[proofplan] We prove directly that $G_{\mathrm{tors}}$ is a subgroup of $G$ under the operation inherited from $G$. The identity is torsion because $e^1=e$. Closure under products uses the abelian hypothesis: if $a^m=e$ and $b^n=e$, then $(ab)^{mn}=a^{mn}b^{mn}=e$. Closure under inverses follows from $(a^{-1})^m=(a^m)^{-1}=e$. [/proofplan] [step:Show that the identity element is torsion] By definition, \begin{align*} G_{\mathrm{tors}}=\{g\in G:\text{ there exists } n\in\mathbb{N}\text{ such that }g^n=e\}. \end{align*} Since $e^1=e$, the identity element $e$ belongs to $G_{\mathrm{tors}}$. Hence $G_{\mathrm{tors}}$ is nonempty. [/step] [step:Use commutativity to prove closure under products] Let $a,b\in G_{\mathrm{tors}}$. By the definition of $G_{\mathrm{tors}}$, there exist $m,n\in\mathbb{N}$ such that \begin{align*} a^m=e \end{align*} and \begin{align*} b^n=e. \end{align*} Since $G$ is abelian, $a$ and $b$ commute. Therefore, for every $r\in\mathbb{N}$, \begin{align*} (ab)^r=a^r b^r. \end{align*} Taking $r=mn$, we get \begin{align*} (ab)^{mn}=a^{mn}b^{mn}. \end{align*} Because $a^{mn}=(a^m)^n=e^n=e$ and $b^{mn}=(b^n)^m=e^m=e$, it follows that \begin{align*} (ab)^{mn}=ee=e. \end{align*} Since $mn\in\mathbb{N}$, this proves $ab\in G_{\mathrm{tors}}$. [guided] Let $a,b\in G_{\mathrm{tors}}$. The goal is to prove that the product $ab$ is again torsion, meaning that some positive power of $ab$ is equal to the identity element $e$. Because $a\in G_{\mathrm{tors}}$, there exists $m\in\mathbb{N}$ such that \begin{align*} a^m=e. \end{align*} Because $b\in G_{\mathrm{tors}}$, there exists $n\in\mathbb{N}$ such that \begin{align*} b^n=e. \end{align*} The natural exponent to try is $mn$, because it is a common multiple of both $m$ and $n$. The abelian hypothesis is exactly what allows us to separate the power of a product. Since $G$ is abelian, $a$ and $b$ commute, so for every $r\in\mathbb{N}$ we have \begin{align*} (ab)^r=a^r b^r. \end{align*} This identity follows by induction on $r$: for $r=1$ it is the identity $(ab)^1=a^1b^1$, and the induction step uses commutativity to move the new factor $a$ past $b^r$. Taking $r=mn$, we obtain \begin{align*} (ab)^{mn}=a^{mn}b^{mn}. \end{align*} Now each factor on the right is the identity. Indeed, \begin{align*} a^{mn}=(a^m)^n=e^n=e \end{align*} and \begin{align*} b^{mn}=(b^n)^m=e^m=e. \end{align*} Therefore \begin{align*} (ab)^{mn}=ee=e. \end{align*} Since $mn$ is a positive integer, the definition of $G_{\mathrm{tors}}$ gives $ab\in G_{\mathrm{tors}}$. [/guided] [/step] [step:Prove closure under inverses] Let $a\in G_{\mathrm{tors}}$. By definition, there exists $m\in\mathbb{N}$ such that \begin{align*} a^m=e. \end{align*} Using the inverse law for powers in a group, \begin{align*} (a^{-1})^m=(a^m)^{-1}=e^{-1}=e. \end{align*} Thus $a^{-1}\in G_{\mathrm{tors}}$. [/step] [step:Conclude that the torsion elements form a subgroup] We have shown that $G_{\mathrm{tors}}$ is nonempty, is closed under the group product inherited from $G$, and is closed under inverses. Associativity of the inherited operation is inherited from the associativity of $G$. Therefore $G_{\mathrm{tors}}$ is a subgroup of $G$, that is, \begin{align*} G_{\mathrm{tors}}\le G. \end{align*} [/step]