Let $M=S^3\setminus K$ be the figure-8 knot complement. The fundamental group of $M$ has a presentation $\pi_1 M=\langle x,y\mid x^{-1}yxy^{-1}x^{-1}y^{-1}xyx^{-1}y^{-1}=1\rangle$. There is a surjective homomorphism $\varphi\colon\pi_1 M\to A_4$ defined by $\varphi(x)=(134)$ and $\varphi(y)=(234)$. Let $G:=\ker\varphi$. It is not hard to show that $G=\langle\langle x^3,y^3,(xy)^2 \rangle\rangle$ (double angle brackets denote normal closure).
Now, the theory of covering spaces ensures that there is a cover $f\colon M_G\to M$, such that $\deg f=12$, $\pi_1 M_G=G$, and the group of Deck transforms is $A_4$. There is then a space $\overline{M_G}$ in which $M_G$ is dense such that $f$ extends to a ramified cover $\overline{f}\colon\overline{M_G}\to S^3$, with ramification over $K$. Here are my questions:
- Can we find a presentation for $G$?
- Can we describe $M_G$ and $\overline{M_G}$ topologically? E.g., by a (simplicial/CW/handle) decomposition?
- How many components does $\overline{f}^{~-1}(K)$ have, and what is the ramification index of $\overline{f}$ around each of these components?