Show that the group is trivial. 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.
 A: 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 $y$.
