On computing the Galois group $\mathrm{Aut}\left(\Bbb Q (\alpha,\omega) / \Bbb Q( \omega + \omega^2 + \omega^4)\right)$ Let $\alpha:= \sqrt[7]{2},\omega:= e^{\frac{2\pi i }{7}}\in \Bbb C$. We set $E:=\Bbb Q(\alpha,\omega)$ and $B:=\Bbb Q(\omega+\omega^2+\omega^4)\leq E$. 
Thus, we have the Tower of Fields
$$\Bbb Q \leq B \leq E.$$
We can prove that $|\mathrm{Aut}(E/\Bbb Q)|=42$ and that $\theta,\sigma\in \mathrm{Aut}(E/\Bbb Q)$, given by 
$$\theta:\alpha \longmapsto \alpha \omega,\ \omega \longmapsto \omega$$
$$\sigma: \alpha \longmapsto \alpha,\ \omega \longmapsto \omega^3.$$
Question: How can we prove that $\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle$?
A first thought is to compute the Galois group $\mathrm{Aut}(E/B)$, but this seems to be extremely hard, because of $B$. Another might be to use somehow the Fundamental Theorem of Galois Theory, so to exploit the relations $[E:B]=|\mathrm{Aut}(E/B)|$ and then find the order of the subgroup $\langle \theta,\sigma^2 \rangle$.
Note that $\theta^i(\alpha)=\alpha\omega^i$ and $\sigma^j(\omega)=\omega^{3^j}$.
 A: You meant $\alpha=\sqrt[7]{2}$. Let $K=\Bbb{Q}(\omega)$.
We get $E/K/B/\Bbb{Q}$ where $E/\Bbb{Q}$ and $K/B$ are Galois.
We know that $Aut(K/\Bbb{Q})=\Bbb{Z/7Z}^\times=\langle \sigma\rangle,Aut(E/K)=\Bbb{Z/7Z}$. It is quite immediate that both mix together to give $Gal(E/\Bbb{Q})=Aff(\Bbb{Z/7Z})$ (the group of affine transformations $x\to ax+b$ which corresponds to $\omega^x\alpha\to \omega^{ax+b}\alpha$).
$B$ is a subfield of $K$ fixed by $\sigma^2$, and it is not $\Bbb{Q}$, so it is the subfield fixed by $\sigma^2$ and hence $ Aut(K/B)=\langle \sigma^2\rangle$. 
$K$ is the subfield of $E$ fixed by $\theta$.
Extend $\sigma\in Aut(K/\Bbb{Q})$ to an element $\sigma'\in Aut(E/\Bbb{Q})$. 

Then it is immediate that $B$ is the subfield of $E$ fixed by $(\sigma')^2$ and $\theta$ which means that $E/B$ is Galois with Galois group $Aut(E/B)=\langle(\sigma')^2,\theta\rangle$.

$[E:B]=[E:K][K:B]=21$ and $Aut(E/B)= \{x\to a^2 x+b\}$.
A: Below please find a few pieces that help you to onoe way of proving the claims.


*

*The calculation here reveals that $B=\Bbb{Q}(\sqrt{-7})$. Therefore $[B:\Bbb{Q}]=2$, and you know that $[E:B]=21$.

*You can easily verify that $\theta$ and $\sigma^2$ leave $B$ fixed pointwise, so they belong to $Gal(E/B)$.

*The calculation showing that $G=Gal(E/\Bbb{Q})$ is generated by $\theta$ and $\sigma$ more specifically describes the Galois group as a semidirect product
$$G\simeq C_7\rtimes C_6,$$ where the normal subgroup $\simeq C_7$ is generated by $\theta$ and the non-normal factor by $\sigma$. This follows also from the relation
$$\sigma\theta\sigma^{-1}=\theta^3$$ that you can verify by calculating their effects on $\alpha,\omega$.

*It is the easy to check that the subgroup generated by $\theta$ and $\sigma^2$ has order $21$ (see reuns's answer for a way of making that explicit), so it must be all of $Gal(E/B)$.



