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It is well known that braid groups and knot groups share many common properties. For example, they have the same $H_1$ and they are both residually finite and (hence) Hopfian. On the other hand, we know that $B_2$ and $B_3$ are isomorphic to the knot groups of unknot and trefoil knot respectively. My question is, for any $n\geq 4$, is there a knot with knot group $B_n$?

If not, how about high dimensional knot $S^n$ in $S^{n+2}$?

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    $\begingroup$ The first question is answered (negatively) here: math.stackexchange.com/questions/2118701/… $\endgroup$ – Igor Rivin Aug 8 '17 at 2:50
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    $\begingroup$ To summarize Igor's link, for $n\ge 4$, $B_n$ contains a copy of $\mathbf{Z}^3$ (use $\sigma_1$, $\sigma_3$, and a generator of the center), while an aspherical 3-manifold group can't contain $\mathbf{Z}^3$ unless it's virtually abelian. $\endgroup$ – YCor Aug 8 '17 at 15:07
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This is a simple addendum to Danny's answer (because I can't use the citation thing in comments): Homology of braid groups has been computed by a number of people, and an extensive survey is:

Vershinin, Vladimir V., Homology of braid groups and their generalizations, Jones, Vaughan F. R. (ed.) et al., Knot theory. Proceedings of the mini-semester, Warsaw, Poland, July 13--August 17, 1995. Warszawa: Polish Academy of Sciences, Institute of Mathematics, Banach Cent. Publ. 42, 421-446 (1998). ZBL0905.20032.

Unless I am misinterpreting the results, there is nontrivial homology in all degrees, so the answer to the OP's second question is negative also.

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This is an incomplete answer to the last question, but perhaps someone can supply the last piece of information to complete it. The question of whether a finitely presented group G is a higher-dimensional knot group was resolved by Kervaire. Necessary and sufficient conditions are (1) G is the normal closure of one element, (2) $H_1(G) \cong {\mathbb Z}$, and (3) $H_2(G) = 0$.

The first two of these hold for the braid groups. I'm sure that the calculation of $H_2(B_n)$ is somewhere in the literature, but I couldn't locate it.

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