If a Galois group has $n$ subgroups of some order $k$, will there always be $n$ intermediate field extensions of order $k$?

Question

I realised today that I don't really understand the entirety of the fundamental theorem of Galois theory. It might be that the way it's phrased in my book confuses me, or it might be the subject itself (Galois theory has seemed like magic on several occasions already).

I was working with $x^3 -7 \in \mathbb{Q}[x]$ and found the Galois group of its splitting field $E$ over $\mathbb{Q}$ to be isomorphic to $S_3$. Now, $S_3$ has three subgroups of order $2$, namely the ones with the identity permutation along with a transposition (the reflections of the triangle). Can I then guarantee that there will exist three distinct intermediate field extensions $\mathbb{Q} \subset K \subset E$ where $[K:\mathbb{Q}] = 2$? If yes, will it generalize?

In my example, $E=\mathbb{Q}(7^{1/3},\omega)$ where $\omega = e^{2\pi i/3}$. I thought I could adjoin $\omega$, $7^{1/3}\omega$ and $7^{1/3}\omega^2$ to $\mathbb{Q}$ because it seems like they would create extensions of order $2$, but I'm not completely sure. But regardless of whether or not I can actually create the extensions, will they exist?

Answer 1

$S_3$ has three subgroups of index 3, so there are three intermediate fields of degree 3, namely, those generated by $\root3\of7$, by $\omega\root3\of7$, and by $\omega^2\root3\of7$.

Answer 2

The other answer covers cubic intermediate fields, so let me cover the quadratic ones. The fundamental theorem of galois theory yields:

$$\# \{\text{ intermediate fields } K \text{ with } [K:\mathbb Q]=2 \}$$ $$=\# \{\text{ subgroups of } S_3 \text{ index } 2 \}$$ $$=\# \{\text{ subgroups of } S_3 \text{ order } 3 \}$$

There is one such subgroup, namely $A_3$, hence there is one quadratic intermediate field, namely $\mathbb Q(\omega)=\mathbb Q(\sqrt{-3})$.