Galois Group of $(x^3-5)(x^2-3)$ I am having some trouble calculating the Galois group (over $\mathbb{Q}$) of $(x^3-5)(x^2-3)$. I can see the splitting field is $F:=\mathbb{Q}(\sqrt[3]{5},\omega,\sqrt{3})=\mathbb{Q}(\sqrt[3]{5},i,\sqrt{3})$, where $\omega$ is a primitive 3rd root of unity, and it has degree 12 over $\mathbb{Q}$. Since the extension is Galois (it is a splitting field extension for a separable polynomial), this makes $|\operatorname{Gal}(F/\mathbb{Q})|=12$ so it is either $S_3 \times S_2$ or $A_4$ as these are the only subgroups of of $S_5$ of order 12. It has a transposition (complex conjugation) so I conclude that it is $S_3 \times S_2$. I found all the subgroups of $S_3 \times S_2$, (all 16 of them), but I can't see to match them up with the intermediate fields. For example, there are three subgroups of order 6 in $S_3 \times S_2$ which should correspond to three field extensions of degree 2. One of them should be $\mathbb{Q}(i)$ since this is an intermediate field with a degree 2 minimal polynomial, but none of the subgroups of order 6 I found have every element fix $i$, so this is a problem. (A subgroup of order 6 has an element of order 3, which should be permuting the roots of $x^3-5$, no?)
Have I gone wrong somewhere? I would appreciate the help.
 A: It may help you to have a general remark on when to recognize that a Galois group splits as a direct product.
Suppose that $E/F$ is Galois with Galois group $G$, and that $G$ factors as a product $G = G_1 \times G_2$.  Then by the Galois correspondence we have Galois
subextensions $E_i/F$ contained in $E/F$, where $E_i = E^{G_i}$.  The fact that
$G_1$ and $G_2$ generate $G$ says that $E_1 \cap E_2 = E^G = F.$  The fact
that $G_1 \cap G_2 = \{1\}$ says that $E_1$ and $E_2$ together generate $E$
(in standard, if slightly flowery, terminology, we say that $E$ is the compositum of $E_1$ and $E_2$).
So to show that $Gal(E/F)$ is a product, we have to write $E$ as the compositum
of two subfields $E_1$ and $E_2$, each Galois over $F$, such  that $E_1\cap E_2 = F$.
In your case $E = \mathbb Q(5^{1/3}, \omega,\sqrt{3})$ can be written as the compositum of $E_1 = \mathbb Q(5^{1/3},\omega)$ and $E_2  = \mathbb Q(\sqrt{3})$.  This gives the product struture on $G$.
A: Let's use the following numbering on the roots of your polynomial: $$\{z_1=\root3\of5,z_2=ωz_1,z_3=ω^2z_1,z_4=\sqrt3,z_5=−\sqrt3\}.$$ Any automorphism $\sigma$ of the splitting field must permute these numbers as they are the zeros of your polynomial. As the splitting field $F$ is generated by them, the automorphism $\sigma$ is fully determined once we know $\sigma(z_j),j=1,2,3,4,5.$ This gives us the usual way of identifying $\sigma$ with an element of $S_5=\operatorname{Sym}(\{z_1,z_2,z_3,z_4,z_5\})$.
But there are further constraints. The numbers $z_1,z_2,z_3$ are zeros of the factor $x^3-5$. As that factor has rational coefficients, all automorphisms must permute these three roots among themselves. Similarly for the remaining pair $z_4,z_5$ as they are the zeros of $x^2-3$. Therefore in the identification of an automorphism with a permutation in $S_5$ of the previous paragraph only the permutations in $\textrm{Sym}(\{z_1,z_2,z_3\})\times \textrm{Sym}(\{z_4,z_5\})$ are allowed. There are 12 such permutations forming a group isomorphic to $S_3\times S_2$. As you had established by other means that $[F:\mathbb{Q}]=12$, we can conclude that all such permutations come from actual automorphisms, and thus the Galois group is isomorphic to $G=S_3\times S_2$.
Let us first calculate the fixed field of the subgroup $H_1=\langle\sigma\rangle$, where $\sigma=(123)(45)$. Here we are given that $\sigma(z_1)=z_2$ and that $\sigma(z_2)=z_3$. As $\sigma$ is an automorphism of fields we get
$$
\sigma(\omega)=\sigma\left(\frac{z_2}{z_1}\right)=\frac{\sigma(z_2)}{\sigma(z_1)}=\frac{\omega^2z_1}{\omega z_1}=\omega.
$$
Therefore $\omega$ is fixed by $\sigma$, and therefore also by any power of $\sigma$.
Thus $\mathbb{Q}(\omega)\subseteq Inv(H_1)$. From Galois theory we know that $[Inv(H_1):\mathbb{Q}]=|G|/|H|=12/6=2$. As $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ it follows that $\mathbb{Q}(\omega)=Inv(H_1)$.
It is fairly clear that the fixed field of the subgroup $H_2\simeq S_3$ that keeps both $z_4$ and $z_5$ fixed is $Inv(H_2)=\mathbb{Q}(\sqrt3)$. 
Let's try the last subgroup $H_3$ of order 6. 
I describe it as follows. We have a homomorphism $f$ from $S_3=Sym(\{z_1,z_2,z_3\})$ to
$S_2=\textrm{Sym}(\{z_4,z_5\})$ that maps a permutation $\alpha\in S_3$ to the identity element $(4)(5)$ (resp. to the 2-cycle $(45)$) according to whether $\alpha$ is an even (resp. odd) permutation. Then
$$
H_3=\{(\alpha,f(\alpha))\in S_3\times S_2\mid \alpha\in S_3\}.
$$
We easily see that $H_3$ is generated by $\beta=(123)$ and $\gamma=(12)(45)$.
Because $\beta$ acts on the set $\{z_1,z_2,z_3\}$ the same way as $\sigma$ above, we see that $\beta(\omega)=\omega$ and also that $$\beta(\sqrt{-3})=\beta(2\omega+1)=2\omega+1=\sqrt{-3}.$$ Because $\beta(z_4)=z_4$, we obviously also have $\beta(\sqrt3)=\sqrt3$. What about $\gamma$? First we get
$$
\gamma(\omega)=\gamma(\frac{z_2}{z_1})=\frac{\gamma(z_2)}{\gamma(z_1)}=\frac{z_1}{z_2}=\omega^2.
$$
As $2\omega=-1+\sqrt{-3}$ and $2\omega^2=-1-\sqrt{-3}$, this implies that $\gamma(\sqrt{-3})=-\sqrt{-3}$.
But we also have
$$
\gamma(\sqrt3)=\gamma(z_4)=z_5=-\sqrt3.
$$
Putting all these bits together we see that
$$
\gamma(i)=\gamma\left(\frac{\sqrt{-3}}{\sqrt3}\right)=\frac{-\sqrt{-3}}{-\sqrt3}=i.
$$
Similarly we see that $\beta(i)=i$. We can then conclude that $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.
