How can we compute the Galois group of a subgroup of a splitting field?

[mathmari]

Hey!

Let $\rho=\sqrt[3]{\frac{1+\sqrt{5}}{2}}$.
We have that $\rho$ is a root of $f(x)=x^6-x^3-1\in \mathbb{Q}[x]$, that is irreducible over $\mathbb{Q}$.
We have that all the roots of $f(x)$ are $\rho, \omega\rho, \omega^2\rho, -\frac{1}{\rho}, -\frac{\omega}{\rho}, -\frac{\omega^2}{\rho}$, where $\omega$ is the cubic root of $1$, $\omega\neq 1$ ($\omega^2+\omega+1=0$).
We have that the splitting field of $f(x)$ over $\mathbb{Q}$ is $L=\mathbb{Q}[\rho, \omega]$.
There are automorphisms $\sigma, \tau\in \mathcal{G}(L/\mathbb{Q})$ such that $\sigma (\rho)=-\frac{\omega}{\rho}, \sigma (\omega)=\omega^2, \tau (\rho)\rho, \tau (\omega)=\omega^2$. We have that the order of $\sigma$ is $6$ and the order of $\tau$ is $2$ and that $\tau\sigma=\sigma^5\tau$. So, $\mathcal{G}(L/\mathbb{Q})\cong D_6$.

Let $E$ be an intermediate extension of $L/\mathbb{Q}$ with $E\neq \mathbb{Q}, L$.
($\mathbb{Q}\subset E\subset L$)

We have the following:

1. The generator of $E$ is $\omega$, the minimal polynomial of the generator is $x^2+x+1$ and $\mathcal{G}(L/E)=\langle \sigma^2, \sigma\tau\rangle$.
   
   How have we found that $\mathcal{G}(L/E)=\langle \sigma^2, \sigma\tau\rangle$ ? (Wondering)
   One of the automorphisms of $\mathcal{G}(L/E)$ is the identity $id_L$.
   
   It holds that $[E:\mathbb{Q}]=\deg (x^2+x+1)=2$, right? (Wondering)
   
2. The generator of $E$ is $\theta=1+\rho-\rho^4+\omega(\rho+\rho^2-\rho^4)$, the minimal polynomial of the generator is $x^3-3x^2-1$ and $\mathcal{G}(L/E)=\langle \sigma^3, \sigma^2\tau\rangle$.
   How have we found that $\mathcal{G}(L/E)=\langle \sigma^3, \sigma^2\tau\rangle$ ? (Wondering)
   Also how can we compute $[E:\mathbb{Q}]$ without using the minimal polynomial? Maybe with the Theorem of Galois Theory that $\mathcal{G}(E/\mathbb{Q})\cong \mathcal{G}(L/E)/ \mathcal{G}(L/\mathbb{Q})$ and so $[E:\mathbb{Q}]=|\mathcal{G}(E/\mathbb{Q})|=\frac{|\mathcal{G}(L/E)|}{|\mathcal{G}(L/\mathbb{Q})|}$ ?
   We have that $|\mathcal{G}(L/\mathbb{Q})|=12$ and to find $|\mathcal{G}(L/E)|$ we have to find all the elements, or not?
   $\mathcal{G}(L/E)=\langle \sigma^3, \sigma^2\tau\rangle=\{\sigma^3, \sigma^6=id, \sigma^2\tau, (\sigma^2\tau)(\sigma^2\tau)=\dots=id\}=\{id, \sigma^3, \sigma^2\tau\}$
   So, $|\mathcal{G}(L/E)|=3$ and so $[E:\mathbb{Q}]=\frac{3}{12}=\frac{1}{4}$ ? That is wrong, isn't it? (Wondering)

 

[mathmari]

At the second case since $\mathcal{G}(L/E)=\langle \sigma^3, \sigma^2\tau\rangle$, it must hold that $\sigma^3 (1-\rho^2+\omega(-\rho-\rho^2+\rho^4))=1-\rho^2+\omega(-\rho-\rho^2+\rho^4)$ and $\sigma\tau (1-\rho^2+\omega(-\rho-\rho^2+\rho^4))=1-\rho^2+\omega(-\rho-\rho^2+\rho^4)$.

We have that and $\sigma^3(\rho)=\rho^2-\rho^5$, $\sigma^3(\omega)=\omega^2$ and $\sigma^2\tau(\rho)=\omega\rho$, $\sigma^2\tau(\omega)=\omega$, right?

I tried to prove the above relations:

$$\begin{align*}\sigma^3\left (1+\rho^2+\omega(\rho+\rho^2-\rho^4)\right )&=1+\left (\sigma^3 (\rho)\right )^2+\sigma^3(\omega)\left (\sigma^3 (\rho)+\left ( \sigma^3(\rho)\right) ^2-\left (\sigma^3(\rho)\right )^4\right ) \\ &=1+\left (\rho^2-\rho^5\right )^2+\omega^2\left ((\rho^2-\rho^5)+\left ( \rho^2-\rho^5\right )^2-\left (\rho^2-\rho^5\right )^4\right ) \\ & =1+\left (\rho^2-\rho^5\right )^2+\omega^2(\rho^2-\rho^5)\left (1+\left ( \rho^2-\rho^5\right )-\left (\rho^2-\rho^5\right )^3\right )
\\ \sigma^2\tau \left (1+\rho^2+\omega(\rho+\rho^2-\rho^4)\right )&=1+\left (\sigma^2\tau (\rho)\right )^2+\sigma^2\tau (\omega)\left (\sigma^2\tau (\rho)+\left ( \sigma^2\tau (\rho)\right) ^2-\left (\sigma^2\tau (\rho)\right )^4\right ) \\ &=1+\left (\omega\rho\right )^2+\omega\left (\omega\rho+\left (\omega\rho\right) ^2-\left (\omega\rho\right )^4\right ) \\ &=1+\omega^2\rho^2+\omega\left (\omega\rho+\omega^2\rho^2-\omega^4\rho^4\right ) \\ &=1+\omega^2\rho^2+\omega\left (\omega\rho+\omega^2\rho^2-\omega\rho^4\right ) \\ &= 1+\omega^2\rho^2+\omega^2\rho+\omega^3\rho^2-\omega^2\rho^4 \\ &=1+\omega^2\rho^2+\omega^2\rho+\rho^2-\omega^2\rho^4 \\ &=1+\rho^2+\omega^2 (\rho^2+\rho-\rho^4)\end{align*} $$

How could we continue at the first relation to get the desired result? (Wondering)