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18 votes

Examples of locally hyperbolic groups

Many examples can be exhibited using a theorem of Gersten: Theorem (Gersten): Let $G$ be a hyperbolic group of cohomological dimension 2. Every finitely presented subgroup $H$ of $G$ is hyperbolic. ...
HJRW's user avatar
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18 votes
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Is solvability semi-decidable?

The Adian–Rabin theorem, although it is not usually stated like this, says that Markov properties are not co-semi-decidable, thus « not being metabelian » or « not being solvable » are not semi-...
E.Rauzy's user avatar
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16 votes

Examples of locally hyperbolic groups

Finite groups. ... ... Fundamental groups of closed connected hyperbolic three-manifolds. Here is a proof of the latter using far too many tools from kleinian groups. Fundamental groups of closed ...
Sam Nead's user avatar
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15 votes

Examples of locally hyperbolic groups

A locally hyperbolic group is in particular coherent (i.e. locally finitely presented), which is already a special property. To add to the examples already given by Sam Nead: ascending HNN extensions ...
Giles Gardam's user avatar
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14 votes
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Commutator problem vs conjugacy/word problem

Denis Osin [Osin, Denis, Small cancellations over relatively hyperbolic groups and embedding theorems, Ann. Math. (2) 172, No. 1, 1-39 (2010). ZBL1203.20031.] proved that every torsion-free countable ...
Ashot Minasyan's user avatar
14 votes
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When are biautomatic groups hyperbolic?

$\DeclareMathOperator\BS{BS}$Since this question goes in several directions, I hope a discursive answer is appropriate. The question fits into an important family of questions in geometric group ...
HJRW's user avatar
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Examples of hyperbolic groups with non-hyperbolic subgroups

It is an open problem to find a coherent hyperbolic group with a finitely generated, non-hyperbolic subgroup. See Wise's survey article for the state of the art on coherent groups. Wise, Daniel T. (3-...
HJRW's user avatar
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13 votes

Is there a one relator group with property (T)?

In fact, much more is true: if $H$ is a finitely generated subgroup of a one-relator group and $H$ has property (T) then $H$ is finite. Indeed, a classical theorem of Brodskii and Howie asserts that ...
HJRW's user avatar
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Are Artin-Tits groups ordered groups?

Your question is answered in Mulholland and Rolfsen's article Local indicability and commutator subgroups of Artin groups. On the one hand, the Artin-Tits group $A(I_2(n))$ is not bi-orderable because ...
AGenevois's user avatar
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12 votes
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Minimum number of relations that must be added to make a group abelian

The question has been answered in the comments by SashaP and Derek Holt, taking the following definition of $c(G)$: Definition 1. Let $G$ be a finitely generated group. We denote by $c(G)$ the ...
Luc Guyot's user avatar
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12 votes

Is It possible to determine whether the given finitely presented group is residually finite with MAGMA or GAP?

It is a consequence of the Adian-Rabin theorem that there is no algorithm that decides, given a finite presentation of a group, whether the group is residually finite. In a similar spirit, it is ...
Giles Gardam's user avatar
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11 votes
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Reference request: Recent progress on the conjugacy problem for torsion-free one-relator groups?

As mentioned in the comments, this is still considered an open problem. I thought I'd flesh out a few aspects. A solution was claimed in 1992 by Juhasz, but it seems to have failed to convince experts....
Carl-Fredrik Nyberg Brodda's user avatar
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Is an HNN extension of a virtually torsion-free group virtually torsion-free?

Yes, here's an example with an HNN over finite index subgroups as requested. It's based on constructing an amalgam of two f.g. virtually free groups, that has no proper finite index subgroups, using ...
YCor's user avatar
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11 votes
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Subgroups of RAAGs vs. subgroups of RACGs

The fundamental group of the (non-orientable) closed surface of Euler characteristic -1 provides a counterexample. On the one hand, it’s a subgroup of index 4 in the reflection group on the right-...
HJRW's user avatar
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10 votes
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Maximal power in a sequence of iterated commutators in the rank two free group

A word in $F_2$ can be represented by a path on the unit square grid on the plane. Now, $w_0$ is a horizontal unit interval, $w_1$ is a vertical unit interval, $w_2$ is a unit square and the image of $...
Jarek Kędra's user avatar
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10 votes

Is an HNN extension of a virtually torsion-free group virtually torsion-free?

