# Compute the subgroup $M_{\{0, -3, \infty\}} \subset M$ consisting of 6 transformations, preserving the set $\{0, -3, \infty\}$

Compute the subgroup $M_{\{0, -3, \infty\}} \subset M$ consisting of 6 transformations, preserving the set $\{0, -3, \infty\}$, together with an explicit isomorphism

$$M_{\{0, -3, \infty\}} = S_3$$

I said that this would be some transformation that will send $\{0, 1, \infty\}$ to $\{0, -3, \infty\}$, i.e $L: z \mapsto -3z$. We therefore have

$$M_{\{0, -3, \infty\}} = L M_{\{0, 1, \infty\}} L^{-1}$$

However I am now unsure as to how to get the answer, with the explicit isomorphism from here.

Thanks

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I think it'd be agreat idea if you explained your symbols. For example, M seems to be some group of Moebius transformations...or not, who knows? Over the reals, the complex...?? – DonAntonio Dec 13 '12 at 20:33
Yes, M is a group of Mobius transforms over the extended complex plane – Kaish Dec 14 '12 at 0:55

There are of course $3!=6$ permutations of the set $S=\{0,w,\infty\}$ (let $w\ne0,\infty\in\hat{\Bbb C}$). Pick one, say the one given by $0\mapsto\infty,w\mapsto0,\infty\mapsto w$ for instance. We can compute the unique $\gamma\in\mathrm{PSL}_2(\Bbb C)$ such that $\gamma$ restricted to $S$ realizes this permutation. So, in particular,

$$\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix},\quad\begin{cases}\gamma(0)=\infty, \\ \gamma(w)=0, \\ \gamma(\infty)=w \end{cases}\implies\begin{cases}b/d=\infty, \\ (aw+b)/(cw+d)=0, \\ a/c=w.\end{cases}$$

The top condition tells us $d=0$, the bottom condition tells us $c\ne0$ so we might as well set $c=1$ (we can normalize $\gamma$ however we want, because we are working in a projective setting, so fixing one of the nonzero entries is possible without loss of generality), which tells us $a=w$. The middle condition says $aw+b=0$ so that $b=-w^2$. Thus,

$$\gamma=\begin{pmatrix}w&-w^2\\1&0\end{pmatrix}:z\mapsto w-w^2/z.$$

Do this type of computation for all six permutations of $\{0,w,\infty\}$ with $w=3$ and you will be set. If you have $M_{0,1,\infty}=\{z,1/z,1-z,1-1/z,1/(1-z),z/(z-1)\}$ already at hand, then you can of course conjugate it by $L:z\mapsto-3z$. For example, $1-z$ conjugated would be $-3(1-z/(-3))$, which can be simplified to $-3+z/3$.

Once you have the elements of $G=M_{0,-3,\infty}$, to find an explicit isomorphism with $S_3$ it suffices to pick a bijection $\{0,-3,\infty\}\cong\{1,2,3\}$ and then pick out how $G$ acts on this set under the bijection.

For instance, with $0\mapsto1,-3\mapsto2,\infty\mapsto3$, our $\gamma$ acts as $1\mapsto3,2\mapsto1,3\mapsto2$ so $\gamma$ acts like $(132)$.

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