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.

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 $F_n$. $\endgroup$ – André Henriques Jan 3 '12 at 14:46