Does a finite-by-(infinite dihedral) group have the form $N\rtimes H$ for a finite normal subgroup $N$ and a cyclic subgroup $H$

Question

Let $G$ be a finite-by-(infinite dihedral) group, i.e. it contains a finite normal subgroup $K$ with $G/K$ isomorphic to the infinite dihedral group.

Question: If $G$ is a finite-by-(infinite dihedral) group, then $G=N\rtimes H$ for a finite normal subgroup $N$ and a cyclic subgroup $H$ of $G$?

What I've tried: Assume that $G$ contains a finite normal subgroup $N$ with $G/K$ isomorphic to the infinite dihedral group. Consider ‎the ‎short ‎exact ‎sequence ‎‎$K‎\rightarrowtail ‎G‎\twoheadrightarrow‎‎ G/K$. If $G/K$ is projective, ‎we ‎have‎ ‎‎$‎‎G= K\rtimes G/K$.‎ Then by the fact that $G/K=\mathbb{Z}_2 \ltimes \mathbb{Z}$, we have that $G=(K\ltimes \mathbb{Z}_2)\rtimes \mathbb{Z}$.

My problem is that I don't know the infinite dihedral group is projective or not.

Answer 1

From a geometric group theory perspective, the way I would suggest approaching this question is from the theory of ends of groups. In particular there is a nice structure theorem for 2-ended groups:

If $G$ is a finitely generated, 2-ended group then $G$ has a properly discontinuous, cocompact action on $\mathbb R$, and exactly one of the following two alternatives occurs:

1. No element of $G$ swaps the two ends of $\mathbb R$; equivalently, there exists a homomorphism $G \to \mathbb Z$ with finite kernel; or
2. Some element of $G$ swaps the two ends of $\mathbb R$; equivalently, there exists a homomorphism $G \to D_\infty$ with finite kernel.

As a corollary of this, if $G$ is finite-by-dihedral then $G$ has 2 ends and some element of $G$ swaps the 2 ends, and therefore there does not exist a homomorphism $G \to \mathbb Z$ with finite kernel.