- Let $p$ be prime.
- Let $\alpha$ be a generator of the finite field $\mathbb{F}_{p^2}$.
- So, $\mathbb{F}_{p^2}=\mathbb{F}_p[\alpha]$.
- Multiplication by $\alpha$ is an $\mathbb{F}_p$-linear operator on $\mathbb{F}_{p^2}$, when $\mathbb{F}_{p^2}$ is regarded as a $2$ dimensional vector space over $\mathbb{F}_p$ with basis $\{1,\alpha\}$.
- Let $A$ be the $2\times 2$ invertible matrix representing the multiplication by $\alpha$.
- Let $e_1,e_2$ be the standard basis for $\mathbb{F}_p^2$.
- Almost by definition, the elements $\{A^ie_1\}_i$ go over all nonzero vectors.
- So, $\mathbb{F}_{p^2}^\times/\mathbb{F}_p^\times$ acts simply transitively, through powers of the matrix $A$, on the $1$ dimensional subspaces of $\mathbb{F}_p^2$ (the projective line $P^1(\mathbb{F}_p)$).
Let's represent a $1$ dimensional subspace by a $2\times 2$ matrix of rank $1$. So, for a $1$ dimensional space represented by the matrix $L$, the action we described is given by $\alpha\cdot L= AL$
Now to my question. By some hints in a paper I'm trying to read, and by trial and error, it seems that there's another action of $\mathbb{F}_{p^2}^\times/\mathbb{F}_p^\times$ on $P^1(\mathbb{F}_p)$.
It seems that the element $\alpha$ can act on $\mathbb{F}_p^2$ by conjugation by $A$ (instead of acting by translation), and that this action is transitive.
Is it true? Why?
In other words:
Does $\{A^i\begin{pmatrix}0&0\\0&1\end{pmatrix}A^{-i}\}_{i=1}^\infty$ go over all $2 \times 2$ matrices of rank $1$ over $\mathbb{F}_p$, up to multiplication by scalars (from $\mathbb{F}_p$)?