How to show that $\langle a,b \mid aba^{-1}ba = bab^{-1}ab\rangle$ is not Abelian? I'd like to show that
$$
G = \langle a,b \mid aba^{-1}ba = bab^{-1}ab\rangle
$$
is non-Abelian.
I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.
Thanks a lot!

Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.
One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group.  There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2b\equiv a+c\pmod{5}$ at each crossing).  Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $\pi_1(S^3-4_1)$ to $D_{2\cdot 5}$, the dihedral group of order $10$, is surjective.
