Denote by $\mathbb{F}_q$ the finite field with $q$ elements, and denote by $\bar{\mathbb{F}}_q$ its algebraic closure. Let $G$ be an affine algebraic group over $\bar{\mathbb{F}}_q$, and let $F$ be a Frobenius morphism of $G$, which defines a $\mathbb{F}_q$-structure $G(\mathbb{F_q})$ on $G$.
QUESTION: Is the order of any element in $G$ finite?
If the answer is YES, I feel very surprised. But it seems that I can prove it as following.
Denote by $\bar{\mathbb{F}}_q[G]$ the ring of regular functions of G. Because $G$ has a $\mathbb{F}_q$-structure defined by $F$, $\bar{\mathbb{F}}_q[G]=\mathbb{F}_q[G]\otimes_{\mathbb{F}_q}\bar{\mathbb{F}}_q$. Let $a_i$ for $1\leq i\leq k$ be generators of $\mathbb{F}_q[G]$. Because there exsits some $n\in\mathbb{Z}^+$ such that $a_i(g)\in\mathbb{F}_{q^n}$ for all $i$, it follows that $F^n(g)=g$. On the other hand, $F^n$ is a Frobenius morphism attached to an $\mathbb{F}_{q^n}$-structure on $G$, and hence the set of fixed points $G^{F^n}$ forms a finite group. Now, $g$ lies in the finite group $G^{F^n}$ and hence has a finite order.