Does every virtually cyclic group $G$ have the form $G=H\ltimes N$? A virtually cyclic group is a group that has a cyclic subgroup of finite index. Every virtually cyclic group in fact has a normal cyclic subgroup of finite index (namely, the core of any cyclic subgroup of finite index), and virtually cyclic groups are therefore also known as cyclic-by-finite groups.
My Question: Let $G=H\ltimes N$, where $H$ is a finite subgroup and $N$ is a normal cyclic subgroup. Then $G$ is a virtually cyclic group. Does every virtually cyclic group $G$ have the form $G=H\ltimes N$, where $H$ is a finite subgroup and $N$ is a normal cyclic subgroup?
What I've tried: ‎Let $G$ be a virtually cyclic group. If ‎$‎‎G$ ‎is ‎finite, ‎then ‎the ‎proof ‎trivially ‎holds. ‎So ‎let's ‎suppose ‎‎$‎‎G$ ‎is ‎infinite.‎   Then $G$ contains an infinite cyclic subgroup $N$ of finite index. So b‎y ‎Lemma 11.4 from 3-manifolds by J. Hempel, ‎‎$‎‎G$ ‎contains a‎ ‎finit‎e normal subgroup ‎$‎‎H$ ‎with ‎‎$‎‎G/H$ ‎isomorphic ‎to ‎either ‎‎$‎‎‎\mathbb{Z}‎$ ‎or ‎‎$‎‎‎\mathbb{Z}_2 * ‎\mathbb{Z}_2‎‎$. If $G/H\cong \mathbb{Z}$, then by considering ‎the ‎short ‎exact ‎sequence ‎‎$H‎\rightarrowtail ‎G‎\twoheadrightarrow‎‎ G/H$ ‎with ‎‎$‎‎G/H$ ‎free, ‎we ‎have‎ ‎‎$‎‎G= H\rtimes G/H$.‎
But I don't know what to do about $G/H\cong \mathbb{Z}_2 * ‎\mathbb{Z}_2(\cong \mathbb{Z}_2 \ltimes \mathbb{Z})$.
My another problem is that $N=G/H$ here is not normal in $G$ (As I mentioned in my question).
 A: Let $G$ be a virtually cyclic group. By definition, $G$ contains a cyclic normal subgroup $N$ of finite index. Since $|G:N|<\infty$, we have that $|C_G (N):Z(C_G (N))|<\infty$. So by a result of Schur (if $Z(G)$ is of finite index in $G$, then the derived subgroup of $G$ is finite), $C_G (N)'$ is finite. Since $G/C_G (N)$ embeds in ${\rm Aut}(N)$ which has order $2$, we have either $C_G (N)=G$ or $|G:C_G (N)|=2$. We consider two following cases:
First case: Let $G=C_G (N)$. Since $G/G'$ is an abelian group with at most one infinite cyclic factor, it has a finite subgroup $H/G'$ with $G/H$ trivial or infinite cyclic. If $G/H$ is trivial, then $G$ is the semidirect product of $H$ and $1$. Otherwise, $G$ is the semidirect product of $H$ and $\mathbb{Z}$ (because the short exact sequence $1\to H\to G\to G/H \to 1$ splits by the fact that $G/H\cong \mathbb{Z}$).
Second case: Let $|G:C_G (N)|=2$. Then $H$ is normal in $G$ and $G/H$ is infinite dihedral. Since the short exact sequence $1\to H\to G\to G/H \to 1$ splits, then $$G=H\ltimes G/H\cong H\ltimes (\mathbb{Z}_2\ltimes \mathbb{Z})\cong (H\ltimes \mathbb{Z}_2)\ltimes \mathbb{Z}.$$
A: No. A counterexample is the group defined by the presentation
$$\langle x,y,z \mid x^2=y^2=1, yz=zy, xy=yx,  xz=zxy \rangle.$$
This group is virtually cyclic, with infinite cyclic subgroup $N =\langle z \rangle$ of index $4$, but $N$ is not normal in $G$. The subgroup $\langle z^2 \rangle$ is normal but has no complement.
This group is a semidirect product $\langle x,y \rangle \rtimes N$, but that's the other way round.
