How to get an expression for this isomorphism Define $\hat{\mathbb{R}} = \mathbb{R}\cup\{{\infty}\}$ and $1/0=\infty,1/\infty=0, \infty/\infty=1,1-\infty=\infty-1=\infty$. Then the six functions $\hat{\mathbb{R}}\to\hat{\mathbb{R}}$ given by $x$, $\frac{1}{x}$, $1-x$, $\frac{1}{1-x}$, $\frac{x}{x-1}$ and $\frac{x-1}{x}$ form a group $G$ with composition $\circ$ as operation.

Show that $S_3$ (the permutation group, $S_3=\{id,(12),(23),(13),(132),(123)\}$) is isomorphic to $G$. Hint: The elements in the latter group permute $\{0,1,\infty\}$.

First of all, as the hint suggests, I should use the $S_3$ as the permutation group of the set $X = \{0,1,\infty\}\subset\hat{\mathbb{R}}$ ? But if so then $G=S_3$ and they're isomorphic trivially.
If not, then I think I should identify each element of $G$ with a element of $S_3$ that permutes $\{1,2,3\}$ similarly. I know how to do that if I identify the elements of the sets they permute. Then I have defined a bijection $\varphi$ between $S_3$ and $G$. Is this right? But then I have to check, I think, $\binom62+6$ cases to show that it is a homomorphism. I'm quite sure it is, but there is a way to get a expression for this isomorphism that allows me to show that it is a homomorphism in a less extensive way?
Any help will be appreciated.
 A: Follow the hint. For each permutation $\sigma$ of $\{0, 1, \infty\}$, can you find an element $g_\sigma\in G$ that acts like $\sigma$?
Then you have a surjective, hence bijective (why?) map $G\rightarrow S_3$. 
Since it is a bijection, the permutation $\sigma$ in turn determines the element $g_\sigma$. Using this, you should be able to prove that your map is a homomorphism i.e. that $g_\sigma g_{\sigma'}=g_{\sigma \sigma'}$ for each $\sigma, \sigma' \in S_3$.
A: The hint pretty much tells you what the proof is going to be — in the sense that proving that the statement of the hint (rather than using it) is true pretty much solves the problem.
What the hint wants to tell you is that each of the given six functions is a permutation on the set $X=\{0,1,\infty\}$. To verify this claim, you need to show that  for any $g\in G$ and for any $x\in X$ we have $g(x)\in X$. (If this were not true, i.e. if for example we had $g(1)=2$ for one of these functions $g$, then that would be a problem — but the hint says it's not gonna happen.)
You verify this in the most straightforward manner. For example, consider $g(x)=1-x$, one of the given functions. Then by plugging in, we can directly verify that:
$$g(0)=1-0=1, \quad g(1)=1-1=0, \quad g(\infty)=1-\infty=\infty.$$
By definition, this $g$ is a permutation of $X$. Moreover, if you (informally speaking) match $\{0,1,\infty\}$ with $\{1,2,3\}$, then you can see that this function $g(x)=1-x$ acts on $X=\{0,1,\infty\}$ just like the permutation $(12)$ acts on $\{1,2,3\}$.
Verifying all six given functions would pretty much complete the proof: you should see that you indeed have all six different permutations of $X=\{0,1,\infty\}$.
