Application of Lang' theorem about finite groups Let $G$ be a connected algebraic group defined over the finite field $\mathbb{F}_q$, and let $F: G \to G$ be the Frobenius morphism.

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*Show that $G^F = \{g \in G\mid g = F(g)\}$ is a finite group;

*For $x\in G^F$, let $Z_G(x) = \{g\in G\mid gx = xg\}$. Show that $Z_G(x)$ is $F$-stable, i.e., $g\in Z_G(x)$ implies $F(g) \in Z_G(x)$;  and if $Z_G(x)$ is connected, $y\in G^F$ is conjugate to $x$ in $G$, then $y$ is conjugate to $x$ in $G^F$.

I have proved 1. and the first half of 2. About the 2nd half

Let $h \in G$, s.t. $hyh^{-1} = x$, hence I want to prove $\{F^n(h)\}_{n \geq 0} \cap G^F \neq \emptyset$.

But I have no idea using the connectivity of $Z_G(x)$ to prove this.
Any help will be appreciate.
 A: I’m not quite sure what the blockquoted statement is about.
Let
$$L_G:G\to G:x\mapsto x^{-1}F(x)$$
be the Lang isogeny. Lang’s theorem then says that if $G$ is smooth and connected (which I assume is what you’re assuming) then $L$ is surjective. Note that $G^F=\ker(L)$.
So, for 2. we want to show that $Z_G(x)$ is $F$-stable if $x$ is in $G^F$. But, this is simple since if
$$gx=xg$$
then applying $F$ we get
$$F(g)F(x)=F(x)F(g)$$
but $F(x)=x$ and so this says
$$F(g)x=xF(g)$$
so $F(g)$ is in $Z_G(x)$. We now want to show that if $Z_G(x)$ is connected and if $y$ in $G^F$ is conjugate to $x$ in $G$ then it’s conjugate to $x$ in $G^F$. But, note that since $y$ is conjugate to $x$ in $G$ we can write
$$x=z^{-1}yz$$
for some $z$ in $G$. But, note that then that $L_G(z)$ is in $Z_G(x)$. Indeed,
$$(z^{-1}F(z))x(F(z)^{-1}z)=z^{-1}F(zxz^{-1})z=z^{-1}F(y)z=z^{-1}yz=x$$
But, since $L_{Z_G(x)}$ is surjective this implies that we can write $L_G(z)=L_G(w)$ for some $w$ in $Z_G(x)$. Let us then note that since
$$z^{-1}F(z)=w^{-1}F(w)$$
that
$$wz^{-1}=F(wz^{-1})$$
so $wz^{-1}\in G^F$. But, note that
$$(wz^{-1})y(wz^{-1})^{-1}=wz^{-1}yzw^{-1}=wxw^{-1}=x$$
where the last equality holds since $w$ is in $Z_G(x)$. Thus, $wz^{-1}$ is in $G^F$ and conjugates $y$ to $x$ as desired.
