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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

 

How many knots have the same group?


Author: Jonathan Simon
Journal: Proc. Amer. Math. Soc. 80 (1980), 162-166
MSC: Primary 57M25
MathSciNet review: 574528
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Abstract: Let K be a knot in $ {S^3},G = {\pi _1}({S^3} - K),n = $ number of prime factors of $ K,\nu (G) = $ number of topologically different knot-complements with group G and $ \kappa (G) = $ number of distinct knot types with group G.

Theorem. If K is prime, then $ \nu (G) \leqslant 2$. If $ n \geqslant 2$, then $ \nu (G) = \kappa (G) \leqslant {2^{n - 1}}$. For each $ n \geqslant 2$, the bound $ {2^{n - 1}}$ is the best possible. For K prime, we still have the conjecture $ \nu (G) = \kappa (G) = 1$. If K is a cable-knot, then $ \kappa (G) \leqslant 2$.


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DOI: http://dx.doi.org/10.1090/S0002-9939-1980-0574528-5
PII: S 0002-9939(1980)0574528-5
Article copyright: © Copyright 1980 American Mathematical Society