Here is the proof of $E^H /F$ is a normal extension $\Leftrightarrow$ $H\lhd Gal(E/F)$.(Please zoom in to see the second picture)
I cannot understand $2$ of the lines of the proof:
$1$:
$2$:
I cannot understand(in the second picture) why it is the fact that $(E^H)^{G/H}=E^G=F$ (here we regard $G/H$ as the image of $G/H$ in $Aut_F(E^H)$) and why is the fact that the degree of $E^{\sigma H\sigma^{-1}}$ is the same as $E^H$.
May I please ask for a more explicit explanation? Thanks a lot!
There is what I get from the comments :
Let $E/N/F$ be a field extension such that $E/F$ is Galois. $N/F$ is a normal extension iff for any $\alpha \in E$ and $\sigma \in Aut(E/F)$, if $\alpha \in N$ then $\sigma(\alpha) \in N$.
Let $G = Aut(E/F)$, $H$ a subgroup, $E^H$ its fixed field.
If $E^H/F$ is normal then for every $\sigma \in G, h \in H$, we have $\alpha \in E^H$ iff $\sigma(\alpha) \in E^H$ so that $h(\sigma(\alpha))=\sigma(\alpha)$. Thus $\alpha =\sigma^{-1}(h(\sigma(\alpha)))$ and $\alpha \in E^{\sigma^{-1} H \sigma}$ iff $\alpha \in E^H$. By the Galois correspondence, $E^{\sigma^{-1} H \sigma} = E^H \implies H = \sigma^{-1} H \sigma$ and $H$ is a normal subgroup of $G$.
Conversely, if $H$ is a normal subgroup of $G$ then $H = \sigma^{-1} H \sigma$ so that $\alpha \in E^H$ iff $\sigma(\alpha) \in E^H$ and hence $E^H/F$ is normal.