Let $G$ be a finite simple group. Suppose that $A, B < G$, $G = AB$ and $A$ is an Abelian group. Is it true that $A \cap B=1 $? Let $G$ be a finite simple group. Suppose that $A$ and $B$ are proper subgroups of $G$, $ G = AB$ and $A$ is an Abelian group. Is it true that $A \cap B=1 $ ?

I checked it with some examples and it seems to be true; but I have no proof for it.
Set $C:= A \cap B$. On the contray suppose that $C \neq 1$. So $C^B \neq 1$.
We have $C^G = C^{AB}$. Since $A$ is Abelian, $C^G = C^B$.
But $ C^G \trianglelefteq G$. Thus $C^B=C^G=G$.
For every $c \in C$ and $b \in B$, we have $c^b = b^{-1}cb \in B$, so $G = C^B \le B$, a contradiction.
Counter-example: Let $G= Z_p$ with $p$ prime number. G is simple and abelian, $G=G\cdot G$ and $G \cap G=G\neq 1$.