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Large scale properties of groups; growth functions; Dehn functions; small cancellation properties; hyperbolicity and CAT(0); actions and representations; combinatorial group theory; presentations
9
votes
3
answers
490
views
Residually solvable Bianchi groups
Let $d$ be a square-free positive integer, and let $\mathcal{O}_d$ be the ring of integers of the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$. Consider the Bianchi group $\Gamma_d = \oper …
6
votes
Accepted
Is there a finitely generated residually finite group with solvable word problem that does n...
A preprint by E. Rauzy appeared today on the arXiv, and gives a negative answer to this question. In other words (if the proof is correct), there exists a f.g. residually finite group with decidable w …
6
votes
0
answers
623
views
Minimum Simple Burger-Mozes Type Group
Burger and Mozes constructed (Burger and Mozes - Lattices in products of trees) infinite, finitely presented, torsion-free simple groups which split as amalgams of two finitely generated free groups o …
1
vote
Accepted
Some questions on a paper of Baumslag and Solitar
I'll write out the answer to your first question (solving the word problem in one-relator groups) for how one might do this in practice. I'll focus on the case $\ell = 2, m=3, p=2$, but you'll hopeful …
2
votes
General properties of free-by-cyclic groups
They are all residually finite.
They are not all subgroup separable/LERF.
They do not all have decidable submonoid membership problem.
Residual finiteness is a result which can be found in the (ap …
7
votes
Subgroup membership problem in simple groups
After some digging, I was able to find that the answer to my question exists: the problem can be undecidable. Rattaggi, in an unpublished manuscript (available here), proved that there exists a finite …
8
votes
2
answers
480
views
Subgroup membership problem in simple groups
Let $G$ be a finitely presented simple group. By Kuznetsov (1958), $G$ has decidable word problem. However, by Scott [1], $G$ may have undecidable conjugacy problem. Is anything known about other deci …
24
votes
Recognizing free groups
As indicated in the comments, it's undecidable in general to take as input a finite presentation of a group and try to output whether or not this group is free or not. This is a direct consequence of …
2
votes
Examples of residually-finite groups
Here's a few examples in line with classical combinatorial group theory.
Though small cancellation groups as a whole have already been mentioned, one important subclass of these are the one-relator g …
9
votes
1
answer
225
views
Yang-Mills algebra and lower central series of surface groups
Here is a connection that I recently noticed, but I haven't quite been able to make sense of. It might follow from well-known facts; apologies, if so. This is quite far from my area.
First, in "Yang-M …
10
votes
Analogous results in geometric group theory and Riemannian geometry?
Here is a very classical example. As stated in the comments, Gromov was an early proponent of importing ideas from geometry to group theory, but already thirty years earlier there was work in this dir …
5
votes
0
answers
290
views
For which classes of metric spaces can we prove that quasi-isometry is an equivalence relati...
Given two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$, a map $\phi \colon (M_1, d_1) \to (M_2, d_2)$ is a large-scale Lipschitz essentially surjective map if there exist constants $A \geq 1, B \geq 0$ …
3
votes
Which groups are LERF?
Polycyclic groups are LERF, by Mal'cev 1948. In particular, all nilpotent and all abelian groups are LERF.
As mentioned in the comments, as not all one-relator groups are residually finite, not all on …
4
votes
Are there any computational problems in groups that are harder than P?
While most other answers have mentioned computational problems related to finitely presented (but generally infinite) groups, there are many problems in finite group theory which are either conjecture …
8
votes
1
answer
348
views
Finite two-relator groups and quotients of knot groups
Let $G$ be a one-relator group $\langle A \mid R = 1 \rangle$. Then clearly $G$ is finite if and only if it is cyclic of finite order, i.e. can be given by a presentation $\langle a \mid a^n = 1 \rang …