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.