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).
Best Answer
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 $$G=H\ltimes G/H\cong H\ltimes (\mathbb{Z}_2\ltimes \mathbb{Z})\cong (H\ltimes \mathbb{Z}_2)\ltimes \mathbb{Z}.$$