Abstract
Two groups have a common model geometry if they act properly and cocompactly by isometries on the same proper geodesic metric space. The Milnor-Schwarz lemma implies that groups with a common model geometry are quasi-isometric; however, the converse is false in general. We consider free products of uniform lattices in isometry groups of rank-1 symmetric spaces and prove, within each quasi-isometry class, residually finite groups that have a common model geometry are abstractly commensurable. Our result gives the first examples of hyperbolic groups that are quasiisometric but do not virtually have a common model geometry. Indeed, each quasi-isometry class contains infinitely many abstract commensurability classes. We prove that two free products of closed hyperbolic surface groups have a common model geometry if and only if the groups are isomorphic. This result combined with a commensurability classification of Whyte yields the first examples of torsion-free abstractly commensurable hyperbolic groups that do not have a common model geometry. An important component of the proof is a generalization of Leighton's graph covering theorem. The main theorem depends on residual finiteness, and we show that finite extensions of uniform lattices in rank-1 symmetric spaces that are not residually finite would give counterexamples.
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