Order of products of elements in a finite Abelian group We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have 
$\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$. 
How I approached this question was by saying let $\operatorname{LCM}(|a|,|b|) = L$. Then if we show that $L$ divides $|ab|$ and $|ab|$ divides $L$ then $|ab| = L$. Showing that $|ab|$ divides $L$ was fine. But then I am having trouble with the second part that is showing $L$ divides $|ab|$. For this, so far I have:
Consider $(ab)^{|ab|} = e = a^{|ab|}b^{|ab|} $  since we know $a \neq b^{-1}$ by assumption then we can conclude that $|a|$ divides $|ab|$ and $|b|$ divides $|ab|$. We know that $L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}$ from here can we conclude that $L$ divides $|ab|$?
 A: 
But then I am having trouble with the second part that is showing |ab| divides L. 

Recall that if $g\in G$ and $g^n=e$, then this implies implies $|g|$ divides $n.$ So $(ab)^L = e \implies |ab|$ divides $L.$
Edit: 
If you are trying to show that $L$ divides $|ab|$, then with respect to your last question, the work you have, knowing that $$L = \frac{|a||b|}{\gcd(|a|,|b|)}$$ tells you $L$ divides $|a||b|$. If you can show it follows that $$L = \frac{|ab|}{\gcd(|a||b|)}$$ then you have shown that $L$ indeed divides $|ab|$.
A: The stated result is wrong
In the (additively written) cyclic group $\Bbb Z/6\Bbb Z$ of order$~6$, the elements $a=1$ and $b=2$ respectively have orders $|a|=6$ and $|b|=3$. Then $L=\operatorname{lcm}(6,3)=6$, but that is not the order of $a+b=3$, since that element has order$~2$. So the statement fails.
This also shows that your argument that $|a|$ and $|b|$ both divide $|ab|$ cannot work. You seem to think that since $a,b$ are not inverses of each other, $a^m$ and $b^m$  are not inverses of each other either, with $m=|ab|$. But in the example $m=2$, and here it is the case that $a^2$ and $b^2$ are inverses of each other (additively written these elements are $2a=2$ and $2b=4$). When you say "by assumption then we can conclude that $|a|$ divides $|ab|$" it is not clear what assumption you are referring to, but your conclusion is wrong.
