Galois Group of $(x^2-a)^2-b$ Let $a,b \in\mathbb{Q},\,\ b>a^2, \,\,f=(x^2-a)^2-b \in \mathbb{Q}$.
Let $L$ be the splitting field of $f$ over $\mathbb{Q}$ and let $f$ be irreducible.
So now I want to calculate the Galois Group.

The roots of $f$ are $\pm \sqrt{a \pm \sqrt{b}}$. Let $\omega_1 = \sqrt{a + \sqrt{b}}, \,\,\omega_2 = \sqrt{a - \sqrt{b}}$.
You can see fast that: $[\mathbb{Q}(\omega_1, \omega_2):\mathbb{Q}]=8$. So this is also the order of the Galois Group.
I found the automorphism to be:
$(\omega_1 \mapsto \omega_1, \,\omega_2 \mapsto \omega_2) = \tau_1 = \text{id}$
$(\omega_1 \mapsto -\omega_1, \,\omega_2 \mapsto \omega_2) = \tau_2$
$(\omega_1 \mapsto -\omega_1, \,\omega_2 \mapsto -\omega_2) = \tau_3$
$(\omega_1 \mapsto \omega_1, \,\omega_2 \mapsto -\omega_2) = \tau_4$
$(\omega_1 \mapsto \omega_2, \,\omega_2 \mapsto \omega_1) = \sigma_1$
$(\omega_1 \mapsto -\omega_2, \,\omega_2 \mapsto \omega_1) = \sigma_2$
$(\omega_1 \mapsto -\omega_2, \,\omega_2 \mapsto -\omega_1) = \sigma_3$
$(\omega_1 \mapsto \omega_2, \,\omega_2 \mapsto -\omega_1) = \sigma_4$
Now I need to find all fields between $K$ and $L$. So first I find the subgroups.
A subgroup must have order $1, 2, 4$ or $8$.
Order 1: $\{\text{id}\}$
Order 2: $\langle\{\tau_2\}\rangle,\langle\{\tau_3\}\rangle,\langle\{\tau_4\}\rangle,\langle\{\sigma_1\}\rangle,\langle\{\sigma_3\}\rangle$
Order 4: $\langle\{\sigma_2\}\rangle,\langle\{\sigma_4\}\rangle$
Order 8: $G$ the Galois Group, so just all elements

But I don't know how to continue from now. How to find all the fields corresponding to these subgroups?
 A: First of all you didn't get all subgroups correct. The subgroup $\langle \sigma_2 \rangle$ and $\langle \sigma_4 \rangle$ are the same because $\sigma_2^3=\sigma_4$, $\sigma_4^2=\sigma_2^2$, $\sigma_4^3=\sigma_2$, so they span the same subgroup.
On top of that your list of subgroups is missing two more subgroups, namely $\langle \tau_2, \tau_3 \rangle$ or equivalently any group that is spanned by two of the non-identity transpositions $\tau_i$ - this will give a subgroup with four elements that is not the same as $\langle \sigma_2 \rangle$. The last missing subgroup is $\langle \sigma_1, \sigma_3 \rangle$ that consists of the elements $\sigma_1, \sigma_3, \tau_3$ and $id$.
Now by the fundamental theorem of Galois Theory each field between $K$ and $L$ corresponds to a fixed field.
The easy ones first: The fixed field of $\{ id\}$ is all $L=\mathbb Q(\omega_1, \omega_2)$. The fixed field of Gal$(L/K)$ is $K=\mathbb Q$.
The order $2$ subgroups correspond to field of order $4$ over $\mathbb Q$. The subgroup $\langle \tau_2 \rangle$ obiously fixes $\omega_2$ so $L^{\langle \tau_2 \rangle} = \mathbb Q(\omega_2)$. Then it is also clear that $L^{\langle \tau_4 \rangle} = \mathbb Q(\omega_1)$. Both fields have degree $4$ over $\mathbb Q$ because $f$ is assumed to be irreducible so it is the minimal polynomial of both $\omega_1$ and $\omega_2$. The subgroup $\langle \tau_3 \rangle$ fixes $\mathbb Q(\omega_1, \omega_2)$ which should also have degree $4$ over $\mathbb Q$ if I am not forgetting something.
For the order $4$ subgroups we have to look for extensions of degree $2$ over $\mathbb Q$. For example $\mathbb Q(\omega_1^2)$ is fixed by the group $\langle \tau_2, \tau_3 \rangle$.
And $\mathbb Q(\omega_1\omega_2)$ is fixed by $\langle \sigma_1, \sigma_3 \rangle$.
This way you can figure out all the fields between $L$ and $K$ - by looking which elements of $L$ are fixed by the homomorphisms in a given group. And you always know that if $[L:K]=n$ and $\#G=k$ you are looking for a field $E$ with $[E:K] = \frac nk$.
