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.
A 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 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 $D_{8}$
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 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 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 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 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 […]
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!