Let $G=\langle x,y|x^{8}=y^{2}=e,yxyx^{3}=e \rangle$. Show that $|G|\leq16$. Assuming that $|G|=16$, find the center of $G$ and the order of xy.

Question

Let $G=\langle x,y|x^{8}=y^{2}=e,yxyx^{3}=e \rangle$. Show that $|G|\leq16$. Assuming that $|G|=16$, find the center of $G$ and the order of xy.

I am having trouble with this proof. I have a hint but do not understand how to piece it together. Any help would be appreciated.

Answer 1

For the first part you can use the commutativity relation $yx=x^{-3}y^{-1}$ to rewrite every element in the group in the form $x^my^n$, with $0\le m\lt8$ and $0\le n\lt2$. This gives $|G|\le16$.

Assuming $|G|=16$, we have $G=\{e,x,x^2,x^3,x^4,x^5,x^6,x^7,y,xy,x^2y,x^3y,x^4y,x^5y,x^6y,x^7y\}$.

For the center, if it had order $8$, the quotient would be cyclic and $G$ would be abelian. So it has order $2$ or $4$. The center has order $2$, and is $\{e,x^4\}$.

It's looking more and more like we have $D_8$.

Finally, the order of $xy$ is $8$. Because $(xy)^8=(xyxy)^4=(xx^{-3})^4=(x^{-2})^4=x^{-8}=e$, and no smaller power gives $e$.