Find $Gal(\mathbb{Q}[\sqrt{13}, \sqrt{17}]/\mathbb{Q})$ and explicitly determine the intermediate subfields Let K be the splitting field of $(x^2-3x-1)(x^2-3x-2)$. Find the Galois group $Gal(K/\mathbb{Q})$ and determine the intermediate subfields explicitly.
I have that $(x^2-3x-1)(x^2-3x-2)$ = $(x-\frac{3}{2}+\frac{\sqrt{13}}{2})(x-\frac{3}{2}-\frac{\sqrt{13}}{2})(x-\frac{3}{2}+\frac{\sqrt{17}}{2})(x-\frac{3}{2}-\frac{\sqrt{17}}{2})$
Then $K = \mathbb{Q}[\sqrt{13}, \sqrt{17}]$. I know that the degree of K is 4, so the cardinality of the Galois group must also be 4. I do not know how to determine the Galois group or explicitly find the subfields. 
 A: Hint: There are only two groups of order $4$: $C_4$ and $C_2 \times C_2$. There are two elements of order $2$ in the Galois group (which?) and so it cannot be cyclic.
A: Up to isomorphism, there are only two groups of order $4$, namely $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ and $\mathbb{Z}_4$. Hence $G = \mathsf{Gal}(f)$ is isomorphic to one of these. 
Note that $\mathbb{Q}(\sqrt{13})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{17})/\mathbb{Q}$ are subextensions of $K$ of degree $2$, and so by the Galois correspondence $G$ has two subgroups of order $2$. This implies that $G \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2$ which has six subgroups:
$$
\{(0,0)\}, \mathbb{Z}_2 \oplus 0, 0 \oplus \mathbb{Z}_2, \langle (1,1) \rangle \text{ and } \mathbb{Z}_2 \oplus \mathbb{Z}_2.
$$
You already know the subextensions $\mathbb{Q}, \mathbb{Q}(\sqrt{13})/\mathbb{Q}, \mathbb{Q}(\sqrt{17})/\mathbb{Q}$ and $K$, so once again by Galois correspondence there is only one subextension left, of degree two over the rationals.
One can make an ansatz and consider $\mathbb{Q}(\sqrt{13}\sqrt{17})$. This is a quadratic extension contained in $K$, to see that it is the one missing we just have to show that it doesn't equal the other quadratic extensions.
If we were to have $\mathbb{Q}(\sqrt{13}) = \mathbb{Q}(\sqrt{13}\sqrt{17})$, in particular we should have $\sqrt{13}, \sqrt{13}\sqrt{17} \in \mathbb{Q}(\sqrt{13})$. However, this implies $\sqrt{17} \in \mathbb{Q}(\sqrt{13})$ which is absurd, as you probably have shown when calculating $[K:\mathbb{Q}]$. The other case is similar.
