Galois group of field extension I was asked to find the Galois group of the extension $\mathbb{Q}(\sqrt[3]{2},\sqrt{2},e^{\frac{2\pi i}{3}})$. Since the degree of the minimal polynomials of $\sqrt[3]{2},\sqrt{2}$ and $e^{\frac{2\pi i}{3}} $ are not relatively prime, I can’t argue that the order of the Galois group is the product of the degree of such polynomials. I would appreciate if anyone could give me an idea on how to proceed.
Thanks.
 A: $\mathbb{Q}(\sqrt 2,\sqrt[3]{2},\zeta_{3}) = \mathbb{Q}(\sqrt 2,\sqrt[3]{2},\zeta_{3},\zeta_{2}) =\mathbb{Q}(\sqrt 2,\sqrt[3]{2},\zeta_{6}) = \mathbb{Q}(\sqrt[6]{2},\zeta_{6}) := \mathbb{F}$
Which is the splitting field of $x^6-2$ : 
Infact, if we have $\alpha  = \sqrt[6]{2}$, $\alpha^{3} = \sqrt 2 , \alpha^{2} = \sqrt[3]{2}$. 
Conversely since gcd$(2,3) = 1 \hspace{0.2cm} \exists a,b \in \mathbb{Z} : 3a+2b = 1$ which leads to $(\sqrt 2)^{a} (\sqrt[3]{2})^{b} = 2^{\frac{a}{2}+\frac{b}{3}} = 2^{\frac{3a+2b}{6}} = \sqrt[6]{2}$
So we discover that $\mathbb{Q}(\sqrt 2,\sqrt[3]{2},\zeta_{3}) = \mathbb{Q}(\sqrt[6]{2},\zeta_{6})$. 
Note $\mathbb{Q}(\sqrt[6]{2})$ has degree $6$ over $\mathbb{Q}$ since $x^{6}-2$ is irreducible thanks to Eistenstein's criterion and $\mathbb{Q}(\zeta_{6})$ over $\mathbb{Q}$ is $\phi(6) = 2$.
The total degree, i.e. the cardinality of the Galois group Gal($\mathbb{F}/\mathbb{Q}$) is $12$ indeed, because if we add $\zeta_{3}$ to $\mathbb{Q}(\sqrt[6]{2})$ we get all the splitting field $\mathbb{F}$; but $\zeta_{3} \not\in \mathbb{Q}(\sqrt[6]{2})$ since $\zeta_{3} \in \mathbb{C} \not\subset \mathbb{Q}(\sqrt[6]{2})$, 
Hence degree it at least $12$, and since it at most $12$ we have the desired result.
To conclude, thanks to fundamental theory of Galois since $\mathbb{Q}(\zeta_{6})/\mathbb{Q}$ is a normal extension Gal($\mathbb{F}/\mathbb{Q}(\zeta_{3}) \triangleleft$ Gal$(\mathbb{F}/\mathbb{Q})$ of index $2$, so it corresponds to a subgroup of order $6$ $H$ in the galois group $G$. 
Now consider $\mathbb{Q}(\sqrt[6]{2})/\mathbb{Q}$, which corresponds to a subgroup K order $2$ in the galois group since it has index $6$. 
To conclude note that $|HK| = \frac{|H| \cdot |K|}{|H| \cap |K|} = 12 = |Gal(\mathbb{F}/\mathbb{Q})|$ Since the intersection is trivial, do you see why ? (If you note carefully we already said that) and because $H$ is normal we have that  $Gal(\mathbb{F}/\mathbb{Q} \cong H \rtimes_{C_{g}} K \cong D_{6}$
Where $C_{g}$ denotes the conjugancy.
