Let $B_{\infty}$ denote the infinite strand braid group. Let $\mathrm{sh}:B_{\infty}\rightarrow B_{\infty}$ be the group homomorphism where $\mathrm{sh}(\sigma_{i})=\sigma_{i+1}$ for all $i>0$.
Define an operation $*$ on $B_{\infty}$ where $$x*y=x\cdot\mathrm{sh}(y)\cdot\sigma_{1}\cdot\mathrm{sh}(x)^{-1}.$$
The operation $*$ satisfies the self-distributivity identity $$x*(y*z)=(x*y)*(x*z).$$
Suppose that $N$ is a normal subgroup of $B_{\infty}$ such that if $x\in N$, then $\mathrm{sh}(x)\in N$. Then $B_{\infty}/N$ is a self-distributive algebra where we set $$(xN)*(yN)=(x\cdot\mathrm{sh}(y)\cdot\sigma_{1}\cdot\mathrm{sh}(x)^{-1})N.$$
If $x\in B_{\infty}$, then let $A_{x}$ be the subalgebra of $(B_{\infty},*)$ generated by $x$. Then $A_{x}$ is freely generated by $x$. Does there exist a non-trivial normal subgroup $N\subseteq B_{\infty}$ where $A_{x}/N$ is a free self-distributive algebra for each $x\in B_{\infty}?$ Does there exist a non-trivial normal subgroup $N\subseteq B_{\infty}$ where $A_{e}/N$ is a free self-distributive algebra? Does there exist a non-trivial normal subgroup $N\subseteq B_{\infty}$ where there exists some $x\in B_{\infty}$ where $A_{x}/N$ is a free self-distributive algebra?
Suppose that $X$ is a set and $T:X^{2}\rightarrow X^{2}$ is a bijective function that satisfies the Yang-Baxter equation: $$(1_{X}\times T)\circ(T\times 1_{X})\circ(1_{X}\times T)=(T\times 1_{X})\circ(1_{X}\times T)\circ(T\times 1_{X}).$$ Define an action of $B_{\infty}$ on $X^{\mathbb{N}}$ by letting $$(x_{1},...,x_{n},...)\cdot\sigma_{i}=(x_{1},...,x_{i-1},T(x_{i},x_{i+1}),x_{i+1},...).$$
An algebra $(X,*,*^{-1})$ is said to be a rack if it satisfies the identities
$x*(y*z)=(x*y)*(x*z)$ and
$x*(x*^{-1}y)=x*^{-1}(x*y)=y$.
If $(X,*,*^{-1})$ is a rack and $T:X^{2}\rightarrow X^{2}$ is defined by $T(x,y)=(x*y,x)$, then $T$ satisfies the Yang-Baxter equation.
If $T:X^{2}\rightarrow X^{2}$ satisfies the Yang-Baxter equation, then let $N_{T}$ denote the normal subgroup of $B_{\infty}$ consisting of all braids $b$ such that $\mathbf{x}\cdot b=\mathbf{x}$. If $(X,*,*^{-1})$ is a rack, and then define $N_{*}=N_{T}$ where $T:X^{2}\rightarrow X^{2}$ is the mapping where $T(x,y)=(x*y,x)$.
Does there exist a finite set $X$ along with a bijective $T:X^{2}\rightarrow X^{2}$ that satisfies the Yang-Baxter equation where $A_{b}/N_{T}$ is freely generated by $bN_{T}$ for all $b\in B_{\infty}$ (or for some $b\in B_{\infty}$ or for $b=e$)?
Does there exist a finite rack $(X,*)$ where $A_{b}/N_{*}$ is freely generated by $bN_{*}$ for all $b\in B_{\infty}$ (or for some $b\in B_{\infty}$ or for $b=e$)?
Does there exist a finite set $X$ along with a bijective $T:X^{2}\rightarrow X^{2}$ that satisfies the Yang-Baxter equation (respectively $T:X^{2}\rightarrow X^{2}$ where $T(x,y)=(x*y,x)$ for all $x,y\in X$) along with a sequence $\mathbf{x}\in X^{\mathbb{N}}$ computable in polynomial time where there is some $c\in B$ with (respectively for all $c\in B$) where $\mathbf{x}\cdot a\neq\mathbf{x}\cdot b$ for distinct $a,b\in A_{c}$?
