2.1.5

We know that $$xyz = 1$$ in the group $$G$$ and therefore $$x$$ must be the inverse of $$yz$$. So $$x$$ satisfies

$$(yz)x = x(yz) = 1$$

and the equation $$yzx = 1$$ holds in $$G$$.

Now take $$z$$ to be the inverse of $$xy$$. Because it is not specified that the group $$G$$ is abelian, it does not follow that $$z$$ is also the inverse of $$yx$$, so the equation $$yxz = 1$$ does not hold in $$G$$.