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 

As has been remarked, the non-existence of intermediate fields is equivalent to $S_{3}$ being a maximal subroup of $S_{4}.$ If not, then there is a subgroup $H$ of $S_{4}$ with $[S_{4}:H] = [H:S_{3}] = 2.$ Now $S_{3} \lhd H$ and $S_{3}$ contains all Sylow $3$-subgroup of $H.$ But $S_{3}$ has a unique Sylow $3$-subgroup, which is therefore normal in $H.$ Hence $H$ contains all Sylow $3$-subgroups of $S_{4}$ as $H \lhd S_{4}$ and $[S_{4}:H] = 2.$ Then since $H$ only has one Sylow $3$-subgroup, $S_{4}$ has only one Sylow $3$-subgroup, a contradiction
(for example, $\langle (123) \rangle$ and $\langle (124) \rangle$ are different Sylow $3$-subgroups of $S_{4}$).
