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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?2012-03-19

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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: http://web.utk.edu/~utkreu/docs/SGmini4.pdf

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    An easier way to see the $H_1$ of knot complement has one generator is to pick a generator in $\pi_1$, this is just a pointed loop. Then when one move the loop "upstairs" or "downstairs" in Wirtinger representation to another generator, two generators differ from a conjugation, but when you abelianize everything, two pointed loops represent the same loop in homology. Do this for all generators in fundamental group, then one sees $H_1$ of knot complement only has one generator.2016-11-01