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This question is mostly idle curiosity.

Recall the following standard terminology: if $P$ is a property of groups, a group $G$ is said to be virtually $P$ if it has a subgroup of finite index which is $P$. First, a preparatory question:

Is there an established name in the literature for invariants of groups which are invariant under passage to finite index subgroups?

For example, I think the quasi-isometry class of a finitely generated group has this property. Somewhat more generally I think the coarse equivalence class of a group has this property.

For now let me call such things "virtual invariants." It is tempting to try to study virtual invariants by studying the universal such thing: namely, start with the category of (possibly finitely generated) groups, then localize at the inclusions of finite index subgroups.

Has anyone computed anything about this localization? In particular, is there a model category structure on (possibly finitely generated) groups with inclusions of finite index subgroups as weak equivalences?

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  • $\begingroup$ Apparently if we toss in quotients with finite kernel then the resulting equivalence relation is called weak commensurability. The corresponding question in this case about model structures also seems interesting. $\endgroup$
    – Qiaochu Yuan
    Commented Feb 4, 2014 at 2:26
  • $\begingroup$ Just a small comment: there's no model category structure on the category of finitely generated groups. A model structure always requires the category to have all limits and colimits. $\endgroup$
    – Akhil Mathew
    Commented Feb 4, 2014 at 2:34
  • $\begingroup$ Groups are called abstractly commensurate if they have isomorphic finite index subgroups. This is not the name for the class of invariants though. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 4, 2014 at 3:21
  • $\begingroup$ It seems then that Qiaochu's "virtual invariants" are something like $\mathcal O(\{\text{equivalence classes for abstract commensurability}\})$, right? I have nothing useful to say towards the posted question. $\endgroup$
    – Theo Johnson-Freyd
    Commented Feb 4, 2014 at 4:14
  • $\begingroup$ This question seems related: mathoverflow.net/q/97759/1345 In fact, for many classes of groups, such as lattices in semisimple groups, the abstract commensurator contains any member of the commensurability class as a subgroup. However, incommensurable groups can have isomorphic commensurators. $\endgroup$
    – Ian Agol
    Commented Feb 4, 2014 at 4:21

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