Elements of Finite Order of an Abelian Group form a Subgroup

Problem 522

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 $G$.

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