Some examples of virtually cyclic groups

Question

The only virtually cyclic groups (ie. groups containing $\mathbb{Z}$ as subgroup of finite index) I really know are : the groups $F \times \mathbb{Z}$, where $F$ is a finite group, and the infinite dihedral group $D_{\infty}$ (isomorphic to $\mathbb{Z}_2 \ast \mathbb{Z}_2$).

But all these groups are finitely presented, just-infinite (ie. their proper quotients are finite) and residually finite (ie. for all element $g$, there exists a morphism $\varphi$ onto a finite group such that $\varphi(g) \neq 1$).

So I am looking for examples of virtually cyclic groups without one of these properties. I only know that there exists a virtually abelian group not just-infinite but without having an explicit example.

As other virtually abelian groups, there is also the generalized dihedral groups $\text{Dih}(G)$ where $G$ is an infinite finitely generated abelian group, but I don't know them really. Are they virtually cyclic ?

NB: The groups I consider are finitely generated.

Answer 1

The existing answer is restricted to virtually-cyclic groups, but more general things can be said: Finite presentability and residual finiteness are both preserved when moving from finite index subgroups to the big group. That is, suppose $H$ is a finite index subgroup of $G$. Then,

• If $H$ is finitely presentable then so is $G$. This can be proven using covering spaces.

• If $H$ is residually finite then so is $G$. The way to prove this is to remember that a group $T$ is residually finite if for each $x\in T$ there exists a subgroup $K_x$ of finite index in $T$ such that $x\not\in K_x$. So, suppose $x\in G$ and we shall find such a finite index subgroup of $G$. if $x\not\in H$ then we are done, by taking $K_x=H$, while if $x\in H$ then there exists some $K_x\in H$ such that $x\not\in K_x$ and $K_x$ has finite index in $H$. As $K_x$ has finite index in $H$ it also has finite index in $G$, as required.

Therefore, every virtually-cyclic subgroup is both finitely presentable and residually finite. So the groups you are looking for do not exist!

Answer 2

In [F.T. Farrell, and L. Jones, The lower algebraic K-theory of virtually infinite cyclic groups, K-theory 9 (1995), 13-30], it is shown that a virtually infinite-cyclic group has the form $F \rtimes_{\alpha} \mathbb{Z}$, where $F$ is a finite group, or it maps onto $D_{\infty}$ with a finite kernel.

Case 1 : $G=F \rtimes_{\alpha} \mathbb{Z}$. First, $G$ is necessary virtually infinite-cyclic since $G=F \mathbb{Z}$ and $F$ is finite. Then, there exists a morphism $\phi_m$ such that the following diagram is commutative :

$$\begin{array}{ccc} \mathbb{Z} & \rightarrow^{\alpha} & \text{Aut}(F) \\ & \searrow{\pi_m} & \uparrow{\phi_m} \\ & & \mathbb{Z}_m \end{array}$$

Where $\text{ker}(\alpha)=m \mathbb{Z}$ and $\pi_m : \mathbb{Z} \to \mathbb{Z}_m$ is the canonical epimorphism. Then $\varphi_m : \left\{ \begin{array}{ccc} F \rtimes_{\alpha} \mathbb{Z} & \to & F \rtimes_{\phi_m} \mathbb{Z}_m \\ (f,p) & \mapsto & (f,\pi_m(p)) \end{array} \right.$ is a morphism. Since for all $k \geq 1$, $\varphi_{km}$ is a morphism from $G$ to the finite group $F \rtimes_{\phi_m} \mathbb{Z}_m$, we deduce that $G$ is residually finite.

Moreover, if $F = \langle X |R \rangle$ is a finite presentation of $F$, then $G= \langle X,z | R,z^nxz^{-n}=\alpha(z^n) \cdot x, x \in X,n \geq 1 \rangle$. Yet, $\text{ker}(\alpha) \neq \{e\}$, otherwise $\text{Aut}(F) \simeq \mathbb{Z}$ whereas $\text{Aut}(F)$ is finite. So there is $r \geq 1$ such that $z^r \in Z(G)$. The presentation above, without repetitions, is in fact finite, so $G$ is finitely presented.

Case 2 : there exists an epimorphism $\varphi : G \twoheadrightarrow D_{\infty}$ with $F=\text{ker}(\varphi)$ finite. If $D_{\infty}= \langle a,b | a^2=b^2=1 \rangle$, let $\alpha, \beta \in G$ such that $\varphi(\alpha)=a$ and $\varphi(\beta)=b$. Set $A= \langle F,\alpha \rangle$ and $B= \langle F, \beta \rangle$.

Let $g \in A$. We can write $g=w(\alpha,f_1,...,f_n)$ with $f_1,...,f_n \in F$. Since $F$ is a normal subgroup of $G$, there exists $g_i \in F$ sucht that $\alpha f_i= g_i \alpha$. So $g= \alpha^n \tilde{w}(f_1,...,f_n,g_1,...,g_n)$. Hence $F$ is sugroup of index 2 in $A$ (and also in $B$).

Then, you can show that the inclusions $A,B \hookrightarrow G$ extend to an isomorphism $A \underset{F}{\ast} B$.

In this case, Baumslag proved that $G$ is residually finite and moreover $G$ is finitely presented since $A$, $B$ and $C$ are finite.

So effectively, a virtually infinite-cyclic group is finitely presented and residually finite.

Answer 3

Let $G$ be a group which is virtually infinite cyclic. Then there is a finite index normal subgroup $N\lhd G$, which is also infinite cyclic.

Since $|Aut(N)|=2$, we have $[G:C_G(N)]\le 2$. Let's consider the case $N$ is central:

Then the transfer map $G\rightarrow N$ is a surjection from $G$ to the free group $N$, which means $G$ splits over $N$, so we can write $G=F\times N$, with $F$ a finite group.

Now if $[G:C_G(N)]=2$, we know from the above that $C_G(N)=F\times N$. Since $F\lhd G$, we can consider $G/F$, which must be torsion. Thus there's an element of finite order $gF\in G/F$ with $g^2F\in NF$, implying $g^2\in F$. In other words, $G/F$ splits as the semidirect product $N\rtimes C_2$, more commonly known as $D_\infty$.