From the relation (1), we have
\[xy^2x^{-1}=y^3.\]
Computing the power of $n$ of this equality yields that
\[xy^{2n}x^{-1}= y^{3n} \tag{3}\]
for any $n\in \N$.

In particular, we have
\[xy^4x^{-1}=y^6 \text{ and } xy^6x^{-1}=y^9.\]
Substituting the former into the latter, we obtain
\[x^2y^4x^{-2}=y^9. \tag{4}\]
Cubing both sides gives
\[x^2y^{12}x^{-2}=y^{27}.\]

Using the relation (3) with $n=4$, we have $xy^8x^{-1}=y^{12}$.
Substituting this into equality (4) yields $x^3y^8x^{-1}=y^{27}$.

Now we have
\begin{align*}
y^{27}=x^3y^8x^{-1}=(x^3y)y^8(y^{-1}x^{-3})=yx^2y^8x^{-2}y.
\end{align*}
Squaring the relation (4), we have $x^2y^8x^{-2}=y^{18}$.
Substituting this into the previous, we obtain $y^{27}=y^{18}$, and hence
\[y^9=e,\]
where $e$ is the identity element of $G$.

Note that as we have $xy^2x^{-1} =y^3$, the elements $y^2, y^3$ are conjugate to each other.
Thus, the orders must be the same. This observation together with $y^9=e$ imply $y=e$.

It follows from the relation (2) that $x=e$ as well.
Therefore, the group $G$ is the trivial group.

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Hi, why “y^2⋅x^3y=y⋅yx^2⋅y” is true?

Dear Aidao Chen,

Thank you for finding a mistake. I totally rewrote the proof.

Please let me know if the proof still contains errors.

Thank you very much for sharing this proof. It is a quite clever proof!