# 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? [closed]

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.

• Try proof by induction. Commented Mar 18, 2019 at 17:10
• A simple induction on $n$ for kids is enough to prove it Commented Mar 18, 2019 at 17:10
• Use induction on $n$.... Commented Mar 18, 2019 at 17:10
• include your own ideas or work and don't use such a long title, down vote from me sadly Commented Mar 18, 2019 at 17:13
• 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 $\;(12)(13)=(132)\neq(123)=(13)(12)\;$ Commented Mar 18, 2019 at 17:15

Hint: Show by induction on $$m$$ that $$xy^m=y^mx$$ and $$x^my=yx^m$$ for any $$m\in\Bbb Z$$.
• @Shot: if that answers your question, you should probably consider "accept" it by clicking the $\checkmark$ button.
$$(xy)^n=x(yx)^{n-1}y=x(xy)^{n-1}y=\ldots=x^k(xy)^{n-k}y^k=\ldots=x^ny^n$$