A book to study about hyperbolic plane, hyperbolic translations, etc. In this paper, page $6$, the authors state the following:

The translations of the hyperbolic plane are defined as products of
  two central symmetries; the set of hyperbolic translations forms a
  sharply transitive set on the hyperbolic plane, the associated loop is
  the classical simple Bruck loop.

I would like to have a refence to study about the boldfaced sentences in the text above.
I would appreciate your help.
 A: These are  deep waters. 
Central symmetries are defined on the first page of
T. Banakh, A. Dudko, D. Repovš
http://new.math.uiuc.edu/math402/public/models/htranslation.html 
http://en.wikipedia.org/wiki/Square_root_of_a_matrix#Square_roots_of_positive_operators 
http://en.wikipedia.org/wiki/Bol_loop 
Using the Poincare disk model, a hyperbolic translation is the Möbius transformation given by matrix
$$ A \; = \;  
 \left(  \begin{array}{rr}
  1 & \alpha  \\
   \bar{\alpha}  & 1  
\end{array} 
  \right)  ,
  $$
or
$$ T_\alpha(z) = \frac{z + \alpha}{\bar{\alpha}z + 1 } \; \; ,  $$
where $\alpha$ is a complex number with $|\alpha| < 1.$ The transformation does take the unit circle to itself as along as $\alpha$ is not itself on the unit circle. However, we also need $T_\alpha(0)$ to be inside the disk, and  $T_\alpha(0) = \alpha.$ 
The matrix for $T_\alpha \circ T_\beta$ has matrix
$$  
 \left(  \begin{array}{rr}
  1 + \alpha \bar{\beta} & \alpha + \beta  \\
   \bar{\alpha} +  \bar{\beta} & 1  +  \bar{\alpha} \beta
\end{array} 
  \right)  .
  $$
This is simply not Hermitian, and other things go wrong.
Given two positive Hermitian matrices,
$$ A \; = \;  
 \left(  \begin{array}{rr}
  1 & \alpha  \\
   \bar{\alpha}  & 1  
\end{array} 
  \right)  ,
  $$
and
$$ B \; = \;  
 \left(  \begin{array}{rr}
  1 & \beta  \\
   \bar{\beta}  & 1  
\end{array} 
  \right)  ,
  $$
as you see above $AB$ or $BA$ are not Hermitian, so the classical Bruck loop operation is
$$  \sqrt{B A^2 B},$$ where the square root is the (unique) positive Hermitian matrix $H$ such that $ H^2 = B A^2 B.$ So, the operation is not commutative or associative, but it satisfies a Bol identity and behaves well as far as inverses. 
EDIT, Saturday, March 24, 12:23 Pacific time: got it, very hard. We get, with the above $A,B$ as written,
$$ B A^2 B =  \left(  \begin{array}{rr}
  | 1 + \bar{\alpha} \beta |^2 + |\alpha + \beta|^2 & 2 (\alpha + \beta) (1 + \bar{\alpha} \beta)  \\
  2 (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})   &   | 1 + \bar{\alpha} \beta |^2 + |\alpha + \beta|^2 
\end{array} 
  \right).   $$
Now, as matrices, we get
$$ \sqrt{B A^2 B} =  \left(  \begin{array}{cc}
  | 1 + \bar{\alpha} \beta |  &  \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ | 1 + \bar{\alpha} \beta |}  \\
   \frac{  (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{ | 1 + \bar{\alpha} \beta |}    &   | 1 + \bar{\alpha} \beta | 
\end{array} 
  \right).   $$
However, as a Möbius transformation, we may divide all four entries by the real number $  | 1 + \bar{\alpha} \beta |,$ and call that transformation $A * B.$ That is,
$$ A * B =  \left(  \begin{array}{cc}
  1  &  \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ | 1 + \bar{\alpha} \beta |^2}  \\
   \frac{  (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{ | 1 + \bar{\alpha} \beta |^2}    &   1 
\end{array}  \right)
   $$ 
or
$$ A * B =  \left(  \begin{array}{cc}
  1  &  \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ ( 1 + \alpha \bar{\beta} ) ( 1 + \bar{\alpha} \beta )    }  \\
   \frac{  (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{( 1 + \bar{\alpha} \beta ) ( 1 + \alpha \bar{\beta} ) }    &   1 
\end{array}  \right)
   $$ 
or
$$ A * B =  \left(  \begin{array}{cc}
  1  &  \frac{ \alpha + \beta }{  1 + \alpha \bar{\beta}      }  \\
   \frac{  \bar{\alpha} + \bar{\beta}}{ 1 + \bar{\alpha} \beta   }    &   1 
\end{array}  \right).
   $$ 
We finally have what Grishkov and Nagy write, using Möbius transformations with both diagonal elements equal to $1$ and the matrix Hermitian in any case, everything is determined by a complex number of modulus less than $1$ in the upper right corner. Put together, we have
$$ A \; = \;  
 \left(  \begin{array}{rr}
  1 & \alpha  \\
   \bar{\alpha}  & 1  
\end{array} 
  \right)  ,
  $$
and
$$ B \; = \;  
 \left(  \begin{array}{rr}
  1 & \beta  \\
   \bar{\beta}  & 1  
\end{array} 
  \right)  ,
  $$
then
$$ A * B =  \left(  \begin{array}{cc}
  1  &  \frac{ \alpha + \beta }{  1 + \alpha \bar{\beta}      }  \\
   \frac{  \bar{\alpha} + \bar{\beta}}{ 1 + \bar{\alpha} \beta   }    &   1 
\end{array}  \right).
   $$ 
With $x= \alpha, y = \beta,$ we  see the Grishkov and Nagy formula at the bottom of page 6, 
$$  x \cdot y = \frac{x + y}{1 + x \bar{y}} = \frac{ \alpha + \beta }{  1 + \alpha \bar{\beta}      }.$$ 
