I do not want to be overly detailed so in case something doth not make sense you can ask me.
Let and let .
You should be familar with the theorem that .
Now, so .
If is an automorphism of then it is completely determined by . Let by the -th cyclotomic polynomial, then it is irreducible and its zeros include (the powers of relatively prime with ). If is an automorphism then it means must be a root of and so . Therefore we have at most automorphism. Since the order of the Galois group is it means that each one defines and automorphism. Therefore, where is defined by i.e. the identity automorphism; where is defined by ; where is defined by ; where is defined by .
Now the subgroups of include and itself. An easy why to see why these are the subgroups is because we proved that the Galois group is isomorphic with , and so you can think of the subgroups of to help you visualize what the subgroups of the Galois group need to be.
To find the intermediate fields we need to use the fundamental theorem of Galois theory. First, so that one was easy. Second, so that one was also easy. Now we get a little more involved. The other three intermediate fields are by Galois theory (by "Galois theory" I mean using the correspondence theorem between subgroups and intermediate subfields).
Notice that if we can find so that and is fixed by (for ) then . To prove this, it is really easy. Notice that with and (by Galois theory again). However, we know that from basic field theory, we conclude that .
By the above paragraph to find we just need to find an element in fixed by that has degree over . The claim is that has degree over . Can you see what ? If you write out the sine's and cosine's you will see that is a complex number so it cannot be in . Likewise is not in . Consider this polynomial is fixed by (just do the calculation). Therefore, is a irreducible polynomial over which contains as roots. Therefore, and . To finally find notice that is fixed by and so it is neither in or which forces .