I know this question has asked and answered before, so my apologies for posting a duplicate query. Instead of the solution, I'm seeking a "nudge" in the right direction to complete the proof on my own. So please, hints or criticism only. Thanks!
"Show that in an Abelian group $G$, the set consisting of all elements of $G$ of finite order is a subgroup of $G$."
Let $H\subset G$ consisting of all elements of a finite order. Then $H$ contains the inverse element $e \in G$, since $e^1 = e$.
Next, let $a \in G$ such that $a^n = e$ for some $n \in \mathbb{Z}$. So $a \in H$, and $a^{-1} \in H$ since $(a^{-1})^n = (a^{n})^{-1} = e^{-1} = e$. Hence the order of $a^{-1}$ is finite, and $H$ is closed under inverses.
Last, let $b^{m} = e \in G$ for some $m \in \mathbb{Z}$. Then $b \in H$ $ab \in G$. To show $ab \in H$, notice that $e = ee = a^{n}b^{m}$ and
$$a^{n}b^{m} = (ab)a^{n-1}b^{m-1}$$
since $G$ is Abelian. So $ab \in H$ and $H$ is closed under the operation of $G$. So $H$ is a subgroup of $G$.
This last part I'm nearly certain is wrong. I've tried to use the fact that $G$ is Abelian to manipulate the equation $$ab = ba$$ using left and right multiplication in as many ways as I can think of to yield a proof that the order of $ab$ is finite. As usual, I'm sure I'm missing something obvious, and would appreciate any advice pointing towards, rather than disclosing, the solution.
Best Answer
HINT: Suppose that $a,b\in H$; you need to show that $ab\in H$. There are positive integers $m,n$ such that $a^m=b^n=e$; use the fact that $G$ is Abelian to show that $(ab)^{mn}=e$ and hence that $ab\in H$. Basically this requires showing by induction that $(ab)^k=a^kb^k$ for $k\ge 0$.