I am solving problems in Dummit and Foote; however, this problem I am not able to do.
Show that if $n=3k$, then $X_{2n}$ has order $6$, and it has same generators and relations as $D_6$ when $x$ is replaced by $r$ and $y$ is replaced by $s$, where $$X_{2n}=\langle x,y\mid x^n=y^2=1, xy=yx^2\rangle.$$
So here is my argument
$$\begin{align} xy = yx^2&\rightarrow xy^2 = yx^2y\\ &\rightarrow x = yxxy = yxyx^2 = yy x^2x^2 = x^4\\ &\rightarrow x^3 = e, \end{align}$$
so $\text{ord}(x) \mid 3k$ and $\text{ord}(x) \mid 3$ so $\text{ord}(x) = 1$ or $\text{ord}(x) = 3$; however, I tried to show that $\text{ord}(x) = 1$ will lead to a contradiction but I can't get that contradiction.
Ok so thanks to Paul Hammer I was able to prove it. All I need to show is that I derive the relations of $D_6$ from $X_{6k}$ and vice versa.
So first I will go from relation of $X_{6k}$ to relations of $D_6$.
We want to know if $xy * (xy^{-1}) = e$, so $xy * (xy^{-1}) = yx^2xy^{-1} = yx^3y^{-1} = e$. The reason we know $x^3 = e$ is because we know $\text{ord}(x) = 3$ or $\text{ord}(x) = 1$, so both cases it will satisfy $x^3 = e$.
Now going the other way we want to know if $xy = yx^{-1}$ $\rightarrow$ $xy = yx^2$.
Well this is same as checking whether $xy*(yx^2)^{-1} = e$.
So we have
$$\begin{align} xy*(yx^2)^{-1}& = xy*x^{-2}y^{-1}\\ & = yx^{-1}x^{-1}x^{-1}y^{-1} \\ &= yx^{-3}y^{-1} \\ &= y*(x^3)^{-1}y^-1\\ & = e, \end{align}$$ and we are done!