A group $G$ is defined to have $F$ if there exists a finite $K(G,1)$. This property is clearly not invariant under quasi-isometry as one can see from the trivial group and $\mathbb{Z}_2$.

My question: Is there also a torsion-free counterexample? Meaning are there two torsion-free, quasi-isometric groups of which one has $F$ and theother one not?

  • 1
    $\begingroup$ A variant of the question, with no torsion issue, is: Let $C$ be the class of f.g. groups admitting a proper continuous action on a contractible (locally compact) topological space with compact quotient. Is $C$ closed under QIs? $\endgroup$
    – YCor
    Jun 16, 2016 at 20:13
  • $\begingroup$ As an analogy, it is known that a torsion-free group having a finite index subgroup with Property FP (=$FP_\infty$ and finite cohomological dimension) then it itself has Property FP. This leaves some hope to a negative answer to the original question (no torsion-free counterexample) and rules out some too naive approaches. $\endgroup$
    – YCor
    Jun 16, 2016 at 20:35

1 Answer 1


This is not an answer but a summary of what I know about this issue:

  1. The property $F_n$ is QI invariant (the groups here are not assumed to be torsion free). You can find a proof either in section 18.2 of

R.Geoghan, "Topological methods in group theory"

or in chapter 6 of my book with Cornelia Drutu.

It is quite possible that some version of the arguments in these proofs will handle your question, it is worth checking.

  1. If a group $G$ has type $FP_n$ and is $n$-dimensional (here and below everything is over ${\mathbb Z}$) then $G$ has type $FP$. In particular, if a group has type $F_\infty$ and has finite cohomological dimension then it has type $FP$.

  2. If a finitely presented group $G$ has type $FL$ then there is a finite $K(G,1)$. This is in Brown's book "Cohomology of groups", Ch. 8.8.

  3. There are no known examples of finitely presented groups which are of type FP but not of type FL.

  4. Putting it all together and assuming that FP=FL, the positive answer to your question would follow from:

Suppose that $G$ admits a finite $K(G,1)$ and $H$ is torsion free quasiisometric to $G$. Then $H$ has finite cohomological dimension.

This is quite likely to be true, since cohomological dimension over ${\mathbb Q}$ is QI invariant (R.Sauer, "Homological invariants and quasi-isometry", GAFA, 2006).


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