Conjugacy classes with linking number 1 in the complement of a knot?

Question

Let $K$ be a knot in $S^3$ and let $lk: \pi_1(S^3 - K) \to H_1(S^3-K;\mathbb{Z})$ denote the abelianization map. Let $1 \in H_1(S^3-K;\mathbb{Z})$ denote a choice of generator.

Is it possible that there are multiple conjugacy classes in $\pi_1(S^3-K)$ that are taken to $1$ by $lk$?

For $K$ the unknot, $lk$ is an isomorphism so certainly not, but I imagine in general we have infinitely many conjugacy classes sent to $1$? I'm afraid I am just bad at proving that things are not conjugate.

Answer 1

I may have an example where multiple conjugacy classes are sent to $1$.

Take $K$ the trefoil knot, its knot group $\pi_1(S^3 - K) \cong B_3$ the braid knot (one can use Wirtinger presentation to see it).
So $B_3 =<a,b \ | \ aba=bab >$ therefore $a$ and $b$ are conjugated (by element $ba$). Thus $lk(a)=lk(b)=1$.

But $b^2 a^2 b^{-3}$ is not conjugated to $b$. I do not knot how to prove it, but Sagemath does, since this code:

B.<a,b>=BraidGroup(3)
print (b^2*a^2*b^(-3)).is_conjugated(b)

return:

false

Since $lk(b^2 a^2 b^{-3})=2+2-3=1$, then you have two different conjugacy classes that maps to $1$.

I am sorry that I used a Sage algorithm blackbox in this answer, but I couldn't figure out how exactly one proves that two elements are not conjugated. I think it is a non trivial matter and it is called Conjugacy Problem. But fortunately there is an algorithm for Knot Groups (the problem is soluble as shown in this article ).