It is easy to construct a counter-example when $H$ and $K$ are not of finite index in $G$. Let $A$ be a finitely presented torsion-free group which is not residually finite (e.g., the Baumslag-...
Ashot Minasyan's user avatar
10 votes
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Is Thompson's group $T$ co-Hopfian?

The answer to the question is no, $T$ is not co-Hopfian, i.e., it does contain proper subgroups isomorphic to itself. Nicolás Matte Bon explained this to me over email (he doesn't use Mathoverflow, ...
Matt Zaremsky's user avatar
10 votes

A "simpler" description of the automorphism group of the lamplighter group

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be the unit group of $R$ and let $\theta$ be the ring automorphism of $R$ ...
Luc Guyot's user avatar
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9 votes
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Largest Hopfian quotient

The answer is no even for finitely generated groups. Here's a construction of a finitely generated residually Hopfian, non-Hopfian group. It is even solvable (actually center-by-metabelian). Denote ...
YCor's user avatar
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9 votes
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Finite presentability of semi-direct product of free group and its commutator subgroup

This group is not finitely presentable. Indeed, write $F$ for the given free group and $F'$ for its derived subgroup. The map $$F\ltimes F'\to F\times F,\quad (f,g)\mapsto (f,fg)$$ is an injective ...
YCor's user avatar
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9 votes
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Subgroup membership problem in simple groups

As another example, the problem of computing the order of an element of the finitely presented simple Brin–Thompson group $2V$ is undecidable by Belk, James; Bleak, Collin, Some undecidability results ...
Giles Gardam's user avatar
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9 votes

Do cyclically presented groups of positive word length four relators satisfy the Tits Alternative?

When $j,|k-j|, |l-j|, |n-l|$ are all distinct, then the presentation complex is a $CAT(0)$ square complex, since the length of any path in the link of the vertex (which is a bipartite graph because of ...
Ian Agol's user avatar
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8 votes

Minimal length presentations of cyclic groups

I can beat that by one. $$C_{173} \cong \langle a,b,c,d,e \mid a^4=b, b^2=c,c^4=d,d^4=e, abcde=1 \rangle.$$ It might be hard to prove minimality. I am sure the best you can do is $O(\log n)$, but it ...
Derek Holt's user avatar
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8 votes

Is It possible to determine whether the given finitely presented group is residually finite with MAGMA or GAP?

To precise Giles Gardam's answer, let me add the following. The Adian-Rabin theorem shows that residual finiteness is undecidable, by showing that no algorithm can stop exactly on non-residually ...
E.Rauzy's user avatar
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8 votes

Finite two-relator groups and quotients of knot groups

Question 1: As mentioned in comments, presentations with the same number of generators and relations are called balanced. The triviality problem for balanced presentations appears to be a question of ...
HJRW's user avatar
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8 votes

Centre of group with deficiency at least two (Progress on Murasugi's conjecture)

All $L^2$-Betti numbers of a finitely generated group $G$ with an infinite amenable normal subgroup are 0, by a result of Gromov. (See Theorem 7.2 of Lück's book $L^2$-Invariants: Theory and ...
jonathan's user avatar
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7 votes
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When is the semidirect product of an elementary abelian group and a cyclic group generated by two elements?

This answer corroborates YCor's claim according to which the conditions $n_i \le 1$ for $i > 1$ and $n_1 \le 2$ on the irreducible modular representations'multiplicities $n_i$, are necessary and ...
Luc Guyot's user avatar
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7 votes
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Is there a name of semidirect product of a group with its automorphism group?

As a first remark, note that if $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ is a finite subgroup and $G$ is torsion-free, then the projection $p: G\rtimes \operatorname{Aut}(G)\to \operatorname{Aut}...
Mark Grant's user avatar
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