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Character table from Schrier-Sims Representation.

One of the paper's I that Kreher sent me that has Cameron as an author had a line that intrigued me greatly:

The group $\text{PSL}(2,q)$ is 3-homogeneous on the projective line when $q$ is a prime power congruent to 3 modulo 4. Therefore, a set of $k$-subsets of the projective line is the block set of a $3-(q+1,k,\lambda)$ design admitting $\text{PSL}(2,q)$ for some $\lambda$ if and only if it is a union of orbits of $\text{PSL}(2,q)$. This Simple observation has led different authors to use this group for constructing 3-designs, see for example [1-3,5,7,8,10]

The part that intrigued me was the part I separated from the rest, "This simple observation." If these authors are calling this observation simple, then there MUST exist an algorithm for taking an arbitrary group acting in some way on an arbitrary number of points and turning it's orbits into designs on those points with . Then I realized that's what we did last semester, but we did it with $\lambda =1$, and we had $k$ in mind before hand, it wasn't arbitrary.

Then, In design theory, Tonchev talked about how he sent people looking for the structure of an automorphism group in a given design, and I remembered that's what Kreher said we should do for the GDD's, and I realized if people have been doing this for 20 years or more, (Kramer Mesner was in '76? so 35 years?) why is there not some computerized way of taking $t-(v,k,\lambda)$ and spitting out some things about it's automorphism group?

I've been trying to read as much as I can about character theory to see where that goes, but I just thought I'd get this written down.

Homework 3.

Proposition.
Let \[ f(x)=\frac{ax+b}{cx+d}\text{ and } g(x)=\frac{Ax+B}{Cx+D} \]be elements in $\text{LF}(2,q)$ the group of linear fractionals over $\mathbb{F}_q$.
Suppose $\det f = \det g = 1$. If $g=f$, show that $a=rA$, $b=rB$, $c=rC$, and $d=rD$ where $r=\pm1$.

Proof.
By the proof of theorem 2 that stated $\text{LF}(2,q)\cong \text{PSL}(2,q)$,
\[
\Phi\colon \text{SL}(2,q)\rightarrow \text{LF}(2,q)
\]
defined such that
\[
\Phi\left(\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\right)=\frac{ax+b}{cx+d} \]
is a homomorphism. Let $\{\pm I\}=Z=\ker \Phi$ and $\alpha \in \text{SL}(2,q)$.. Then $\zeta\colon \text{SL}(2,q)/Z\rightarrow \text{LF}(2,q)$ defined as
\[
\zeta\colon \alpha Z\mapsto \Phi(\alpha)
\]
is an isomorphism by the first law of isomorphisms. $\det f = \det g = 1$ imply that
\[ \left[\begin{array}{cc}a&b\\c&d\end{array}\right],\left[\begin{array}{cc}A&B\\C&D\end{array}\right] \in \text{SL}(2,q)
\]
and $f=g$ imply that the two matrices above are in the same coset by the isomorphism of $\zeta$. Thus either $a=A,b=B,c=C,d=D$ or $a=-A, b=-B,c=-C,d=-D$, as each coset contains only two elements.