Any tεH(2) may be expressed as the composite of at most three h-inversions.
If |κ| = 1, then the map r(z) = κz* is a reflection in a line through 0.
Proof of Lemma
Note first that the formula for r involves z*, so r is indirect.
Let O, X, K denote the points with complex coordinates 0, 1 and κ respectively.
Observe that r maps O to O and X to K.
Let L be the bisector of angle XOK, and let rL be reflection in L.
Then rL maps O to O and X to K, so that r-1orL
fixes O and X.
Since the composite is direct, Lemma 1 of the euclidean group page
r-1orL is the identity, so r = rL.
Note that, if L is through O, then reflection in L extends to an h-inversion,
which we also refer to as rL.
Proof of Theorem H2
Suppose first that t is direct, and let γ = t(0).
By the Origin Lemma, there is an h-inversion s mapping γ to 0.
Then u = sot is indirect, and maps 0 to 0.
By Theorem H1, as u(0) = 0, u(z) = κz*, with |κ| = 1.
Then, by the Lemma, u = rL.
Now we have t = s-1ou = sorL,
(as s-1 = s and u = rL)
so t is the composite of two h-inversions.
Now suppose that t is indirect, and let r0 denote the h-inversion
corresponding to reflection in the real axis.
Then v = r0ot is direct, so is a composite of two inversions,
and hence t = r0-1ov = r0ov can be written as a composite of three.
Note. We often use the fact
that an h-inversion h has
order 2, so that h-1 = h.