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I know that the Galois group $f(t)=(t^3-2)(t^3-3)$ over $\mathbb{Q}$ is $Gal(\mathbb{Q}(2^{\frac{1}{3}},3^{\frac{1}{3}},\xi)/\mathbb{Q})$ (where $\xi$ is the 3rd root of unity) and its order is 18. I also know that this group should be some non-abelian subgroup of $S_3 \times S_3$. (here, $S_n$ is the symmetric group)

But I don't know the exact shape of this Galois group.

How can we know this group?

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    $\begingroup$ Take an elementary abelian group of order 9, then add an element of order two, that acts via inversion. $\endgroup$ – Steve D Mar 7 '16 at 5:59
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Given that the Galois group acts on the roots this gives us intution for defining automorphisms: let $\sigma, \tau, \eta : \mathbb{Q}(2^{1/3},3^{1/3},\xi) \to \mathbb{Q}(2^{1/3},3^{1/3},\xi)$, $\sigma(2^{1/3}) = \xi2^{1/3}$, $\tau(3^{1/3}) = \xi3^{1/3}$, $\eta(\xi) = \xi^2$. By composing such maps we generate a group of order $18$ so this must be the whole group. Its easy to see this group is not abelian (as you are aware).

$\sigma$ and $\tau$ generate an abelian subgroup of order $9$, which you know will be normal (index $2$). You can construct an exact sequence $$\langle \sigma, \tau \rangle \to \textrm{Gal}(\mathbb{Q}(2^{1/3},3^{1/3},\xi)/\mathbb{Q}) \to \langle \eta \rangle$$ and from here you'll be able to exhibit the Galois group as a direct product or a semi-direct product (hint: direct product of abeian groups is abelian).

Here's a more mechanical approach. By the Sylow theorems and other tricks from Group theory we can identify all non-abelian groups of order 18: $$D_{18}, \hspace{1mm} S_3 \times \mathbb{Z}/3\mathbb{Z}, \hspace{1mm} (\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}$$ Now look at the orders of the elements in each group and it should be clear which group you can identify the Galois group with.

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  • $\begingroup$ Thank tou for your kind reply. From your remark, I am comvincing that the Galois group should be $(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}$. Because the exact sequence is splitting exact sequence. But I am also wondering whether many of Galois group arise the semi direct product like in this way when the Galois extension admit some normal sub extension. Anyway, thank you for your answering! $\endgroup$ – user29422 Mar 8 '16 at 5:35

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