This is a question about one step in the proof for the following theorem:
Let $F$ be a field, and let $G \leq\mathrm{Aut}(E)$. If $\sigma \in \mathrm{Aut}(E)$ fixes $E^G$, then $\sigma \in G$.
The proof goes by contradiction:
Suppose $\sigma \notin G$, then $E^G$ is fixed by $G \cup \{\sigma\}$. By previous lemmas in the book, $[E:E^G] = |G| = n$, and $[E:E^{G \cup \{\sigma\}}] \geq n+1$, which contradicts $[E:E^{G \cup \{\sigma\}}] \leq [E:E^{G}]$.
I am not clear with the last step where the contradiction is derived. From my understanding, $E^{G \cup \{\sigma\}}$ is fixed by more things in $\mathrm{Aut}(E)$, so must be contained in $E^G$. Hence the tower goes like $E/E^G/E^{G \cup \{\sigma\}}$. The inequality seems to be the other way. Could anyone help with this?
You are assuming that $ \sigma $ fixes $ E^G $, so in fact $ E^G = E^{G \cup \{ \sigma \}} $. So $ [E : E^G] = [E : E^{G \cup \{ \sigma \}}] $ , and using the previously proven result in the book, we deduce that $ |G| = |G \cup \{\sigma\}| $. Hence, $ \sigma \in G $.