# Hyperbolic groups from Dehn functions

Hyperbolic groups may be defined as finitely generated groups admitting a linear Dehn function. I wonder whether it is possible to prove most of the classifical properties of hyperbolic groups in this context, probably using van Kampen diagrams. For instance, using this point of view:

Is it possible to prove that a hyperbolic group cannot contain $\mathbb{Z}^2$?

Is it possible to prove that the size of the finite subgroups of a hyperbolic group is bounded?

• The answer to both questions is yes, by using the theorem that linearity of the Dehn function implies other formulations of hyperbolicity such as existence of $\delta \ge 0$ such that all geodesic triangles in the Cayley graph are $\delta$-thin. One can proceed from there, for example, to the classification of isometries, and from there one can deduce that the centralizer of every element contains a cyclic subgroup of finite index. Hence a hyperbolic group does not contain $\mathbb{Z}^2$. However, I am guessing that's not what you have in mind? – Lee Mosher Mar 4 '16 at 14:31
• Indeed, I know the usual proofs of these two facts, and I would be interested in alternative proofs which do not use this formulation of thin triangles. – Seirios Mar 4 '16 at 14:36
• I don't know of any alternate proof like that. If you assumed in addition that your $\mathbb{Z}^2$ subgroup was undistorted in both metric and area then you can get a very short proof, but that's a very strong hypothesis, and the existence of examples of distorted subgroups of hyperbolic groups makes it an unreasonable hypothesis. – Lee Mosher Mar 4 '16 at 14:39
• I thibk it is a little easier to prove that a linear Dehn function implies that geodesic bigons in the Cayley group are uniformly thin, than it is to prove this property for geodesic triangles,. That's enough to prove that the group is biautomatic, and then you can use the results in Gersten and Short's paper about rational subgroups of biautomatic groups to show that, if the group contained ${\mathbb Z}^2$, then it would contain a rational, and hence undistorted, subgroup ${\mathbb Z}^k$ for some $k \ge 2$, contradicting linear Dehn function. – Derek Holt Mar 4 '16 at 15:37
• I think it should not be too difficult to prove that, if the Cayley graph is hyperbolic, then the associated Dehn function is linear. Therefore, if we are able to prove that the bigons in the Cayley graph are uniformly thin, then it is sufficient to conclude thanks to a result of Papasoglu implying the hyperbolicity of the Cayley graph. But I do not know Gersten and Short's paper, so I will take a look. – Seirios Mar 4 '16 at 19:38