Let $G=\langle x,y\mid x^3=y^3=(xy)^3=1\rangle$.
Show that $A\rtimes\langle t\rangle\cong G$ where $t^3=1$ and $A=\langle a\rangle\times\langle b\rangle$ is an abelian group isomorphic to $\mathbb{Z^2}$ and $t$ is action on $A$ with this relation$a^t=b$ and $b^t=a^{-1}b^{-1}$
Tip:
The group $\langle xyx,x^{2}y\rangle$ is a normal abelian subgroup of $G$.
I was trying to solve this problem using von Dyck's Theorem but I have no idea where I would map the generators $x,y$ to. Would really appreciate if someone showed me how to solve this problem .
Best Answer
You need to put more effort into solving the problem yourself, so I am just going to provide some hints to get you started. First show that $(xyx)(x^2y) = (x^2y)(xyx)$. That proves that the subgroup in the hint is abelian.
To show it's normal, show that $x(xyx)x^{-1} = x^2y$ and $x(x^2y)x^{-1} = (xyx)^{-1}(x^2y)^{-1}$. So now we can see that we need to put $a=xyx$, $b=x^2y$, and $t=x^{-1}$
Now, putting $A = \langle a,b \rangle$, it is easy to see that $\langle A,x \rangle = G$, so we have shown that $A \unlhd G$ and clearly $|G:A| \le 3$.
To complete the proof, prove that the semidirect product ${\mathbb Z}^2 \rtimes \langle t \rangle$ with the specified action satisfies the relations of the presentation of $G$.