Galois Groups and intermediate field extensions I'm stuck in the structure of Galois group of a field extension over $\mathbb{Q}$.
I have the polynomial $fg=(x^{6}-4)(x^{4}-3)\in\mathbb{Q}[x]$. Is easy to find that the extension over $\mathbb{Q}$ given by this polynomial is $E=\mathbb{Q}(\sqrt[3]{2},i,\sqrt[4]{3})$. So $[E:\mathbb{Q}]=24$; as this extension is normal and separable we have that $[E:\mathbb{Q}]=|G=Gal(E:\mathbb{Q})|=24$. 
Justify that exist 3 intermediate fields between $\mathbb{Q}$ and $E$ is not hard (because by Sylow's Theorem if $s_{2}=1$ then exists $H\triangleleft G$, and so $E^{H}|\mathbb{Q}$ is normal of degree 3, but the only subextension with this property is $\mathbb{Q}(\sqrt[3]{2}:\mathbb{Q})$ that is not normal.)
Justify that exists only one $3$-Sylow subgroup and determinate the fixed field $E^{H}$ is where the problems start to me. I know that by Sylow's Theorem $s_{3}$ (where this is the 3-Sylow subgroups of G) should be 1 or 4. I don't know how to discriminate $s_{3}=4$, because I don't know what this implies (all that comes to my mind is that there are only one extension of order 8, $\mathbb{Q}(i,\sqrt[4]{3})$. I'm not sure if this reasoning is correct.)
And the last thing that I don't know: justify the existence of a normal subgroup $K\subset G$ so $E^{K}=E_{f}\cap E_{g}$ and that $G/K$ is abelian. What comes to my mind is that this is related to the structure of $G$ (it can be written as the direct product of two subgroups).
My try: $E_{f}\cap E_{g}=\mathbb{Q}(\sqrt{3}i)$. So I have to find one $K$ that $E^{K}=\mathbb{Q}(\sqrt{3}i)$ and that $G/K$ is abelian.
Any hint to continue is appreciated.
Thanks for all the answers!
 A: For justifying the existence of a unique Sylow-$3$ subgroup, note that such a subgroup is unique if and only if it is normal, and a normal Sylow-$3$ subgroup would correspond to a Galois subextension $ K/\mathbf Q $ with degree $ 8 $. You already know a degree $ 8 $ subextension of $ E/\mathbf Q $, namely $ \mathbf Q(\sqrt[4]{3}, i)/\mathbf Q  $, so all you need to show is that this subextension is Galois. This trivially follows from it being the splitting field of $ X^4 - 3 $.
For the last problem, you know that $ [E_f : \mathbf Q] = 6 $ and $ [E_g : \mathbf Q] = 8 $, and yet their compositum has degree $ 24 $. This means $ [E_f \cap E_g : \mathbf Q] = 2 $, and in fact armed with this result you can determine $ E_f \cap E_g = \mathbf Q(\sqrt{-3}) $. The subgroup $ K \subset G $ corresponding to this subfield has all of the required properties.
Since the extensions $ E_f $ and $ E_g $ become linearly disjoint over $ E_f \cap E_g $, you can in fact see from this result that there is a short exact sequence 
$$ 0 \to C_3 \times C_2 \times C_2 \to G \to C_2 \to 0 $$
A: Your first justification is fine.
For the second one we in fact have $E_f \cap E_g = \mathbb{Q}(\sqrt{-3})$. Now $E / \mathbb{Q}(i)$ is a Galois extension with $\lvert \text{Gal}(E/\mathbb{Q}(\sqrt{-3})) \rvert = [E : \mathbb{Q}(\sqrt{-3}) ] = 12$ and so $K:=\text{Gal}(E/\mathbb{Q}(\sqrt{-3})) \leq G$ is a subgroup of index $2$ and hence it is normal in $G$. The quotient $G/K$ has order $2$, so it is abelian.
However, this does not mean that $G$ is a direct product of $K$ and $G/K$. For example, the dihedral group $D_{6}$ has a subgroup $H =\{1, r, r^2\}$, but is not a direct product of $H$ and $D_6/H$ - it is not even abelian.
