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This question already has an answer here:

Show that the following group is identity:

$$G=\langle x,y,z \mid xyx^{-1}=y^{2}\, , \, yzy^{-1}=z^{2}\, , \, zxz^{-1}=x^{2} \rangle.$$

This group is its own derived group. So all I get is group is perfect. Am I correct? What else I missed? Should I use tietze transformation, I am not getting anywhere with them.

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marked as duplicate by user147263, PVAL-inactive, user99914, dustin, Johanna Mar 7 '15 at 3:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Well, it's relatively straightforward to show that no proper normal subgroup can contain an $x_i$. Don't know if that helps, though. $\endgroup$ – Avi Steiner Feb 5 '15 at 6:48
  • $\begingroup$ Also, (again just tossing out an idea) notice that cyclically permitting the $x_i$'s gives an automorphism of the group. So, all the $x_i$'s have to have the same order. Moreover, if they're finite order, that order must be odd. $\endgroup$ – Avi Steiner Feb 5 '15 at 14:25
  • $\begingroup$ The same question is here although there is no answer so far. I added a bounty to that question, so you may want to check there every once and a while and see if it gets any responses. $\endgroup$ – Paul Plummer Feb 5 '15 at 21:50
  • $\begingroup$ Do you know any reference where you found this question. I saw it in previous year question paper of my institution. $\endgroup$ – Bhaskar Vashishth Feb 5 '15 at 21:53
  • $\begingroup$ I found the linked question in the "Related" section that is to the right of the questions with links to other questions. Your the one that posted the question which is where I found it. "Some presentations of the trivial group" by Miller and Schupp could be relevant but I don't see an obvious way to extend the technique to this presentation. Where did you find the question? $\endgroup$ – Paul Plummer Feb 5 '15 at 22:02
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Here is a proof-

$G=\langle x,y,z \mid xyx^{-1}=y^{2}\, , \, yzy^{-1}=z^{2}\, , \, zxz^{-1}=x^{2} \rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.

So consider the subgroup generated by $H=\langle x,y \rangle$ and show that $H$ is solvable. For that consider $H_1=\langle y \rangle < H$ and it is easily seen that $H_1 \unlhd\ H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1\ \cong\ \langle x \rangle$ which is abelian and hence $H$ is solvable.

Now only thing remains is to check that $H=G$, i.e. $z \in \langle x,y \rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.

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