AGenevois
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2 answers
25 votes
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Is the intersection of two subgroups, defined below, always trivial?
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32 votes

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 ...

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1 answers
13 votes
875 views
Examples of hyperbolic groups
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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 ...

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2 answers
7 votes
289 views
Examples of non-cubulated hyperbolic groups
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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 (...

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2 answers
7 votes
439 views
Ends of finitely generated torsion groups
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 ...

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2 answers
11 votes
547 views
Constraints on the homology of amenable groups
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 &&&...

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1 answers
10 votes
625 views
CAT(0) groups that does not act on CAT(0) cubical complex
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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 ...

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1 answers
7 votes
337 views
Abelianization of mapping class groups $\Gamma_{g,n}$
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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\{ ...

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1 answers
7 votes
199 views
Hyperbolic groups and spaces of negative curvature
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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 ...

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1 answers
4 votes
147 views
Conjugating generators in free groups
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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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1 answers
10 votes
493 views
Parabolic subgroups of relatively hyperbolic and CAT(0) groups
10 votes

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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2 answers
8 votes
301 views
Contractible Rips complex from non-hyperbolic group
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9 votes

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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4 answers
7 votes
539 views
Roots in Thompson's groups
9 votes

About Question 3, the answer is "no" 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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2 answers
5 votes
620 views
3-manifold with fundamental group $\mathbb Z$
9 votes

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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1 answers
3 votes
165 views
Group acting on a CAT(0) cube complex then acting also on a tree
9 votes

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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1 answers
7 votes
335 views
Properties (T) and (FA)
8 votes

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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2 answers
8 votes
214 views
Is the automorphism group of free group of rank two relatively hyperbolic?
8 votes

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 \...

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2 answers
8 votes
398 views
Why does not a closed 3-manifold modelled on SL(2,R) admit a metric of nonpositive curvature?
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....

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2 answers
10 votes
384 views
Sequence of epimorphisms of residually finite groups stabilizes
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 ...

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3 answers
4 votes
639 views
Is Higman's group a free product?
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 ...

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3 answers
6 votes
594 views
Bass-Serre theory textbook
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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 ...

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1 answers
6 votes
300 views
Easy example of an infinite simple group with an embedding into a finitely presented group
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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 ...

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2 answers
10 votes
431 views
A question about conjugacy in Higman's group
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....

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3 answers
10 votes
297 views
Subgroups of RAAGs vs. subgroups of RACGs
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 ...

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3 answers
16 votes
576 views
Group with non-trivial center containing trivially-intersecting copies of itself
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 ...

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2 answers
13 votes
493 views
Cantor-Bernstein for quasi-isometric embeddings?
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 ...

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2 answers
9 votes
432 views
Braid groups and Kazhdan's property (T)
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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2 answers
16 votes
388 views
What is the largest known Dehn function of f.p. subgroup of a f.p. group with quadratic Dehn function?
6 votes

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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3 answers
7 votes
442 views
Dimension of classifying space of a group
6 votes

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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2 answers
18 votes
641 views
Why are Thompson's groups called $F$, $T$ and $V$?
6 votes

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

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2 answers
7 votes
392 views
Relative/acylindrical hyperbolicity of free-by-cyclic groups
6 votes

In addition to Henry's answer, I would like to mention that the acylindrical hyperbolicity of free-by-cyclic groups is fully characterised in the recent preprint Acylindrical hyperbolicity of ...

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