# How to show that $G\cong\mathbb Z_3$

I am working on this one:

Let $G=\langle x,y|xyx^{-1}y^2=1, yx^{-1}yx^2y=1\rangle$. Show that $G\cong\mathbb Z_3$

What I have done for this is to consider subgroup $H=\langle x\rangle$ and to find the index of $H$ in $G$ by doing Todd-Coxeter Algorithm. I did this algorithm and find $3$ right cosets for the subgroup $H$ in the group. That means to me that $[G:H]=3$. I am not sure this made the group to be cyclic, because I know some non abelian semi-direct product. Thanks for helping me in this problem.

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With your preliminary results, it seems necessary to show that $H$ is trivial, right? –  Hagen von Eitzen Oct 1 '12 at 10:05
Perhaps. But if I could conclude $G$ is cyclic, since $[G:G']=3$ then the group would be $\mathbb Z_3$. –  B.S. Oct 1 '12 at 10:14
Does that mean it is easy to see that $G'=H$? –  Hagen von Eitzen Oct 1 '12 at 10:18
I don't know. I could just find $[G:H]=3$ and the order of $$\frac{G}{G'}=\langle x,y|xyx^{-1}y^2=1, yx^{-1}yx^2y=1, [x,y]=1\rangle\cong \langle x,y|x=1,y^3=1,[x,y]=1\rangle$$ –  B.S. Oct 1 '12 at 10:23
With $H=\langle x\rangle$ you have already shown that $[G:H]=3$. You have also shown that $[G:G']=3$. Since $x=1$ appears as relation in your expression for $G/G'$, you have $H\subseteq G'$ and thus $G'=H=\langle x\rangle$. Therefore $yxy^{-1}x^{-1}\in H$ and thus $yx=x^ky$ for some $k\in\mathbb Z$. But then $$G=\langle x,y\mid xyx^{-1}y^2=1, yx^{-1}yx^2y=1, yx=x^ky\rangle.$$ We conclude $yx^{-1} = x^{-k}y$, i.e.$1=xyx^{-1}y^2=x^{1-k}y^3$. But then $yx=x^ky=xx^{k-1}y=xy^4$. Hence $1=yx^{-1}yx^2y=yx^{-1}xy^4xy=yx^{-1}xxy^{16}y=xy^{21}$. With this we see at least that $G$ is cyclic (generated by $y$) and is in fact presented by $$G=\langle y\mid y^{-21}yy^{21}y^2=1, yy^{21}yy^{-42}y=1\rangle =\langle y\mid y^3=1, y^{-18}=1\rangle=\langle y\mid y^3=1\rangle\cong\mathbb Z/3\mathbb Z.$$
Sorry for backing but, do you think we could find a cyclic subgroup with index 1 in $G$? Thanks and sorry again. –  B.S. Oct 1 '12 at 12:03
$y$ has order 3. –  i. m. soloveichik Oct 1 '12 at 13:21