Knot group

Remarks

The complement $\R^3 \setminus K$ is path-connected, so the group does not depend on $x_0$ up to isomorphism. If $K$ and $K'$ are equivalent, an ambient isotopy gives a homeomorphism of their complements, so their knot groups are isomorphic. Removing a point at infinity does not change it: $\pi_1(S^3 \setminus K) \cong \pi_1(\R^3 \setminus K)$. For a tame knot, the abelianization of the knot group is $\Z$. Example: the complement of the unknot deformation retracts onto the union of a $2$-sphere and a circle that meet in one point. So its knot group is $\Z$, generated by a loop that goes once around the unknot. For the trefoil knot, the knot group has the presentation $\langle x, y \mid xyx = yxy \rangle$; it is not abelian, so the trefoil is not equivalent to the unknot. Non-example: the fundamental group of the knot $K$ itself is not the knot group. It is $\Z$ for every knot, since $K$ is homeomorphic to a circle, so it distinguishes no knots.