Let $\text{GL}_2(\mathbb{F}_p)$ act on $\mathbb{F}_p^2$ ($2$-vectors with entries in $\mathbb{F}_p$) by matrix multiplication:
$$\mathbf{M}\in\text{GL}_2(\mathbb{F}_p),\;\mathbf{v}\in\mathbb{F}_p^2:\quad \mathbf{M}(\mathbf{v})=\mathbf{Mv}$$ I would like to prove the following claim:
If $\mathbf{A}\in\text{GL}_2(\mathbb{F}_p)$ has $\text{ord}\,\mathbf{A}=p$ then there exists non-zero $\mathbf{w}\in\mathbb{F}_p^2$ such that $\mathbf{A}(\mathbf{w})=\mathbf{w}$
The hint given is to use the orbit-stabiliser theorem. I think the idea is to show that for some $\mathbf{w},\;p$ cannot divide $|\text{orb}(\mathbf{w})|$ which in turns means $\mathbf{A}\in\text{stab}(\mathbf{w})$ which gives us what we want. But I cannot seem to make the initial argument. Help?
Let $G$ be the subgroup generated by $A$. Then, $G$ acts on $\mathbb{F}_{p}^{2}$. Note that the size of the orbit of each element must divide the order of $G$, which is $p$. Hence, the orbit of each element must be $p$ or $1$.
But note that $(0, 0) \in \mathbb{F}_{p}^{2}$ has a singleton orbit. Hence, we must have another orbit of size 1 because otherwise, the sum of the sizes of the orbits could not be $p^{2}$, the size of $\mathbb{F}_{p}^{2}$ (as $p$ would then divide the size of $\mathbb{F}_{p}^{2}$ but not the sum of the size of the orbits.
Hence, we must have a $w \in \mathbb{F}_{p}^{2}$ such that $G \cdot w = \{w\}$ as desired.