# amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?

I heard from someone that the following problem is an open question.

(Open Problem 1)For a countable discrete group $G$, suppose it does not contain any Baumslag-Solitar subgroups $BS(m,n):=\langle x,y|xy^mx^{-1}=y^n\rangle$ and it admits a finite $K(G,1)$, is it a hyperbolic group?

I could not find the relevent stuff on this problem, so I am wondering whether the following is known.

(My question)For a countable discrete amenable group $G$, suppose it does not contain any Baumslag-Solitar subgroups $BS(m,n):=\langle x,y|xy^mx^{-1}=y^n\rangle$ and it admits a finite $K(G,1)$, is it a virtually cyclic group?

Note that Yes for problem 1 implies Yes for my question" since $G$ is amenable and hyperbolic iff $G$ is virtually cyclic(?)

Any references or comments are welcome!

## migrated from math.stackexchange.comNov 18 '14 at 13:25

This question came from our site for people studying math at any level and professionals in related fields.

• What do you mean by a finite $K(G, 1)$? You mean that $K(G, 1)$ can be modeled by a finite CW complex? – Qiaochu Yuan Nov 18 '14 at 1:02
• Anyway, this seems like it would be better on MO. – Qiaochu Yuan Nov 18 '14 at 1:09
• I guess the main expectation for the "open problem 1" is that no, there are examples, they are just hard to find. If so, your question is probably even harder. Or else you expect a positive answer for amenable groups, but it would sound difficult as well. Very few structural results for groups are known even under a strong assumption such as finite $K(G,1)$. – YCor Nov 18 '14 at 13:36
• Another comment: the assumption "does not contain any Baumslag-Solitar is equivalent to "does not contain any solvable Baumslag-Solitar", i.e., does not contain $BS(1,n)$ for any $n\ge 1$ (including $BS(1,1)\simeq\mathbf{Z}^2$). The reason is that $BS(m,n)$ for $\min(|m|,|n|)\ge 2$ contains a copy of $\mathbf{Z}^2$. This is not a hard fact, but it can sound surprising (at least two experts told me that $BS(2,3)$ does not contain $\mathbf{Z}^2$). – YCor Nov 18 '14 at 13:50
• Note for a finite dim. $K(G,1)$, virtually cyclic is equivalent to infinite cyclic (or trivial I suppose), since there can be no torsion. – Ian Agol Nov 18 '14 at 19:01

Yes, this problem is as open as Problem 1. There are no feasible obstructions to counter-examples, the main obstacle is lack of tools: All known constructions of non-hyperbolic groups with finite $K(G,1)$, yield groups which contain some $BS(p,1)$, $p\ge 1$.