The Strong Affine Fixed Point Theorem
If G ≠ {e} is a finite subgroup of A(2),then either
(1) there is a unique point fixed by every element of G, or
(2) G = {e,s^{1}oros}, where r is reflection in a line, and sεA(2).
Proof
By the Affine Fixed Point Theorem, there is a point P fixed by every element of G.
Suppose that Q ≠ P is also fixed by all elements of G. Then, by Lemma 2, each
element of G fixes all points of the line L through P and Q. There is a euclidean,
and hence affine transformation s' which maps L to the line Y : y=0.
By Lemma 3, the elements of H = s'Gs'^{1} fix the points of Y.
Now suppose that h ≠ e is in H. Since h fixes all of Y, it fixes (0,0), so that
h(x,y) = (ax+by,cx+dy) for some real a,b,c,d. Now, h fixes (x,0) for all real x,
so we have (ax,cx) = (x,0) for all x. Thus, a = 1 and c = 0 and we see that
h(x,y) = (x+by,dy).
As H is conjugate to G, it has the same order, n say. Then h^{n} = e. After some
calculation, we see that h^{n}(x,y) = (x+Ky,d^{n}y) for some K. Since h^{n} = e, we
have d^{n} = 1, so d = ±1. If d = 1, then the calculation is easier, and we get
h^{n}(x,y) = (x+nby,y), so we must have b = 0. This means that h = e, which
contradicts an earlier assumption. Thus, h(x,y) = (x+by,y) for some b. It is
easy to check that such h have order 2.
Finally, suppose that h' ≠ e is also in H. As it fixes Y, it must have the form
h'(x,y) = (x+b'y,y). Then h" = hoh'has the form h"(x,y) = (x+(b'b)y,y).
But h" is in H, so h"^{n} = e. As before, we must have (b'b) = 0, so h' = h.
Thus H = {e,h}.
Now H = s'Gs'^{1}, so G = s'^{1}Hs' =
{e,s'^{1}ohos'}. If h(x,y) = (x,y), then this
is reflection in Y, and we are done. Otherwise h(x,y) = (x+by,y). The matrix A
associated with this clearly has eigenvalues 1 and 1, so there is a nonsingular
matrix M with MAM^{1} = D =diag(1,1). If we let s" be the affine transformation
s"(x) = Mx, then r(x) = s"ohos"^{1}(x) = Dx
i.e. is reflection in Y. Then we have
G = {e, (s"os')^{1}oro(s"os')} =
{e,s^{1}oros}, with s = s"os'.

