How to show that $\langle a,b \mid aba^{-1}ba = bab^{-1}ab\rangle$ is not Abelian?

Question

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!

Answer 1

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.

Answer 2

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.