Prove that we have an isomorphism from the group generated by $1/x$, $(x-1)/x$ to the symmetric group $S_3$

Question

Prove that the group generated by $1/x$, $(x-1)/x$ is isomorphic to the symmetric group $S_3$.

What I already did is generate the elements $f \circ g=x/(x-1)$, $g \circ f=1-x$, $f^2=x$, $f\circ g \circ f=1/(1-x)$. It seems to be reasonable to assign $\phi(1/x)=(12)$ and $\phi((x-1)/x)=(123)$ since $(1/x)^2=x$ and $((x-1)/x)^3=x$. And through this to assign to every function a permutation.

But how do I show that I cannot generate more elements with the two base functions $1/x$ and $(x-1)/x$?

Answer 1

Call the group $G$. Show that

$$\begin{align} \varphi:G&\to D_3,\\ (x-1)/x &\mapsto a,\\ 1/x &\mapsto b, \end{align}$$

defines an isomorphism between $G$ and the dihedral group

$$D_3\cong\langle a,b\mid a^3, b^2, bab=a^{-1}\rangle$$

of order six. (Doing this could entail showing $G$ satisfies the relations in the presentation, so that $|G|$ is bounded above by $|D_3|$, since $D_3$ would then map onto $G$.)

Use the well-known fact that $D_3\cong S_3$.