Order of a product in an abelian group. Suppose $G$ is a finite abelian group and has two element $a$ and $b$, such that $\circ(a)=m$ and $\circ(b)=n$ and $lcm(m,n)\neq m,n$. Is it true that $\circ(ab)=lcm(m,n)$? Thanks in advance.

But it is true that if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$ and $\operatorname{gcd}(m, n)=1$ then $\operatorname{ord}(ab)=mn=\operatorname{lcm}(m, n)$.
In general, one can say the following:

Proposition. If $G$ is a finite abelian group, and $\operatorname{ord}(a)=m$, and $\operatorname{ord}(b)=n$. Then
  $$\frac{mn}{\gcd(m, n)^2}\mid \operatorname{ord}(ab) \mid
 \frac{mn}{\gcd(m, n)}=\operatorname{lcm}(m, n)$$

It looks like you have shown $\frac{mn}{\gcd(m, n)^2}\mid \operatorname{ord}(ab)$ as you indicated in the comments. Very nice! 
For the proof of the above fact, see the article
Dieter Jungnickel, On the Order of a Product in a Finite Abelian Group. Mathematics Magazine, Vol. 69, No. 1 (Feb., 1996), pp. 53-57
