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 $E\neq \mathbb{Q}, L$.
($\mathbb{Q}\subset E\subset L$)
The generator of $E$ is $\sqrt{-15}$, the minimal polynomial of the generator is $x^2+15$ and $\mathcal{G}(L/E)=\langle \sigma\rangle$.
The generator of $E$ is $\theta=1-\rho^2+\omega(-\rho-\rho^2+\rho^4)$, the minimal polynomial of the generator is $x^3-3x^2-1$ and $\mathcal{G}(L/E)=\langle \sigma^3, \sigma\tau\rangle$.
The generators of $E$ are $\sqrt{-15},\theta$ and $\mathcal{G}(L/E)=\langle \sigma^3\rangle$.
I want to show that in the first and third case $\mathcal{G}(L/E)$ is a normal subgroup of $\mathcal{G}(L/\mathbb{Q})$ and that in the second case it is not a normal subgroup.
I have done the following:
- We have the roots of $X^2+15$ are simple, so the polynomial is separable. So, $E$ is the splitting field of a separable polynomial. Therefore, the extension $E/\mathbb{Q}$ is Galois. From a theorem we get that $\mathcal{G}(L/E)\triangleleft \mathcal{G}(L/\mathbb{Q})$.
How can we show it in the other two cases?
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EDIT:
At the second case since $\mathcal{G}(L/E)=\langle \sigma^3, \sigma\tau\rangle$, 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 $\sigma^3(\rho)=\rho^2-\rho^5$, $\sigma^3(\omega)=\omega^2$ and $\sigma\tau(\rho)=-\frac{\omega}{\rho}$, $\sigma\tau(\omega)=\omega$, right?
I tried to prove the above relations:
$$\begin{align*}\sigma^3\left (1-\rho^2+\omega(-\rho-\rho^2+\rho^4)\right )&=1-\left (\sigma^3 (\rho)\right )^2+\sigma^3(\omega)\left (-\sigma^3 (\rho)-\left ( \sigma^3(\rho)\right) ^2+\left (\sigma^3(\rho)\right )^4\right ) \\ &=1-\left (\rho^2-\rho^5\right )^2+\omega^2\left (-(\rho^2-\rho^5)-\left ( \rho^2-\rho^5\right )^2+\left (\rho^2-\rho^5\right )^4\right ) \\ & =1-\left (\rho^2-\rho^5\right )^2+\omega^2(\rho^2-\rho^5)\left (-1-\left ( \rho^2-\rho^5\right )+\left (\rho^2-\rho^5\right )^3\right ) \\ \sigma\tau \left (1-\rho^2+\omega(-\rho-\rho^2+\rho^4)\right )&=1-\left (\sigma\tau (\rho)\right )^2+\sigma\tau (\omega)\left (-\sigma\tau (\rho)-\left ( \sigma\tau (\rho)\right) ^2+\left (\sigma\tau (\rho)\right )^4\right ) \\ &=1-\left (-\frac{\omega}{\rho}\right )^2+\omega\left (\frac{\omega}{\rho}-\left (-\frac{\omega}{\rho}\right) ^2+\left (-\frac{\omega}{\rho}\right )^4\right )\\ &=1-\frac{\omega^2}{\rho^2}+\omega\left (\frac{\omega}{\rho}-\frac{\omega^2}{\rho^2}+\frac{\omega^4}{\rho^4}\right ) \\ &=1-\frac{\omega^2}{\rho^2}+\omega\left (\frac{\omega}{\rho}-\frac{\omega^2}{\rho^2}+\frac{\omega}{\rho^4}\right ) =1-\frac{\omega^2}{\rho^2}+\frac{\omega^2}{\rho}-\frac{1}{\rho^2}+\frac{\omega^2}{\rho^4} \\ &=1+\frac{\omega^2}{\rho}-\frac{1+\omega^2}{\rho^2}+\frac{\omega^2}{\rho^4} =1+\frac{\omega}{\rho^2}+\omega^2\frac{\rho^3+1}{\rho^4} \\ &=1+\frac{\omega}{\rho^2}+\omega^2\frac{\rho^6}{\rho^4} =1+\omega(\rho^5-\rho^2)^2+(-\omega-1)\rho^2 \\ &= 1+\omega(\rho^{10}-2\rho^7+\rho^4)-\omega\rho^2-\rho^2= 1+\omega(\rho^4-2\rho+\rho^4)-\omega\rho^2-\rho^2 \\ &=1+\omega(2\rho^4-2\rho)-\omega\rho^2-\rho^2=1+2\omega\rho(\rho^3-1)-\omega\rho^2-\rho^2 \end{align*}$$
How could we continue at both relations to get the desired result?