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**dori1123** Let $\displaystyle G \leq Gal(K/F)$ be a subgroup of the Galois group of the extension $\displaystyle K/F$ and suppose $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$ are generators for $\displaystyle G$. Show that the subfield $\displaystyle E/F$ is fixed by $\displaystyle G$ if and only if it is fixed by the generators $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$.

I think the forward direction is clear. Since $\displaystyle G$ fixes $\displaystyle E$ and since the generators $\displaystyle \sigma_1, \sigma_2, ..., \sigma_n$ are in $\displaystyle G$, then these generators must fix $\displaystyle E$.

I am not sure how to prove the backward direction. Some help please.