Proof of Theorem I1

 Theorem I1 I+(2) = M(2) Lemma For c ε C, I+(2) and M(2) each contain the element jc given by jc(z) = 1/(z-c), z ≠ c, ∞, jc(c) = ∞, jc(∞) = 1. Proof To see that it is in I+(2), observe that jc = i1or0ot-c. The result for M(2) is clear from the formula for jc. reflection in the real axis: r0(z)=z*, z ≠ ∞, r0(∞)=∞ translation : t-c(z)=z-c, z ≠ ∞ t-c(∞)=∞ Proof of Theorem I1 We show that each of I+(2) and M(2) is generated by the elements of S+(2) and the jc, so the groups are equal. From the ∞ Theorem S(2) is the subgroup of I(2) which fixes ∞ Taking just the direct elements of each, we have S+(2) is the subgroup of I+(2) which fixes ∞. Now suppose that tεI+(2) is such that t(∞) = c ≠ ∞. As jc(c) = ∞, s = jcot fixes ∞. Thus, by the previous paragraph, sεS+(2). Hence, since t = jc-1os, I+(2) is generated by S+(2) and the jc. For M(2), we have a similar argument. For the first part, since m with m(z) = (αz+β)/(γz+δ) fixes ∞ if and only if γ = 0. Then m(z) = (αz+β)/(δ) = (α/δ)z+(β/δ) εS(2). Since each element of S+(2) arises in this way, S+(2) is the subgroup of M(2) which fixes ∞. Finally, just as for I+(2), we can show that M(2) is generated by S+(2) and the jc.