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Are there known examples of knot groups that do not surject onto any alternating group? I would be a little surprised if the answer is negative.

Update: To reflect the interesting part of the question, I'd prefer to avoid the unknot and the "abelianization" in the obvious way. So, the target alternating group should be considered as $A_n$ $(n > 3)$. That is, as suggested by YCor, does there exist some nonabelian knot group $G$ such that for every $n > 3$, the alternating group $A_n$ is not a quotient of $G$?

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  • $\begingroup$ Since the meaning of "any" might be ambiguous, the question is, if I'm correct: does there exist some know group $G$ such that for every $n\ge 3$, the alternating group $\mathrm{Alt}_n$ is not a quotient of $G$. Right? $\endgroup$
    – YCor
    Commented May 17 at 16:51
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    $\begingroup$ The heuristic model of Dunfield-Thurston suggests that the probability a fundamental group of "random" $3$-manfold surjects onto $A_n$ tends to a (positive) limiting probability as $n \rightarrow \infty$ (arxiv.org/pdf/math/0502567), so it seems exceedingly unlikely for hyperbolic knots. OTOH Torus knots have fundamental group $\langle x,y |x^p=y^q \rangle$ and I suspect it should be elementary to prove that this group has many alternating quotients outside the degenerate cases. $\endgroup$
    – user523984
    Commented May 17 at 18:48
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    $\begingroup$ Re: the previous comment: it's not hard to show a $p$-cycle and a $q$-cycle (where $p<q$ are both odd) generate $A_q$. $\endgroup$
    – Steve D
    Commented May 17 at 22:42
  • $\begingroup$ @YCor There might be some misunderstandings when I said “abelianization” I was referring to the exclusion of the unknot. Similar to the nontrivial knot…I’ll use the phrase “unknot” then. $\endgroup$
    – Shijie Gu
    Commented May 18 at 10:35
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    $\begingroup$ Probably you should say "non-abelian knot group", because "nontrivial group" usually means not reduced to the unit. $\endgroup$
    – YCor
    Commented May 18 at 15:15

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The unknot and the hopf link have abelian knot groups. See here. Hence they can't surject onto an $A_n$ for $n\gt3.$

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    $\begingroup$ :D Of course this doesn’t really answer the interesting version of the question. $\endgroup$
    – HJRW
    Commented May 17 at 17:52

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