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.

 
LoadingAdd to solve later

Proof.

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.


LoadingAdd to solve later

More from my site

  • A Simple Abelian Group if and only if the Order is a Prime NumberA Simple Abelian Group if and only if the Order is a Prime Number Let $G$ be a group. (Do not assume that $G$ is a finite group.) Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.   Definition. A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]
  • Dihedral Group and Rotation of the PlaneDihedral Group and Rotation of the Plane Let $n$ be a positive integer. Let $D_{2n}$ be the dihedral group of order $2n$. Using the generators and the relations, the dihedral group $D_{2n}$ is given by \[D_{2n}=\langle r,s \mid r^n=s^2=1, sr=r^{-1}s\rangle.\] Put $\theta=2 \pi/n$. (a) Prove that the matrix […]
  • Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$ Let $D_8$ be the dihedral group of order $8$. Using the generators and relations, we have \[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\] (a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$. Prove that the centralizer […]
  • Non-Abelian Simple Group is Equal to its Commutator SubgroupNon-Abelian Simple Group is Equal to its Commutator Subgroup Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.   Definitions/Hint. We first recall relevant definitions. A group is called simple if its normal subgroups are either the trivial subgroup or the group […]
  • Every Cyclic Group is AbelianEvery Cyclic Group is Abelian Prove that every cyclic group is abelian.   Proof. Let $G$ be a cyclic group with a generator $g\in G$. Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.) Let $a$ and $b$ be arbitrary elements in $G$. Then there exists […]
  • Group Generated by Commutators of Two Normal Subgroups is a Normal SubgroupGroup Generated by Commutators of Two Normal Subgroups is a Normal Subgroup Let $G$ be a group and $H$ and $K$ be subgroups of $G$. For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$. Let $[H,K]$ be a subgroup of $G$ generated by all such commutators. Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup […]
  • Prove that a Group of Order 217 is Cyclic and Find the Number of GeneratorsProve that a Group of Order 217 is Cyclic and Find the Number of Generators Let $G$ be a finite group of order $217$. (a) Prove that $G$ is a cyclic group. (b) Determine the number of generators of the group $G$.     Sylow's Theorem We will use Sylow's theorem to prove part (a). For a review of Sylow's theorem, check out the […]
  • Every Finitely Generated Subgroup of Additive Group $\Q$ of Rational Numbers is CyclicEvery Finitely Generated Subgroup of Additive Group $\Q$ of Rational Numbers is Cyclic Let $\Q=(\Q, +)$ be the additive group of rational numbers. (a) Prove that every finitely generated subgroup of $(\Q, +)$ is cyclic. (b) Prove that $\Q$ and $\Q \times \Q$ are not isomorphic as groups.   Proof. (a) Prove that every finitely generated […]

You may also like...

3 Responses

  1. Aidao Chen says:

    Hi, why “y^2⋅x^3y=y⋅yx^2⋅y” is true?

    • Yu says:

      Dear Aidao Chen,

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

      Please let me know if the proof still contains errors.

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Group Theory Problems and Solutions
The Product of Distinct Sylow $p$-Subgroups Can Never be a Subgroup

Let $G$ a finite group and let $H$ and $K$ be two distinct Sylow $p$-group, where $p$ is a prime...

Close