Theorem E2


Proof Suppose that t is an isometry with t(0) =b and t(1) =g.
Suppose that s maps z to az + b.
Then s(0) = b, s(1) = a + b.

The idea is to construct a map with the same effect as t on 0 and 1.

We must check that s is an isometry. This is easy since s(z)  s(w) = (az+b)(aw+b) = a(zw) = zw. 
The case where s(z) =az*+b 
A simple calculation shows that r = s^{1}°t fixes 0 and 1. By Lemma 1, r is the identity or r_{x} (reflection in the xaxis). Hence t = s or s°r_{x}, so has the required form.
 By Lemma 1.
s^{1}°t is more or less determined. 
euclidean group 