Find order of $xy$ provided $x^2=e, y^3=e$ and $yxy=xy^2x$ Let $G$ be a group and $x,y \in G$.  Suppose $x^2=e$, $y^3=e$ and $yxy=xy^2x$. Find the order of $xy$.
I have no idea what to do. I am suspecting $(xy)^6=e$ (order 6) but I don't know how to solve it.
 A: When in doubt, look at the successive powers of $xy$ and try to simplify.  Here, we have:
$$
xy\\
(xy)^2=xyxy=x(yxy)=x(xy^2 x)=x^2y^2x=y^2x\\
(xy)^3=xy(y^2x)=xy^3x=xx=e
$$
So, in fact, $(xy)^3=e$, which means that the order of $xy$ is (at most) $3$.
A: Recall that $x^2=1$ and $y^3=1$. Then we have the following.
$$\begin{align*}
yxy&=xy^2x\\
\Rightarrow
xyxy&=y^2x\\
\Rightarrow
xyxyx&=y^2\\
\Rightarrow
xyxyxy&=y^3\\
\Rightarrow (xy)^3&=1
\end{align*}$$
This gives an upper bound. I will leave you to prove that this is, in fact, a lower bound (so long as there are no other relators). To get you started, consider a regular Tetrahedron, or equivalently the following graph.
$\hskip2in$
(Indeed, it seems there are precisely five groups with this property, and only two of these are such that $x\neq 1 \neq y$).
A: Another way to organize the argument, though ultimately the same as user1729's answer: The first two equations tell you that $x=x^{-1}$ and $y^2=y^{-1}$. Plugging those into the third equation, you get $yxy=x^{-1}y^{-1}x^{-1} =(xyx)^{-1}$. That is, $yxy$ is the inverse of $xyx$, meaning that their product (in either order) is $e$. So $xyxyxy=e$, which means that $xy$ has order $3$ or $1$.
