Let $E/F$ be a Galois extension, and let $B$ be an intermediate field between $E$ and $F$. Let $H$ be the subgroup of $Gal(E/F)$ that maps $B$ into itself (but does not necessarily fix $B$). Prove that $H$ is the normalizer of $Gal(E/B)$ in $Gal(E/F)$.

I approached this by showing inclusion in two directions. I have managed to show that $H$ is a subgroup of the normalizer of $Gal(E/B)$ but am unsure how to show the other direction. I think I would need to show that an element of the normalizer maps $F$ into $F$, but don't know how to do this. Any suggestion is greatly appreciated! (So I would like to show that the image of F under an element of the normalizer is a subset of F)


Assuming that the extensions (and hence also the Galois groups) are all finite.

Let us denote the Galois groups by $G=\operatorname{Gal}(E/F)$ and $K=\operatorname{Gal}(E/B)$. So all the automorphisms $\tau\in K$ fix $B$ elementwise.

  1. Show that if $\sigma\in G$, then all the automorphisms in $\sigma K\sigma^{-1}$ fix the field $\sigma(B)$ elementwise.
  2. Conclude that, given $\sigma\in G$, the Galois correspondence links the subgroup $\sigma K\sigma^{-1}$ with the intermediate field $\sigma(B)$. In other words, $\operatorname{Gal}(E/\sigma(B))=\sigma K\sigma^{-1}$, and consequently also the fixed field of $\sigma K\sigma^{-1}$ is equal to $\sigma(B).$
  3. Conclude that $\sigma(B)=B$, iff $\sigma$ normalizes $K$.

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