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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
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A virtually cyclic group is always finitely generated. –  tomasz Jul 11 '12 at 14:04
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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
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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
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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

2 Answers 2

up vote 1 down vote accepted

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
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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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