"Given that the automorphism group of $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$ is isomorphic to $\mathbb{Z}_2 \oplus\mathbb{Z}_2 \oplus\mathbb{Z}_2$, determine the number of subfields of $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$ that have degree 4 over $\mathbb{Q}$."
Looking at $\mathbb{Z}_2 \oplus\mathbb{Z}_2 \oplus\mathbb{Z}_2$, it is easy to see that every nontrivial element generates a subgroup of order 2, and thus having index 4. Considering that $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$ isn't the splitting field of a polynomial over $\mathbb{Q}$ we are not able to apply the Fundamental Theorem of Galois Theory here to obtain a one-to-one correspondence between subgroups of the Galois group and subfields of our extension field. However, is the existence of appropriate index subgroups in this case enough to give us these subfields of order 4?
Generally, when we're unable to use the FToGT, what basic facts do we still know to help us out?
Lastly, are these subfields always just the fixed fields corresponding to each subgroup of the Galois group?