# is a group $G$, that admits finite $k(G, 1)$ and has no Baumslag-Solitar subgroups, necessarily hyperbolic?

This is the first question asked in Bestvina's article "Questions in Geometric Group Theory". Does anyone know if there has been any progress made on this problem? Is the question answered if $G$ is instead automatic and contains no subgroups isomorphic to $\mathbb{Z} \times \mathbb{Z}$? If the first question is open, can someone clarify what is meant by "admit finite $k(G, 1)$" (is $G$ the fundamental group of a space with trivial higher homotopy groups or is the action of $G$ on the space properly discontinuous)? Any helpful references would be great.

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I guess you consider $\mathbb{Z}^2$ as a Baumslag-Solitar group too. The answers are still not known. "Finite $K(.,1)$ is a finite CW-complex whose fundamental group is your group, and whose universal cover is contractible. –  Mark Sapir Nov 29 '11 at 13:21
I am wondering what happens for free by $\mathbb{Z}$-groups. When do they have these properties ? Can one easily see that they are hyperbolic then? Something is known about when they are CAT(0)... –  HenrikRüping Nov 29 '11 at 13:49
@Henrik: There is a paper Kapovich, Ilya Mapping tori of endomorphisms of free groups. Comm. Algebra 28 (2000), no. 6, 2895–2917. It addresses these questions. As far as I know, the results of this paper are still the best known. –  Mark Sapir Nov 29 '11 at 13:59
@Henrik: For mapping tori of free group automorphisms, which I believe are the ones you are asking about, the result is an immediate consequence of Bestvina and Feighn's combination theorem. –  HJRW Nov 29 '11 at 15:59
This is unknown even for CAT(0) groups (where it reduces to no ZxZ subgroups). A recent result: msp.warwick.ac.uk/agt/2011/11-03/agt-2011-11-059s –  Ian Agol Nov 29 '11 at 17:23