PRIME KNOTS ARE DETERMINED UP TO MIRROR IMAGE BY THE FINITE QUOTIENTS OF THEIR GROUPS

Let K ⊂ S³ be a prime knot and J ⊂ S³ be any knot. We prove that if the knot groups π₁(S³ \ J) and π₁(S³ \ K) have the same finite quotients, then J is equivalent to K or to its mirror image. Neither the primeness of K nor the mirror-image ambiguity can be removed. For hyperbolic knots, this follows from Xu’s theorem that lattices in PSL₂(ℂ) are profinitely rigid among lattices, and for graph knots, it is due to Wilkes. The remaining case is that of satellite knots whose exteriors contain a hyperbolic JSJ piece. For these, the reduction in Xu’s proof—that the A-polynomial of a prime knot is a profinite invariant up to mirror image—shows that an arbitrary isomorphism of profinite completions can be arranged to preserve the meridian and longitude: of the knot itself when the root of the JSJ tree is hyperbolic, and of the companion when the root is a cable space. The marked recognition theorem of Biswas and Singh then completes the proof. We give a self-contained account of the argument.