Let $B_n$ denote the braid group on $n$ strands. The knot group of the trefoil knot is isomorphic to $B_3$. Are there other knots $K_n$ such that the knot group of $K_n$ is $B_n$?
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No. Knot complements are aspherical by the sphere theorem, so the cohomology of knot groups coincides with that of the knot complement, which vanishes outside degree 1. Now we have the following calculation, due to Fuks, of the Poincare polynomial of pure braid groups: $$\sum_{i=0}^\infty \text{rk} H^i(P_n, \Bbb Z) t^i = \prod_{j=1}^{n-1} (1 + jt).$$ In particular, for $n > 3$, the pure braid group has cohomology in degrees 3 and higher. But if $B_n$ were the fundamental group of a knot complement, then a finite (degree $n!$) cover of this 3-manifold would have fundamental group $P_n$. But that's preposterous, since a noncompact 3-manifold cannot have nontrivial cohomology in degrees 3 and higher.
Alternatively starting at $B_6$ there are evident $\Bbb Z^3$ subgroups of the braid groups, and an aspherical noncompact 3-manifold cannot have fundamental group $\Bbb Z^3$ for the same cohomological reasons as before (but now you can calculate it yourself, since $\Bbb Z^3$ is the fundamental group of $T^3$). But then $B_4$ and $B_5$ need special arguments. You can find $\Bbb Z^3 \subset B_4$ in a somewhat more complicated way: it's generated by $\sigma_1$, $\sigma_3$, and $(\sigma_1 \sigma_2 \sigma_3)^4$.