AGenevois
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Your question is related to a famous conjecture: Kervaire Conjecture: Given a non-trivial group $H$ and an element $g \in H \ast \mathbb{Z}$, the quotient $(H \ast \mathbb{Z} ) / \langle \!\langle ... View answer 1 answers 13 votes 875 views Accepted answer 28 votes Below are some sources of hyperbolic groups. Of course, the list is far from being exhaustive. Groups defined by generators and relations: Finitely generated free groups, as their Cayley graphs are ... View answer 2 answers 7 votes 289 views Accepted answer 12 votes If a group$G$satisfies Kazhdan's property (T), then any action of$G$on a CAT(0) cube complex has a global fixed point. See Niblo and Roller's article Groups acting on cubes and Kazhdan's Property (... View answer 2 answers 7 votes 439 views 12 votes A direct proof. Following Yves' comments above, it is possible to give an easy proof of the fact that an infinitely-ended group must contain an infinite-order element, which implies that it cannot be ... View answer 2 answers 11 votes 547 views 11 votes Abels' groups provide simple examples of solvable groups that are finitely presented but not of type$F$. Given a ring$R$, define the group $$A_n(R):= \left\{ \left( \begin{array}{ccccc} 1 &amp;&amp;&... View answer 1 answers 10 votes 625 views Accepted answer 11 votes Many CAT(0) groups cannot act geometrically on CAT(0) cube complexes. For instance: CAT(0) groups satisfying Kazhdan's property (T), eg. uniform lattices in simple Lie groups of higher rank or in ... View answer 1 answers 7 votes 337 views Accepted answer 10 votes The following statement can be found in Section 5 of Low-dimensional homology groups of mapping class groups: a survey: Theorem: Let g \geq 1. Then$$H_1(\Gamma_{g,r}^n,\mathbb{Z}) \simeq \left\{ ... View answer 1 answers 7 votes 199 views Accepted answer 10 votes As already mentioned in the comments, it is still unknown whether hyperbolic groups are CAT(-1) or even CAT(0). A related question is: Let$G$be a hyperbolic group (endowed with a finite generated ... View answer 1 answers 4 votes 147 views Accepted answer 10 votes 1. About injectivity: Yves already answered about injectivity in the comments. Below is an alternative argument which works more generally for free products: Proposition 1: Let$G=A_1 \ast \cdots \...

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Peripheral subgroups of relatively hyperbolic CAT(0) groups are indeed CAT(0) themselves. In fact, more is true: Morse subgroups of CAT(0) groups are CAT(0) themselves. Definition. Given a finitely ...

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Another source of Cayley graphs with contractible Rips complexes comes from Helly graphs. Proposition: Rips complexes of uniformly locally finite Helly graphs are contractible. See Lemma 5.28 and ...

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About Question 3, the answer is &quot;no&quot; because Thompson's groups $F$, $T$, and $V$ act properly on CAT(0) cube complexes (as proved by D. Farley, see MR1978047 and MR2136028). Indeed, a ...

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See for instance Kawauchi's article A classification of compact 3-manifolds with infinite cyclic fundamental groups. However, by looking at the review on mathscinet: The author provides a somewhat ...

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There exist plenty of groups satisfying Serre's property (FA), meaning that any action on a simplicial tree has a global fixed point, which act nicely on a CAT(0) cube complex. It includes: Many ...

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Here is a point of view which justifies why Property $(FA)$ is a very particular case of Property $(T)$. First, Chatterji-Drutu-Haglund proved that: Theorem: A discrete group has $(T)$ iff all its ...

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Here is more direct and elementary argument. Lemma: $\mathrm{Aut}(W_3)$ and $\mathrm{Aut}(\mathbb{F}_2)$ are isomorphic, where $W_3$ denotes the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \... View answer 2 answers 8 votes 398 views 8 votes I am not sure that this is what the OP is looking for, but here is a justification of the fact that the unit tangent bundle of a hyperbolic surface cannot be endowed with a nonpositively curved metric.... View answer 2 answers 10 votes 384 views 8 votes In the same vein as dodd's answer, a counterexample can also be deduced from the second Houghton group$H_2$, which is defined as the group of bijections$L^{(0)} \to L^{(0)}$that preserves adjacency ... View answer 3 answers 4 votes 639 views 8 votes Here are some details about constructing normal subgroups and some properties of negative curvature of groups (such as splitting as a free product). The connection between these two subjects comes ... View answer 3 answers 6 votes 594 views Accepted answer 8 votes As mentioned by Andy Putman in the comments, the classical (and probably the best) references are Serre's book Trees and Scott and Wall's paper Topological methods in group theory. Serre's approach is ... View answer 1 answers 6 votes 300 views Accepted answer 8 votes I think D. L. Johnson's article Embedding some recursively presented groups should answer your question. The abstract is: We seek to illustrate the Higman Embedding Theorem by finding actual ... View answer 2 answers 10 votes 431 views 8 votes Below is a geometric argument based on the action of Higman's group on its natural CAT(0) square complex. (Of course, Yves' argument is more elementary, but I find this alternative viewpoint enjoyable.... View answer 3 answers 10 votes 297 views 7 votes I finally found a elementary example: $$BS(1,-1):= \langle x,y \mid yxy^{-1}=x^{-1} \rangle.$$ It embeds into$\mathbb{Z} \oplus \mathbb{D}_\infty$, which itself embeds in the right-angled Coxeter ... View answer 3 answers 16 votes 576 views 7 votes Here is an idea to construct Thompson-like groups with non-trivial centers. The construction depends on an arbitrary group$G$we fix once for all. Labelled strand diagrams. In the same way that every ... View answer 2 answers 13 votes 493 views 6 votes 1) LAMPLIGHTER GROUPS As mentioned by Yves, lamplighter groups over$\mathbb{Z}$provide counterexamples thanks to Eskin, Fisher, and Whyte's work. Other counterexamples are given by lamplighter ... View answer 2 answers 9 votes 432 views 6 votes As suggested by Will Sawin in the comments, I took a look at Bekka, de la Harpe and Valette's book. The claim is a straightforward consequence of the following statement: Theorem. (Artin) For every$k ...

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As mentioned in the comments, Sapir and Olshanskii recently proved in Algorithmic problems in groups with quadratic Dehn functions that: Theorem. For every recursive function $f$, there exist ...

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Famous examples come from the so-called Bestvina-Brady groups. Given a simplicial graph $\Gamma$, define the right-angled Artin group $A(\Gamma)$ as \langle \text{$u$ vertex of $\Gamma$} \mid [u,v]=...

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Matt Zaremsky's justification of &quot;$F$&quot; for &quot;free&quot; can also be found in Ross Geoghegan's review MR1239554 on Mathscinet. I also took a look at McKenzie and Thompson's article An ...