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A good way to show that a subspace is undistorted is to give a coarse Lipschitz retraction of the whole space onto that subspace. This question is about a failure of the converse.

Let $G$ be a finitely generated group, and fix a word-metric $d$ on $G$. Say that $G$ has the "cyclic coarse retracts property (CCRP)" if for every $g\in G$, either:

(1) there is a coarsely Lipschitz coarse retraction $f:G\to\langle g\rangle$, or

(2) $\langle g\rangle$ is distorted.

(In (1), I just mean a map $f$ so that there is a constant $C$ such that $d(f(g^n),g^n)\leq C$ for all $n$ and $d(f(h),f(k))\leq Cd(h,k)+C$ for all $h,k\in G$. I don't ask that $f$ is a homomorphism.)

Note that whether or not $G$ has the CCRP is independent of the choice of finite generating set. Examples of groups with CCRP include: torsion groups (trivially), hyperbolic groups, CAT(0) groups, Baumslag-Solitar groups, the Heisenberg group (I think), etc.

What are some examples of finitely presented groups $G$ that do not have the CCRP? I'd like to have a list of a few examples to see how this property fails. A characterization of groups which have this property would be even better!

(Note: one can construct finitely generated examples, but the ones I am aware of are lacunary hyperbolic. Since finitely presented lacunary hyperbolic groups are hyperbolic, these non-CCRP groups can't be finitely presented.)

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    $\begingroup$ Olshanskii (1997) proved that any recursively presented f.g. group has an undistorted embedding into a finitely presented group (the same statement without "undistorted" is Higman's theorem). Since clearly CCRP passes to undistorted subgroups, constructing a f.p. group without CCRP reduces to proving the existence of a recursively presented f.g. group that fails CCRP. I expect you can easily arrange your lacunary hyperbolic example to be recursively presented. $\endgroup$ – YCor Nov 29 '17 at 15:10

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