Quartic Equation having Galois Group as $S_4$ Suppose $f(x)\in \mathbb{Z}[x]$ be an irreducible Quartic polynomial with Galois Group as $S_4$. Let $\theta$ be a root of $f(x)$ and set $K=\mathbb{Q}(\theta)$.Now, the Question is:
Prove that $K$ is an extension of degree $\mathbb{Q}$ of degree 4 which has no proper Subfields?
Are there any Galois Extensions of $\mathbb{Q}$ of degree 4 with no proper sub fields.
As i have adjoined a root of irreducible quartic, I can see that $K$ is of degree $4$ over $\mathbb{Q}$.
But, why does there is no proper subfield of $K$ containing $\mathbb{Q}$.
suppose $L$ is proper subfield of $K$, then $L$ has to be of degree $2$ over $\mathbb{Q}$. So, $L$ is Galois over $\mathbb{Q}$. i.e., $L$ is normal So corresponding subgroup of Galois group has to be normal.
I tried working in this way but could not able to conclude anything from this. 
any help/suggestion would be appreciated.
Thank You 

Convert the question to a problem in permutation groups.
Let $F$ be the splitting field of 
$$f(x)=(x-\theta_1)(x-\theta_2)(x-\theta_3)(x-\theta_4)$$
with $\theta=\theta_1$.
We were given that the Galois group realizes all the 24 permutations of the roots $\theta_i,i=1,2,3,4.$ Therefore
$$
\operatorname{Gal}(F/\mathbb{Q}(\theta))=\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})
$$
contains automorphisms realizing all the six permutations of the other roots.
Galois correspondence then means that the claim is equivalent to:

There are no subgroups $H$ properly between $\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})$ and $\operatorname{Sym}(\{\theta_1,\theta_2,\theta_3,\theta_4\})$.

In other words, this is equivalent to proving that the obvious copy of $S_3$ inside $S_4$ is a maximal subgroup. Have you seen that? If not, can you prove it?
