Group of push-out I'm looking at a question regarding the push-out of $\mathbb{Z}$ to itself via the $\times2$ and $\times3$ homomorphisms. As I see it, the group will be $\{x,y\mid x^3=y^2\}$. But the questions asks to prove that it is $\{x,y\mid xyx=yxy\}$, with a hint to consider $xy$ and $yxy$.
I guess that I can switch to generators that are combinations of $x$ and $y$, such as the suggested ones, but I still can't get the presentation that I'm after.
Any suggestions on how I can prove this result?

Note that in $\{x,y\mid xyx=yxy\}$ you have $(xy)^3=xyxyxy=yxyyxy=(yxy)^2.$ You can show that the homomorphism $\Phi:\{x,y\mid x^3=y^2\} \to \{x,y\mid xyx=yxy\}$ defined by $\Phi(x)=xy$ and $\Phi(y)=yxy$ is an isomorphism.
