, themselves, commutators. $\endgroup$ – Gerry Myerson Oct 21 '12 at 23:12
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    $\begingroup$ This is problem 3.11 in Isaacs's CT book. Problem 3.10 is a great help. $\endgroup$ – user641 Oct 22 '12 at 2:58
  • $\begingroup$ Thank Gerry for pointing my mistakes and thanks also to Steve for pointing me out the problem. I will try to solve it and ask you later if I could not then. $\endgroup$ – 9999 Oct 22 '12 at 19:37
  • $\begingroup$ @Steve : Yes, using the hints to Problem 2.12 and Problem 3.10, we can deduce this problem easily. Thanks again to Steve. $\endgroup$ – 9999 Oct 22 '12 at 19:50
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    $\begingroup$ $g=[a,b]=a^{-1}b^{-1}ab$ for some $a,b\in G$. Then $g^k=k^{-1}a^{-1}b^{-1}abk=k^{-1}a^{-1}kk^{-1}b^{-1}kk^{-1}akk^{-1}bk=[a^k,b^k]$. Just kidding. Anyway, I just answered what is evidently problem 2.12 over here math.stackexchange.com/questions/218302/… if that would help you. $\endgroup$ – Alexander Gruber Oct 22 '12 at 22:35

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