# Prove $G \cong \langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$

Let $D^3= D \times D \times D$ where $D = D_\infty$ where we see $D$ as the group generated by $\mathbb{Z}$ and element $0^*$ of order $2$ such that $0^*n0^*=-n$ for all $n \in \mathbb{Z}$. Letting $n0^*$ be $n^*$ gives the following multiplication: $$nk=n+k \\ nk^*=(n+k)^* \\ n^*k=(n-k)^* \\ n^*k^*=(n-k)$$
Now let $G$ be the subgroup of $D^3$ generated by $a=(1,0^*,0^*)$ and $b=(0^*,1,1^*)$. Then I have to show that $G \cong \langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$

Let $F(x,y)$ be the free group on $x$ and $y$, and we take a map $\phi : F(x,y) \to G$ such that $x \to (1,0^*,0^*)$ and $y \to (0^*,1,1^*)$. This is a homomorphism and also onto, so it is an epimorphism.

Now we claim that $ker(\phi)=R^F$, where $R^F$ is the normal closure of $R$ in $F$ where $R$={$x^{-1}y^2xy^2, y^{-1}x^2yx^2$} $\subset F$.

Now $R^F \subseteq ker(\phi)$ is clear as $R \subset ker(\phi)$ but to prove $ker(\phi) \subseteq R^F$, if I let $x^{k_1}y^{l_1}x^{k_2}y^{l_2}\dots x^{k_s}y^{l_s}x^{k_{s+1}}$ be a word in $ker(\phi)$, how can I put it in $R^F$?

## 1 Answer

Your notation for $D$ with its mixture of additive and multiplicative is horrible, but fortunately I think I can ignore that!

Let $H$ be the group $\langle x,y \mid y^{-1}x^2y=x^{-2}, x^{-1}y^2x=y^{-2} \rangle$. A good general strategy for this type of problem is to find a normal form for elements of $H$. If you can show that none of your normal form words in $H$ lie in $\ker \phi$, then you have proved that $\phi$ is an isomorphism, and you are done.

Now, from the relations of $H$, you can see that $N = \langle x^2,y^2\rangle$ is an abelian normal subgroup of $H$, and you can move all even powers of $x$ and $y$ to the right of a word in the generators. You can also replace $x^{-1}$ by $xx^{-2}$ and thereby reduce every word to the form $wx^{2k}y^{2l}$ for $k,l \in {\mathbb Z}$, where $w$ is an alternating positive word in $x,y$.

Now, since $(yx)^2N = (xy)^{-2}N$, $xyxN= y^{-1}(yx)^2N = y(xy)^{-2}N$, $yxyN = x^{-1}(xy)^2N=x(xy)^2N$, and $yxN = x^{-1}y^{-1}(yx)^2N=xy(xy)^{-2}N$, we can write any element of $H$ in the form $w(xy)^{2j}x^{2k}y^{2l}$, for $j,k,l \in {\mathbb Z}$, where $w$ is one of the four words $1,x,y,xy$. (In fact $(xy)^2$ commutes with $x^2$ and $y^2$ and $\langle (xy)^2,x^2,y^2 \rangle$ is normal in $H$, but I don't think you need to prove that.)

So now you just need to check that none of these normal form words lie in $\ker \phi$.

• This an example of torsion free non unique product group, but to prove it is torsion free, I should consider $M=\langle (xy)^2,x^2,y^2 \rangle$, and then show that $x,y, xy$ are not torsion in $G/M$, but then $G/M \cong <x,y\ |\ x^2=y^2=1, (xy)^2=1>$ but is not it presentation for $D_2$ i.e. $\mathbb{Z}_2$ Can you please tell me how to generally go for proving torsion free, given a group presentation – Bhaskar Vashishth Jun 2 '15 at 2:09
• But $x$, $y$ and $xy$ clearly are torsion elements of $G/M$. You have to show that they are not torsion elements of $G$, which is clear because none of $x^2$, $y^2$ or $(xy)^2$ are torsion elements. Also, $G/M$ is a Klein $4$-group so I don't know why you wrote i.e. ${\mathbb Z}_2$. – Derek Holt Jun 2 '15 at 8:53
• Ohh yeah I am sorry, $D_2$ is klein 4 group. I am having trouble understanding that why you wrote so easily that $x^2, y^2, (xy)^2$ is not torsion elements. Suppose if they were torsion, with say order of $x^2=n$ then there their order of images (=$1$) in $G/M$ would divide their original orders i.e. $1\ |\ n$. (so no contradiction here) . – Bhaskar Vashishth Jun 2 '15 at 9:31
• Their images under $\phi$ are not torsion elements so they cannot be either. – Derek Holt Jun 2 '15 at 13:57