Let $L = \mathbb{Q}(\sqrt{2}, \sqrt{3},\sqrt{5})$.
(a) Given a basis for $L/\mathbb{Q}$.
(b) List the elements of $Gal(L/\mathbb{Q})$ and show that $|Gal(L/\mathbb{Q})| = [L/\mathbb{Q}]$. Which known group is isomorphic to $Gal(L/\mathbb{Q})$?
(c) For each of the following fields $K_{i}$, determine the subgroup $H_{i}$ of $Gal(L/\mathbb{Q})$ corresponding to $K_{i}$ by the Galois correspondence: $$K_{1} = \mathbb{Q}(\sqrt{10}),\quad K_{2} = \mathbb{Q}(\sqrt{6} + \sqrt{15}),\quad K_{3} = \mathbb{Q}(\sqrt{2} + \sqrt{5}),\quad K_{4} = \mathbb{Q}(\sqrt{30}).$$
I already showed the items (a) and (b). If I'm correct, in the item (b), the elements of $Gal(L/\mathbb{Q})$ are the automorphisms that permutes $\sqrt{2},\sqrt{3},\sqrt{5}$ and their inverse. But, I don't know how to use it for find the subgroups correspondents. Can anybody help me?
As an example, consider $K_2$. An element of the Galois group fixes $K_2$ pointwise iff it fixes $\alpha=\sqrt 6+\sqrt{15}$. The images of $\alpha$ under the Galois group are the four numbers $\pm\sqrt6\pm\sqrt{15}$. So the subgroup fixing $\alpha$ has order $8/4=2$, and consists of the identity and one more element $\sigma$. Bearing in mind that for each of the square roots $\sqrt2$, $\sqrt3$ and $\sqrt5$, $\sigma$ either fixes it or negates it, the $\sigma$ in question must have $\sqrt2\mapsto-\sqrt2$, $\sqrt3\mapsto-\sqrt3$ and $\sqrt5\mapsto-\sqrt5$.