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.



Analogues of the dihedral group
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 $\Gamma$.)


