finding the number of sub fields such that $(K : Q) = 2$ Consider the polynomial $f(x) = x^5 - 4x + 2$. Let $L$ be the complex splitting field of $f(x)$ over $\mathbb{Q}$. I want to find the number of subfields $K$ of $L$ such that $(K : \mathbb{Q}) = 2$.
I believe there is only one such subfield and that is $S_{5}$. I am not sure how to prove it though.
 A: You can easily see that $f(x) = x^5 -4x + 2$ is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion with $p = 2$. Also, by using calculus you can check that $f(x)$ has three real roots and two complex conjugate roots.

Then you can use the following standard result.
Proposition Let $f(x) \in \mathbb{Q}[x]$ be irreducible over $\mathbb{Q}$ and of prime degree $p$. If $f(x)$ has exactly two nonreal roots in $\mathbb{C}$ then the Galois group of $f(x)$ is isomorphic to $\mathbb{S}_p$.
Thus we conclude that the Galois group $G$ of $f(x) = x^5 -4x +2$ is $G \cong \mathbb{S}_5$, as you correctly mention in your question.
Then, by the Galois correspondence, any intermediate subfield $\mathbb{Q} \subset K \subset L$ of degree $[K:\mathbb{Q}] = 2$ corresponds to a subgroup $H \leq G$ of index $[G:H] = 2$. And then it is known from basic group theory that the only index $2$ subgroup of $\mathbb{S}_5$ is $\mathbb{A}_5$. Thus there is exactly one intermediate subfield $\mathbb{Q} \subset K \subset L$ of degree $2$.
