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I am studying for my Galois theory final for tomorrow (and I'm really getting burned out), I need help with the following question:

Galois group $G$ of $f(x)=(x^3-2)(x^5-1)$ over $\mathbb{Q}$.

Let $\alpha = \sqrt[3]{2}$ and let $\omega_3$ be the primitive cubed root of unity, and $\omega_5$ the primitive $5$th root of unity. The splitting field for $f$ is $E=\mathbb{Q}(\alpha,\omega_3, \omega_5)$. We know that $E:\mathbb{Q}$ is separable since $\text{char}\mathbb{Q}=0$ and that it is normal since it is a splitting field. Hence $E:\mathbb{Q}$ is Galois. So $|G|=|E:\mathbb{Q}|$. It is easy for me to show that $\text{Gal}((x^3-2)/\mathbb{Q}) \cong S_3$ and that $\text{Gal}((x^5-1)/\mathbb{Q}) \cong C_4$.

But how can I determine $\text{Gal}(f)$? How can I find the size of the extension $|E:\mathbb{Q}|$? I know it is divisible by $6$ and $4$.

Edit: I see people mentioning the result about cartesian products. I have not seen this result. Given that this is a past exam question I would be interested in a proof that does not use the result

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  • $\begingroup$ Do you know the degrees of the cyclotomic extensions? If you know that $[\Bbb{Q}(\zeta_{15}):\Bbb{Q}]=8$, and that the extension is abelian, then you are basically done. $\endgroup$ May 26, 2019 at 18:46
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    $\begingroup$ The splitting fields of $x^3-2$ and $x^5-1$ are linearly disjoint, so $G$ is the product of their Galois groups. $\endgroup$ May 26, 2019 at 18:46
  • $\begingroup$ @JyrkiLahtonen I know both of those facts. But why is that the case? $\endgroup$ May 26, 2019 at 18:47
  • $\begingroup$ @LordSharktheUnknown hm, never saw that results. The past papers I'm doing are from a different lecturer so seems like it was not covered this year. $\endgroup$ May 26, 2019 at 18:48
  • $\begingroup$ The extension generated by both $\omega_3$ and $\omega_5$ is the same extension you get by adjoining $\omega_{15}$. Do you see why? Then $8$ is coprime to $3$. That is enough to give you the degree. I first thought that you need to use the fact that the cyclotomic extension is abelian to deduce that $\root3\of2\notin\Bbb{Q}(\omega_{15})$, but that was an error. $\endgroup$ May 26, 2019 at 18:50

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The Galois group of $f=gh$ is equal to the cartesian product of the Galois group of $g$ and the Galois group of $h$ iff the splitting field of $g$ over $\mathbb{Q}$ is disjoint with the splitting field of $h$ over $\mathbb{Q}$. I.e. you have to determine whether $\mathbb{Q}(\omega_3,\sqrt[3]{2})\cap\mathbb{Q}(\omega_5)=\mathbb{Q}$. If shown, you have that Gal$(\mathbb{Q}(\omega_3,\sqrt[3]{2},\omega_5)/\mathbb{Q})\cong S_3\times C_4$.

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