Note that $x^5 - 1 = (x - 1)(x^2 - 4x + 1)(x^2 + 5x + 1)$ in $\mathbb{Z}/19\mathbb{Z}$. Suppose $A$ is a 2 by 2 matrix with coefficients in $\mathbb{Z}/19\mathbb{Z}$ satisfying $A^5 - I = 0$. Then the minimal polynomial of $A$ divides $x^5 - 1$. As the characteristic polynomial of a 2 by 2 matrix has degree 2, the minimal polynomial (which divides the characteristic polynomial) has degree 1 or 2. So the minimal polynomial (and characteristic polynomial) of $A$ must be $x^2 - 4x + 1$ or $x^2 + 5x + 1$ (if the minimal polynomial is $x - 1$ then it is just the identity, with order 1).
If we let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we can manually compute coefficients such that its characteristic polynomial $x^2 - (a + d)x + ad - bc$ equals one of the two polynomials above. Is there a way to show that matrices with such characteristic polynomials exist without doing this?
