Knot Group and the Unknot Hi I am stuck in trying to show that given a knot $K$ such that the knot group $\pi_1(K)=\mathbb Z$ then $K\simeq U$.
I tried to use the fact that the infinite cyclic cover is the universal cover but I don't know if that is the right way.
Any hints would be appreciated.
 A: Note: I am assuming the OP is referring to the usual knot group $\pi_1(S^3\setminus K)$. That's the only way his second sentence makes sense anyway.
There is a curve on the boundary torus which is null-homotopic in the complement, since the inclusion of the boundary into the complement is $\mathbb Z\oplus\mathbb Z\to \mathbb Z$ on $\pi_1$. This means that the boundary torus is not incompressible. So there is a simple closed curve on the torus that bounds a disk into the complement. Furthermore, the only such curve which is null-homologous is the longitude. Therefore, the longitude is null-homotopic and by the disk theorem bounds an embedded disk. Thus the knot is the unknot.
A: You need some tools of 3-manifold topology --- the loop theorem and the generalized Schönflies theorem --- in order to prove this (assuming you mean $\pi_1(S^3 \setminus K)$. 
Consider the manifold $M$ obtained from $S^3$ by removing an open solid torus neighborhood of $K$, so $\partial M$ is homeomorphic to $T^2$. The inclusion induced homomorphism 
$$\mathbb{Z}^2 \approx \pi_1(T^2) \mapsto \pi_1(M) \approx \mathbb{Z}
$$
is not injective. 
Applying the loop theorem, there exists a tamely embedded 2-disc $D \subset M$ such that $D \cap \partial M = \partial D$ and that circle is nontrivial in the torus $\partial M$. It follows that the boundary of a regular neighborhood of $D \cup \partial M$ is a tamely embedded 2-sphere $S^2 \hookrightarrow M \subset S^3$. 
Applying the generalized Schönflies theorem, that 2-sphere is standard, after composing with a self-homeomorphicm of $S^3$ we may assume that this is the usual embedding of $S^2$ in $S^3$. 
From this it follows that $K$ is the unknot.
A: For any knot $K$ you always have $\pi_1(K)=\Bbb Z$ because any knot is homeomorphic to the circle. The major trouble comes when one considers the complement of the knot and depending where is embedded. 
