From Abstract Algebra Third Edition (Foote, Dummit). Section 1.3 Question 17.
Let $X_{2n}$ be the group represented by $\langle x,y \mid x^n = y^2 = 1, xy = yx^2\rangle $. Prove that if $n = 3k$, then $X_{2n}$ has order $6$.
Here is my attempted proof.
It is sufficient to show that $|x| = 3$, because then $X_{2n} = \{1,x,x^2,y,yx,yx^2\}$ which has order 6. Consider the following
$$\begin{equation}\begin{aligned} yx &= yxy^2 \\ &= y(xy)y \\ &= y(yx^2)y \\ &= y^2x^2y \\ &= x^2y \\ &= yx^4 \end{aligned}\end{equation}\tag{2}\label{eq2}$$
So we have $yx = yx^4$. By the cancellation laws, $x = x^4 \implies x^3 = 1 \implies |x| \leq 3$.
$|x| \neq 1$ because $x$ is not the identity. $|x| \neq 2$ because then $X_{2n} = \{1, x, y, yx\}$ which is of order $4 \neq 3k$. Then |x| is of order 3.
Q.E.D.
The reason I feel uncertain of this proof is that I didn't use the hypothesis that $n=3k$ until the very end of the proof, and I was able to find a bound on the order of $x$ without knowing anything about the size of the group, using only the relations provided. I feel like if I did make a mistake, it has something to do with my assumptions of what $X_{2n}$ looks like, for example maybe $x=y$ for some values of n. The book did mention the "collapsing" (pg. 26) of relations, which might be what is going on.
You’ve merely asserted that $x$ is not the identity. But if $n$ was not a multiple of $3,$ $x$ would be the identity.
One way to show $x$ is not the identity is to find a group $G$ with $|G|=6$ and $x,y\in G$ such that:
$$x^{3k}=1,y^2=1,xy=yx^2.$$
There is only one non-commutative group of order $6,$ so...