Problem source

Let $\Omega=\exp(\mathrm{i}\pi/3).$ Prove that $\Omega^{2}-\Omega+1=0.$

		Two transformations, $R$ and $T$, of the complex plane are defined
		by 
		\[
		R:z\longmapsto\Omega^{2}z\qquad\mbox{ and }\qquad T:z\longmapsto\dfrac{\Omega z+\Omega^{2}}{2\Omega^{2}z+1}.
		\]
		 Verify that each of $R$ and $T$ permute the four point $z_{0}=0,$
		$z_{1}=1,$ $z_{2}=\Omega^{2}$ and $z_{3}=-\Omega.$ Explain, without
		explicitly producing a group multiplication table, why the smallest
		group of transformations which contains elements $R$ and $T$ has
		order at least 12. 

		Are there any permutations of these points which cannot be produced
		by repeated combinations of $R$ and $T$?

Solution source

$R(0) = 0$, 
$R(1) = \Omega^2 1 = \Omega^2$, 
$R(\Omega^2) = \Omega^4 = -\Omega$, 
$R(-\Omega) = -\Omega^3 = 1$

$T(0) = \frac{\Omega^2}1 = \Omega^2$,
$T(1) = \frac{\Omega + \Omega^2}{2\Omega^2+1} = \frac{2\Omega - 1}{2\Omega-1} = 1$
$T(\Omega^2) = \frac{\Omega^3 + \Omega^2}{2\Omega^4+1} = \Omega \frac{\Omega^2+\Omega}{-2\Omega+1} = \Omega \frac{2\Omega-1}{-2\Omega+1} = - \Omega$
$T(-\Omega) = \frac{-\Omega^2 + \Omega^2}{-2\Omega^3+1} = \frac{0}{3} = 0$

Thinking of $R$ and $S$ as elements of $S_4$, we have that $R = (234), S = (134)$, we can also construct $RS = (14)(23), R^2S = (12)(34),  RSR^2S = (13)(24)$. Therefore we have the subgroups $\{e, (234), (243)\}$ of order $3$ and the subgroup $\{e, (12)(34), (13)(24), (14)(23) \}$ of order $4$. By Lagrange's theorem this means that both $3$ and $4$ divide the order of the group, therefore the group has order divisible by $12$ (and therefore is at least $12$).

Yes, we cannot produce any odd permutation, for example $(12)$ cannot be produced. (Since all our generators are even permutations).