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

Group Theory Problems and Solutions in Mathematics

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.

 
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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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