I asked here that: does $\mathbb{Z}\oplus G$, where $G$ is a finite abelian group, has only finitely many retracts? The answer was yes. Now I've tried to prove that the number of retracts of $\mathbb{Z}\rtimes G$ is also finite, where $G$ is a finite group. My try is based on the argument of here. My question is that does the argument still holds for $\mathbb{Z}\ltimes G$?
What I've tried: The notation $+$ is used for the group action of all groups throughout this argument. Let $G$ be a finite group with identity element $e$ and $H$ be an arbitrary retract of $G\ltimes \mathbb{Z}$. Then there exists a normal subgroup $N$ of $G\ltimes \mathbb{Z}$ such that $G\ltimes \mathbb{Z}=H\ltimes N$. Write each element of $G\ltimes \mathbb{Z}$ as $(g,z)$. We show that either $H$ or $N$ must be of the form $J \ltimes \{ 0\}$, where $J$ is a subgroup of $G$. In fact, if $(g_1,m)\in H$ and $(g_2,n)\in N$ where $m\neq0$ and $n\neq0$, then $(e,mn|G|)=n|G|(g_1,m)=m|G|(g_2,n)\in H\cap N$ which is a contradiction.
First, let's suppose $H=J \ltimes \{ 0\}$. Writing $N\cap(G\ltimes \{ 0\})$ in the form $F \ltimes \{ 0\}$, we have that $G$ is the semidirect product of $F$ and $J$, i.e. $G=J\ltimes F$.
Now, we assume that $N=J \ltimes \{ 0\}$. Since all elements of $N$ have the second coordinate as being $0$, and since $(e,1)=h+n$ for some $h\in H$ and $n\in N$, such a value of $h$ must have the second coordinate as being $1$. Thus $H$ must have at least one element of the form $(g,1)$. Also, for such a choice of $g$, we have that $H$ is the semidirect product of $\langle (g,1)\rangle$ and $F' \ltimes \{ 0\}$, where $F' \ltimes \{ 0\}=H\cap(G\ltimes \{ 0\})$. Hence, $G\ltimes \mathbb{Z}$ has only finitely many retracts.
Recall that a retract of $G$ I mean a subgroup $H$ of $G$ such that there exists a normal subgroup $N$ with $G=N\rtimes H$.
If $G$ is infinite dihedral, then every subgroup of order 2 is a retract. And there are infinitely many. However, there are only two conjugacy classes.
In general, let $G$ be virtually cyclic. Then its finite subgroups have bounded order. So its infinite retracts have finite bounded index. Since there are finitely many subgroups of given index, we deduce there are finitely many retracts of finite index (this actually applies in every finitely generated group with a maximal finite subgroup "finite radical").
In a virtually cyclic group, infinite index subgroups are finite. So we have to count finite retracts.
By a result of Wall, $G$ is either finite-by-cyclic or finite-by-dihedral. If $G$ is finite-by-cyclic, then it has finitely many finite subgroups.
Now suppose that $G$ is finite-by-dihedral. Let $W$ be its finite radical. Then maximal finite subgroups of $G$ have index 2 over $W$, and there are two conjugacy classes of such subgroups, and each has finitely many subgroups. Hence there are finitely many retracts up to conjugacy (but possibly infinitely many).