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.
On
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.
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