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A virtually-$\mathbb{Z}$ group $G$ admits either a epimorphism onto $\mathbb{Z}$ or a epimorphism onto $D_\infty$.

So what happens if one replaces $\mathbb{Z}$ by another group $F$ (like the free group or $\mathbb{Z}^n$). My first guess would be that any virtually-$F$ group $G$ maps surjectively onto one of the groups

$F\rtimes H$, where $H\le $Aut$(F)$ is any finite subgroup.

This is clear in the case where $G$ is semidirect product of $F$ and a finite group $K$; one can simply take $H$ to be the image of $K\rightarrow $ Aut $ (F) $.

So my questions are:

1) Is it still true that there is an epimorphism even in the non-split case?

2) Is it also true that two groups $F\rtimes H_1$ and $F\rtimes H_2$ (with $H_i\subset $ Aut $(F)$ finite) cannot surject onto each other unless $H_1$ and $H_2$ are conjugated (in which case the groups are isomorphic) ?

I doubt that this is true for all groups $F$ but maybe one can find sufficient conditions that guarantee this.

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I am skeptical that this works for virtually free groups. Virtually infinite cyclic groups are 2-ended and so act on the real line by simplicial maps. This means the group surjects to Z or to the infinite dihedral group. A general virtually cyclic group acts on a tree with finite vertex and edge stabilizers. I don't see how you are guaranteed the epimorphism you want, but I leave counterexamples or proving me wrong to the experts on this site. – Benjamin Steinberg Jan 3 '12 at 14:43
"A virtually-$\mathbb{Z}$ group $G$ admits either a epimorphism onto $\mathbb{Z}$ or...". Let's think a bit about which $\mathbb{Z}$ we are talking about. For example, the group $\frac 1 2 \mathbb Z$ (subgroup of $\mathbb Q$) is virtually $\mathbb{Z}$ but it doesn't admit any surjective isometry onto $\mathbb{Z}$. This point might sound pedantic since $\mathbb{Z}$ and $\frac 1 2 \mathbb Z$ are isomorphic... but I bet that it makes a difference in the case of the free group $F_n$. – André Henriques Jan 3 '12 at 14:46
Virtually free groups include free products of finite groups. What do you want to surject those on? – André Henriques Jan 3 '12 at 16:32
So does this mean that there is no set of virtually-$F$ groups $S$ such that each virtually $F$-group surjects exactly onto one of the groups from $S$. It certainly means that taking semidirect products with finite subgroups of $Aut(F)$ is not enough... – HenrikRüping Jan 4 '12 at 15:28
There's another interpretation of the surject $\mathbb{Z}$ or $D_\infty$ result. One may interpret these as graph-of-groups, an HNN extension of the trivial group or an amalgamated product. A nice theorem of Stallings generalizes this to virtually free group. In fact, a virtually free group is a graph of finite groups. If you fix the rank of the free group $F_n$, then I think one should be able to use this to derive a structure theorem for virtually $F_n$ groups with is analogous to the virtual $\mathbb{Z}$ case. – Ian Agol Feb 26 '12 at 16:48
up vote 6 down vote accepted

Here's an idea for a proof that the modular group $\Gamma=\mathbb{Z}/2*\mathbb{Z}/3$, which is, of course, virtually free, doesn't surject a group of the form $F\rtimes H$. I don't have time to work out the details.

First, I think it's plausible that the only reduced, non-trivial graph-of-groups decomposition for $\Gamma$ is the one that realises it as the free product given above, which I will denote by $\mathcal{H}$. One idea for the proof is as follows. If $\mathcal{G}$ is any such graph of groups with fundamental group $\Gamma$, then the generators $a$ of order two and $b$ of order 3 can be taken to lie in distinct vertices of $\mathcal{G}$; this leads to an obvious immersion of graphs of groups $\mathcal{H}\to\mathcal{G}$. But $\mathcal{H}$ carries the whole of $\Gamma$, so the immersion is actually an isomorphism.

Remark: If the above is true, then it should be known, and there should be a reference out there somewhere.

Now suppose that $f:\Gamma\to F\rtimes H$ is a surjection. Then $F\rtimes H$ can be decomposed as a graph of groups $\mathcal{J}$, and this induces a graph of groups decomposition $\mathcal{J}'$ for $\Gamma$, by setting $\mathcal{J}'_v=f^{-1}\mathcal{J}_v$ for each vertex $v$ and $\mathcal{J}'_e=f^{-1}\mathcal{J}_e$ for each edge $e$. But, by the previous claim, $\mathcal{J}'=\mathcal{H}$. As $\mathbb{Z}/2$ and $\mathbb{Z}/3$ are both simple, it follows that $f$ is an isomorphism. But $\Gamma$ is not of the form $F\rtimes H$. (To see this, note, for instance, that all the torsion in $F\rtimes H$ is conjugate into a singe finite subgroup $H$; this is not the case in $\Gamma$.)

