The question is Exercise 3.89 from Advanced Modern Algebra, Rotman.
Stochastic group $\sum (2,\mathbb{F}_4) < \text{GL}_2(\mathbb{F}_4)$ is defined as following:
$\sum (2,\mathbb{F}_4) = \left\{ \begin{pmatrix} a & b \\ 1-a & 1-b \end{pmatrix} \bigg| a,b \in \mathbb{F}_4, a \neq b \right\}$
$A_4$ is the alternating subgroup of $S_4$
My attempt:
Based on Exercise 3.19 of the same book. I proved $\sum (2,\mathbb{K}) \approx \text{Aff}(1,\mathbb{K})$ where the affine group $\text{Aff}(1,\mathbb{K}) < \text{GL}_2(\mathbb{K})$ is defined by:
$\text{Aff}(1,\mathbb{K}) = \left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \bigg| a,b \in \mathbb{K}, a\neq 0 \right\}$
and the isomorphism $\varphi$ is defined by: $\varphi: \begin{pmatrix} a & b \\ 1-a & 1-b \end{pmatrix} \mapsto \begin{pmatrix} a-b & b \\ 0 & 1 \end{pmatrix}$
Then I aimed to form an isomorphism $\phi: A_4 \to \text{Aff}(1, \mathbb{F}_4)$. In fact, I got a concrete function as the following:
$\phi((1)) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \phi((12)(34)) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \phi((13)(24)) = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}, \phi((14)(23)) = \begin{pmatrix} 1 & a^2 \\ 0 & 1 \end{pmatrix}$ $\phi((123)) = \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}, \phi((243)) = \begin{pmatrix} a & 1 \\ 0 & 1 \end{pmatrix}, \phi((134)) = \begin{pmatrix} a & a \\ 0 & 1 \end{pmatrix}, \phi((142)) = \begin{pmatrix} a & a^2 \\ 0 & 1 \end{pmatrix}$ $\phi((132)) = \begin{pmatrix} a^2 & 0 \\ 0 & 1 \end{pmatrix}, \phi((234)) = \begin{pmatrix} a^2 & a^2 \\ 0 & 1 \end{pmatrix}, \phi((143)) = \begin{pmatrix} a^2 & 1 \\ 0 & 1 \end{pmatrix}, \phi((124)) = \begin{pmatrix} a^2 & a \\ 0 & 1 \end{pmatrix}$
My questions are
Is there any easier method to solve this problem?
Can we further generalize this conclusion to something like $\sum (2,\mathbb{F}_{p^n}) \approx A_{p^n}$
Thank you all