# Does handle reduction determine braid equivalence in quotients of braid groups?

Let $B_{n}$ denote the $n$-strand braid group. Let $G$ be a group generated by elements $x_{1},...,x_{n-1}$. Let $N$ be the smallest normal subgroup of $B_{n}$ such that the mapping $x_{1}\mapsto\sigma_{1}N,.,..,x_{n}\mapsto\sigma_{n}N$ extends to a homomorphism $L_{2}$. Let $L_{1}:B_{n}\rightarrow B_{n}/N$ be the quotient mapping. Let $W_{n}=\{w_{0},...,w_{n-1}\}$ and let $\textrm{br}:W_{n}^{*}\rightarrow B_{n},\textrm{gr}:W_{n}^{*}\rightarrow G$ by the homomorphisms defined by $\textrm{br}(w_{i})=\sigma_{i},\textrm{br}(w_{i}^{-1})=\sigma_{i}^{-1},\textrm{gr}(w_{i})=x_{i},\textrm{br}(w_{i}^{-1})=x_{i}^{-1}$

Consider the following game.

Play $0$: Alice goes first and she chooses a word $w_{0}\in W_{n}^{*}$ with $\textrm{br}(w_{0})N=N$.

Play $2n+1$. Bob chooses a word $w_{2n+1}$ where $\textrm{gr}(w_{2n})=\textrm{gr}(w_{2n+1})$.

Play $2n+2$. Alice chooses a word $w_{2n+2}$ which is a left handle reduction of $w_{2n+1}$ (and hence $\textrm{br}(w_{2n+2})=\textrm{br}(w_{2n+1})$).

If there is a word $w_{n}$ where $w_{n}=e$, then Bob wins. Otherwise Alice wins.

This game is a closed game so one of the players has a winning strategy.

For which groups with generators does Alice (Bob) have a winning strategy? I would like to see examples of groups with generators where Alice has a winning strategy and of groups with generators where Bob has a winning strategy.

My main motivation for this question is that if Bob has a winning strategy, then the group $B_{n}/N$ would very likely have a word problem which is solvable in a similar way to how the handle reduction solves the word problem for braid groups.

• It's undubitably true that braid groups $B_n$ have finitely presented quotients with unsolvable word problem. (More precisely, pure braid groups $PB_n$, which are only of finite index in braid groups, have free quotients. A little work is needed to get from there to quotients of $B_n$ with unsolvable word problem, but it can certainly be done.) So any attempt to solve the word problem in finitely presented quotients of braid groups is doomed to fail. – HJRW Nov 22 '16 at 12:53