The Galois group of a specific polynomial $ f(x) = x^6-2x^3+2 \in Q[x] $ Hello all I was given this question in Algebra asking me to show a polynomial's Galois group has both a subgroup and quotient group isomorphic to $S_3$ the three symmetric group.
The polynomial in question over Q is $ f(x) = x^6-2x^3+2 \in Q[x] $
I know this polynomial is irreducible over Q via Eisenstein with p=2
This I know so the polynomial is irreducible over a field Q with char zero therefore it is separable, so it has a Galois group for its splitting field.
I also know this is the minimal polynomial of $ \sqrt[3]{1+i} $ over Q. I cannot seem to proceed and actually prove that the Galois group has both a subgroup and quotient groups isomorphic to $ S_3 $, I also know the Galois group is a subgroup of the six symmetric group $S_6$ but cannot use any of this information to prove the claim. Help appreciated thanks all

Getting late here, so a quick sketch (or extended hint) with steps only.


*

*Show that the zeros of $f(x)$ are the three cube roots of $1+i$ and the cube roots of $1-i$.

*Let $L$ be the splitting field of $f(x)$ over $\Bbb{Q}$. Given that the ratio $(1+i)/(1-i)$ is a primitive fourth root of unity show that $L$ contains a twelfth root of unity $\zeta$.

*If $w$ is one of the cube roots of $1+i$, show that $L=\Bbb{Q}(w,\zeta)$, and conclude that $[L:\Bbb{Q}]=12$.

*Because $L$ contains a cube root of $(1+i)$ and a cube root of $(1-i)$ it also contains a cube root of $(1+i)(1-i)=2$. Why does this imply that $L$ contains the splitting field $M$ of $x^3-2$?

*It is well known that $Gal(M/\Bbb{Q})\cong S_3$. Why does this imply that $Gal(L/\Bbb{Q})$ has a quotient group $\cong S_3$?

*Show that $L$ is the splitting field of $x^3-(1+i)$ over $K=\Bbb{Q}(i)$. Why does this imply that $Gal(L/\Bbb{Q})$ has a subgroup $\cong S_3$?


If you want to play with a realization of $Gal(L/\Bbb{Q})$ as a subgroup of $S_6$ you can use the numbering of zeros of $f(x)$:
$$
\begin{aligned}&x_1=w, &&x_2=w\zeta^4, &&x_3=w\zeta^8,\\
 &x_4=\overline{w}=w\zeta^{11},&&x_5=x_4\zeta^4=w\zeta^3,&&x_6=x_4\zeta^8=w\zeta^7.
\end{aligned}$$
The conjugates of $\zeta$ are $\zeta^k,\gcd(k,12)=1$. But there is the relation $w^3=1+i=1+\zeta^3$ that the automorphisms must respect. So if you specify the image $\sigma(w)$ under an automorphism $\sigma$ to one of the roots, then this leaves only two options for $\sigma(\zeta)$ (instead of all four).
