Theorem AS1
The affine symmetry group of E_{0} is E_{0}(2) = {t : t(x) = Ax, with A orthogonal}.
Proof
We know that the elements of E_{0}(2) are symmetries of E_{0}.
It remains only to show that there are no affine symmetries.
Suppose that t is an affine symmetry.
We know that affine transformations map centres to centres, so t(O) = (O).
Now let P be any point on E_{0}. Since t maps E_{0} to
E_{0}, t(P) = Q on E_{0} and so
t maps OP to OQ, a segment of the same length, i.e. t(OP) =OP = 1.
For any segment AB, there is a radius OP parallel to AB. Then, since t is affine,
it preserves the ratio of the parallel segments AB, OP. Thus,
t(AB) = AB.
Since AB is any segment, we see that t is an isometry, so tεE(2).
Since t
maps E_{0} to itself, it belongs to E_{0}(2).