Extras:


*

*$\Bbb{Q}(\omega)/\Bbb{Q}$ is Galois with an abelian Galois group isomorphic to $\Bbb{Z}/7\Bbb{Z}^*\simeq C_6$ generated the restriction of $\sigma$. Therefore the intermediate field $B$ must also be Galois over $\Bbb{Q}$.

*You see that 
$$\sigma(\omega+\omega^2+\omega^4)=\omega^3+\omega^6+\omega^5=-1-(\omega+\omega^2+\omega^4)$$ implying that $\sigma$ does not fix $B$ element wise but $\sigma^2$ does. This also implies that $[B:\Bbb{Q}]=2$ without the calculation relating $B$ to $\sqrt{-7}$.

A: Taking into account Jyrki's answer, I tried to collect the pieces. Please, have a look.
(1) By the first lines of his answer: If $[B:\Bbb{Q}]=2$, then 
$$[E:\Bbb Q]=[E:B][B:\Bbb Q] \implies 42=[E:B]\cdot 2 \iff [E:B]=21=|\mathrm{Aut}(E/B)|.$$ Since $B=\Bbb Q(i\sqrt 7)=\langle1,i\sqrt 7\rangle$, we can see that $\theta(b)=\sigma^2(b)=b,\ \forall b\in B$. Hence, 
$$\theta,\sigma^2 \in \mathrm{Aut}(E/B)\implies \langle \theta,\sigma^2 \rangle \subseteq \mathrm{Aut}(E/B).$$ But also we have $|\langle \theta,\sigma^2\rangle|=|\langle \theta\rangle | \cdot |\langle \sigma^2\rangle|=7 \cdot 3 =21=|\mathrm{Aut}(E/B)|$. Therefore, $$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle.$$

(2) By the extras: We observe that 
$$\theta(\omega+\omega^2+\omega^4)=\omega+\omega^2+\omega^4,\\
\sigma^2(\omega+\omega^2+\omega^4)=\dotsb =\omega+\omega^2+\omega^4, $$ 
so, $\omega+\omega^2+\omega^4 \in \mathrm{Fix}(\langle \theta,\sigma^2 \rangle) $.
Then, we have the tower of fields 
$$\Bbb Q \leq B=\Bbb Q (\omega+\omega^2+\omega^4) \leq \mathrm{Fix}(\langle \theta,\sigma^2 \rangle).$$
But, 
\begin{align*}
|G: \langle \theta,\sigma ^2 \rangle |=[\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q] & \iff \frac{|G|}{|\langle \theta,\sigma^2\rangle|}=[\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q] \\
& \iff \frac{42}{21}=[\mathrm{Fix}(\langle \theta, \sigma^2\rangle):\Bbb Q] \\
& \iff [\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q]=2.
\end{align*}
So, $B=\Bbb Q$ or $B=\mathrm{Fix}(\langle \theta,\sigma^2\rangle)$. But $\omega+\omega^2+\omega^4 \notin \Bbb Q$. So, 
$$B=\mathrm{Fix}(\langle \theta,\sigma^2\rangle) \implies [B:\Bbb Q]=2.$$ Therefore, 
$$[E:\Bbb Q]=[E:B][B:\Bbb Q] \iff [E:B]=21 \iff |\mathrm{Aut}(E/B)|=21.$$
Now $[\underbrace{\Bbb Q(\omega+\omega^2+\omega^4)}_{B}:\Bbb Q]=2 < \infty \iff \omega+\omega^2+\omega^4$ is algebraic over $F$ and therefore it determines uniquely the basis of $B$. So, for showing that $\theta,\sigma^2 \in \mathrm{Aut}(E/B)$, it sufficies to show that $\theta,\sigma^2 $ leave $\omega+\omega^2+\omega^4$ fixed, and indeed that happens. So, $\langle \theta,\sigma^2 \rangle \subseteq \mathrm{Aut}(E/B)$. Also, like (1), $|\langle \theta,\sigma^2 \rangle|=21=|\mathrm{Aut}(E/B)|$ and eventually 
$$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2 \rangle,$$
as wanted.
