Later Lawson (1998) gave another more direct way to prove maximality by first showing that there is a g in S sending D onto the disk D, | z | > 1. In fact if x ∈ ∈ S ∖ ∖ H ¯ ¯ , then there is a small disk D 1 in D such that xD 1 lies in D. Then for some h in H, D 1 = hD. Similarly yxD 1 = D for some y in H. So g = yxh lies in S and sends D onto D. It follows that g fixes the unit disc D so lies in SU(1,1). So g lies in S. If t lies in H then tgD contains gD. Hence g − − 1 t − − 1 g ∈ ∈ H ¯ ¯ . So t lies in S and therefore S contains an open neighbourhood of 1. Hence S = SL(2, C).