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!

Put $\xi = \frac{1}{2}(3 + \sqrt{5})$. The map $G \to GL_2(\mathbb{R})$ defined by
\begin{align*}
a &\to \begin{pmatrix}1 & 0 \\ 0 & \xi\end{pmatrix} & 
b &\to \begin{pmatrix}1 & 1 \\ 0 & \xi\end{pmatrix} 
\end{align*}
is well-defined with nonabelian image.
