I am studying for my topology qualifying exam. I am stuck on showing if there is a surjective homomorphism from the fundamental group of the knot complement of $6_1^2$ shown below to the dihedral group of order $6$.
Assuming that I have calculated the fundamental group correctly, I have the presentation
$$\pi_1(\mathbb{R}^3 - 6_1^2) = \langle a,b \mid ababab = bababa \rangle$$
and, of course,
$$D_6 = \langle r,s \mid r^3 = s^2 =1, rs = sr^{-1}\rangle $$ I'm really stuck on finding a homomorphism that makes sense (assuming that a surjective one does in fact exist). So let $\phi:\pi_1(\mathbb{R}^3 - 6_1^2) \to D_6$ be our homomorphism. I've tried $\phi(a) = r$ (which means that $\phi(a)^2 = r^2 = r^{-1})$ and $\phi(babab) = s$. I know that I need $\phi(ababab) = \phi(bababa)$ in order for this to be a homomorphism. But the mapping above gives
$$rsr = \phi(a)\phi(babab)\phi(a) = \phi(abababa) = \phi(bababaa) = \phi(babab)\phi(a^2) = \phi(babab)\phi(a)^2 = sr^2 = sr^{-1}$$
Thus,
$$rsr = sr^{-1} \implies rsr^2 = s \implies rsr^{-1} = s \implies rs = sr$$
which isn't true.
Another mapping I've had suggested to me is $\phi(a) = r$ and $\phi(bababa^{-1}b^{-1}a^{-1}b^{-1}a^{-1}b^{-1})$ = s but then
$rs = \phi(a)\phi(bababa^{-1}b^{-1}a^{-1}b^{-1}a^{-1}b^{-1})) = \phi (abababa^{-1}b^{-1}a^{-1}b^{-1}a^{-1}b^{-1}) = \phi(1) = \phi(bababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1}a^{-1}) = \phi(bababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1})\phi(a^{-1}) = sr^{-1}$
Now $rs = sr^{-1}$ but since a homorphism must map the identity $\pi_1(\mathbb{R}^3 - 6_1^2)$ to the identity of $D_6$, this would imply that $rs = id_{D_6}$, which is a contradiction.
Please note that I have minimal knot theory knowledge and am looking for an explanation using algebra. Thank you for your consideration.
Update: I think I may have found a homomorphism: $\phi(a)=s$, $\phi(b)=rs$, so $$\phi(ababab)=\phi(a)\phi(b)\phi(a)\phi(b)\phi(a)\phi(b)=s(rs)s(rs)s(rs)=r^{−3}=id$$ and $$\phi(bababa)=\phi(b)\phi(a)\phi(b)\phi(a)\phi(b)\phi(a)=(rs)s(rs)s(rs)s=r^3=id$$ so $\phi(ababab)=\phi(bababa)$ and it's nontrivial. And it's surjective since we can generate D6 with s and rs. Does this look right?