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

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?

• Get your facts straight: The order of the subgroup is equal to $[E:K]$, not to $[K:\mathbb Q]$. – MooS Jun 9 '16 at 10:04
• @MooS First of all, the exercise I worked with asked me how many fields $K$ existed such that $[K:\mathbb{Q}] = 2$. Secondly, I didn't claim anything, I asked if there existed such fields. Thirdly, the fundamental theorem in my book includes $[K:\mathbb{Q}]$, not only $[E:K]$, so I should think there is some link there. – Auclair Jun 9 '16 at 10:08
• I would expect you to figure out the link between $[K:\mathbb Q]$ and $[E:K]$ yourself. Their connection is the most basic proposition about finite field extension. – MooS Jun 9 '16 at 10:30

$S_3$ 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 $\omega^2\root3\of7$.
• Thanks. So instead I want to look for subgroups of $S_3$ of index $2$, which is only $A_3$. So there should exist only one such field extension $K$ with $[K:\mathbb{Q}] = 2$? – Auclair Jun 9 '16 at 10:31
• Yes, and that's the extension by $\omega$. – Gerry Myerson Jun 9 '16 at 12:40
$$\# \{\text{ intermediate fields } K \text{ with } [K:\mathbb Q]=2 \}$$ $$=\# \{\text{ subgroups of } S_3 \text{ index } 2 \}$$ $$=\# \{\text{ subgroups of } S_3 \text{ order } 3 \}$$
There is one such subgroup, namely $A_3$, hence there is one quadratic intermediate field, namely $\mathbb Q(\omega)=\mathbb Q(\sqrt{-3})$.