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Let $G := F \rtimes H = f(\Gamma)$. Since $G$ is virtually free, it is a fundamental group of a graph of groups with finite vertex groups. But (like $\Gamma$) $G$ has finite abelianization, so $G$ can be defined by free products with amalgamation of a finite number of finite groups. There must be known results about minimal generator numbers of free products with amalgamation in terms of the generator numbers of the factors. I would expect the fact that $G$ is 2-generated to imply that it is a free product with amalgamation of two cyclic groups, which should lead rapidly to a contradiction. – Derek Holt Jan 4 '12 at 9:15
'There must be known results about minimal generator numbers of free products with amalgamation in terms of the generator numbers of the factors'. Some care is needed here! For example, the free group of rank two admits very complicated graph-of-group decompositions. See, for instance, the examples on page 454 of Bestvina and Feighn's Inventiones paper 'Bounding the complexity of simpliciai group actions on trees'. – HJRW Jan 4 '12 at 10:37
You are right that care is needed! It is not hard to construct examples with $d(A*_CB) < d(A),d(B)$ with $A$ and $B$ finite, which surprised me! I can see how to do that with $d(A*_CB) \le 4$ and $d(A),d(B)$ arbitrarily large. I expect it is possible with $d(A*_CB) = 2$, but I haven't managed that yet! – Derek Holt Jan 4 '12 at 13:48
I think an easier argument gives something stronger about $C_2\ast C_3$. Namely, if $K$ is the (non-trivial) kernel of some surjection, then either $K$ has torsion (and thus the quotient is abelian), or $K$ is torsion free. But every torsion-free subgroup of $C_2\ast C_3$ is contained in the commutator ($F_2$), and there are no non-trivial quotients of $F_2$ which are both non-abelian and free. – Steve D Nov 14 '14 at 15:05
@SteveD, that's a nice idea. You're right that $F_2/K$ can't be free, but it could still be virtually free, so I don't see how to finish the argument... – HJRW Nov 14 '14 at 22:12

Actually, for free abelian groups something does in fact happen along the lines of what you ask. It is just that the semidirect product idea is a little bit of a red herring.

In the $Z^n$ case, what happens is that any virtually $Z^n$ group surjects onto one of the finitely many $n$-dimensional Euclidean crystallographic groups, with finite kernel. This is basically the content of the Bieberbach theorems. In the rank 2 case, that's 17 different Wallpaper groups to surject onto. In rank 3, that's 219 different Space groups.

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It looks like the answer to your first question is no for both of the two classes of groups $F$ that you ask about, free groups and free abelian groups.

${\mathbb Z}_2 * {\mathbb Z}_3$ is virtually $F_2$ (free of rank 2), with quotient ${\mathbb Z}_6$, but it does not map onto a semidirect product of $F_2$ by a finite group, basically because its finite subgroups have order at most 3.

The group $\langle x,y \mid y^{-1}xy=x^{-1} \rangle$ has a subgroup $\langle x,y^2 \rangle$ of index 2 isomorphic to ${\mathbb Z}^2$, but it does not surject onto a semidirect product of ${\mathbb Z}^2$ with a finite group.

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Oh - you got there just ahead of me! Could you elaborate on `...basically because its finite subgroups have order at most 3.'? – HJRW Jan 3 '12 at 19:40
Sorry, that's not so clear as I thought! Your argument should work. – Derek Holt Jan 3 '12 at 23:31

Let me elaborate on the comment I made in the case of a virtually free group. Suppose $G$ is a virtually $F_n$ group, $F_n \leq G$ a finite-index free group of rank $n$. By Stallings' theorem, $G$ is a graph of finite groups. In particular, there is a tree $\mathcal{T}$ and an action of $G$ on $\mathcal{T}$, such that $\mathcal{T}/G$ is a finite graph. The action of $F_n$ on $\mathcal{T}$ is faithful and free, since the stabilizer of any vertex in the $G$ action on $\mathcal{T}$ is finite. Consider the kernel $K=ker\{ G \to Aut(\mathcal{T})\}$. Then $K$ is finite since it stabilizes any vertex of $\mathcal{T}$. Then we should think of $\Lambda=\mathcal{T}/(G/K)$ as an orbispace, in the sense of Haefliger, where we attach to each cell of $\Lambda$ the stabilizer in $G/K$ of a lift of the cell to $\mathcal{T}$. Also, $F_n\hookrightarrow G/K$, since $F_n \leq G$ acts faithfully on $\mathcal{T}$. The quotient $\mathcal{T}/F_n$ is a finite graph $\Gamma$ with $b_1(\Gamma)=n$ and no vertices of degree 1. There are only finitely many topological types of such graphs, and only finitely many orbispace covers $\Gamma \to \Lambda$. Thus, there are only finitely many possibilities for $G/K$. These groups are maybe the analogues of $\mathbb{Z},D_\infty$ in your question, or of the Bieberbach groups in Mosher's answer in the $\mathbb{Z}^n$ case.

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Ian - I think you mean $b_1(\Gamma)\leq n$. – HJRW Feb 26 '12 at 20:53
@HW: No, I mean $b_1(\Gamma)=n$. I forgot to state before that $K$ is finite - does this make sense now? The action of $F_n$ on $\mathcal{T}$ is free and faithful, so the quotient graph should have fundamental group isomorphic to $F_n$, hence $b_1(\Gamma)=n$. – Ian Agol Feb 26 '12 at 23:21
Ian - of course you're right. I had misread $\Gamma$ as $T/G$. I also wrote something silly in an earlier comment here, which I have now deleted. – HJRW Feb 27 '12 at 16:12

Let me add another class of examples: virtual hyperbolic surface groups, by which I mean groups that have some $\pi_1(S_g)$ as a finite index subgroup, where $S_g$ is the closed, oriented surface of genus $g \ge 2$. Each virtual hyperbolic surface group $\Gamma$ surjects, with finite kernel, onto the fundamental group of a closed hyperbolic 2-orbifold (predecessors to Haefliger's orbispaces mentioned above by Agol). And if you fix $g$ then there are only finite many such orbifold groups which are the targets of a virtual $\pi_1(S_g)$ group.

This class of examples comes from quasi-isometric rigidity of the class of hyperbolic surface groups, a theorem of Gabai and of Cannon and Jungreis, which says that any finitely generated group quasi-isometric to the hyperbolic plane (which includes virtual hyperbolic surface groups) has a surjection as described above for $\Gamma$.

One can mine the theory of quasi-isometric rigidity for other classes of examples.

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