Theorem N1
Suppose that P and Q are distinct points and that L is the line PQ.
(1) If gεN(2) fixes P and Q, then g = e or g = rL.
(2) If h,kεN(2) have the same effect on P and Q, then k = h or k = horL.
proof
(1) Let R be any point on L.
By (6), R is uniquely determined by n(P,R) and n(Q,R).
As g is an isometry, g(R) = R, and so g fixes L.
Now suppose that R is not on L.
The circles C, centre P, radius n(P,R), and D, centre Q, radius n(Q,R) meet
in R. By (9) they meet at most once more. But the n-reflection rL fixes
P and Q and is an isometry, so that R' = rL(R) is another intersection.
As g is also an isometry fixing P and Q, g(R) lies on C and D. Thus we see
that g(R) = R or g(R) = R' = rL(R), the only intersections.
To finish the proof we have to show that, either g(R) = R for all R, or
g(R) = R' for all R. Note that, for R on L, R = R', so we look at points which
do not lie on L.
Suppose that we can find R, S not on L with g(R) = R' and g(S) = S.
Now R and R' lie on opposite sides of L, by the definition of rL.
Then one of the segments RS, R'S cuts L as one pair {R',S}, {R,S} lie on
the same side. But g maps RS to R'S and fixes L. Thus, if RS meets L at
the point T, so does R'S, and vice versa. This contradiction completes the
proof of (1).
(2) Suppose that h,k have the same effect on P and Q.
Then h-1ok fixes P and Q. (If k(P) = P', then h(P) = P', so h-1(P') = P).
By (1) h-1ok = e or h-1ok = rL.
(2) follows by premultiplying by h.
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