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It is known that in a topologically finitely-generated profinite group, every subgroup of finite index is open. (See this paper.) If $G$ is a finitely-generated residually finite group, then the prof …

If I understand the question correctly, you are concerned that there might be hidden relations between the elements of $T$. Moreover, since the images of $T$ are linearly independent in the abelianiz …

As far as I can tell, a solution to this problem has not appeared in the literature. Unless I'm mistaken, the best partial result was obtained by Liousse in this 2008 paper, where it is proven by exam …

It is possible to use closed strand diagrams to check whether a given element of $F$, $T$, or $V$ has an $n$th root. Before discussing this algorithm, I'd like to say a little bit about the situation …

Francesco Matucci, James Hyde and I have just posted an arXiv preprint with a solution to this problem. We prove that $\mathbb{Q}$ embeds in the group $\overline{T}$ of piecewise-linear homeomorphism …

Well, every action of $F$ corresponds to a subgroup $H\leq F$ in the standard way. Specifically, the "standard" action on the interval corresponds to the stabilizers of various points in the interval …

Here is a construction for a group similar to the braided Thompson group $BV$ that ought to have this property. Define the $n$th ribbon group to be the semidirect product $$ R_n = \mathbb{Z}^n \rtime …

It is known that there exists a finitely presented group $H$ with solvable word problem that has a finitely generated subgroup $K$ whose subgroup membership problem is unsolvable. For example, Mikhail …

My understanding is that the situation has not changed much. A group of mathematicians at Binghamton University had been investigating Shavgulidze's argument, and they found a flaw which Shavgulidze …

Here's a case where $G$ and $H$ can be conjugate. First some notation: given a sequence $\{k_n\}$ of positive integers, let $[k_1,k_2,\ldots]$ denote the permutation $$(1,\ldots,k_1)(k_1+1,\ldots,k_ …

Thompson's group $F$ is totally ordered. See here for a description of all possible bi-orderings. Indeed, all diagram groups are totally orderable (see here). The pure braided Thompson group $BF$ a …

Just to make the method as concrete as possible, I'll compute the growth rate for the fundamental group $G$ of a surface of genus two. The Cayley graph of $G$ is the 1-skeleton of a tiling of the hyp …

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 F …

The answer is already no for $\mathbb{Z}$, assuming the question is whether this holds for every meaure. Let $n\in\mathbb{N}$, and let $$ S \,=\, \{(a_1,\ldots,a_n)\in\mathbb{Z}^n \mid \gcd(a_1,\ldot …

If we place no restrictions on $k$, then this is precisely the class of finite groups that are generated by involutions. In particular, if $G$ is any finite group of order $n$, then in the left regul …