Consider a $(p,q)$ torus knot $K$ in 3-dimensional Euclidean space $\mathbb R^3$ where $p,q \geq 2$ and $\operatorname{GCD}(p,q)=1$. 


Let $\operatorname{Isom}(\mathbb R^3,K)$ be the isometries of $\mathbb R^3$ that preserve $K$. 


It's a fairly standard argument using theorems about uniqueness of Seifert fiberings to prove that it's impossible for $\operatorname{Isom}(\mathbb R^3, K)$ to contain subgroups isomorphic to both $\mathbb Z\_p$ and $\mathbb Z\_q$. Of course, $\operatorname{Isom}(S^3,K)$ can and does for the standard embeddings of torus knots in $S^3$. In some sense the core issue is that when this does happen, the $\mathbb Z\_p$ and $\mathbb Z\_q$ subgroups of $\operatorname{Isom}(S^3,K)$ have disjoint fixed point sets. 


My question: is there a reasonably elementary proof $\operatorname{Isom}(\mathbb R^3, K)$ does not contain subgroups isomorphic to both $\mathbb Z\_p$ and $\mathbb Z\_q$ that *avoid* the use of Seifert-fiber space techniques? I'm particularly interested if any "quantum topology" invariants can make this kind of symmetry argument. I thought a little about this, at least I'm not seeing how one could use the Alexander polynomial. 



Torus knots in Euclidean space -- a symmetry argument
The answer no. Neither quantum invariants nor the Alexander polynomial sees the difference between a knot in the three sphere and a knot in Euclidian three space. In the case of the Alexander polynomial the missing point does not interfere with the first homology of the infinite cyclic cover. In the case of quantum invariants, it is often proved that the coefficients of the vector assigned to the the knot complement by whatever TQFT you are working with are given by evaluations of one of the standard knot polynomials, which are the same for knots in the sphere and knots in Euclidian space.


No matter how you present the torus knot there is a subgroup of the homeomorphism group of the complement of the $(p,q)$-torus knot ( in the sphere) that is isomorphic to $\mathbb{Z}\_p\times \mathbb{Z}\_q$.
So you cannot rule it out with topological invariants that don't distinguish between knots in $\mathbb{R}^3$ and knots in $S^3$. 


