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Suppose that $E/F$ is a Galois extension, and $F_1\subset F_2$ are subfields of $E/F$ which correspond in the Galois correspondence to $H_1$ and $H_2$.

If $H_2 \unlhd H_1$, then is $F_2$ Galois over $F_1$? If so, how to prove this? (bonus: how about the converse?)

Context: I am trying to find an example of fields $\mathbb Q \subset F_1 \subset F_2 \subset F_3 $ such that each extension is Galois with the exception that $F_2$ is not Galois over $\mathbb Q$. I thought it would be a nice idea to try out $F_1 = \mathbb Q(i\sqrt 2)$, $F_2 = \mathbb Q((1+i)\sqrt[4] {2})$, $F_3 = \mathbb Q(i,\sqrt[4]{2})$. These correspond to $H_1 = \langle \sigma ^2,\tau \sigma^3 \rangle$, $H_2 = \langle \tau \sigma^3 \rangle$ and $H_3 = \langle \sigma^4 \rangle$ in $\text{Gal}(\mathbb Q(i,\sqrt[8]{2})/\mathbb Q)$, where:

$$\sigma: \begin{cases} \sqrt[8]{2} \mapsto \zeta \sqrt[8]{2} \\ i \mapsto i \end{cases}, \tau: \begin{cases} \sqrt[8]{2} \mapsto \sqrt[8]{2} \\ i \mapsto -i \end{cases}$$

and $\zeta = e^{i\pi/4}$.

Now these satisfy: $F_3$ is Galois over $\mathbb Q$ (hence over everything), $H_1 \unlhd G$ (so $F_1$ Galois over $\mathbb Q$), $H_2 \not \unlhd G$ (so $F_2$ is not Galois over $\mathbb Q$). If the thing in the yellow box is true, then I'm done because $H_2 \unlhd H_1$.

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    $\begingroup$ For the example, how about $F_1=\mathbb{Q}(\sqrt{2})$, $F_2=F_1(\sqrt[4]{2})$ and $F_3=F_2(i)$ being the Galois closure of $F_2$? These are all quadratic extensions of the field below (so Galois), but $F_2/\mathbb{Q}$ is not Galois. $\endgroup$ – Matt B Aug 11 '17 at 15:31
  • $\begingroup$ @MattB of course that would be much easier. $\endgroup$ – Cauchy Aug 11 '17 at 15:56
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We have $\operatorname{Gal}(E/F_1) \cong H_1$ and $E^{H_2} = F_2$ (which does not depend on whether we consider $H_2$ as a subgroup of $H_1$ or $\operatorname{Gal}(E/F)$), so under the Galois correspondence for $E/F_1$, $F_2$ corresponds to $H_2$. Thus $H_2 \unlhd H_1$ iff $F_2/F_1$ is Galois.

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