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I have an exercise which involves groups. It's probably common, but I don't know how to solve it. It sounds like this:

Let (G, $\cdot$) be a group and x, y $\in$ G. Show that ord(xy) = ord(yx).

where ord(x) is the order of x. I just tried to write $xy\cdot xy \cdot \space ...\space\cdot xy = x \cdot (yx) \cdot (yx) \cdot\space ...\space \cdot (yx) \cdot y$, letting x and y outwards. I don't know if this is a good step and I don't know how to continue. I would be glad if someone could help me.

Uncle Sam
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  • Note: all you need do is remark that $xy$ and $yx$ are always conjugates, as $yx = x^{-1},xy,x$ – lulu Feb 15 '17 at 13:27

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Hint: $xy\cdot xy \cdot \space ...\space\cdot xy = y^{-1} \cdot (yx) \cdot (yx) \cdot\space ...\space \cdot (yx) \cdot y$, so the elements $xy$ and $yx$ are conjugate. Prove that conjugate elements have the same order and use $(h^{-1}gh)^i=h^{-1}g^ih$.

Nicky Hekster
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