Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]

Multiplying the equality by $yx$ from the right, we obtain
\begin{align*}
yx&=(xy)(xy)(yx)\\
&=xyxy^2x\\
&=xyx^2 \quad (\text{ since } y^2=1)\\
&=xy \quad (\text{ since } x^2=1).
\end{align*}
Thus we obtain $xy=yx$ for any elements $x, y \in G$. Thus the group $G$ is an abelian group.

Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group
Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.
(a) Prove that $T(A)$ is a subgroup of $A$.
(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]

Order of the Product of Two Elements in an Abelian Group
Let $G$ be an abelian group with the identity element $1$. Let $a, b$ be elements of $G$ with order $m$ and $n$, respectively.
If $m$ and $n$ are relatively prime, then show that the order of the element $ab$ is $mn$.
Proof.
Let $r$ be the order of the element […]

Prove a Group is Abelian if $(ab)^3=a^3b^3$ and No Elements of Order $3$
Let $G$ be a group. Suppose that we have
\[(ab)^3=a^3b^3\]
for any elements $a, b$ in $G$. Also suppose that $G$ has no elements of order $3$.
Then prove that $G$ is an abelian group.
Proof.
Let $a, b$ be arbitrary elements of the group $G$. We want […]

Prove a Group is Abelian if $(ab)^2=a^2b^2$
Let $G$ be a group. Suppose that
\[(ab)^2=a^2b^2\]
for any elements $a, b$ in $G$. Prove that $G$ is an abelian group.
Proof.
To prove that $G$ is an abelian group, we need
\[ab=ba\]
for any elements $a, b$ in $G$.
By the given […]

Order of Product of Two Elements in a Group
Let $G$ be a group. Let $a$ and $b$ be elements of $G$.
If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.
Proof.
We claim that it is not true. As a […]

Elements of Finite Order of an Abelian Group form a Subgroup
Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,
\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]
Prove that $H$ is a subgroup of $G$.
Proof.
Note that the identity element $e$ of […]

If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse
Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.
Show that $x$ is not conjugate to $x^{-1}$.
Proof.
Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. […]

Finite Group and a Unique Solution of an Equation
Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]
We give two proofs.
Proof 1.
Since $m$ and $n$ are relatively prime […]