# Some examples of virtually cyclic groups

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.

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In fact, $G \lhd G \rtimes \mathbb{Z}_2$ and $G \rtimes \mathbb{Z}_2 / G \simeq \mathbb{Z}_2$, so $G$ is a subgroup of finite index in $G \rtimes \mathbb{Z}_2$. So $\text{Dih}(G)=G \rtimes \mathbb{Z}_2$ is virtually cyclic iff $G$ is virtually cyclic. –  Seirios Jul 11 '12 at 13:38
A virtually cyclic group is always finitely generated. –  tomasz Jul 11 '12 at 14:04
Also, how is $F\times \mathbf Z$ (for nontrivial $F$) just-infinite? $F$ is nontrivial and it is absolutely not of finite index! –  tomasz Jul 11 '12 at 14:10
I think that what you're actually definining is not virtually cyclic groups (every finite group is such), but virtually infinite- cyclic groups, which is another thing. Check this. –  DonAntonio Jul 11 '12 at 14:33
The groups I consider are indeed the virtually infinite-cyclic groups. With this keyword I found some interesting results! –  Seirios Jul 11 '12 at 15:22

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!

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I asked my question a long time ago, and the answer I gave is just so complicated! Of course I agree with you, the answer is elementary; user641 pointed it out in comments, but the accepted answer clearly should include this argument, thank you. –  Seirios Oct 30 '14 at 14:43
@Serios I didn't spot the comment of user641 (aka SteveD)! Yes, I didn't want to detract from the other answers, and I wouldn't have posted this answer if the post had not recently been bumped. But I felt that the current answers were just a bit too complicated (through trying to valiantly classify all such groups), especially if someone was reading this and they were new to infinite groups. –  user1729 Oct 30 '14 at 20:29
Could you add a hint on how one could prove the first statement using covering spaces? I can only think of a proof in the converse direction, i.e. for the statement: If $G$ is finitely presentable and $H$ is a finite index subgroup, then $H$ is finitely presentable. –  Jim Apr 9 at 13:42

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.

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As tomasz said, not all virtually infinite-cyclic groups are just-infinite, but with the above classification, you can show that a virtually infinite-cyclic group has finitely many infinite quotients. –  Seirios Jul 12 '12 at 20:28
By the way the description that you give for the class of virtually infinite cyclic groups goes back quite a bit further than the paper of Farrell and Jones that you cite. I know that it goes back at least to Stallings Ends Theorem: a virtually infinite cyclic group is the same as a 2-ended group, which were completely described by Stallings. –  Lee Mosher Jul 28 '12 at 20:44
What is the point of this long answer? Virtually finitely presented implies finitely presented, and virtually cyclic groups are clearly residually finite! RF means the identity subgroup is separable, and separability passes to finite extensions. –  user641 Aug 15 '12 at 19:05
Indeed, but it would be more difficult to show that a virtually infinite cyclic group has finitely many infinite quotients without using the classification given above (thank you for the proof!). –  Seirios Aug 23 '12 at 15:36

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 $D_\infty$.

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