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The knot group (the fundamental group of the complement of a knot) of the unknot is $\mathbb{Z}$ and the Hopf link is $\mathbb{Z}^2$, so those are knots (links) with Abelian knot group but are there any more?

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At least for a knot, what would that say about the Wirtinger representation of the knot group? – Neal Mar 19 '12 at 14:51

Here's a partial answer:

At least for knots, you're very limited in what can appear. The first homology group $H_1$ of a knot complement is $\mathbb{Z}$, which can be proved from Alexander Duality (see Hatcher, Algebraic Topology, section 3.3). Now the first homology group is always the abelianization of the fundamental group $\pi_1$, so if $\pi_1$ were abelian, it would have to be $\mathbb{Z}$.

This forces the knot to be the unknot, but I believe the reason for this is difficult. Here's a reference that quotes this result:

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