Let $F_{q}$ be a finite field of order $q=p^{r}$ ($p$ odd) and let $V$ be a $3$-dimensional vector space over $F_{q}$. Consider the subgroup $\Omega(3,q)$ of $SO(3,q)$., where we are picking the standard basis $e,f,d$ which satisfies $(e,e)=(f,f)=(e,d)=(f,d)=0$ and $(e,f)=(d,d)=1$.
I would like to show that for any element $A\in \Omega(3,q)$ $$sp(A)=sp(A^{p}),$$ where $sp$ denotes the spinor norm and $A^{P}$ denotes raising each entry of $A$ to the $p$-th power.
I have tried to write $A$ as a product of reflections but I am not sure how to write $A^{p}$ as a product of reflections using that so that I can then deduce the spinor norm of $A^{p}$. Any help with this method, or another method would be appreciated.
Thanks!
Maybe I'm missing something, but since the spinor norm is a morphism of groups $$ \mathrm{sp}\colon \mathrm{SO}(3,q)\rightarrow \mathbf F_q^\star/\mathbf F_q^{\star2} $$ one has $$ \mathrm{sp}(A^p)=\mathrm{sp}(A)^p=\mathrm{sp}(A) $$ in $\mathbf F_q^\star/\mathbf F_q^{\star2}$, since $p$ is odd.