Could someone look over my solution to this exercise in Artin's algebra text?
Assume that the equation $xyz = 1$ holds in a group $G$. Does it follow that $yzx = 1$? That $yxz = 1$?
Solution. It follows that $yzx = 1$. We have: \begin{align*} xyz = 1 &\iff yz = x^{-1}*1 = x^{-1} \\ &\iff yzx = x^{-1}x = 1. \end{align*} It does not follow that $yxz = 1$. For a contradiction, suppose it does. Then, we have: \begin{align*} yxz = 1 \iff y^{-1} (yxz) z^{-1} = y^{-1} * 1 * z^{-1} \iff x = y^{-1}z^{-1}. \end{align*} Plugging into $xyz = 1$: \begin{align*} xyz = 1 & \iff y^{-1} z^{-1} yz = 1\\ & \iff y(y^{-1}z^{-1}yz)\\ & \iff y\\ & \iff z(z^{-1}yz)\\ &\iff zy = yz = zy. \end{align*} So $zy = yz$ for all $z, y \in G$. This contradicts the fact that not all groups are abelian. Hence, $yxz = 1$ holds if and only if $G$ is abelian, but not generally.
The first part is OK but the second part is wrong.
The question is actually ambiguous. It isn't clear whether the condition $xyz=1$ holds for all $x,y,z$ or just for some $x,y,z$.
In your answer to the second part, you wrote (emphasis mine):
You seem to have implicitly taken the premise to mean $xyz=1$ for all $x,y,z\in G$. But in this case we should have $x11=1$ for all $x\in G$, i.e. $G=\{1\}$. Hence $yxz\equiv1$ and your answer is wrong.
If you take the premise to mean that $xyz=1$ for some $x,y,z\in G$, then your answer is incomplete. You have correctly shown that the condition $xyz=1=yxz$ implies that $yz=zy$. However, this does not lead to a contradiction because in every non-abelian group, there do exist $x,y,z$ that satisfy all these conditions (such as when $x=y=z=1$). So, in order to give a negative answer to the question, you should at least explain why there is some non-abelian group $G$ that contains some $x,y,z$ such that $xyz=1$ but $yz\ne zy$.
This can be done by picking a pair of non-commuting elements $y,z\in G$ and let $x=(yz)^{-1}$, or by exhibiting a concrete counterexample, such as $x=\bigl(\begin{smallmatrix}1&2\end{smallmatrix}\bigr),\,y=\bigl(\begin{smallmatrix}2&3\end{smallmatrix}\bigr)$ and $z=\bigl(\begin{smallmatrix}1&2&3\\ 3&1&2\end{smallmatrix}\bigr)$ in the symmetry group $S_3$ with composition from left to right.