This is a standard fact, part of the Fundamental Theorem of Galois Theory.
Assuming separability throughout, let’s take this as a definition of $E$ being Galois over a field $k$: that every $k$-morphism $\varphi$ of $E$ into a separably closed extension $\Omega$ of $E$ actually sends $E$ into $E$.
So let’s suppose that $F_2$ is Galois over $F_0$ and that $\text{Gal}(F_2/F_1)
$ is a normal subgroup of $\text{Gal}(F_2/F_0)$. Now let $\varphi$ be an $F_0$-morphism of $F_1$ into $\Omega$. It can be extended to $\bar\varphi\colon F_2\to\Omega$, by a standard theorem, and by the hypothesis that $F_2$ is Galois over $F_0$, it sends $F_2$ to $F_2$, and thus is an element of $\text{Gal}(F_2/F_0)$. I want to show that $\bar\varphi(F_1)\subset F_1$, by showing that any element $\bar\varphi(x))$ of this field is fixed under $\text{Gal}(F_2/F_1)$. Now let $g$ be any element of this group. I want $g(\bar\varphi(x))=\bar\varphi(x)$. But by normality of the subgroup, $g\circ\bar\varphi=\bar\varphi\circ g'$, for an (perhaps other) element $g'$ of the subgroup. But since $g'$ leaves $x$ fixed, we have $g(\bar\varphi(x))=\bar\varphi(x))$, as desired.
This argument is probably unnecessarily wordy; I’m sure you can improve it.
$\newcommand{Gal}{\operatorname{Gal}}$Preliminaries:
- $\operatorname{irr}(\zeta, \Bbb Q) = \Phi_{12}(X) = X^4-X^2+1$
- $\zeta = \dfrac{\sqrt3+i}2$
- $\sqrt3 = \zeta + \zeta^{-1}$
- $i = \zeta + \zeta^5$
- $\Gal(\Bbb Q(\zeta)/\Bbb Q) = C_2 \times C_2$
- Subfields are $\Bbb Q$, $\Bbb Q(\sqrt3)$, $\Bbb Q(i)$, $\Bbb Q(\sqrt{-3})$, $\Bbb Q(\zeta)$.
We wish to determine $\Gal(\Bbb Q(\zeta, \alpha) / \Bbb Q(\zeta))$, so we wish to factorize $X^{12}-2$ in $\Bbb Q(\zeta)$. We claim that it is irreducible.
Since $\Bbb Q(\zeta)$ has all the $12$th roots of unity, and $\Bbb Z[\zeta]$ is a Euclidean domain, it suffices to show that $2$ is neither a square nor a cube.
It is clear that $2$ is not a cube, since $N_{\Bbb Q(\zeta)/\Bbb Q}(2) = 16$ is not a cube.
That $2$ is not a square, i.e. $\sqrt 2 \notin \Bbb Q(\zeta)$, is clear, since none of the quadratic subfields of $\Bbb Q(\zeta)$ contain $\sqrt 2$.
Since $X^{12} - 2 \in \Bbb Q(\zeta)[X]$ is irreducible, we can know that $[\Bbb Q(\zeta,\alpha) : \Bbb Q(\zeta)] = 12$ and $\Gal(\Bbb Q(\zeta,\alpha) : \Bbb Q(\zeta)) = C_{12}$. Thus we have an exact sequence:
$$1 \to \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q(\zeta)) \to \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q) \to \Gal(\Bbb Q(\zeta)/\Bbb Q) \to 1$$
i.e.
$$1 \to C_{12} \to \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q) \to C_2 \times C_2 \to 1$$
Now $G = \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q) = \langle \rho, \sigma, \tau \rangle$, where:
- $\rho(\zeta) = \zeta$, $\rho(\alpha) = \zeta \alpha$
- $\sigma(\zeta) = \zeta^5$, $\sigma(\alpha) = \alpha$
- $\tau(\zeta) = \zeta^{11}$, $\tau(\alpha) = \alpha$
It is clear that $N = \langle \rho \rangle$ is the normal subgroup $\Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q(\zeta))$ with $H = \langle \sigma, \tau \rangle$ being its complement, thus $G = N \rtimes H$.
To determine the action, we need to compute $\sigma \rho \sigma^{-1}$ and $\tau \rho \tau^{-1}$:
- $\sigma \rho \sigma^{-1} \alpha = \sigma \rho \alpha = \sigma (\zeta \alpha) = \zeta^5 \alpha$
- $\tau \rho \tau^{-1} \alpha = \tau \rho \alpha = \tau (\zeta \alpha) = \zeta^{11} \alpha$
So $\sigma \rho \sigma^{-1} = \rho^5$ and $\tau \rho \tau^{-1} = \rho^{11}$, and that is the group action that determines the semidirect product. The group presentation is:
$$G = \langle \rho, \sigma, \tau \mid \rho^{12} = \sigma^2 = \tau^2 = (\sigma \tau)^2 = 1, \sigma \rho \sigma^{-1} = \rho^5, \tau \rho \tau^{-1} = \rho^{11} \rangle$$
And another way of presenting the group is $C_{12} \rtimes \operatorname{Aut}(C_{12})$ with the identity homomorphism.
The problem with your attempt is that, while every element of $G$ must send $\zeta$ and $\alpha$ to one of their respective conjugates, not every such pair of conjugates must correspond to an element of $G$.
In other words, we have a set-theoretic injection $G \to \{\text{conjugates of $\zeta$}\} \times \{\text{conjugates of $\alpha$}\}$ which you withou justification assumed to be surjective.
Best Answer
Note that $\tau^2$ maps $\eta$ to $\eta^4=\eta^{-1}$. So you might want to consider $\eta+\eta^{-1}$, which is contained in $\mathbb{Q}(\eta)$, and is fixed by both $\tau^2$ and $\sigma$. Now it's just a question of determining if it is the complete fixed field, or just a subfield thereof. But this should get you started.