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

Sponsored Links

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

Sponsored Links

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?

Please Login to Comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

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