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,
\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]

Note that the identity element $e$ of $G$ has order $1$, hence $e\in H$ and $H$ is not an empty set.

To show that $H$ is a subgroup of $G$, we need to show that $H$ is closed under multiplications and inverses.

Let $a, b\in H$.
By definition of $H$, the orders of $a, b$ are finite.
So let $m, n \in \N$ be the orders of $a, b$, respectively:
We have
\[a^m=e \text{ and } b^n=e.\]

Then we have
\begin{align*}
(ab)^{mn}&=a^{mn}b^{mn} && \text{since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^ne^m=e.
\end{align*}

This implies that the order of $ab$ is at most $mn$, hence the order of $ab$ is finite.
Thus $ab\in H$ for any $a, b\in H$.

Next, consider any $a\in H$. We want to show that the inverse $a^{-1}$ also lies in $H$.
Let $m \in \N$ be the order of $a$.
Then we have
\begin{align*}
(a^{-1})^m=(a^m)^{-1}=e^{-1}=e.
\end{align*}
This implies that the order of $a^{-1}$ is also finite, and hence $a^{-1}\in H$.

Therefore we have proved that $H$ is closed under multiplications and inverses.
Hence $H$ is a subgroup of $G$.

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 […]

Finite Group and Subgroup Criteria
Let $G$ be a finite group and let $H$ be a subset of $G$ such that for any $a,b \in H$, $ab\in H$.
Then show that $H$ is a subgroup of $G$.
Proof.
Let $a \in H$. To show that $H$ is a subgroup of $G$, it suffices to show that the inverse $a^{-1}$ is in $H$.
If […]

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 […]

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 […]

If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group
Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.
Proof.
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 […]

Abelian Normal Subgroup, Intersection, and Product of Groups
Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$.
(That is, $A$ is a normal subgroup of $G$.)
If $B$ is any subgroup of $G$, then show that
\[A \cap B \triangleleft AB.\]
Proof.
First of all, since $A \triangleleft G$, the […]

Abelian Group and Direct Product of Its Subgroups
Let $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers.
Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.
Hint.
Consider […]