What Are the Limitations of Rigid Motion in a Hilbert Plane?

[dismo]

Working in a Hilbert plane, show that any rigid motion that fixes at least three noncollinear points must be the identity.

I am certain that I can claim that:
(i) any translation of the plane will fix none of the points
(ii) any rotation will fix a single point
(iii) any reflection will fix only the points on the line about which the plane is reflected

The trouble is I don't know how to prove that no composition of these could fix only three points in the plane...

Where do I go next?

 

[VKint]

I'm unfamiliar with the axioms of a Hilbert plane. If this question were posed in the context of an ordinary Euclidean plane, however, this is how I would approach it. Rigid motions preserve distances (i.e., the distance between points f(a) and f(b) is the same as that between a and b, where f is a rigid motion). Given a two points A and B and a third point X not on \overline{AB}, there exists exactly one other point Y \neq X such that d(A,Y) = d(A,X) and d(B,Y) = d(B,X) (where d(\; , \;) denotes the distance function). Furthermore, Y = r_{AB}(X), the reflection of X about \overline{AB}. If X is on \overline{AB}, then X is the unique point in the plane satisfying these equations. (Neither of these assertions is hard to prove.)

Let f be a rigid motion fixing the three noncollinear points A,B,C. From the last observation, we know that f fixes all of \overline{AB}, \overline{AC}, and \overline{BC}. Let X be a point not on any of these lines. Suppose f(X) \neq X; then, by distance conservation, we must have f(X) = r_{AB}(X) = r_{AC}(X) = r_{BC}(X) simultaneously, a contradiction since (by assumption) \overline{AB} \neq \overline{AC} \neq \overline{BC}.

 

[Doom of Doom]

A Hilbert Plane is just a Euclidean Plane, but without the Parallel Axiom and the Circle–Circle Intersection Property.

So, yeah, your proof works.