Product of two Abelian group My Question :

$G=AB$, where $A,\, B$ are abelian groups. Are $A$ and $B$ normal?

The problem is a very basic, and very much possibly duplicate of an existing question in this community.
But my question actually might have gone a different way......

I post this today. When I check it, I found in the final part that $[a,b]^{a'} = [a, b^{a'}] \in [A,B]$ may imply $ b^{a’}\in B,\,\forall a’\in A,\, b\in B $. So $B$ is invariant under conjugations by elements of $A$, and $B$ is hence normal, since $G=AB$. The same is true for $[a,b]^{b'} $.

I know my statement is wrong, but what have I misunderstood? Any help is sincerely appreciated! 

Not necessarily. For example, the Klein bottle group (that is, the fundamental group of a Klein bottle) is given by $G=\langle a, b; a^{-1}ba=b^{-1}\rangle$. Here, $A=\langle a\rangle$ and $B=\langle b\rangle$. Clearly, $G=AB$. However, $A$ is not normal in $G$.
This group can be viewed as $\mathbb{Z}\rtimes\mathbb{Z}$, where the action of one infinite cyclic group on the other is to negate it: $b\mapsto b^{-1}$. If the action was trivial then we would have a direct rather than semidirect product, so $G=\mathbb{Z}\times\mathbb{Z}$, which is clearly abelian so all subgroups are normal.

The above is rather complicated. For a simpler example, consider the smallest non-abelian group. This is also the smallest counter-example to your question.
