Knot group is $\mathbb{Z}$ iff $K$ is the unknot

Question

Let $K \subset S^3$ be a knot; we call knot group the fundamental group of the complementary of $K$ in $S^3$. I've come across the fact that the only knot whose knot group is isomorphic to $\mathbb{Z}$ is the unknot; it is easy to check that, if $K$ is the unknot, than its knot group is infinite cyclic, but I don't know how to prove the converse implication (I'm concerning only about knots, I know that the same statement is false for links with more than one component).

Answer 1

The converse implication requires a deep theorem, namely the loop theorem. I suspect that a detailed proof can be found in textbooks on 3-manifolds or on knots, such as Rolfsen's book. But here are a few details.

Letting $N(K)$ be a solid torus neighborhood of $K$, and letting $M$ be $S^3 - \text{interior}(N(K))$, it follows that $T = \partial N(K) = \partial M$ is homeomorphic to the torus $S^1 \times S^1$. The manifold $M$ is a deformation retract of $S^3 - K$ and so $\pi_1 M \approx \mathbb Z$. It follows that the homomorphism $\pi_1 T \mapsto \pi_1 M$ that is induced by the inclusion map $T \hookrightarrow M$ is not injective. This exactly matches the hypothesis of the Loop Theorem, and from the conclusion of that theorem one obtains an embedded disc $D \subset M$ such that $D \cap T = \partial D$.

Now one needs that the inclusion $\partial D \hookrightarrow N(K)$ is isotopic to $K$ in the solid torus $N(K)$. The only other alternatives are that $\partial D$ is null homotopic in $N(K)$, or it is homotopic to a nontrivial iterate of $K$, and in each case one would then use a Mayer-Vietoris argument to get an absurd computation for $H_1(S^3)$, which should of course be trivial.

It follows that $\partial D \cup K$ bounds an annulus $A \subset N(K)$. The union $D \cup A$ is therefore an embedded disc in $S^3$ with boundary $K$.