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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. –  Qiaochu Yuan Jun 26 '12 at 7:13
    
@QiaochuYuan: But for the first part $0\neq ad-bc=k^2$ and for the second it isn't. What is confusing me? –  B. S. Jun 26 '12 at 7:16
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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 $\mathbb{F}_{q^2}$. –  Qiaochu Yuan Jun 26 '12 at 7:21
    
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. –  Jyrki Lahtonen Jun 26 '12 at 7:57

1 Answer 1

up vote 1 down vote accepted

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) –  Jack Schmidt Jun 26 '12 at 14:16
    
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. –  B. S. Jun 26 '12 at 14:17
    
@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. –  Jack Schmidt Jun 26 '12 at 14:20
    
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. –  Jack Schmidt Jun 26 '12 at 14:23
    
$q^2$ is an odd wherein $q=p^f$ with $p$ is an odd prime number and $f=|Aut(GF(q^2))|$ –  B. S. Jun 26 '12 at 14:24

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