Let F/K be a field extension with galois group G. The extension need not be galois. Let H be any finite subgroup of G. Let E be the intermediate extension fixed by H, H′ in our notation. By construction, E is fixed by H, and F/E is galois.
Let J be the subgroup of G that fixes E, thus J contains H. In other words, there may be more automorphisms in G that fix E, besides those in H. As we move from automorphism groups to fixed fields, or from fields back to groups, the dimension/index can only decrease. We know that H is finite, and the dimension of 1′ over H′ can be no larger. thus F/E is finite, a finite galois extension. Its dimension equals the size of its galois group, namely |J|. If H is a proper subgroup of J, it corresponds to a proper intermediate extension between E and F. Yet H fixes E, a contradiction. Therefore H = J. Our subgroup H is indeed the galois group for the galois extension F/E.
If H has order n, adjoin n indeterminants to the field K and call the resulting field F. Let H act on the indeterminants by left translation. This produces a group of automorphisms that is isomorphic to H. Apply the previous theorem and H becomes the galois group for a galois extension.
The above is rather contrived. We are more interested in galois groups over the rationals, and other specific fields that are used in algebraic number theory.
Finite fields, for instance, were dealt with in the previous page. Every finite extension of a finite field is galois, with a cyclic galois group, and every cyclic group acts as a galois group for an extension of Zp, or any other finite field for that matter. (A field extension of dimension n is galois, with cyclic galois group, which must have order n.) So it's cyclic groups all around.