Find intermediate fields of $\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i)$ This is the problem I am facing:

Compute the intermediate fields of the extension $K | \mathbb{Q}(i)$
  where $K = \mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i)$ and find the
  intermediate fields $M$ such that $M | \mathbb{Q}(i)$ is a Galois
  extension.

What I have done:
I start by thinking about the grade of the extension $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(i)]$
$[\mathbb{Q}(\sqrt{3},i) \, : \, \mathbb{Q}(i)] = 2$ because the irreducible polinomial of $\sqrt{3}$ over $\mathbb{Q}(i)$ is $x^2 -3$ (since $\sqrt{3} \notin \mathbb{Q}(i)$
$[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(\sqrt{3},i)] = 3$ because the irreducible polinomial of $\sqrt[3]{2}$ over $\mathbb{Q}(\sqrt{3},i)$ is $x^3 -2$ (*)
Hence, $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(i)] =[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(\sqrt{3},i)] * [\mathbb{Q}(\sqrt{3},i) \, : \, \mathbb{Q}(i)] = 3 * 2 = 6$
Now, a $\mathbb{Q}(i) $-base for $\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i)$ is $\{ 1, \sqrt[3]{2} \, \xi , \sqrt[3]{2} \, \xi^2, \sqrt{3} ,\sqrt{3}  \sqrt[3]{2} \, \xi , \sqrt{3} \sqrt[3]{2} \, \xi^2 \}$
The non-linear elements whose images fix every $\sigma \in \textrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i))$ are:
$$ \sqrt{3} \to \pm \sqrt{3}$$
$$ \sqrt[3]{2} \to \{\sqrt[3]{2}, \sqrt[3]{2} \, \xi, \sqrt[3]{2} \, \xi ^2 \}$$
The extension $\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i)$ is a Galois extension (it is normal(splitting field of $(x^3-2)(x^2-3)$, separable (char($\mathbb{Q}(i)$) = 0) and finite) so $| \textrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i)) | = 6$ which means that all the options are valid.
I am stuck here, how can I determine the Galois group, and the normal subgroups (and then, how to extract the matching fields for those subgroups?)
(*) How to show justify this?
 A: To address the (*):
To show that $f(x) = x^3 - 2$ is irreducible over $\mathbb{Q}(\sqrt{3}, i)$, it suffices to show that none of its roots are in this field (since $\deg(f) \leq 3$).  Indeed, $\sqrt[3]{2}, \ \omega\sqrt[3]{2}$, and $\omega^2\sqrt[3]{2}$ are not elements of $\mathbb{Q}(\sqrt{3}, i)$.  If they were, we would have a tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\sqrt{3}, i)$, which would imply $\Big[ \mathbb{Q}(\sqrt{3}, i):\mathbb{Q} \Big] = \Big[ \mathbb{Q}(\sqrt{3}, i): \mathbb{Q}( \sqrt[3]{2}) \Big] \cdot \Big[ \mathbb{Q}(\sqrt[3]{2}):\mathbb{Q} \Big]$.  Contradiction!  Look at the degrees of the extensions: $3$ does not divide $4$.

To address the crux of the question:
You're close!  You've done a good bit of the work already.
Remember that the only groups of order $6$ are $\mathbb{Z}_6$ and $S_3$.  However, you've found $6$ automorphisms and none of them generate the entire group.  Therefore, it must be the case that $\operatorname{Gal}(K/\mathbb{Q}(i)) \cong S_3$.
Next, the fundamental theorem of Galois theory tells us that there is a one-to-one correspondence between subgroups of $\operatorname{Gal}(K/\mathbb{Q}(i))$ and the intermediate fields between $K$ and $\mathbb{Q}(i)$ that they fix.  Let's list the subgroups of $S_3$.  They are:

*

*$1$ copy of the trivial subgroup

*$3$ copies of $\mathbb{Z}_2$

*$1$ copy of $\mathbb{Z}_3$

*The whole group

You should check this yourself!  Hint: Lagrange's theorem tells us that the proper subgroups are of prime order, and hence are cyclic.  First, the trivial subgroup corresponds to the entire field $K$, and the whole group corresponds to the base field $\mathbb{Q}(i)$.
Next, let's call the two 'generator' automorphisms that you found $\phi$ and $\tau$ such that $\phi$ sends $\sqrt{3} \mapsto \sqrt{3}$ while fixing the other basis elements, and likewise $\tau$ sends $\sqrt[3]{2} \mapsto \sqrt[3]{2}\xi$.
You should check that the copies of $\mathbb{Z}_2$ are:

*

*$\{\text{id}, \phi \}$

*$\{\text{id}, \tau \phi \}$

*$ \{\text{id},  \tau^2 \phi \} $
And the copy of $\mathbb{Z}_3$ is $\{\text{id}, \tau, \tau^2\}$.
Some proper intermediate fields between $K$ and $\mathbb{Q}(i)$ are $\mathbb{Q}(i, \sqrt[3]{2})$, $\mathbb{Q}(i, \sqrt{3})$, $\mathbb{Q}(i, \sqrt[3]{2}\xi)$, and  $\mathbb{Q}(i, \sqrt[3]{2}\xi^2)$.  How do we know this is a complete list?  Keeping in mind that $\displaystyle \xi = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$, I'll leave it to you match the subgroups with the intermediate fields they fix.
Finally, to determine which intermediate fields are Galois over $\mathbb{Q}(i)$, simply apply the fact than an intermediate field $E$ is Galois if and only if its corresponding subgroup is a normal subgroup of $\operatorname{Gal}(K/\mathbb{Q(i)})$.  What are the normal subgroups of $S_3$?
