Suppose $G$ is a group, such that there exist two its normal abelian subgroups $A$ and $B$, satisfying the condition $G = AB = \{ab | a \in A, b \in B\}$. Is $G$ always abelian?
The statement is true if the intersection of A and B is trivial, due to the obvious fact that for each $a \in A$ and $b \in B$ we have $[a, b]$ lying in $A \cap B$ (because both subgroups are normal).
It is also quite easy to conclude from that, that $G$ is always metabelian.
However, I do not know how to proceed further.
Any help will be appreciated.
No, $G=AB$ with $A,B$ abelian subgroups implies that $G$ is metabelian by Ito's Theorem, but not that $G$ is abelian, see here:
A question about the proof of Ito's theorem (Metabelian)
Commutator property in product of groups
Even if we assume that $A,B$ are normal subgroups, $G$ need not be abelian. Consider non-abelian groups where all proper subgroups are normal.