Is Abelian Group Subset consisting of elements of finite order a Subgroup? 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.    

Here is another take.
Consider $G_n = \{ x \in G : x^n = e \}$. It is easy to prove that $G_n$ is a subgroup of $G$.
Moreover, $G_m \subseteq G_n$ when $m \mid n$. In particular, $G_m$ and $G_n$ are contained in $G_{mn}$.
The subgroup you want is the torsion subgroup of $G$, denoted by $\operatorname{tor}(G)$.
It follows from the definiton that $\operatorname{tor}(G)=\displaystyle\bigcup_{n=0}^{\infty} G_n$.
It now follows that $\operatorname{tor}(G)$ is a subgroup because every group operation you need to perform happens in one of the $G_n$. In particular, if $a\in G_m$ and $b\in G_n$, then $ab \in G_{mn}$.
