1 problem found
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: \(R(0) = 0\), \(R(1) = \Omega^2 1 = \Omega^2\), \(R(\Omega^2) = \Omega^4 = -\Omega\), \(R(-\Omega) = -\Omega^3 = 1\) \(T(0) = \frac{\Omega^2}1 = \Omega^2\), \(T(1) = \frac{\Omega + \Omega^2}{2\Omega^2+1} = \frac{2\Omega - 1}{2\Omega-1} = 1\) \(T(\Omega^2) = \frac{\Omega^3 + \Omega^2}{2\Omega^4+1} = \Omega \frac{\Omega^2+\Omega}{-2\Omega+1} = \Omega \frac{2\Omega-1}{-2\Omega+1} = - \Omega\) \(T(-\Omega) = \frac{-\Omega^2 + \Omega^2}{-2\Omega^3+1} = \frac{0}{3} = 0\) 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).