If Generators $x, y$ Satisfy the Relation $xy^2=y^3x$, $yx^2=x^3y$, then the Group is Trivial

Problem 554

Let $x, y$ be generators of a group $G$ with relation
\begin{align*}
xy^2=y^3x,\tag{1}\\
yx^2=x^3y.\tag{2}
\end{align*}

Prove that $G$ is the trivial group.

Proof.

Let $e$ be the identity element of $G$.
We compute
\begin{align*}
&xy^2x\stackrel{(1)}{=} y^3x^2=y^2\cdot yx^2\stackrel{(2)}{=}y^2\cdot x^3y=y\cdot yx^2 \cdot y \stackrel{(2)}{=}y\cdot x^3y \cdot y\6pt] &=yx^2 \cdot xy^2 \stackrel{(2)}{=}x^3y\cdot xy^2\stackrel{(1)}{=}x^3y\cdot y^3x=x^3y^4x. \end{align*} Canceling the leftmost and the rightmost x, we obtain \[y^2=x^2y^4.

Canceling $y^2$, we get
$e=x^2y^2.$ Hence we have
$y^2=x^{-2}.$ Substituting this into the relation (1) yields that
$x^{-1}=yx^{-1},$ and thus $y=e$.
Then it follows from the relation (2) that $x^2=x^3$, and hence $x=e$.

Therefore the generators $x, y$ must be the identity element.
Thus, $G$ is the trivial group $\{e\}$.

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