Residually Finite Braid Group

Question

In Braid Groups of Kassel, Turaev, it mentions that $\mathcal{B}_n$ is a residually finite group. The definition that they give as a residually finite group is a group $G$ such that for each $g\in G-\{e_G\}$ ($e_G$ the identity of $G$), there exists an homomorphism $f$ to a finite group $H$ such that $f(g)\neq e_H$. My question is:

How can I obtain the group $H$ for a given element $g\in \mathcal{B}_n$ and the homomorphism that fulfills this?

I hope you can help me. Nice Holidays.

Someone explained this to me once. Something along the lines of: "$B_n$ is a subgroup of $\text{Aut}(F_n)$, which is much easier to show is residually finite." I don't remember how to do either half of that argument. — , Dec 15 '15 at 19:18
Well if $g$ is not a pure braid, the obvious quotient map to the symmetric group will do. For pure braids, I assume you'll want to comb the braid. So use the fact that $P_n=F_n\rtimes P_{n-1}$ where $F_n$ is the free group on $n$ generators (which is residually finite) and then use induction — Dan Rust, Dec 15 '15 at 19:18

score 2 · Accepted Answer · edited Apr 13 '17 at 12:58

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If the braid is not pure, then you can detect it by a homomrphism to the symmetric group. So this reduces to showing that the pure braid group $P_n$ is residually finite. $P_n$ is an iterated semi-direct product of free groups. (This is called combing the braid.) The result now follows because free groups are residually finite. https://mathoverflow.net/questions/20471/why-are-free-groups-residually-finite

edited Apr 13 '17 at 12:58

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answered Dec 15 '15 at 19:18

Cheerful Parsnip

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By the Symmetric Group which one you mean? $S_n$ for the group $\mathcal{B}_n$? I'm looking for an explicit example of this homomorphisms between a $\mathcal{B}_n$ to that $H$ group mentioned in the question. Can be this possible? – iam_agf Dec 15 '15 at 19:55
@MonsieurGalois The group H depends on the element. So, yes, for many elements you can use the homomorphism to $S_n$, but it won't work on the pure braid subgroup, when you have to start using different homomorphisms to other finite groups. – Cheerful Parsnip Dec 16 '15 at 00:37
For the group $P_n$ which group(s) can be used? – iam_agf Dec 16 '15 at 00:51
@MonsieurGalois: $P_n$ is an iterated semidirect product of free groups. In the link I gave, the groups $H$ are constructed for free groups. – Cheerful Parsnip Dec 16 '15 at 00:53

score 2 · Answer 2 · edited Jan 03 '16 at 18:30

Another possibility is to embed $\mathcal{B}_n$ into the automorphism group $\mathrm{Aut}(\mathbb{F}_n)$ of the free group $\mathbb{F}_n$. Now, Baumslag gave a very short prove of the fact that, for any finitely generated residually finite group $G$, $\mathrm{Aut}(G)$ is also residually finite. The conclusion follows from the residual finiteness of finitely generated free groups (see for example the beautiful proof of Stallings in Topology of finite graphs), since a subgroup of a residually finite group is clearly residually finite itself.

For more details, see Basic results on braid groups and the references therein.

Residually Finite Braid Group

2 Answers2