A Klein Page

Proof of Theorem I1

Theorem I1
I⁺(2) = M(2)

Lemma
For c ε C, I⁺(2) and M(2) each contain the element j_c given by

j_c(z) = 1/(z-c), z ≠ c, ∞,
j_c(c) = ∞,
j_c(∞) = 1.
Proof
To see that it is in I⁺(2), observe that j_c = i₁or₀ot_-c.
The result for M(2) is clear from the formula for j_c.

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 j_c, 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 j_c(c) = ∞, s = j_cot fixes ∞.
Thus, by the previous paragraph, sεS⁺(2).
Hence, since t = j_c^-1os,
I⁺(2) is generated by S⁺(2) and the j_c.
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 j_c.

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Theorem I1 I⁺(2) = M(2)

Lemma For c ε C, I⁺(2) and M(2) each contain the element j_c given by j_c(z) = 1/(z-c), z ≠ c, ∞, j_c(c) = ∞, j_c(∞) = 1. Proof To see that it is in I⁺(2), observe that j_c = i₁or₀ot_-c. The result for M(2) is clear from the formula for j_c.	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 j_c, 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 j_c(c) = ∞, s = j_cot fixes ∞. Thus, by the previous paragraph, sεS⁺(2). Hence, since t = j_c^-1os, I⁺(2) is generated by S⁺(2) and the j_c. 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 j_c.