Groups of order 8 proof

Question

I understand the solution to these questions I was just wanting to confirm that the solution to Q10 excludes the possibility that $y =x^2$ ( and hence the proof is not complete) as it uses the solution to 9b).

Answer 1

The way $y$ is defined is as an element of $G-\langle x\rangle$; that is, it is not in the subgroup generated by $x$. Since $x$ is of order $4$, this subgroup's underlying set is $\{e,x,x^2,x^3\}$.

So $y\neq e,x,x^2,$ or $x^3.$

They skip over the fact that $\langle x\rangle\neq G$, but $x$ is of order $4$ and $G$ is of order $8$, so this should be obvious. Other than that, the proof looks complete, using Q9 only to show there is an element of order $4$ and for the case of $G\cong D_8$.