Proving property of Fibonacci sequence with group theory

Call the group $G = \left\{A \in (\mathbb{Z}/p\mathbb{Z})^{2\times 2} \ \middle\vert\ \det A = \pm 1\right\}$.

First, we count the order of $G$. We have elements $\left(\array{a & b \\ c & d}\right)$, with $a,b,c,d \in \mathbb{Z}/p\mathbb{Z}$ and $ad - bc \equiv \pm 1$ ($\equiv$ means congruent modulo $p$). Let us first count the elements with determinant one. $\mathbb{Z}/p\mathbb{Z}$ is a field, which means that every element except $0$ has a multiplicative inverse. Therefore, $ad - bc \equiv 1$ is equivalent to $(d \equiv 0\ \land\ c \equiv -b^{-1})\ \lor\ (a \not\equiv 0 \land d \equiv a^{-1}(1 + bc))$. Counting the possibilities gives $$ \underbrace{(d \equiv 0\ \land\ c \equiv -b^{-1})}_{\text{$a$: $p$ choices, $b$: $p - 1$ choices}}\ \lor\ \underbrace{(a \not\equiv 0 \land d \equiv a^{-1}(1 + bc))}_{\text{$a$: $p - 1$ choices, $b,c$: $p$ choices}} $$ for a total of $p(p-1) + (p-1)p^2 = p(p+1)(p-1) = p(p^2 - 1)$ elements with determinant $1$. By symmetry (swapping $a$ with $b$ and $d$ with $c$) there has to be an equal number of elements with determinant $-1$. So the order of $G$ is $|G| = 2p(p^2 - 1)$.

Now, let $M = \left(\array{1 & 1 \\ 1 & 0}\right)$. You already know that $M^n = \left(\array{F_{n+1} & F_{n} \\ F_{n} & F_{n-1}}\right)$. For $M$, as for any element in a finite group, we know that $M^{|G|} = I$ (consider the cyclic subgroup $\left<M\right>$ generated by $M$ and use Lagrange's theorem). This fact translates to $$ \left(\array{F_{2p(p^2 - 1) + 1} & F_{2p(p^2 - 1)} \\ F_{2p(p^2 - 1)} & F_{2p(p^2 - 1) - 1}}\right) \equiv \left(\array{1 & 0 \\ 0 & 1}\right) \mod{p}.$$ The off-diagonal components of this equation are what you wanted to show.

The matrix $\left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]$ has the property that its powers encode Fibonacci numbers. It lives in the group of $2 \times 2$ matrices $\bmod p$ with determinant $\pm 1$ (a certain subgroup of $GL_2(\mathbb{F}_p)$). If you count the size of this group...