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Anyone who studies Permutation Groups will be encountering the following definition:

A group $G$ acting on a set $\Omega$ is said to be “Sharply m-Transitive” iff $$\forall (a_1,a_2…,a_m) , (b_1,b_2…,b_m) \in \Omega^{m};\ ∃! g \in G , a_i^g=b_i, 1\leq i\leq m$$

While reviewing my written notes in the class about group $PGL_2(q)$, I have faced to this matter that $PGL_2(q)$ acting on set $\Omega=GF(q)\cup\{\infty\}$ is sharply $3-$transitive. The idea for judging that is as follows:

Since

$PGL_2(q)=\{f|f:\Omega\longrightarrow\Omega, f(z)=\frac{az+b}{cz+d},ad-bc\neq 0; a,b,c,d\in GF(q)\}$ so we have

$PGL_2(q)_{\infty}=\{f|f:\Omega\longrightarrow\Omega, f(z)= az+b ,a\neq 0; a,b\in GF(q)\}$,

$PGL_2(q)_{\infty,0}=\{f|f:\Omega\longrightarrow\Omega, f(z)= az ,a\neq 0; a\in GF(q)\}$ and

$PGL_2(q)_{\infty,0,1}=\{f|f:\Omega\longrightarrow\Omega, f(z)= z\}=\{id\}$.

Knowing that $|PGL_2(q)|=q(q^2-1)$, we get $| PGL_2(q)_{\infty}|=q(q-1)$ and then

$| PGL_2(q)_{\infty,0}|=q-1$.Now, since $|PGL_2(q):PGL_2(q)_{\infty}|=q+1=|\Omega|$ then $PGL_2(q)$ is

acting transitively on $\Omega$; and because of having $|PGL_2(q))_{\infty,0}:PGL_2(q)_{\infty}|=q=|\Omega-\{\infty\}|$

therefore $PGL_2(q)$ is acting $2-$transitively on $\Omega$. Finally, it was concluded in the class that

since $|PGL_2(q)_{\infty,0,1}|=1$; $|PGL_2(q)|$ is acting 3-transitively on $\Omega$. I should confess, I can't

reach to the last result and above definition is hopeless for me. My question is "Why the group is

sharply 3-transitive on $\Omega$. Thanks and sorry for my long question here.

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Your definition of sharply m-transitive is a little bit off: you need to assume the $a_i$ are distinct, and the $b_i$ are distinct. In other words you cannot have $a_1 =a_2$ (unless also $b_1=b_2$). –  Jack Schmidt Jun 20 '12 at 17:38
    
If you can send (∞,0,1) to any (x,y,z) then you can send (a1,a2,a3) to (∞,0,1) and then to (b1,b2,b3). You can just give a formula for the (∞,0,1) to (x,y,z) case. –  Jack Schmidt Jun 20 '12 at 17:40
    
@JackSchmidt: Yes that is right. we kave $a_i\neq a_j$ when $i\neq j$. The same is held for $b_i$. Thanks. –  B. S. Jun 21 '12 at 5:50

2 Answers 2

up vote 5 down vote accepted

To show that a group is acting sharply $3$-transitively, you need to show that it acts $3$-transitively and that the stabilizer of one (and hence of any) triplet of distinct points is trivial. Since you know that $|PGL_2(q)_{\infty,0,1}|=1$, you need to show just that the action is $3$-transitive. For that, it is sufficient to show that the action of the stabilizer of two points, i.e. $PGL_2(q)_{\infty,0}$, is transitive on $\Omega\setminus\{\infty,0\}$. Indeed, if $x,y\in GF(q)\setminus\{0\}$, then take $a=yx^{-1}$ and you have $f(z)=az$ such that $f(x)=y$.

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Thanks Dennis for your answer. Indeed, you made me criteria for what I was looking for. Thanks again. –  B. S. Jun 20 '12 at 17:42

Just to point out, this is just basic algebraic manipulation. You solve a few systems of linear equations. The inclusion of infinity requires a little projective fiddling, but it is just a few extra cases to check.

If $f(z) = \frac{az+b}{cz+d}$ then $f(\infty) = \frac{a}{c}$, $f(0) = \frac{b}{d}$ and $f(1) = \frac{a+b}{c+d}$. Hence you are solving a system of three linear equations in four variables.

If $x,y,z \neq \infty$, set $a=x$, $b=\frac{xy-yz}{z-y}$, $c=1$, and $d=\frac{x-z}{z-y}$.

If $x=\infty$, set $c=0$ and solve for the rest. If $y=\infty$, set $d=0$ and solve for the rest. If $z=\infty$, set $c=-d$ and solve for the rest.

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Thanks for noting me that. –  B. S. Jun 20 '12 at 18:14

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