Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 3 / 7D / Solution

By transitivity, choose $h\in G$ with $hy=x$. If $gy=y$, thenThus $hg{h}^{-1}\in B$. This proves the conjugation assertion directly from the transitive group action.

For the complex special linear group in dimension two, the Möbius transformation of a matrix $g$ acts by $z\mapsto (az+b)/(cz+d)$, with $ad-bc=1$. It fixes infinity exactly when $c=0$, so its stabilizer subgroup isWe distinguish the cases to avoid dividing by a vanishing coefficient.

If $c\ne 0$, infinity is not fixed, and the finite fixed points solveNeither root is the pole $-d/c$: substituting that value gives $(ad-bc)/c=1/c\ne 0$. The fixed-point set is thereforewith the two entries coinciding when the discriminant is zero. The discriminant identity uses $(a-d{)}^2+4bc=(a+d{)}^2-4$.

If $c=0$ and $a\ne d$, the fixed points are $\{\infty ,b/(d-a)\}$. If $c=0$, $a=d$ and $b\ne 0$, only infinity is fixed. If $c=0$, $a=d$ and $b=0$, the determinant condition gives $g=\pm I$ and every point is fixed. Both central matrices induce the identity Möbius transformation.

Every case has a fixed point, because a quadratic over the complex numbers has a root. The action is transitive: any finite $z_0$ is sent to infinity byand infinity already needs no change. Applying the first argument to a fixed point proves every element of $S{L}_2(\mathbb{C})$ is conjugate to an element of $B$.