I'm studying the following Lemma:
Let the group $G = AB$ be the product of two cyclic subgroups $A$ and $B$. Then $G$ is supersoluble and abelian-by-finite.
Proof: Using a preliminary lemma we know that at least one of the factors $A$ and $B$ contains a non-trivial normal subgroup of $G$. Hence $G$ has a cyclic non-trivial normal subgroup. Since the hypothesis is inherited by every homomorphic image of $G$, it follows that $G$ is hypercyclic, and hence even supersoluble, as it is finitely generated. In particular, $G$ is nilpotent-by-finite. If one of the factors $A$ and $B$ is finite, $G$ is obviously abelian-by-finite. Thus we may suppose that $A$ and $B$ are both infinite cyclic. If $A \cap B \neq 1$, then $G/ (A \cap B)$ is finite, and so $G$ is abelian-by-finite. If $A \cap B = 1$, it follows from a previous Corollary that the Fitting subgroup of $G$ is abelian, and so $G$ is abelian-by-finite. $\square$
First question: how can I proof that hypercyclic and finitely generated imply supersoluble?
Second question: In the last part of the proof we take the Fitting subgroup of $G$ to show that $G$ is abelian-by-finite. How do we know that the Fitting has finite index in $G$?
For Question 1, $G$ has a normal cyclic subgroup $N \ne 1$ with $N \le A$ or $N \le B$. So in $G/N$ one of the cyclic factors $AN/N$ and $BN/N$ is smaller than the factors $A$ and $B$ in $G$ (or finite if originally infinite). So it follows by a downward induction argument on $(|A|,|B|)$ that $G$ is supersolvable.
You have already answered Question 2 earlier in the proof.