(The question has been edited. It was pointed out in the comments that $\Gamma_G$ could be a surface group, thought of as a finite extension of another surface group, in which case $G$ is finite.)

Let $\pi$ be the fundamental group of a closed orientable surface, and let $G$ be a subgroup of $\mathrm{Out}(\pi)$.
Associated to $G$ is an extension \begin{equation} 1 \to \pi \to \Gamma_G \to G \to 1 \end{equation} which sits inside the sequence \begin{equation} 1 \to \mathrm{Inn}(\pi) \to \mathrm{Aut}(\pi) \to \mathrm{Out}(\pi) \to 1. \end{equation}

If there is a finite $K(G,1)$, it is not difficult to see that there is a finite $K(\Gamma_G,1)$. A colleague and I are interested in a converse.

If $G$ is torsion-free and there is a finite $K(\Gamma_G, 1)$, is there a finite $K(G,1)$?

I am aware of Wall's theorem that a finitely presented group $\mathcal{G}$ has a finite $K(\mathcal{G},1)$ if and only if it is of type (FL), meaning that there is a finite resolution of $\mathbb{Z}$ by finitely generated free $\mathcal{G}$-modules. A group $\mathcal{G}$ is of type (FP) if there is a finite resolution of $\mathbb{Z}$ by finitely generated projective $\mathcal{G}$-modules. As far as I know, there is still no known group of type (FP) that is not of type (FL).

I imagine that careful examination of the Hochschild-Serre spectral sequence may tell you that $G$ is of type (FP), but I didn't pursue this as I'd like to get all the way to (FL).

Is there some geometric construction I'm missing?

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    $\begingroup$ Could you explain why the assumption on $\Gamma_G$ implies that $G$ is torsion-free? (Which is necessary for having a finite classifying space). $\endgroup$ Oct 30, 2012 at 2:53
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    $\begingroup$ It doesn't imply it, does it? If a finite group $G$ acts freely on a closed surface that is a $K(\pi,1)$ then the orbit space is a closed surface that is a $K(\Gamma_G,1)$. $\endgroup$ Oct 30, 2012 at 3:17
  • $\begingroup$ Oops. I was thinking that $\pi$ is a maximal surface subgroup, but it isn't. I'll edit. $\endgroup$ Oct 30, 2012 at 3:26
  • $\begingroup$ Richard: you might want to mention that $\Gamma_G$ is necessarily a subgroup of $Aut(\pi)$ that contains $Inn(\pi)$, so $G$ is simply $\Gamma_G/Inn(\pi)$ (this is an exercise in Brown's book at the end of Chapter 4). Also I am puzzled why you expect $G$ to have a finite classifying space, e.g. there are Rips type constructions in which the middle group (in the extension) has a finite classifying space, and the quotient group does not; of course, the kernel there isn't a surface group. $\endgroup$ Oct 30, 2012 at 3:38
  • $\begingroup$ Igor, that's a good suggestion. And I'm not sure what I expect the answer to be. $\endgroup$ Oct 30, 2012 at 3:43


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