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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.

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    I made your link clickable. Hopefully this is the way you wanted it to be. See the current source code for the syntax. Roll back, if you wanted something else. Qiaochu's comment fully answers your question IMHO.2012-06-26

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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.

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    Oh, I haven't read it, but Rotman's chapter 9, pages 281-286 and then 272-281 looks really clear and easy.2012-06-26