Order of $ g= \big(\begin{smallmatrix} \ 1 & 1 \\ 1 & 0 \end{smallmatrix}\big)\in GL_2(\mathbb F_3)\;. $

Question

Let\begin{align*} g= \begin{pmatrix} \ 1 & 1 \\ 1 & 0 \end{pmatrix}\in GL_2(\mathbb F_3)\;. \end{align*} Its minimal polynomial is $P_g(X)=X^2-X-1$ which divides $X^8-1$ in $\mathbb F_3[X]$, hence $g^8=1$ and so $\operatorname{ord}(g)\le8$. How can I prove that $\operatorname{ord}(g)$ is exactly equal to $8$?

Thank you all!

Answer 1

Just compute $g^{4}$, and see that it is different from $1$.

Or, better still, note that $x^{2} - x - 1$ does not divide $x^{4} - 1 = (x-1)(x+1)(x^{2} + 1)$.

This is of course because if $g^{8} = 1$, then the order of $g$ divides $8$.