Galois group of $(x^3-2)(x^2-2)$ over $\mathbb{Q}$

I want to find the Galois group over $\mathbb{Q}$ of $(x^3-2)(x^2-2).$

I already know that the Galois group over $\mathbb{Q}$ of $f(x) = x^3-2$ is isomorphic to $S_3$ and there are four intermediate fields between the splitting field of $f$ and $\mathbb{Q}$: $\mathbb{Q}(\sqrt{2}),\,\mathbb{Q}(\omega),\,\mathbb{Q}(\omega\sqrt{2}),\,\mathbb{Q}(\omega^2\sqrt{2}),$ where $\omega$ is a primitive $3^\mathrm{rd}$ root of unity. It's easy to verify that $\sqrt{2}$ does not belong in one of these intermediate fields, so by the fundamental theorem, $\mathbb{Q}(\sqrt{2})\not\subset\mathbb{Q}(\sqrt{2},\omega).$ Now we have that $$[\mathbb{Q}(\sqrt{2},\omega,\sqrt{2}) : \mathbb{Q}]=12.$$ This is where I'm stuck. I don't know how to conclude what the Galois group of the polynomial is, other than it is some subgroup of $S_6$ of order $12$.

• First, show that your extension field is a splitting field of $x^6-2$ over $\mathbb{Q}$. Then see this: math.stackexchange.com/questions/1575214/… – Sungjin Kim Aug 5 '16 at 22:50
• Actually you should be able to realize this as a subgroup of $S_5$ because your polynomial only has five zeros. Anyway, the imaginary part of the eighth root of unity found a nice way of doing it. Also, this question is quite similar to this. It doesn't seem to matter, which square root you adjoin in addition to the cube roots (ok, $\sqrt{-3}$ would be an exception). – Jyrki Lahtonen Aug 5 '16 at 23:11

Well, you know what the generating automorphisms of the Galois group are, namely $\sigma, \tau, \rho$ which act as follows on the generators of the splitting field: $$\sigma(\sqrt{2}) = \omega\sqrt{2}; \sigma(\omega) = \omega; \sigma(\sqrt{2}) = \sqrt{2}$$ $$\tau(\sqrt{2}) = \sqrt{2}; \tau(\omega) = \omega^{2}; \tau(\sqrt{2}) = \sqrt{2}$$ $$\rho(\sqrt{2}) = \sqrt{2}; \rho(\omega) = \omega; \rho(\sqrt{2}) = -\sqrt{2}$$ Now, we compute $$(\sigma\rho)(\sqrt{2}) = \omega\sqrt{2}; (\sigma\rho)(\omega) = \omega; (\sigma\rho)(\sqrt{2}) = -\sqrt{2}$$ It is easy to see that $\sigma\rho$ has order $6$ (why?). Further, $(\sigma\rho)^{-1} = \rho^{-1}\sigma^{-1} = \rho\sigma^{2}$. Finally, note that $\sigma$ and $\rho$ commute, as well as $\tau$ and $\rho$, and $\sigma, \tau$ obey the relations $\sigma\tau = \tau\sigma^{2}$. Now, your group $G$ cannot be abelian, and must contain $S_{3}$. A good candidate for $G$ is thus $D_{6}$, which is nonabelian of order $12$ and contains an isomorphic copy of $D_{3} \cong S_{3}$. Let $\beta = \sigma\rho$. Then $$\beta\tau = \sigma\rho\tau = \sigma\tau\rho = \tau\sigma^{2}\rho = \tau\rho\sigma^{2} = \tau\beta^{-1}$$ Furthermore, $\beta^{4} = \sigma$, $\beta^{3} = \rho$, so $G$ has the presentation $\langle \beta, \tau \mid \beta^{6} = \tau^{2} = e, \beta\tau = \tau\beta^{-1}\rangle \cong D_{6}$.