Let $G$ be an infinite, virtually cyclic group, i.e., $G$ has an infinite cyclic subgroup $H$ of finite index (or equivalently, it contains a finite index infinite cyclic normal subgroup $N$).
By a retract of $G$ I mean a subgroup $K$ of $G$ such that there exists a normal subgroup $L$ with $G=K\ltimes L$.
Question: Is there a finite bound on the indexes of infinite retracts of an infinite virtually cyclic group $G$?
What I've tried: Let $K$ be an infinite retract of $G$. By definition, there is a normal subgroup $L$ with $G=K\ltimes L$. I don't know how to show that $L$ is a finite subgroup of $G$. If so, then the index of $K$ in $G$ becomes finite. Since finite subgroups of $G$ have bounded order (in fact, the order of the finite quotient group $G/N$ is a finite bound on the orders of finite subgroups of $G$ (see here)), then its infinite retracts have finite bounded index.
Suppose that $G$ is virtually cyclic with infinite cyclic normal subgroup $N$ and $|G:N|$ finite. Suppose also that $G = K \ltimes L$.
Now $|L:L \cap N| \le |G:N|$ is finite. So, if $L$ is infinite, then so is $L \cap N$, and hence $|N:L \cap N|$ is finite, and so $|G:L \cap N|$ is finite, and so $|G:L|$ is finite. But $G/L \cong K$, so $K$ is finite.
Hence, if $K$ is infinite, then $L$ must be finite.