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Can someone provide me with an example of a satellite knot with symmetry group which is neither cyclic nor dihedral?

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Makoto Sakuma has a nice paper in this area: M. Sakuma "Uniqueness of symmetries of knots" Math. Z. 192 225--242 (1986) – Ryan Budney May 19 '14 at 20:40

If by the symmetry group of a knot you mean the group of isometries of $S^3$ leaving the knot invariant, then this can only be cyclic or dihedral, apart from the special case of torus knots which can have an $O(2)$ group of symmetries. By restricting symmetries to the knot itself one gets a homomorphism from the symmetry group of the knot to the symmetry group of a circle, which can only be cyclic or dihedral. The kernel of this restriction map must be zero by the Smith Conjecture which says that a nontrivial periodic homeomorphism of $S^3$ cannot fix a nontrivial knot pointwise. (The conjecture was proved by combining work of several people following Thurston's breakthrough work on hyperbolic structures on Haken manifolds.)

There might be other less rigid definitions of the symmetry group of a knot. For example one could take the mapping class group of the pair $(S^3,K)$ for a knot $K\subset S^3$. For satellite knots this group will generally be infinite due to Dehn twists along incompressible tori in the knot complement. One could define a finite "supersymmetry" group by taking the quotient of the mapping class group in which twists along tori are factored out. This can be larger than cyclic or dihedral since there could be for example rotational symmetries of solid tori containing the knot and bounded by incompressible tori. Examples for this are easy to find, such as the Whitehead double of the figure eight knot.

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