Surjection $\pi_1 (S^3 - 4_1) \to D_n$ Let $4_1$ be the figure-of-eight knot. Its knot group $\pi_1 (S^3 - 4_1) $ has a presentation $$ \langle a,b \mathrel| aba^{-1}bab^{-1}=bab^{-1}a \rangle .$$
I want to prove that $4_1$ is not the unknot by defining a surjective group homomorphism to a nonabelian group. I read here that the dihedral group $D_6 \cong \langle x,y \mathrel| x^3= y^2 =1 ,\ yxy^{-1}=x^{-1} \rangle$ should work, but I cannot find a surjective map which is group homomorphism. What should this map be?
 A: Unfortunately, this is not possible.  I would assume that Lee Mosher misspoke there, and as is pointed out, was probably just thinking of the trefoil.
To show there is no map from $\pi_1 = \pi_1(S^3 - 4_1)$ to the dihedral group $D_6$, we will use the idea of $n$-coloring. A  knot $n$-coloring is an assignment of $n$ colors, $\{0,1, \ldots, n-1\}$ to the arcs of a diagram, so that at each crossing, we have $a+b \equiv 2c \mod n$, where $a,b$ are the colors of the under arcs and $c$ is the color of the over arc. 
If we pick $n=3$, we get the classic tricolorability invariant.  But this can be generalized, see wikipedia.  $G$-colorability is defined for any group $G$, if there is a homomorphism from  $\pi_1$ to $G$.  Taking $G$ to be the dihedral group, $D_{2n}$ we get back the classic $n$-coloring I mentioned above.  
So, if there was a map from $\pi_1$ to $D_6$, then we would be able to 3-color the figure 8 knot, but we cannot, see again wikipedia.  The figure 8 knot is 5-colorable though, so there is a map from $\pi_1$ to $D_{10}$.  Of course, if the goal is to show that the figure 8 knot is not trivial, then we have done that already if you show it is 5-colorable.  But finding the map to $D_{10}$ is still interesting in its own right.
Hope this helps.
A: For the sake of future exposition, I will include a full argument to my question above: as @N. Owad pointed, the explicit surjection can be found using Fox $n$-coloring. 
It turns out the $4_1$ is 5-colored, see here 2.3 (i). On the other hand, 3.10 gives an explicit bijection between $n$-colorings of $K$ and group homomorphisms $\pi_1 (S^3 - K) \to D_n$: if the dihedral group of order 10 has a presentation $$D_5 = <x,y | x^5 = y^2=1, yxy^{-1}= x^{-1}>,$$
then the given $5$-coloring corresponds to the group homomorphism $$a \mapsto yx^3 \qquad , \qquad b \mapsto yx.$$
It is easily shown that such a map is surjective, so we found a surjective group homomorphism from the knot group of $4_1$ to a nonabelian group, so $4_1$ cannot be the unknot. 
