Use the Orbit-Stabiliser theorem to deduce that $g\in $ Stab$(\mathbf x)$.

Question

Let $\mathbb F_p$ be the set of (residue classes of) integers mod p, and let $$G=\{\begin{pmatrix} a & b\\c & d \end{pmatrix} : a,b,c,d \in \mathbb F_p, ad-bc \ne 0\}$$ and let $X$ be the set of 2-dimensional column vectors with entries in $\mathbb F_p$.

Let $g \in G$ be an element of order p. By considering the group action given by $\begin{pmatrix}\begin{pmatrix} a & b\\c & d \end{pmatrix}, \begin{pmatrix} x\\y \end{pmatrix}\end{pmatrix} \mapsto \begin{pmatrix} ax+by\\cx+dy \end{pmatrix}$ use the Orbit-Stabiliser theorem to show that there exists $x,y \in \mathbb F_p$, not both zero, with $$g\begin{pmatrix} x\\y \end{pmatrix}=\begin{pmatrix} x\\y \end{pmatrix}$$.

Deduce the $g$ is conjugate in G to $\begin{pmatrix} 1 & 1\\0 & 1 \end{pmatrix}$.

I got the order of $G$ as $p(p-1)^2(p+1)$ and that for $\mathbf x \ne \mathbf 0$, orb$(\mathbf x)=X \space \backslash \{\mathbf 0\}$, so by the orbit-stabiliser theorem $|$Stab$(\mathbf x)|=p(p-1)$ but I just can't see where to go from here or how to use this result at all.

Any help with this would be very useful, thank you

Answer 1

Let $g \in G$ be an element of order $p$ and let $H:= \langle g \rangle \cong \mathbb{Z}/p \mathbb{Z}$ be the subgroup generated by $g$. Set $Y := X \backslash \{0\}$. Consider the corresponding action of $H$ on $Y$ and now use the orbit stabilizer theorem.

What does it tell you about the order of each orbit/stabilizer of elements $y \in Y$ under the action of $H$?

Recall that $Y$ is a disjoint union of its orbits (under $H$). Now you may finish easily.