Proof of Theorem H2

 Theorem H2 Any tεH(2) may be expressed as the composite of at most three h-inversions. Lemma 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 shows that 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 Let tεH(2). 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.