Take the <parabolic element of PSL2(R)>
$$
A=\begin{pmatrix}1&1\\0&1\end{pmatrix},
\qquad z\longmapsto z+1.
$$
It fixes no point in $\mathbb H$. A <Geodesic in the Poincare half-plane model> is determined by its unordered pair of ideal endpoints: either two real numbers or one real number together with $\infty$. Translation adds one to every finite endpoint, so no such pair is invariant. Therefore \b[this $A$ preserves neither a point nor a hyperbolic line in $\mathbb H$.]

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 14G / d / Solution

{title2=$|\operatorname{tr}A|=2,\quad A\ne\pm I$}

= Parabolic Möbius transformation
{synonym}

A nonidentity real <Möbius transformation> represented by $A\in\mathrm{SL}_2(\mathbb R)$ with $|\operatorname{tr}A|=2$. It has one ideal boundary fixed point and is conjugate to a nonzero translation on the <Poincare half-plane model>.

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 Incoming links: Parabolic element of PSL2(R)