Isomorphism type of the Galois group $f=(x^2-2)(x^3-3)$.
Let $K$ be the splitting field of $f$ over $\mathbb{Q}$.
a) Determine the degree of extension of $K$ over $\mathbb{Q}$.
b) Determine the isomorphism type of the Galois group of $K$ over $\mathbb{Q}$.
I have part (a), where I have shown that $K=\mathbb{Q}(2^{(1/2)}, 3^{(1/3)}, 3^{(1/2)}i)$ and hence the degree of extension is $12$.
I am not sure how to figure out part (b).
 A: Part of the difficulty is that $f$ is not irreducible, so a lot of the standard tricks won't work. We can solve this problem by splitting $f$ into its two irreducible subcomponents,
$$g(X) = X^2-2\\h(X) =X^3-3$$
Let $K_g$ and $K_h$ be the splitting fields of $g$ and $h$ over $\mathbb Q$ respectively. You should be able to work out the Galois groups of $K_g$ and $K_h$ as extensions of $\mathbb Q$.
Now $K$ is exactly the compositum $K_g\cdot K_h$, and we can now use the following theorem:

Theorem: Suppose that $K$ is a field, and $L,L'$ are Galois extensions of $K$. If $L\cap L' = K$, then
  $$\mathrm{Gal}(L\cdot L'/K) \cong \mathrm{Gal}(L/K) \times \mathrm{Gal}(L'/K)$$

The proof of this theorem involves showing that the restriction map
$$\mathrm{Gal}(L\cdot L'/K) \cong \mathrm{Gal}(L/K) \times \mathrm{Gal}(L'/K)\\\sigma\mapsto(\sigma\vert_L,\sigma\vert_{L'})$$
is an isomorphism. The map is certainly injective, since if $\sigma$ acts trivially on $L$ and $L'$, then it acts trivially on $L\cdot L' = K$. The fact that $L\cap L' = K$ allows us to show that it is surjective.
A: Let me proceed in a similar, but not identical, fashion to Mathmo. Let $L$ be the splitting field of the irreducible polynomial $x^{3}-3 \in \mathbb{Q}[x]$ and let $M$ be the splitting field of the irreducible polynomial $x^{2}-2 \in \mathbb{Q}[x]$. I am not sure how much Galois Theory you have covered, but ${\rm Gal}(K/L)$ and ${\rm Gal}(K/M)$ are both normal subgroups of ${\rm Gal}(K/\mathbb{Q}).$ Any automorphism of $K$ which fixes both $L$ and $M$ (elementwise), fixes every root of $f$, so is the trivial automorphism. On the other hand, by the fundamental theorem of Galois Theory ( which applies since $L$ and $M$ are each splitting fields in their own right, so Galois extensions of $\mathbb{Q}$), we have 
${\rm Gal}(M/\mathbb{Q}) \cong {\rm Gal}(K/\mathbb{Q})/{\rm Gal}(K/M)$ and ${\rm Gal}(L/\mathbb{Q}) \cong {\rm Gal}(K/\mathbb{Q})/{\rm Gal}(K/L)$ . Hence we have 
(looking at the group orders and using various group isomorphism theorems) that ${\rm Gal}(K/\mathbb{Q}) \cong 
{\rm Gal}(K/M) \times {\rm Gal}(K/L) \cong {\rm Gal}(M/\mathbb{Q}) \times {\rm Gal}(L/\mathbb{Q})$. You can probably figure out what these factors are (but just in case, the rightmost factor group is not Abelian).
A: The splitting field of $f(x)=(x^2-2)(x^3-3)$ is $K=\mathbb{Q}(\sqrt{2}, \sqrt[3]{3}, \omega)$, where $\omega$ is a primitive cubic root of the unity. Consider the subfields $L_1=\mathbb{Q}(\sqrt{2})$ and $L_2=\mathbb{Q}(\sqrt[3]{3}, \omega)$ of $K$.  Both extensions $L_1/\mathbb{Q}$ and $L_2/\mathbb{Q}$ are Galois extensions, since $L_1$ is the splitting field of $x^2-2$ over $\mathbb{Q}$ and $L_2$ is the splitting field of $x^3-3$ over $\mathbb{Q}$.
Hence, the extensions $L_1/\mathbb{Q}$ and $L_2/\mathbb{Q}$ are normal. 
Galois Theory tells us, that the subgroups $\text{Gal}(K/L_1)$ and $\text{Gal}(K/L_2)$ are normal subgroups of $\text{Gal}(K/\mathbb{Q})$. 
The order of $\text{Gal}(K/\mathbb{Q})$ is $[K:\mathbb{Q}]=12$, the order of $\text{Gal}(K/L_1)$ is $[K:L_1]=6$ and the order of $\text{Gal}(K/L_2)$ is $[K:L_2]=2$. Is is easy to see that the intersection $\text{Gal}(K/L_1)\cap\text{Gal} (K/L_2)$ equals the trivial subgroup with one element. Hence, $\text{Gal}(K/\mathbb{Q})$ is isomorphic to the external direct product $\text{Gal}(K/L_1)\times\text{Gal}(K/L_2)$. 
The group $\text{Gal}(K/L_2)$ is isomorphic to the cyclic group $\mathbb{Z}_2$.
The group $\text{Gal}(K/L_1)$ is isomorphic to the symmetric group $S_3$. (For the last statement it is enough to prove the easy fact that $\text{Gal}(K/L_1)$ is non abelian of order 6.)
Finally, $\text{Gal}(K/\mathbb{Q})\cong S_3\times \mathbb{Z}_2$. 
