42

A typical example arising in algebraic geometry is the following.
Take a scheme $X$ of finite type over $\mathbb{C}$. Then there are (at least) two concepts of fundamental group for $X$: the topological fundamental group $\pi_1^{top}(X)$, i.e. the fundamental group of the underlying topological space $X(\mathbb{C})$ with the analytic topology, and the étale ...

38

Yes, the fundamental group of the Hawaiian earring $\pi_1(\mathbb{H},b_0)$ is an important group which is sometimes called the free sigma product $\#_{\mathbb{N}}\mathbb{Z}$. Its is often defined in purely algebraic terms as a group of "transfinite words" in countably many letters. In many ways this group behaves like the non-abelian version of the Specker ...

27

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 simplicial trees.
If $\varphi$ is an atoroidal automorphism of a free group $\mathbb{F}_n$, then the extension $\mathbb{F}_n \rtimes_\varphi \mathbb{Z}$ is ...

24

The conjecture is indeed strictly weaker than $\mathrm{P = NP}$, in the sense that it follows from $\mathrm E=\Sigma^E_2$, which is not known to imply $\mathrm{P = NP}$. Of course, we cannot prove this unconditionally with current technology, as it would establish $\mathrm{P\ne NP}$.
Here, $\mathrm E$ denotes $\mathrm{DTIME}(2^{O(n)})$, $\mathrm{NE}$${}=\...

answered Aug 6 '18 at 12:29

Emil Jeřábek

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24

An earlier reference for groups with this property is
J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.
There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 \subset E_1 \subset E_2 \subset \cdots$, ...

gr.group-theory computational-complexity geometric-group-theory computational-group-theory word-problem

22

I think this is a great question, as there is still a need for an authoritative reference about (word-)hyperbolic groups. Since the textbook doesn't exist, I'd like to take the question in a slightly different direction by listing some of the material I think it should cover. (This is inevitably a personal and biased account.) I'll try to include the best ...

22

Yes. Jan Okninski showed that $$\begin{bmatrix} 1 & 1 \\ 0 &1\end{bmatrix}\ \text{and}\ \begin{bmatrix} 1 & 0\\ 1 & 0\end{bmatrix}$$ generate a semigroup of intermediate growth. Details can be found in Nathanson. The growth was estimated there to be like the Hardy-Ramanujan estimate of the partition function. An exact asymptotic growth ...

answered Mar 27 at 14:37

Benjamin Steinberg

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21

(Probably the question would be more suitable for MathSE)
The answer is no. Well, the trivial group is a counterexample. Also finite groups are counterexamples. So first and for all you should have specified that you assume the group to be infinite. But then the answer is still no.
In a finitely generated group, an element $g$ is called distorted if $\lim |...

21

Bridson and Vogtmann proved a much stronger result. From the abstract: 'If $m$ is less than $n$ then [the image of] a homomorphism $\mathrm{Aut}(F_n)\to\mathrm{Aut}(F_m)$ can have cardinality at most 2.'

19

Assuming the assertions your claim are true, it follows that $\mathrm{GL}_n(\mathbf{Z})$ is generated by 2 elements for all $n$; for $n$ odd (your missing case), the matrices $-s_1,s_3$ indeed generate $\mathrm{GL}_n(\mathbf{Z})$.
This follows from the following three facts:
1) Let $p$ be prime and $C_p=\langle c\rangle$ the cyclic group of order $p$. If $...

19

Yes. The following answer is inspired by Andy Putman's comment. Let $N_\infty(G)$ be the subgroup generated by elements of infinite order, in a group $G$.
Every non-elementary hyperbolic group $G$ with trivial finite radical has a useful property, which bears the ridiculous name "Pnaive", namely that for every finite subset $F$ there exists a $x$ such that ...

19

As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In
Isoperimetric and Isodiametric Functions of Groups,
Mark V. ...

gr.group-theory computational-complexity geometric-group-theory computational-group-theory word-problem

answered Jan 23 '19 at 18:09

Benjamin Steinberg

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18

A better question is:
Given a group $\Pi$, is there a compact manifold $M$ with non-negative Ricci curvature such that $\pi_1(M)=\Pi$?
The answer is given in "On fundamental groups of manifolds of nonnegative curvature" by Wilking.
Here is the main result:
(source: psu.edu)

answered May 8 '13 at 15:54

Anton Petrunin

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18

Thompson's group F is an example. It's finitely presented and, according to this paper of Ken Brown, the integral homology is free abelian of rank 2 in every positive dimension.

