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Let $E$ be a Galois extension of $F$ with Galois group $G$, and let $L$ be the fixed field of a subgroup $H$ of $G$. Show the automorphism group of $L/F$ is $N/H$ where $N$ is the normalizer of $H$ in $G$.

In my opinion, consider the map $Gal(E/E^N)=N\rightarrow Aut(L|F)$ with element $f$ $\rightarrow$$f|_L$, I can see $N/H$ isomorphic with the image in $Aut(L|F)$. But I don't know how to prove it is a surjection.

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Take $g$ in $Aut(L|F)$, and suppose $g$ doesn't fix $E^N$. Extend it to $\sigma:E\rightarrow E$ and let $P$ be the group generated by $<\sigma, Gal(E/E^N)>$. $H$ is normal in $P$ because the extension $L/E^P$ is normal, but that is impossible because $N$ is properly contained in $P$.

This should help you to conclude your result.

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