Presentation $\langle x,y\mid x^3=y^3=(xy)^3=1\rangle $ with normal abelian subgroup.

Question

Let $G$ have the presentation $G=\langle x,y\mid x^3=y^3=(xy)^3=1\rangle $ and let $S=\{xyx,x^2y\}$, $H=\langle\, S\,\rangle $. Prove that $H$ is a normal abelian subgroup of $G$.

That the elements of $S$ commute is easy to see. And then $H$ is abelian. As to $H$ being normal, I think it is enough to prove that given $s\in S$, $x^{-1}sx \in H$ and $y^{-1}sy \in H$. But how do I do it? I have tried but I failed.

Answer 1

I will denote the inverse of an element by a prime. So for instance $x':=x^{-1}=x^2$.

We have for the two generators $s=xyx$ and $t=x^2y$ the relations

• $st=(xyx)(x^2y)=xyx^3y=xyy=xy^2$, and
• $ts=(x^2y)(xyx)=x(xy)(xy)(xy)y^2=xy^2$
• so $st=ts$, and
• its inverse is $yx^2$ because of $(xy^2)(yx^2)=x(y^3)x^2=xx^2=1$.

Then there are following relations of conjugation with $x,y$: $$ \begin{aligned} x'sx &= x^2\ xyx\ x = yx^2=(st)^{-1}\in \langle S\rangle = H\ , \\ y'sy &= y^2\ xyx\ y =yx^2\ (xy)(xy)(xy)=yx^2\in H\ , \\ x'tx &= x^2\ x^2y\ x = xyx = s\in H\ , \\ y'ty &= y^2\ x^2y\ y = (y^2x)(xy^2)=t'(st)\in H\ . \end{aligned} $$ Note: If we want to show $xsx'\in H$ explicitly, we may use $xsx'=x'x'sxx=x'(x'sx)x$, etc.