Supposing we have a group $(G,*)$ (Which we don't know the elements of).
It is given that for each $x, y \in G$, exists $x*y*x=y$.
That's the only information we receive, How can we prove that group G is an abelian group?
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5As a hint, consider what happens when you take y to be the identity element. – Matthew Towers Aug 12 '20 at 10:08
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1Do you mean $y^{-1}$ with "$y*$" ? – Peter Aug 12 '20 at 10:51
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@Peter , No. It was a misclick due to me being on the phone instead on the PC. Anyways, I received the answer I needed. – Max Ilyouchenko Aug 12 '20 at 12:09
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If $xyx=y$ for each pair $x,y\in G$ then in particular this is true if we take $y=e$, where $e\in G$ is the netural element. We obtain $x^2=e$ for any $x\in G$, i.e. every element of $G$ is of order $2$. And such groups are abelian as you can read here.
freakish
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Thank you very much for your quick answer! It really helped me and I finally answered the question. – Max Ilyouchenko Aug 12 '20 at 10:13