$ [x,y]=y, [y,z]=z, [z,x]=x$ implies $x=y=z=e$ in a group

Asked Mar 24 '20 at 07:35

Active Mar 25 '20 at 11:02

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This is a question about commutators. I'm reading Lang's Algebra, and in the section about free groups, Lang asserts that in a group $G$, if $x,y,z\in G$ satisfies $$ [x,y]=y, [y,z]=z, [z,x]=x$$ where $[x,y]=xyx^{-1}y^{-1}$ is the commutator of $x$ and $y$, then we must have $x=y=z=e$.

How does this hold? From the above relations, we obtain $xy=y^2x, yz=z^2y, zx=x^2z$, but I can't see how to proceed next

edited Mar 25 '20 at 11:02

asked Mar 24 '20 at 07:35

user302934

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Are you sure it's not $[x,y]=x^{-1}y^{-1}xy$? – Gerry Myerson Mar 24 '20 at 07:45
@GerryMyerson Yes. The notation can be found in p.69 of the book. – user302934 Mar 24 '20 at 07:58
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You wrote "where $[x,y]=xyx^{-1}y^{1}$". This has a typo, right? – Dietrich Burde Mar 24 '20 at 10:25
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Of course we also get the trivial group if $[x,y]$ is interpreted as $x^{-1}y^{-1}xy$ and the proof is much easier in that case! – Derek Holt Mar 24 '20 at 17:52
@DietrichBurde Yes it is thanks – user302934 Mar 25 '20 at 11:03

$ [x,y]=y, [y,z]=z, [z,x]=x$ implies $x=y=z=e$ in a group

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