Proof of Lemma 1

Lemma 1
Suppose that t is an isometry which fixes the points O(0,0) and X(1,0). Then either
  1. t is the identity, or
  2. t = rx (reflection in the x-axis).

Let P be the point (x,y), and suppose that t(P) = Q, the point (u,v). As t fixes O, and is an isometry, |OP| = |OQ|.
Squaring, we get x2 + y2 = u2 + v2.
Similarly, |XP| = |XQ|, so (x-1)2 + y2 = (u-1)2 + v2.
Combining these, we get -2x + 1 = -2u + 1, so that u = x.

Let Y be the point (0,1), and let Y' = t(Y). Since t is an isometry, |OY'| = 1.
As OY is perpendicular to OX and t fixes OX, OY' is perpendicular to OX.
Thus Y' is on the y-axis.
As |OY'| = 1, Y' = Y or Y- = (0,-1).

In the first case, we have |PY| = |QY|, so, as above, x2 + (y-1)2 = u2 + (v-1)2,
and we deduce that v = y.
Thus, in this case (u,v = (x,y), so t is the identity.

In the second case, we have |PY| = |QY-|, and we deduce that v = -y.
Now we have (u,v) = (x,-y), the reflection of (x,y) in the x-axis.

euclidean group page