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If $x,y$ are element of the finite group $G$ such that $xy=yx$, then is the equation $(xy)^n=x^ny^n$ necessarily true?


I know that if $x^ny^n=(xy)^n$ then $x$ and $y$ commute, but I am not sure about the converse though.

Any tips are welcome.

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    $\begingroup$ Try proof by induction. $\endgroup$ Commented Mar 18, 2019 at 17:10
  • $\begingroup$ A simple induction on $n$ for kids is enough to prove it $\endgroup$
    – Bernard
    Commented Mar 18, 2019 at 17:10
  • $\begingroup$ Use induction on $n$.... $\endgroup$ Commented Mar 18, 2019 at 17:10
  • $\begingroup$ include your own ideas or work and don't use such a long title, down vote from me sadly $\endgroup$
    – AlvinL
    Commented Mar 18, 2019 at 17:13
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    $\begingroup$ What you say you know is false: for example, in $\;S_3\;$ , we have that $\;(12)^6(13)^6=((12)(13))^6\;$ , yet of course $\;(12)(13)=(132)\neq(123)=(13)(12)\;$ $\endgroup$
    – DonAntonio
    Commented Mar 18, 2019 at 17:15

2 Answers 2

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Hint: Show by induction on $m$ that $xy^m=y^mx$ and $x^my=yx^m$ for any $m\in\Bbb Z$.

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    $\begingroup$ That answered my question. Thank you. $\endgroup$
    – Shot
    Commented Mar 18, 2019 at 17:13
  • $\begingroup$ You're welcome, @Shot :) $\endgroup$
    – Shaun
    Commented Mar 18, 2019 at 17:15
  • $\begingroup$ @Shot: if that answers your question, you should probably consider "accept" it by clicking the $\checkmark$ button. $\endgroup$
    – user9464
    Commented Mar 18, 2019 at 21:47
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$$(xy)^n=x(yx)^{n-1}y=x(xy)^{n-1}y=\ldots=x^k(xy)^{n-k}y^k=\ldots=x^ny^n$$

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