The two functions $f(x)=\frac{1}{x}$ and $g(x)=\frac{x-1}{x}$ generate, with the operation of function composition, a group $G$ of functions. Prove that this group is isomorphic to the group $S_3$.
I found a solution which is pretty tedious, so I wonder if there are other ways to prove the above statement. So here is my solution:
With composition we can generate the functions:
$f(f(x))=x$, $g(g(x))=\frac{1}{1-x}$, $g(f(x))=1-x$ and $f(g(x))=\frac{x}{x-1}$
To prove that no more functions can be generated by using the functions above, we need to make a table ($*$ denotes function composition): $$ \begin{array}{c|c|c|c} * & x & 1-x & \frac{1}{x} & \frac{x}{x-1} & \frac{x-1}{x} & \frac{1}{1-x} \\ \hline x & x & 1-x & \frac{1}{x} & \frac{x}{x-1} & \frac{x-1}{x} & \frac{1}{1-x} \\ \hline 1-x & 1-x & x & \frac{x-1}{x} & \frac{1}{1-x} & \frac{1}{x} & \frac{x}{x-1} \\ \hline \frac{1}{x} & \frac{1}{x} & \frac{1}{1-x} & x & \frac{x-1}{x} & \frac{x}{x-1} & 1-x \\ \hline \frac{x}{x-1} & \frac{x}{x-1} & \frac{x-1}{x} & \frac{1}{1-x} & x & 1-x & \frac{1}{x} \\ \hline \frac{x-1}{x} & \frac{x-1}{x} & \frac{x}{x-1} & 1-x & \frac{1}{x} & \frac{1}{1-x} & x \\ \hline \frac{1}{1-x} & \frac{1}{1-x} & \frac{1}{x} & \frac{x}{x-1} & 1-x & x & \frac{x-1}{x} \end{array} $$
Hence no more functions can be generated. After some construction work, it seems that the following function is an isomorphism: $$ \phi(x)=() $$ $$ \phi(1-x)=(1,2) $$ $$ \phi(\frac{1}{x})=(1,3) $$ $$ \phi(\frac{x}{x-1})=(2,3) $$ $$ \phi(\frac{x-1}{x})=(1,2,3) $$ $$ \phi(\frac{1}{1-x})=(1,3,2) $$
To check wether this is an isomorphism or not, we need to make a similar table with $S_3$: $$ \begin{array}{c|c|c|c} * & () & (1,2) & (1,3) & (2,3) & (1,2,3) & (1,3,2) \\ \hline () & () & (1,2) & (1,3) & (2,3) & (1,2,3) & (1,3,2) \\ \hline (1,2) & (1,2) & () & (1,2,3) & (1,3,2) & (1,3) & (2,3) \\ \hline (1,3) & (1,3) & (1,3,2) & () & (1,2,3) & (2,3) & (1,2) \\ \hline (2,3) & (2,3) & (1,2,3) & (1,3,2) & () & (1,2) & (1,3) \\ \hline (1,2,3) & (1,2,3) & (2,3) & (1,2) & (1,3) & (1,3,2) & () \\ \hline (1,3,2) & (1,3,2) & (1,3) & (2,3) & (1,2) & () & (1,2,3) \end{array} $$ If we express this table in terms of $\phi$ we get: $$ \begin{array}{c|c|c|c} * & \phi(x) & \phi(1-x) & \phi(\frac{1}{x}) & \phi(\frac{x}{x-1}) & \phi(\frac{x-1}{x}) & \phi(\frac{1}{1-x}) \\ \hline \phi(x) & \phi(x) & \phi(1-x) & \phi(\frac{1}{x}) & \phi(\frac{x}{x-1}) & \phi(\frac{x-1}{x}) & \phi(\frac{1}{1-x}) \\ \hline \phi(1-x) & \phi(1-x) & \phi(x) & \phi(\frac{x-1}{x}) & \phi(\frac{1}{1-x}) & \phi(\frac{1}{x}) & \phi(\frac{x}{x-1}) \\ \hline \phi(\frac{1}{x}) & \phi(\frac{1}{x}) & \phi(\frac{1}{1-x}) & \phi(x) & \phi(\frac{x-1}{x}) & \phi(\frac{x}{x-1}) & \phi(1-x) \\ \hline \phi(\frac{x}{x-1}) & \phi(\frac{x}{x-1}) & \phi(\frac{x-1}{x}) & \phi(\frac{1}{1-x}) & \phi(x) & \phi(1-x) & \phi(\frac{1}{x}) \\ \hline \phi(\frac{x-1}{x}) & \phi(\frac{x-1}{x}) & \phi(\frac{x}{x-1}) & \phi(1-x) & \phi(\frac{1}{x}) & \phi(\frac{1}{1-x}) & \phi(x) \\ \hline \phi(\frac{1}{1-x}) & \phi(\frac{1}{1-x}) & \phi(\frac{1}{x}) & \phi(\frac{x}{x-1}) & \phi(1-x) & \phi(x) & \phi(\frac{x-1}{x}) \end{array} $$
If we compare it to the first table we see that $\phi(f*g)=\phi(f)*\phi(g)$ $\forall f,g \in G$. Since $\phi$ is bijective it's an isomorphism. And therefore, $G$ and $S_3$ are isomorphic.
Is there a way to avoid making those tables? It seems pretty intuitive after trial and error that $\phi$ is an isomorphism, but how can we prove it formally in a neat and elegant way? Thank you!
A faster way is to note that both groups act on the set $\{0, 1, \infty\}$. The group $S_3$ does as a permutation representation, which is pretty clear. The second one does by viewing this a subset of $\mathbb{R} \cup \{\infty\}$.
If you then look at the group $G$ of functions, then you can see pretty clearly that the function $f(x) = \frac{1}{x}$ acts as the permutation which swaps $0, \infty$, and the function $g(x)$ acts as the 3-cycle $0\to \infty\to 1 \to 0$.
Playing around with these views should help you see a better way to prove this.