Order of an orbit of Frobenius action on a algebraically closed field of characteristic p

Question

Consider the action of the Frobenius homomorphism $F^{2}:\,\overline{\mathbb{F}_{q}}\rightarrow\overline{\mathbb{F}_{q}},\,x\rightarrow x^{q^{2}}$ over $\overline{\mathbb{F}_{q}}$ . Let $s=\left\{ s_{1},\ldots,s_{k}\right\}$ be an orbit of this action such that $s=\tilde{s}$ , where $\tilde{s}:=\left\{ s_{1}^{-q},\ldots,s_{k}^{-q}\right\}$ . What about the order of the orbit $s$ ? I suspect that it has to be odd. Thanks in advance for any suggestions or helps.

Answer 1

Your hunch is correct.

Assume that $s$ is the orbit of the element $t\in\overline{\Bbb{F}_q}$. The assumption implies that $t^{-q}\in s$, so there exists an integer $m$ such that $$ t^{-q}=(F^2)^m(t)=t^{q^{2m}}. $$ Raising both sides of this to power $q^{2m}$ shows that $$ \begin{aligned} t^{q^{4m}}&=(t^{-q})^{q^{2m}}\\ &=(t^{q^{2m}})^{-q}\\ &=(t^{-q})^{-q}\\ &=t^{q^2}. \end{aligned} $$ Therefore $t^{q^{4m-2}}=t$. If we denote $\Phi=F^2$ this means that the iterate $\Phi^{2m-1}(t)=t$. Therefore the size of the orbit $s$ is a factor of $2m-1$, i.e. an odd integer.