---
tag: 2Y52Q00MGE
kind: definition
name: Knot group
msc: [57K10, 57M05]
defines: AQVHM3FP0D
uses: [QYEAZTYGPT, S1XK3HXHFX]
sources:
  - url: https://en.wikipedia.org/wiki/Knot_group
---

## Definition

Let $K \subset \R^3$ be a [[knot|QYEAZTYGPT]], and let $x_0$ be a point of $\R^3 \setminus K$.
The *knot group* of $K$ is the [[fundamental group|S1XK3HXHFX]]
$$\pi_1(\R^3 \setminus K, x_0).$$

## 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|1TQ06R4Q66]] 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|VYTYSVHN4A]], 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.