18

A usually hard problem in Algebraic Geometry is computing the fundamental group of the complement $U=\mathbb{P}^n - V$, where $V$ is a reduced hypersurface.
At least in principle, the computation of $\pi_1(U)$ can be carried out by using Seifert-Van Kampen theorem; the tricky part is that the relations in the presentation depend not only on the ...

answered Jan 26 '17 at 16:47

Francesco Polizzi

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18

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 ...

17

Dehn gave at least three solutions of the conjugacy problem for surface
groups, which can be found in my translation in the book Papers on Group
Theory and Topology (Springer 1986), Papers 2, 4, and 5.
The first is based on an idea of Poincaré: lifting a curve to the universal
cover, which is the disk model of the hyperbolic plane, replacing it by the
...

answered Jul 17 '13 at 6:09

John Stillwell

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17

(1) Bridson showed that if a mapping class group of a surface (of genus at least 3) acts on a CAT(0) space, then Dehn twists act as elliptic or parabolic elements. This implies that the mapping class groups of genus $\geq 3$ are not CAT(0) (Edit: as pointed out by Misha in the comments, this was originally proved by Kapovich and Leeb, based on an observation ...

gt.geometric-topology mg.metric-geometry geometric-group-theory mapping-class-groups alexandrov-geometry

17

This is not exactly an answer to the question, but is instead essentially a comment that was way too long for the comment space.
The OP mentioned that he doesn't have a good sense for the shapes of Følner sets in amenable groups of exponential growth. It seems that it might be helpful to give an example.
The simplest amenable group with exponential growth ...

17

For some reason, the question seems to be asking for an algebraic (number- or ring-theoretical) justification for residual finiteness (and implicitly LERF, though in fact the correct statement is that every hyperbolic group is residually finite if and only if they are all QCERF, as proved by Agol--Groves--Manning). Recall that residual finiteness means that ...

17

No.
Let $G$ be a finite subgroup of $GL_n(\mathbb R)$ containing $W(E_8)$ as a subgroup. Because $G$ is compact, $G$ must preserve a symmetric positive definite form on $\mathbb R^8$. Since $W(E_8)$ preserves a unique such form, it must be that one.
Let $H$ be the largest subgroup of $G$ generated by reflections. Then $H$ contains $W(E_8)$ and thus is an ...

16

My own interest in automatic groups has been principally algorithmic, and I believe that this was Thurston's original motivation for studying them - they provided a method for carrying out practical computations in a variety of interesting groups with negative curvature. Once a (geodesic) automatic structure has been computed, you can compute the growth ...

16

It's undecidable.
Lemma: a group $G$ is nontrivial if and only if the free product $H=G\ast\mathbf{Z}$ has an infinitely generated derived subgroup.
Proof: assume $G$ finitely generated and nontrivial. The kernel $N$ of the canonical epimorphism $H\to\mathbf{Z}$ is isomorphic to $G^{\ast\mathbf{Z}}$ and hence is infinitely generated. Since $H/[H,H]$ is ...

16

This is Corollaire 36, Chapitre 8 in the (French) book "Sur les Groupes Hyperboliques d’après Mikhael Gromov", Editors: Etienne Ghys, Pierre de la Harpe, Progress in Mathematics 83. Birkhäuser Boston (1990)

16

Chloe Perin in her thesis answered the same question for all torsion-free hyperbolic groups. She classified all the subgroups $H$ of a torsion-free hyperbolic group $G,$ that are elementary submodels of $G.$ If $G$ is a f.g. free group, H must be a non-abelian free factor. If $G$ is a surface group the classification of elementary submodels $H$ of $G$ is ...

16

Let $S$ and $T$ be generating sets for $G$, and suppose that $X_\phi$ is $T$-connected (i.e., spans a connected subgraph in the Cayley graph of $G$ with respect to $T$.) Let $[X_\phi]_S$ be the subgraph of $\text{Cay}(G,S)$ spanned by $X_\phi$. We must show that $[X_\phi]_S$ is connected.
Claim 1: For any $n\in\mathbb{Z}$, the set $\{g\in G:\phi(g)\geq n\}$ ...

16

No: let $Q$ be a non-abelian group of order 8. Then the standard lamplighter groups $(\mathbf{Z}/2\mathbf{Z})\wr\mathbf{Z}$ (which is RF) and the wreath product $Q\wr\mathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(\mathbf{Z}/2\mathbf{Z})\wr\mathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(\mathbf{...

15

Up to equivalence there is only one exponential type of growth in which case the answer is trivial. In polynomially bounded growth the answer follows from various old theorems (notably of Wolf and Gromov): the possible growths are $n^d$ for $d$ non-negative integer and things are classified. What's remaining is intermediate growth, and the first examples ...

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