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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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    The condition on the coefficients $a, b, c, d$ is exactly the same on the second part as on the first part.2012-06-26
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    @QiaochuYuan: But for the first part $0\neq ad-bc=k^2$ and for the second it isn't. What is confusing me?2012-06-26
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    Ah, I misread slightly. This is not a problem, though; for a fixed value of $ad - bc$ the number of matrices is the same (they are cosets of $\text{SL}_2$ in $\text{GL}_2$) so it remains to verify that the number of nonzero squares is equal to the number of nonzero non-squares, which follows from the description of the multiplicative group of $\mathbb{F}_{q^2}$.2012-06-26
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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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    Is M related to Zassenhaus's notation for near fields $\mathfrak{M}$? Maybe $M(q^2)$ is the group associated to the projective plane over the Dickson near field $\mathfrak{M}=K_{q^2,2}$? (using Zassenhaus's notation)2012-06-26
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    Shouldn't $...\cup \sigma\PSL(2,q^2) \cup \sigma\tau\PSL(2,q^2)$ be $...\cup \tau\PSL(2,q^2) \cup \sigma\tau\PSL(2,q^2)$. I am reflecting why I thought that the second part is a group. New insight came to my mind. Thanks Jack for the answer. Thanks.2012-06-26
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    @Babak: yup, fixed. no problem. $M(q^2)$ is standard notation. I just meant that if $q$ is not prime, then the groups $\PYL$ and $\PEL$ are much larger than the ones I describe, so I'd have to make up new notation for my smaller groups.2012-06-26
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    I think of M as "the other Zassenhaus group" (besides PGL, PSL, and the Suzuki groups). There is also a group called M from p-groups which I think of as "the other p-group with a maximal cyclic subgroup". I think the two uses of M are unrelated.2012-06-26
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    $q^2$ is an odd wherein $q=p^f$ with $p$ is an odd prime number and $f=|Aut(GF(q^2))|$2012-06-26
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    @Babak: yup, that is the standard requirement on $q$. $\PSL(2,q^2) \cup \sigma\PSL(2,q^2)$ does not have a name when $q$ is not prime, but it has many of the same basic properties as $\PEL(2,q^2)$ (and is equal when $q$ is prime).2012-06-26
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    Sorry Jack for your time, but may I ask to inform me a good book covering issues like you noted kindly here? Thanks2012-06-26
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    @Babak: sure. The discussion in Huppert-Blackburn Finite Groups vol III, Chapter IX is standard and was pretty readable for me. Dixon–Mortimer 7.6 (especially 7.6.2 and page 242, last paragraph) discusses this, but doesn't have many answers. $$ {} $$ Gorenstein's Finite Groups Chapter 13 is probably NOT useful to you. Tsuzuku's Finite Groups and Finite Geometries may be helpful (you've asked theorem 4.5.4), but only if you start to hear words like "collineation" and "finite projective plane". And if you do start to hear those words, M. Hall's Theory of Groups chapter 20 is easy but long.2012-06-26
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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