Let $T_1$ be the standart trefoil knot, embedded in $\mathbb R^3$. Then, one can easily give a simple Wirtinger presentation of $\pi_1(\mathbb R^3 \setminus T_1)$ by $\langle a,b,c | a = bcb^{-1}, b=cac^{-1}, c=aba^{-1} \rangle$. Killing one generator and substituting the corresponding, superflous relation into the other relators, we get the reduced presentation $\langle a,b| a = baba^{-1}b^{-1}, b= abab^{-1}a^{-1} \rangle$. Knowing that $\pi_1(\mathbb R^3 \setminus T_1)$ is isomorphic to the braid group $B_3$, I want to simplify it even further, so that $\pi_1(\mathbb R^3 \setminus T_1) = \langle x,y|x^3 = y^2 \rangle$. Now two obvious candidates are $x = ba$ and $y= bab$. Unsing the above relations, one can immediately compute that $a = yx^{-1}$ and $x^3=y^2$. However, I was so far unable to express $b$ in terms of $x$ and $y$. Can anybody help me out here, it might be something really obvious I am missing....
Further, let $T_1,....T_n$ be a collection of disjoint trefoils with $n \geq 2$ in $\mathbb R^3$ that also have disjoint images under a regular projection of the corresponding link. Then, one can construct the connected sum $T := T_1 \# .....\# T_n$, which is now again a knot in $\mathbb R^3$. Then, one can compute a group presentation of $\pi_1(\mathbb R^3 \setminus T)$ in terms of $n + 1$ generators and $n$ relators, given by $\langle x_i,y| x_i^3 = (x_i x_1^{-1}yx_1^{-1} x_i)^2\rangle $ for $i=1,..n$.
Question: Does there exist a presentation of $\pi_1(\mathbb R^3 \setminus T)$ with less than $n+1$ generators?
EDIT: I found a somewhat simpler presentation, given by $\langle a,b_i: ab_ia = b_iab_i \rangle$ for $i=1,...,n$. Is it clear that this presentation cannot be reduced further ?
It might very well be that a solution to the first question is very helpful for answering the second one. Any help is appreciated
Richard Weidmann proved in his paper "On the rank of amalgamated products and product knot groups" that if $\mathcal K$ is a knot that is the connected sum of $n$ (non-trivial) knots, then $rank(\pi_1(\mathbb S^3 \setminus \mathcal K)) \geq n+1$. In my case, this implies that $rank(\pi_1(\mathbb R^3 \setminus T)) = n+1$.