Let $F$ be a field and $G$ be a finite subgroup of the multiplicative group $F^{\times}$. Then $G$ is cyclic.

A finite extension $K / F$ is simple if and only if it has only finitely many sub-extensions (i.e. there are only finitely many subfields of $K$ containing $F$).

If $K / F$ is finite and separable, then $K / F$ is simple.

Consider $F = \mathbb{F}_{2}(t_{1}, t_{2}) = \left\{ \frac{a(t_{1}, t_{2})}{b(t_{1}, t_{2})} \mid a(t_{1}, t_{2}), b(t_{1}, t_{2}) \in \mathbb{F}_{2}(t_{1}, t_{2}), b(t_{1}, t_{2}) \neq 0 \right\}$ and $K = F(\sqrt{ t_{1} }, \sqrt{ t_{2} })$. Then $K / F$ is finite and inseparable. $$ [K : F] = [F(\sqrt{ t_{1} }, \sqrt{ t_{2} }) : F(\sqrt{ t_{1} })][F(\sqrt{ t_{2} }) : F] = 2 \cdot 2 = 4. $$ Any $\alpha \in K$ can be written as $\alpha_{1} + \alpha_{2}\sqrt{ t_{1} } + \alpha_{3}\sqrt{ t_{2} } + \alpha_{4}\sqrt{ t_{1}t_{2} }$ with $\alpha_{1}, \dots, \alpha_{4} \in F$, and $$ \alpha^{2} = \alpha_{1}^{2} + \alpha_{2}^{2}t_{1} + \alpha_{3}^{2}t_{2}^{2} + \alpha_{4}^{2}t_{1}t_{2} \in F. $$ Hence, $[F(\alpha) : F] \leq 2$ and $F(\alpha) \neq K$ for any $\alpha \in F$. $K / F$ is not simple.

Since every $\sqrt{ t_{1} } + a(t_{1}, t_{2})\sqrt{ t_{2} }$ generate a degree $2$ extension of $F$ contained in $K$, with $a(t_{1}, t_{2})$ varying in $\mathbb{F}_{p}[t_{1}, t_{2}]$, these give infinitely many sub-extensions of $K / F$.

A character $\chi$ of a group $G$ with values in a field $L$ is a homomorphism from $G$ to the multiplicative group $L^{\times}$, i.e. a function $\chi : G \to L^{\times}$ such that $\chi(g_{1})\chi(g_{2}) = \chi(g_{1}g_{2})$ for all $g_{1}, g_{2} \in G$.