$G$ finite Abelian, $a,b\in G$ what can we say about $|\langle a,b\rangle|$? I tried to answer the following question:
Let $G$ be a finite Abelian group and $a,b \in G$. Let 
$$ \langle a,b \rangle = \{a^i b^j \mid i,j \in \mathbb Z \}$$
What can we say about $|\langle a,b \rangle|$ in terms of $|a|,|b|$?
My answer:
$|\langle a,b \rangle| = lcm(|a|,|b|)$.
But it's not so clear to me how to prove this claim. For example, this subgroup of $G$ is not cyclic. If it was it would be obvious. But maybe my claim about the order is wrong. 

Is my claim wrong? If not, how to prove it rigorously?

 A: We can say $|\langle a,b\rangle|$ is a divisor of $|a|.|b|$ and divisible by $|a|$ and $|b|$. 
Consider the group $\mathbb{Z}_4\times \mathbb{Z}_2.$ Consider $a=(1,0)$ and $b=(1,1)$. Then $|\langle a,b\rangle|=|G|=8$ and is not the lcm of the orders of $a$ and $b$ (which are $4$ for both).
We can say this: since $G$ is abelian, $\langle a,b\rangle=\langle a\rangle . \langle b\rangle$, and there is well known formula for order of such product:
$$|\langle a,b\rangle| = |\langle a\rangle.\langle b\rangle|=\frac{|\langle a\rangle|.|\langle b\rangle|}{|\langle a\rangle \cap  \langle b\rangle|}$$
A: As stated in the comment your claim is wrong. In general you have :
$$max(|a|,|b|)\leq |\langle a,b\rangle|\leq |a||b| $$
The left inequality being proven by considering that $\langle a\rangle$ and $\langle b\rangle\subseteq \langle a,b\rangle$. The equality is obtained if and only if $a\in \langle b\rangle$ or $b\in \langle a\rangle$.
The right inequality is obtained by considering the obvious surjective morphism of groups :
$$\langle a\rangle\times\langle b\rangle\rightarrow  \langle a,b\rangle$$
$$(a^k,b^l)\mapsto a^kb^l $$
One can see that we get equality if and only if $\langle a\rangle \cap\langle b\rangle=\{1_G\}$ so exactly when the intersection is trivial (in which the surjective morphism is also injective and leads to an isomorphis of groups). 
Finally one can actually relate $m:=lcm(|a|,|b|)$ to the group $\langle a,b\rangle$. You can show (it is an interesting exercise) that $m$ is the biggest order of an element of the group $\langle a,b\rangle$. In other words, $m$ is the exponent of $\langle a,b\rangle$.
One last thing (following ZoeH's comment), by the first inequality argument $|a|$ and $|b|$ divide $|\langle a,b\rangle|$ so that $lcm(|a|,|b|)$ divides $|\langle a,b\rangle|$ and $|\langle a,b\rangle|$ divides $|a||b|$ since the application in the second inequality argument is a group morphism.
