Asking about $M(q^2)$ and its order I am doing some handy calculation for showing that $M(q^2)$ is a group acting on set $\Omega =GF(q^2)∪\{\infty\}$ $3-$transitively wherein $q^2$ is odd in the way Dennis Gulko showed. So I need the order of  $M(q^2)$.  $M(q^2)$ has two parts:

$M(q^2)=\{f|f:z\rightarrow\frac{az+b}{cz+d},0\neq ad-bc=k^2\}∪\{f|f:z\rightarrow\frac{az^q+b}{cz^q+d},0\neq ad-bc\neq k^2\}$

and $a,b,c,d\in GF(q^2), z\in\Omega$. I know the first part as $PSL_2(q^2)$ and its order is $\frac{1}{2}q^2(q^4-1)$. But I am stumped about the second part. According to my class notes; it has no certain name (?) and its order is the same as the first part(?). I also wrote that $M(q^2)$ is a subgroup of $P\Gamma L_2(q^2)$. I am asking kindly about bold line above. Thanks for you time.
 A: The second part, $$
\newcommand{PYL}{\operatorname{P\Gamma L}}
\newcommand{PEL}{\operatorname{P\Sigma L}}
\newcommand{PGL}{\operatorname{PGL}}
\newcommand{PSL}{\operatorname{PSL}}
\newcommand{Aut}{\operatorname{Aut}}
\newcommand{Alt}{\operatorname{Alt}}
\newcommand{Sym}{\operatorname{Sym}}
\left\{ f \in \PYL(2,q^2) ~\middle|~~f:z\mapsto\frac{az^q+b}{cz^q+d},0\neq ad-bc\neq k^2\right\}$$
is not a group, just a coset.  Let's pretend $q$ is prime so that I don't have to make up any non-standard notation.  Divide $\PYL(2,q^2)$ into cosets over $\PSL(2,q^2)$.  The cosets have representatives $1$, the Frobenius automorphism $\sigma:z\mapsto z^q$, the diagonal element $\tau:z \mapsto \zeta z$ where $\zeta$ is a primitive $q+1$st root of unity, and of course $\sigma\tau$ the combination of the last two.
The group $\PYL(2,q^2)/\PSL(2,q^2)$ is elementary abelian of order 4, and so it has three non-identity proper subgroups, each generated by a single element: $\sigma$, $\tau$, or $\sigma\tau$.
Using the lattice isomorphism theorem (subgroups of a quotient correspond to subgroups of the original containing the kernel), we get the following subgroups:
$$\begin{array}{rcl}
\PEL(2,q^2) & = & \PSL(2,q^2) \cup \sigma\PSL(2,q^2) \\
\PGL(2,q^2) & = & \PSL(2,q^2) \cup \tau\PSL(2,q^2) \\
M(q^2)      & = & \PSL(2,q^2) \cup \sigma\tau\PSL(2,q^2), \text{ as well as} \\
\PSL(2,q^2) & = & \PSL(2,q^2) \text{ and } \\ 
\PYL(2,q^2) & = & \PSL(2,q^2) \cup \sigma\PSL(2,q^2) \cup \tau\PSL(2,q^2) \cup \sigma\tau\PSL(2,q^2)
\end{array}$$
When $q=3$ we get a particularly important version of this that you'll want to know about at some point:
$$\begin{array}{rcl}
\PSL(2,9) & \cong & \Alt(6) \text{ the alternating group of degree 6 } \\
\PEL(2,9) & \cong & \Sym(6) \text{ the symmetric group of degree 6 } \\
\PGL(2,9) & = & \PGL(2,9) \\
M(9) & \cong & M_{10} \text{ the Mathieu group of degree 10 } \\
\PYL(2,9) & \cong & \Aut( \Alt(6) ) \text{ the automorphism group of the alternating and symmetric groups }
\end{array}$$
I believe the "M" is not an abbreviation for "Mathieu". Huppert–Blackburn (Vol 3, XI.1.3 p. 163) does not name it.
