Prove that the group $G$ with generators $x,y,z$ and relations $z^y=z^2$, $x^z=x^2$, $y^x=y^2$ has order $1$ 
Prove that the group $G$ with generators $x,y,z$ and relations $z^y=z^2$, $x^z=x^2$, $y^x=y^2$ has order $1$.

This is a problem on Page 56 of Derek J.S. Robinson's A Course in the Theory of Groups (GTM 80). I think $z^y$ means the result of $y$ acting on $z$, and may be defined as $y^{-1}zy$.
Suppose that $F$ is a free group generated by $x,y,z$. The epimorphism $\pi: F \rightarrow G$ has its kernel $K$ generated by $z^yz^{-2}$, $x^zx^{-2}$, $y^xy^{-2}$. How to prove $K=F$? Or, how to prove $x,y,z \in K$? I've tried but didn't find the right way.
Thank you very much.

EDIT: I changed the steps to be somewhat simpler.
Since this is tagged homework, I will only give hints (I hope!).


*

*Look at $z^{yx}$, and write it in two fairly different ways; use the equality between these two words to write $x$ as a word in $y$ and $z$.

*Now look at $y^x$, and write it in two different ways; use the equality between these two words to show $x$ is a power of $z$.

*Use the fact $x$ is a power of $z$ to write $z^x$ in two different ways, and show $x$ is trivial; conclude your group is trivial.
