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Recall that a group $G$ is of type F if there exists a compact $K(G,1)$.

There are many examples of groups which are not of type F but which are virtually of type F, that is, they have finite-index subgroups of type F. For example, $SL(n,\mathbb{Z})$, mapping class groups, automorphism groups of free groups, etc.

In all the examples I am aware of, the problem comes down to torsion (a group of type F must be torsion-free), and the required finite-index subgroup is just one that has no torsion.

Question: Does anyone know of a torsion-free group that is not of type F but is virtually of type F?

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It is conjectured that no such example exists. It is conjectured that $G$ is of type $F$, but it is hard to prove a group is type $F$ without explicitly exhibiting a classifying space. The class of groups of type $FP$ is a well-behaved proxy. Moreover, it is conjectured that every finitely presented group of type $FP$ is actually of type $F$. It is known that a finitely presented group of type $FP$ has a classifying space which is "finitely dominated" meaning that it is a retract in the homotopy category of a finite complex, which is just as good for most homotopical purposes. I will prove that if a torsion-free group $G$ has a finite index subgroup $H$ of type $FP$, then $G$ is itself of type $FP$. Applying this to your situation, since $H$ is finitely presented, so is $G$, so the conjecture implies that $G$ is of type $F$. It is possible that the hypothesis that $H$ is of type $F$ allow an unconditional proof, but I do not expect it.

$G$ has type $FP$ if the trivial $G$-module $\mathbb Z$ has a finite length resolution by finitely generated projective $G$-modules. Type $F$ is equivalent to a finite length resolution by finitely generated free $G$-modules, plus finite presentation. The advantage of type $FP$ is that it is the intersection of two properties that can be treated separately, namely finite dimension and finite generation of cohomology $FP_\infty$. These can be interpreted as the existence of two resolutions, one of finite length and other by finite modules. The two properties can be combined to produce a resolution that is simultaneously finitely generated and finite length, but only if one allows projective modules and not just free modules, so type $F$ does not decompose this way.

The easy part is finite generation of homology. If $G$ contains a finite index subgroup $H$, then the intersection of the conjugates of $H$ is a finite index normal subgroup $N$. The Hochschild-Serre spectral sequence for the group extension $N\to G\to G/N$ allows us to combine the $FP_\infty$ properties of $N$ and $G/N$ to conclude that $G$ is $FP_\infty$.

The other part is finite homological dimension. By assumption $H$ has a finite dimensional classifying space with universal cover $EH$. Consider the space of $H$-equivariant maps from $G$ into $EH$. That is, intertwining the left action of $H$ on $G$ and the only action of $H$ on $EH$. This leaves the right action of $G$ on $G$, which makes this a $G$-space, a candidate for $EG$. For the quotient to be a model of $BG$, we need the space to be contractible and the action to be free. This space is isomorphic to $EH^{[G:H]}$, hence contractible and finite dimensional. The size of the isotropy is bounded by the index. So the isotropy is a finite subgroup of the torsion-free group $G$, so actually $G$ acts freely on this space. This shows that $G$ has a finite dimensional classifying space. In fact, the dimension of $G$ is the same as the dimension of $H$, although that does not come out of this construction.

Reference: Ken Brown's book, Chapter VIII or his notes.

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  • $\begingroup$ Bestvina and Brady disproved the conjecture $F=FP$ (Inventiones Math. 1997). Their examples are certain normal subgroups of 2-dimensional RAAGs: The subgroups they construct are $FP_2$ (and, hence, $FP$ since the groups are 2-dimensional) but are not finitely-presented. $\endgroup$ – Misha Aug 23 '14 at 4:28
  • $\begingroup$ Still, in Sarah's question finite presentability is not an issue, and thus from FP plus finite presentability we can deduce F, answering negatively the question (torsion-free + virtually F implies F). $\endgroup$ – YCor Aug 23 '14 at 12:56
  • $\begingroup$ I edited to include finite presentation. As YCor says, the BB examples cannot be used for this question and the current conjecture FP+fp=F would apply. $\endgroup$ – Ben Wieland Aug 23 '14 at 16:14
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    $\begingroup$ @Misha - I don't think the Bestvina--Brady examples come from 2-dimensional RAAGs. Indeed, they come from RAAGs defined by the 1-skeleton of a flag triangulation of a homology sphere. $\endgroup$ – HJRW Aug 24 '14 at 19:36
  • $\begingroup$ I asserted some conjectures, but I didn't quite isolate them. The obstruction for a finitely presented finite dimensional $FP_\infty$ group to have a finite classifying space is for the $G$-module $\mathbb Z$ to be trivial in the reduced $K$-group $\bar{K}_0(\mathbb ZG)$. The $K$-theory generalization of one of the Kaplansky conjectures says that this $K$-group is zero (even if the group is just torsion-free). I don't know if this conjecture has a name, other than the $K$-theory (variant) Kaplansky conjecture. This is subsumed by the monster Farrell-Jones conjecture. $\endgroup$ – Ben Wieland Mar 22 at 18:39

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