Fixed fields for Galois groups

I need some help with a Galois problem.

I need to find the splitting field for

$\displaystyle x^7 - 1$ over Q, then find its Galois group, then the subgroups of the Galois group, and then the fixed fields for the subgroups.

Now I thought I understood this, but I got into trouble at the end, which started to make me realize I must have been doing something wrong earlier as well.

The roots of this are obviously (besides 1)

$\displaystyle w, w^2, w^3, w^4, w^5, w^6, w=e^{2\pi i/7}$

and $\displaystyle w+w^2+w^3+w^4+w^5+w^6=0, w^7=1$

Which gives me $\displaystyle Q(e^{2\pi i/7})$ as the field extension, which a Galois group of order 6 isomorphic with Z6 and Z7*, all the elements of which are generated by setting w to one of the other roots. This gives me two subgroups, one with 2 elements and one with 3 elements.

I was completey unable to figure out how to find the fixed fields for these subgroups in any meaningful way. And then I started to think my Galois group wasn't even correct. Most of the permutations seemed to change Q, which means they aren't correct, though I thought this was the standard technique with the cyclotomics (but I'm really new to this stuff).

Now, now I'm thinking that there must be something I'm missing. I see a few opitions here. One is that $\displaystyle cos(2\pi/7)$ can be expressed in terms of a cube root, which would mean I only need to add that cube root and i to Q to get the rigth field, which gives me the same isomorphy anyway.

But this isn't some "obvious" cosine to me, so a question is how would i recognize that this was a root like that if it were true without a table (given an arbitrary cyclotomic)?

The other idea I had was that there might be some obvious relations between various of the roots, like being conjugates of each other or something like that, which would seem to have to be the case now that I think about it ... I hadn't thought about that before, but obviously if one complex number is a root, the conjugate must also be a root, which gives me 3 roots and their conjugates. I guess the question is will that make everything cancel the way it should wihtout me needing to figure out if the cosine is a cube root?

Any comments, help would be appreciated.

Thanks.