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Let $G$ be a group. Let $x,y,z \in G$ such that $[x,y]=y$, $[y,z]=z$, $[z,x]=x$ (the commutators; $[x,y]=xyx^{-1}y^{-1}$). Prove that $x=y=z=e$.

I tried to show it by proving that $zx^mz^{-1}=x^{2m}$ with induction. Therefore, if the order of $x$ is even, we can take $zx^{|x|/2}z^{-1}=x^{|x|}=e$ and thus, $x^{|x|/2}=e$ proving it.

However, what if the order of $x$ is infinite (or even just odd)? I don't know what to do about that cases.

Thank you very much in advance!

Shaun
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Micheal21
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  • It might help to look at the presentation $$\langle a,b,c\mid [a,b]=b, [b,c]=c, [c,a]=a\rangle.$$ – Shaun Jun 12 '20 at 15:53
  • @Shaun I don't really see how it helps. I probably would like to show it equals the trivial set, but I don't see how. Could you clarify it maybe a little bit? – Micheal21 Jun 12 '20 at 16:03
  • @user158834 The question has surely been asked here before, but unfortunately the question you link to asks for a specific kind of proof so isn't necessarily relevant (if it had an answer then that would be fine though!). – user1729 Jun 12 '20 at 16:10
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    I've closed this as a duplicate. Every group satisfying your conditions is a homomorphic image of the group shown to be trivial in the linked question, so your question easily follows from it. – user1729 Jun 12 '20 at 16:16
  • @user1729: Did you remove my comment? What makes you think you have the right to do that? The exercise in Serre does not have a solution in Serre. The proof can be easily found on the Web. – markvs Jun 12 '20 at 16:25
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    @user158834 That wasn't me. Only moderators can delete comments. (I presume it was deleted automatically, as it was auto-generated [I guess?] and about closing as a duplicate.) – user1729 Jun 12 '20 at 16:28
  • Another duplicate https://math.stackexchange.com/questions/2394006/prove-the-higman-group-is-infinite-but-a-similar-one-is-trivial – markvs Jun 12 '20 at 16:28

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We have $$xyx^{-1}y^{-1}=y$$ so $$xyx^{-1}=y^2$$ Similarly $$yzy^{-1}=z^2$$ and $$zxz^{-1}=x^2$$ Note also that since $xyx^{-1}y^{-1}=y$, we have $$yx^{-1}y^{-1}=x^{-1}y$$ so $$yxy^{-1}=y^{-1}x$$ Thus $$yzxz^{-1}y^{-1}=z^2y^{-1}xz^{-2}=y^{-1}xy^{-1}x$$ hence $$yzxz^{-1}y^{-1}x^{-1}=y^{-1}xy^{-1}$$ so that $$yzxz^{-1}x^{-1}y^{-2}=y^{-1}xy^{-1}$$ so that $$yzxz^{-1}x^{-1}=y^{-1}xy$$ Thus $$yzxz^{-1}x^{-1}yzxz^{-1}x^{-1}=x$$ so that $$yzxz^{-1}x^{-1}yzxz^{-1}=e=yx^2yx^{-1}yx^2=yxy^3x^2$$ so $$x^{-1}y^{-1}=y^3x^2$$ so that $$xy^4x^2=x$$ and hence $$xy^4x=e$$ so that $$xxy^4xx^{-1}=xy^8x=e$$ so that $$xy^4xx^{-1}y^{-8}x^{-1}=xy^{-4}x^{-1}=y^{-8}=e$$ It follows that at least one of $y,y^2,y^4,y^8$ is the identity, and since each of these can be obtained by conjugating $y$ by some power of $x$ it follows that they all are. It is easy from there to see that $x=e$ and $z=e$ as well.

Matt Samuel
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