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Questions tagged [combinatorial-group-theory]

7 questions from the last 365 days
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5 votes
2 answers
448 views

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

In 1965, Murasugi [1] conjectured that any finitely presented group with deficiency at least two has trivial centre. The year before, he had proved it true for one-relator groups, and in [1] he proved ...
Carl-Fredrik Nyberg Brodda's user avatar
8 votes
1 answer
349 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 \...
Carl-Fredrik Nyberg Brodda's user avatar
4 votes
1 answer
266 views

Are (group theoretic) Markov properties on groups with decidable word problems, decidable?

(Link to SE duplicate: https://math.stackexchange.com/questions/4959071/are-group-theoretic-markov-properties-on-groups-with-decidable-word-problems) The Adian-Rabin theorem says that if a property of ...
Perry Bleiberg's user avatar
7 votes
1 answer
502 views

Are Artin-Tits groups ordered groups?

We consider Artin-Tits groups of two generators $(I_2(n))$. Are these groups ordered groups?
navashree chanania's user avatar
3 votes
1 answer
165 views

When the fundamental group of subgraph of groups embeds?

Given a connected graph of groups $\mathcal G$ (where edge maps are embeddings), by a subgraph we mean a graph of groups obtain by omitting some vertices, some edges, and replacing the remaining ...
tomasz's user avatar
  • 1,338
5 votes
1 answer
284 views

Word length in the surface groups

I want to know if there are some results about the title of this question. Let $G$ be an orientable closed surface group with genus $n$ greater than 1. We know it has a canonical presentation. $$G=\...
keqiyehuopo's user avatar
4 votes
0 answers
603 views

Show me that I have not simplified the proof of the Adian-Rabin theorem

Let $G$ be a group with presentation $\langle x_1,x_2...,x_m|R \rangle$ and let $G'=G \ast \langle y_0 \rangle$. Now define $y_i=y_{0}x_i$. Notice that $G'=\langle y_0,y_1,...y_m\mid R'\rangle$ for ...
Perry Bleiberg's user avatar