Post #1815111

@hallasurvivor This argument is a bit over the top, but it's the first thing I thought of. Let G be the unit norm quaternions. Then -1 generates an order 2 subgroup which is the center of G. We have that i^2 = -1, and hence for any g in G, g i g^(-1) is also a square root of -1. We have a Lie group action of G on itself by conjugation, and so the orbit of i consists of square roots. But this orbit is a subspace homeomorphic to G/Stab(i). Stab(i) is a closed subgroup of G and hence an embedded Lie subgroup. Thus, since G is connected, if it is not G itself, G/Stab(i) is uncountable. But j is not in Stab(i). You can also show that G/Stab(i) is homeomorphic to S^2 with a bit more work. Essentially, the adjoint representation Ad: G -> GL(g) will show that Stab(i) is the inverse image of the elements of Ad(G) = SO(3) which fix the subspace spanned by i (I'm identifying g with the imaginary quaternions), so rotations with some fixed axis of rotation. But this is the stabiliser for the natural action of SO(3) on S^2, hence G/Stab(i) is homeomorphic to S^2.