Any relation between degree of intermediate field and order of corresponding subgroup? Let $\mathbf{K}$ be a finite normal extension of a field $\mathbf{F}$. For any intermediate field $\mathbf{E} : \mathbf{F} \le \mathbf{E} \le \mathbf{K}$, is there any correspondence between the degree of $\mathbf{E}$ over $\mathbf{F}$ and the order of the corresponding subgroup $\mathrm{G}(\mathbf{K}/\mathbf{E})$ of the full Galois group of $\mathbf{K}$ over $\mathbf{F}$? 
I was working on an example of finding the Galois group of $x^4 - 2$ and all intermediate fields, and I wondered if this were the case or not. Of course the Galois correspondence shows that each such $\mathbf{E}$ has a corresponding subgroup $\mathrm{G}(\mathbf{K}/\mathbf{E})$, but I was wondering if there was any connection between the degree of the intermediate field and and the order of its corresponding subgroup.
 A: Here is one of the key building blocks of Galois theory.

(Artin's Lemma) For any field $K$ and finite group $G$ of automorphisms of $K$, the extension $K/K^G$ is a finite Galois extension and $\operatorname{Aut}(K/K^G) = G$.

For a proof see e.g. Proposition 70 of these notes.  As I point out there, it seems to be common to state the result under the additional hypothesis that
$K/K^G$ has finite degree.  In fact this is superfluous, although the proof requires some additional care.  My source for this was Serge Lang's wonderful Algebra text, which often presents classical results in sharpened form.
Now suppose $K/F$ is a finite Galois extension with $G = \operatorname{Aut}(K/F)$.  Let $F \subset E \subset K$ be an intermediate field, and let $H = \operatorname{Aut}(K/E)$.  Then we have
$E = K^H$
and thus
$[E:F] = [K:F]/[K:E] = [G:H]$.
So the answer to your question is almost yes.  However, a finite field extension $K/F$ is Galois iff it is normal and separable (the latter is automatic in characteristic zero and hence is sometimes deemphasized in undergraduate courses).   What you want can fail if we only assume normality: let $p$ be a prime number, let $K = \mathbb{Z}/p\mathbb{Z}(t)$ and $F = \mathbb{Z}/p\mathbb{Z}(t^p)$: then $K/F$ is normal, not separable and has $[K:F] = p$.  Taking $E = K$ we have $G = H = \{1\}$ and thus
$[E:F] = p > 1 = [G:H]$.
A: Since you are talking about the Galois correspondence, I assume your extension $\mathbf K$ over $\mathbf F$ is also separable.
Hint: $\mathbf K/\mathbf E$ is a Galois extension, as $\mathbf E$ is defined as a fixed subfield of $\mathbf K$ by a finite automorphism group. This implies that the degree of $\mathbf K$ over $\mathbf E$ is the order of $G(\mathbf K/\mathbf E)$.
You should be able to figure out the degree of $\mathbf E$ over $\mathbf F$ from this.
A: Yes, if the original extension $K/F$ is Galois:
Let $H$ be a subgroup of $\text{Gal}(K/F)$. The Galois correspondence gives us an intermediate field $F \leq E \leq K$, where $E$ is the fixed field of $H$, that is:
$$
E = \{x \in K \text{ | } \sigma(x) = x \text{ for all } \sigma \in H \}
$$
It is a well-known result of Galois theory that an extension $K/E$ is Galois if, and only if, $E$ is the fixed field of $\text{Aut}(K/E)$, and that under these circumstances, $\left|\text{Gal}(K/E)\right| = [K:E]$
Thus, in the case above, we in fact have $\left|H\right| = [K:E]$, and so also $[E:F] = \left|G:H\right|$, where $G$ is the full Galois group of $K/F$
