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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.

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    "$g$ 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.2012-10-21
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    This is problem 3.11 in Isaacs's CT book. Problem 3.10 is a great help.2012-10-22
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    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.2012-10-22
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    @Steve : Yes, using the hints to Problem 2.12 and Problem 3.10, we can deduce this problem easily. Thanks again to Steve.2012-10-22
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    $g=[a,b]=a^{-1}b^{-1}ab$ 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 http://math.stackexchange.com/questions/218302/rational-conjugacy-class if that would help you.2012-10-22
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    So now that you can solve the problem, you should post a solution. Later, you can accept it.2012-10-23

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