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.

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.

The post If Generators $x, y$ Satisfy the Relation $xy^2=y^3x$, $yx^2=x^3y$, then the Group is Trivial first appeared on Problems in Mathematics.

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 either the trivial group or $G$ itself.

Proof.

$(\implies)$ If $G$ is a simple abelian group, then the order of $G$ is prime.

Suppose that $G$ is a simple abelian group. Then $G$ is a nontrivial group by definition.

We first show that $G$ is a finite group.
Let $g\in G$ be a nonidentity element of $G$. Then the group $\langle g \rangle$ generated by $g$ is a subgroup of $G$. Since $G$ is an abelian group, every subgroup is a normal subgroup.

Since $G$ is simple, we must have $\langle g \rangle=G$. If the order of $g$ is not finite, then $\langle g^2 \rangle$ is a proper normal subgroup of $\langle g \rangle=G$, which is impossible since $G$ is simple.
Thus the order of $g$ is finite, and hence $G=\langle g \rangle$ is a finite group.

Let $p$ be the order of $g$ (hence the order of $G$).
Seeking a contradiction, assume that $p=mn$ is a composite number with integers $m>1, n>1$. Then $\langle g^m \rangle$ is a proper normal subgroup of $G$. This is a contradiction since $G$ is simple.

Thus $p$ must be a prime number.
Therefore, the order of $G$ is a prime number.

$(\impliedby)$ If the order of $G$ is prime, then $G$ is a simple abelian group.

Let us now suppose that the order of $G$ is a prime.

Let $g\in G$ be a nonidentity element. Then the order of the subgroup $\langle g \rangle$ must be a divisor of the order of $G$, hence it must be $p$.

Therefore we have $G=\langle g \rangle$, and $G$ is a cyclic group and in particular an abelian group.
Since any normal subgroup $H$ of $G$ has order $1$ or $p$, $H$ must be either trivial $\{e\}$ or $G$ itself. Hence $G$ is simple. Thus, $G$ is a simple abelian group.

The post A Simple Abelian Group if and only if the Order is a Prime Number first appeared on Problems in Mathematics.

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 itself.
• The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators $[a,b]=a^{-1}b^{-1}ab$ for $a,b\in G$.

The commutator subgroup $D(G)=[G,G]$ is a normal subgroup of $G$.
For a proof, see: A condition that a commutator group is a normal subgroup.

Proof.

Note that the commutator subgroup $D(G)$ is a normal subgroup.
Since $G$ is simple, any normal subgroup of $G$ is either the trivial group $\{e\}$ or $G$ itself. Thus we have either $D(G)=\{e\}$ or $D(G)=G$.
If $D(G)=\{e\}$, then for any two elements $a,b \in G$ the commutator $[a,b]\in D(G)=\{e\}$.

Thus we have
\[a^{-1}b^{-1}ab=[a,b]=e.\] Therefore we have $ab=ba$ for any $a,b\in G$. This means that the group $G$ is abelian, which contradicts with the assumption that $G$ is non-abelian.
Therefore, we must have $D(G)=G$ as required.

The post Non-Abelian Simple Group is Equal to its Commutator Subgroup first appeared on Problems in Mathematics.