Why $X_{2n}$ when $n=3k$ has order $6$?

Question

Let $X_{2n}$ be the group whose presentation is$\langle x,y\,|\,x^n=y^2=1, xy=yx^2\rangle$. From $x=xy^2$, it is seen that $x^3=1$, hence $X_{2n}$ has at most $6$ elements. I have to show that if $n=3k$, then $X_{2n}$ has exactly $6$ elements.

I can't see where I am having problem if I assume $x=1$.

Answer 1

The elements $x=(1,2,3)$ and $y=(2,3)$ of $S_3$ satisfy the three relations when $n=3k$, and $S_3$ has order $6$ and is generated by $x$ and $y$. So $|X_{2n}| \ge 6$ when $n=3k$.

Also, the relations can be used to write any element of $X_{2n}$ in the form $x^ay^b$, the fact that $x^3=y^2=1$ in $X_{2n}$ proves that $|X_{2n}| \le 6$ for all $n$.

So $|X_{2n}|=6$ when $n=3k$.

Answer 2

You know that $y^2=1$, to show that $x^3=1$, notice that $xyxy=x(x^2)=x^3$, also $xyxy=yx^2yx^2=yx(xyx)x=yxyx^4=x^6$, so $x^3=1$. From here on, you can show that the 6 elements are: $$1,x,x^2,y,xy,yx$$ and that multiplying either of these yields no new element.