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In Serge Lang: Algebra it is stated on page 69 that for a group $G$ that, if for $x,y,z\in G$ $y=[x,y]:=xyx^{-1}y^{-1}, z=[y,z], x=[z,x]$ is satisfied, we have $x=y=z=e$. I see that $y=[x,y]$ directly implies $y=x^{-n}y^{2^n}x^n$ for $n>0$, but I need a hint why it is true for $n=0$.

Edit: As pointed out in the comments: the relation should be $y^n=x^{-1}y^{2^n}x$.

Zikrunumea
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    "but I need a hint why it is true for $n=0$" For $n = 0$, we get $x^{-0}y^{2^0}x^0 = y^1 = y$. Is that really what you were looking for? That doesn't seem to be very helpful, as it is true regardless of any commutator relations. – Arthur Mar 27 '19 at 13:12
  • I cannot deduce that $y=x^{-n}y^{2^n}x^n$ for $n=0$. – Zikrunumea Mar 27 '19 at 13:24
  • I just deduced it for you in my previous comment. It's almost trivial. And I am not sure how it would help. – Arthur Mar 27 '19 at 13:26
  • Ah, okay, you are right. I mixed it up with the relation $y^n=x^{-1}y^{2^n}x$, which also holds. – Zikrunumea Mar 27 '19 at 13:30

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