Let $\displaystyle E $ be an algebraic extension of a field $\displaystyle F $. Let $\displaystyle S = \{\sigma_{i}: i \in I \} $ be a collection of automorphisms of $\displaystyle E $ such that every $\displaystyle \sigma_i $ leaves each element of $\displaystyle F $ fixed. Show that if $\displaystyle S $ generates a subgroup $\displaystyle H $ of $\displaystyle G(E/F) $, then $\displaystyle E_S = E_H $.

The group of automorphisms of a field is cyclic and has the Frobenius automorphism as a natural generator. So $\displaystyle H $ is cyclic. Thus the elements left fixed by $\displaystyle S $ are the same as the elements left fixed by $\displaystyle H $?