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Two groups have a common model geometry if they act properly and cocompactlyby isometries on the same proper geodesic metric space. The Milnor-Schwarzlemma implies that groups with a common model geometry are quasi-isometric;however, the converse is false in general. We consider free products of uniformlattices in isometry groups of rank-1 symmetric spaces and prove, within eachquasi-isometry class, residually finite groups that have a common modelgeometry are abstractly commensurable. Our result gives the first examples ofhyperbolic groups that are quasi-isometric but do not virtually have a commonmodel geometry. Indeed, each quasi-isometry class contains infinitely manyabstract commensurability classes. We prove that two free products of closedhyperbolic surface groups have a common model geometry if and only if thegroups are isomorphic. This result combined with a commensurabilityclassification of Whyte yields the first examples of torsion-free abstractlycommensurable hyperbolic groups that do not have a common model geometry. Animportant component of the proof is a generalization of Leighton's graphcovering theorem. The main theorem depends on residual finiteness, and we showthat finite extensions of uniform lattices in rank-1 symmetric spaces that arenot residually finite would give counterexamples.

Action rigidity for free products of hyperbolic manifold groups