\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Let $\ell$ be a hyperbolic line in ${\bf H}$, and let $m$ and $n$
be elements of ${\rm M\ddot{o}b}({\bf H})$ satisfying $m(\ell) =
\ell = n(\ell)$.  Prove or give a counter-example to the following
claim: $m$ and $n$ commute; that is, $m\circ n =n\circ m$.
\medskip

{\bf Answer}

Consider $\ell$ the positive imaginary axis, so that
stab$_{\rm{M\ddot{o}b}(\bf{H})}(\ell)=G_{\ell}$ is generated by
$\ell_{a}(z)=az\ (a>0), k(z)=\ds\frac{-1}{z}$ and $B(z)=-\bar{z}$.

\begin{itemize}
\item
$\ell_a \circ B(z)=-a\bar{z}$ and $B \circ \ell_{a}(z)=-a\bar{z}$
(since $a \in \bf{R}$) and so $\ell_a, B$ commute
\item
$\ell_a \circ k(z)=\ds\frac{-a}{z}$ and $k \circ
\ell_{a}(z)=\ds\frac{-1}{(az)} \ne \ell_{a}\circ k(z)$
\end{itemize}

and so $G_{\ell}$ is not a commutative group.

\end{document}
