Let $\pi_1(K)$ be the fundamental group of the knot $K$. Let it be generated by two meridians $a$ and $b$ (what is always the case for a two-bridge knot). Consider the normaliser $N \langle a^2, b^2 \rangle$. The $\pi$-group $O(K)$ of $K$, according to Boileau and Zimmermann, in this case is $O(K) \cong \pi_1(K)/ N\langle a^2, b^2 \rangle$.
The proof by Boileau and Zimmermann then goes like that: in fact they show that for a two-bridge knot $O(K)$ is dihedral, once the knot is non-trivial, and if $O(K)$ is dihedral, then the knot is two-bridge and non-trivial.
In fact, if $\pi_1(K)$ is generated by $a$ and $b$, the group $O(K)$ could be either dihedral (finite or infinite) or cyclic of order $2$. The cyclic one corresponds to the trivial knot (B. and Z. refer to the proof of Smith's Conjecture).
The authors use the theorem by Hodgson and Rubinstein in the case of $O(K)$ is finite dihedral. Also they use Thurston's orbifold geometrisation to show that the double cover $V$ of the $\pi$-orbifold along $K$ will be a lens space. Then the theorem by H. and R. applies.
If $O(K)$ is infinite dihedral, B. and Z. consider the double cover $V$ of the $\pi$-orbifold along $K$ again, and deduce that it's $S^1\times S^2$ making use of Smith's conjecture and the $\mathbb{Z}_2$-invariant sphere theorem. A paper by Tollefson (Involutions on $S^1\times S^2$ and other 3-manifolds, Trans. AMS, (183), 139-152 (1973)) says that $K$ must be a trivial two-component link in this case. I don't know how Tollefson obtains this result.
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