Suppose $T \in \mathrm{Isom}^{+}(\mathbb{H})$ . With out loss of generality we may assume that $T$ fixes two points $P,P′$ on the imaginary axis $i\mathbb{R}$. Now let $Q \in \mathbb{H}$. Since $d(P,Q) = d(P,T(Q))$, the point $T (Q)$ lies on the circle radius $d(P, Q)$ centre $P$ . Likewise $T (Q)$ lies on the circle radius $d(P′,Q)$ centre $P′$. Since these two circles are also Euclidean circles, they intersect in precisely two points. (One on each side of the imaginary axis.) Clearly, one of these two points is $Q$.
Now it is told to me that since $T$ preserves orientation that intact $T$ it must be that $T(Q) = Q$. But I cannot realise this truth.
Any help would be much appreciated! Thanks!
Here $\mathbb{H}$ is the hyperbolic plane with standard metric, and $\mathrm{Aut}(\mathbb{H})$ is the group of orientation preserving conformal bijections from $\mathbb{H}$ to itself.
$\mathrm{Isom}^{+}(\mathbb{H})$ is the group of orientation preserving isometries of the hyperbolic plane. Where by orientation preserving I mean that $f \in \mathrm{Isom}^{+}(\mathbb{H})$ if it is a differentiable function when viewed as a map from $\mathbb{R}^2 \mapsto \mathbb{R}^2$, and the determinant of this derivative is non-zero everywhere.