Suppose $p$ is a prime. Show that every element of $GL_2(F_p)$ has order dividing either $p^2 — 1$ or $p(p — 1)$.
My thoughts - Can assume that an element of $GL_2(F_p)$ has one of the canonical forms:
- Different roots so two jordan blocks.
- Repeated Root with either one or two jordan blocks.
Where can I go from there? Not sure what I can say about the order of the matrices.
If the characteristic polynomial has roots in $F_p$, then you are in one of the situations described.
If $J =\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ then $J^{p-1}=\begin{bmatrix} a^{p-1} & 0 \\ 0 & b^{p-1} \end{bmatrix}=I_2$.
If $J=\begin{bmatrix} a & 1 \\ 0 & a \end{bmatrix}$ prove that $J^p$ is diagonal, and hence $J^{p(p-1)}=I_2$.
The other situation is if the characteristic polynomial doesn't have roots in $F_p$. Since the characteristic polynomial is quadratic, there exists an extension $F_p \hookrightarrow K$ of degree 2 where the characteristic polynomial has a root, and hence two roots.
Show that the two roots are different (hint: derivative) and hence, in this algebraic extension the Jordan form is $$$J =\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix} \Rightarrow \\J^{p^2-1}=\begin{bmatrix} \alpha^{p^2-1} & 0 \\ 0 & \beta^{p^2-1} \end{bmatrix}=I_2 \,.$$