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The braid group $ B_1 $ is trivial and the braid group $ B_2 $ is isomorphic to $ \mathbb{Z} $.

The braid group $B_3$ has the property that its central quotient (i.e., $B_3 / Z(B_3)$) is isomorphic to the modular group $\mathrm{PSL}(2,\mathbb{Z})$. The modular group is known to be isomorphic to the free product of $(\mathbb{Z}/2\mathbb{Z}) \ast (\mathbb{Z}/3\mathbb{Z})$.

The Braid group $ B_4 $ surjects onto $ B_3 $, so there is also a quotient of $ B_4 $ which is the free product $(\mathbb{Z}/2\mathbb{Z}) \ast (\mathbb{Z}/3\mathbb{Z})$.

I'm curious about the extent to which this is true for braid groups $B_n$ for $n \geq 5 $.

Question 1: For which $n \geq 5$ is it true that there exists two non-trivial groups $G$ and $H$ such that a quotient of $ B_n $ is isomorphic to $G \ast H$?

I'm also curious about a weaker version of this statement where we allow amalgamation over some shared finite subgroup.

Question 2: For $n \geq 5$ is it true that there are groups $G$ and $H$ with common finite proper subgroup $A$ such that $B_n/Z(B_n)$ is isomorphic to $G \ast_A H$?

This question is a follow-up to Are the central quotients of braid groups non-trivial free products?

Ian Gershon Teixeira
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1 Answers1

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By work of Culler and Vogtmann, if $n \ge 5$ then for every action on a tree $T$ of the $n$-strand braid group $B_n$ the tree $T$ has a line that is invariant under the action of $B_n$, and every element of the action is a translation; one says in this case that $B_n$ has property $F\mathbb R$.

However, suppose that there is a homomorphism $B_n \mapsto G*H$ where $G,H$ are nontrivial free products. The group $G*H$ acts minimally on its Bass-Serre tree $T$, and so by composition $B_n$ acts minimally on $T$ --- meaning that no proper subtree of $T$ is invariant by the action.

Now there are two cases, each of which leads to a contradiction.

  1. If either of $G$ or $H$ has cardinality $\ge 3$ then the tree $T$ has infinitely many ends, and so the action on $T$ of $G*H$, and hence also of $B_n$, has no invariant line at all, contradicting that $B_n$ has property $F\mathbb R$.
  2. If both of $G$ and $H$ both have cardinality $2$ then $T$ is a line but the action contains reflections, also contradicting that $B_n$ has property $F\mathbb R$.
Lee Mosher
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