Periodicity of power of $2\times 2$ matrix $E,A,A^2,\ldots,\bmod p$ s.t. $\det{A}=1$.

Question

I want to prove that periodicity of power of $2\times 2$ matrix $$E,A,A^2,\ldots, \bmod p$$ s.t. $\det{A}=1$ is less or equal to $2p$. Supposee $p$ is a prime number. From Cayley–Hamilton theorem, we have $$A^2=\mathrm{Tr}(A)A-I.$$ Thus, $A^n$ can be expressed as $A^n=uA+vE.$ Since $A^n\neq O$, the order of $\langle A \rangle$ is less or equal to $p^2-1$. But this is still far from $2p$. How to prove this?

Answer 1

Suppose $\alpha, \bar\alpha$ as eigenvalues.

(i) If $\alpha\neq\bar\alpha$

Then, $\mathrm{det}(A)=1\Leftrightarrow N(\alpha)=1$. By Frobenius map, $\alpha^{p+1}=N(\alpha)=1$. Thus, periodicity is less or equal to p+1.

(ii) If $\alpha=\bar\alpha$,

$N(\alpha)=1\Leftrightarrow \alpha=\pm 1$. Then, $$A=S^{-1}\begin{pmatrix} \alpha & 1 \\ 0 & \alpha \end{pmatrix}S$$ or $$A=S^{-1}\begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}S.$$

For latter case, the period is $1$ or $2$. For former case, the period is $p$ or $2p$.