It's, by now, more or less well known that residual finiteness is not a quasi-isometry invariant for finitely generated groups (see here for an example). Thus the following question makes sense:

Question: given a finitely generated group $G$, is there a finitely generated group $H$ such that $H$ is residually finite and $G$ and $H$ are quasi-isometric? That is, does the class of groups quasi-isometric to $G$ always contain residually finite groups?

Acknowledging this question is fairly open-ended, I would be glad with any reference in this direction. Thanks in advance. The following tries to motivate the above via some C*-algebraic questions.

Motivation (and guess): Should the above be known true, then one would be able to manually build so-called quasi-diagonalizing projections for the reduced group C*-algebra of $G$, that is, finite rank projections $p_n \in \mathcal{B}(\ell^2 G)$, with $p_n \rightarrow 1$ in the strong operator topology and $$ ||p_n \lambda_g - \lambda_g p_n|| \rightarrow 0 $$ for every $g \in G$, where $\lambda$ is the left regular representation of $G$. Indeed, note that if $H$ is QI to $G$ and res. finite and amenable, then, by work of Orfanos there are projections $q_n$ for $H$ as above, and those can be pullback-ed to projections $p_n$ for $G$. Thus, since this C*-question is still open, my guess is that the original question is open as well.

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    $\begingroup$ The answer is certainly no, and the question is whether a counterexample (a f.g. group not QI to any RF f.g. group) is already known. The limit is that QI rigidity is known for quite few groups. Possibly Kac-Moody groups (these are simple finitely presented CAT(0) groups) or non-RF Baumslag-Solitar groups would be workable candidates. Also for most non-RF groups I can imagine, there's no natural candidate for a RF group QI to it. $\endgroup$ – YCor Apr 15 '20 at 9:39
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    $\begingroup$ The quasi-isometry class of higher Baumslag-Solitar groups is known (White, Papasoglu and others) and contains no r.f. groups. $\endgroup$ – user6976 Apr 15 '20 at 10:34
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    $\begingroup$ @MarkSapir I googled on White's paper before posting my first comment. Could you provide a reference? I'm aware of an unpublished paper of Whyte, which is known to contain mistakes in its claim on QI-classification (I don't know whether it affects your assertion). $\endgroup$ – YCor Apr 15 '20 at 11:09
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    $\begingroup$ PS it's Whyte, not White, we both misspelled... it's Kevin Wkyte, Coarse bundles arXiv link. Nevertheless I can find in this paper a statement describing arbitrary groups quasi-isometric to $\mathrm{BS}(2,3)$, so I'd be happy to hear about a reference, if any. $\endgroup$ – YCor Apr 15 '20 at 20:54
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    $\begingroup$ @YCor I think you're looking for [Whyte, K. The large scale geometry of the higher Baumslag-Solitar groups. Geom. Funct. Anal. 11 (2001), no. 6, 1327–1343.]. Theorem 5.1 in there is the following: a f.g. group $\Gamma$ is qi to $BS(2,3)$ iff it is a graph of virtual $\mathbb{Z}$s which is neither commensurable to $F_n \times \mathbb{Z}$ nor virtually solvable. $\endgroup$ – Carl-Fredrik Nyberg Brodda Apr 16 '20 at 9:10

Take any finitely-presented group $G$ with undecidable word problem. Then $G$ is not quasi-isometric to any finitely generated group with decidable word problem, in particular, to any residually-finite group. (Note that finite presentability and decidability of the WP are quasi-isometry invariant. The latter is because quasi-isometries preserve the equivalence class of the Dehn function and WP is for a finitely-presented group decidable iff the Dehn function is recursive.)

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    $\begingroup$ Since it's implicit, you're using that finitely presented residually finite groups have solvable word problem, and that among finitely presented groups, to have solvable word problem is QI-invariant. Nice answer! $\endgroup$ – YCor Apr 15 '20 at 16:26

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