Any Grad students or Profs out there that can explain this..Galois Group related.?

Question

IF G is a cyclic group, there exist exactly one intermediate field I of degree k, for each integer k dividing [K : F].

Well, anyway I have to prove that. here is my problem. Our textbook hasn't expalined in any way what a Cyclic Galois Group is at all.
I do know that a Cyclic Group consists of all the powers of the generator, and that every subgroup of cyclic group is cyclic, but it's really not helping me apply this to Galois groups.
Anyway if you know of this in any way or can point me in the right direction or what theorms to look at go for it. My brain is fried and I give up for now.

Answer 1

If K is a Galois extension field of F, the you know that the degree of the Galois group is [K:F]. Thus, G is a cyclic group of degree [K:F].

By the fundamental theorem of Galois theory, every subgroup of G corresponds to a field intermediate between F and K and vice versa. If H is a subgroup, then the corresponding subfield is the stabilizer of H. So the question boils down to what the subgroups of a cyclic group are.

But subgroups of cyclic groups are cyclic with order dividing the order of G. In fact, there is exactly one subgroup of order k for each k dividing the order of G. Furthermore, the index of a subgroup in G is the degree of the extension from F to the subfield corresponding to that subgroup.

Answer 2

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Answer 3

I didn't do so hot in modern algebra. But perhaps this will help:

http://en.wikipedia.org/wiki/Galois_group