Proof of Theorem I1

Theorem I1
I+(2) = M(2)

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.
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 ≠ ∞,
  • translation :
    t-c(z)=z-c, z ≠ ∞
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.

return to inversive pages