Theorem H2
Any tεH(2) may be expressed as the composite of at most three hinversions.
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 r_{L} be reflection in L.
Then r_{L} maps O to O and X to K, so that r^{1}or_{L}
fixes O and X.
Since the composite is direct, Lemma 1 of the euclidean group page shows that
r^{1}or_{L} is the identity, so r = r_{L}.
Note that, if L is through O, then reflection in L extends to an hinversion,
which we also refer to as r_{L}.


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 hinversion 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 = r_{L}.
Now we have t = s^{1}ou = sor_{L},
(as s^{1} = s and u = r_{L})
so t is the composite of two hinversions.
Now suppose that t is indirect, and let r_{0} denote the hinversion
corresponding to reflection in the real axis.
Then v = r_{0}ot is direct, so is a composite of two inversions,
and hence t = r_{0}^{1}ov = r_{0}ov can be written as a composite of three.

Note. We often use the fact
that an hinversion h has
order 2, so that h^{1} = h.
