Let $f $ be an irreducible quartic over a field $K $ of characteristic zero, $G $ the Galois group of $f $, and $u $ a root of $f $. Show that there is no field properly between $K $ and $K (u) $ if and only if $G=A_4$ or $G=S_4$.
Here is my work so far:
I first got the group lattice structures of $S_4$ and $A_4 $from on-line. As $f$ is irreducible over a field of char $0, f $is separable and hence $G (F/K)$ is Galois where $F$ is a splitting field of $K $. Hence by Fundamental Theorem it suffices to show there is no proper subgroup between $G=G (F/K)$ and $G (F/K (u))$ iff $G=A_4$ or $G=S_4.$ We know that $G (F/K (u))$ is Galois. As $f$ is quartic and $f$ is irreducible, we have that the minimal polynomial of $u $ is $f$ itself. If $G $ is isomorphic to $S_4$ then since $G (F/K (u)) $ fixes $u $ it injects into $S_3$. If $G(F/K (u)) $ is isomorphic to $S_3$ then there's no subgroup properly between them. It cannot be isomorphic to a group of order two since it (viewed as a subgroup of $S_3$) cannot leave one of the three (non-$u $) roots fixed since it's Galois. I am not sure why it being isomorphic to $A_3$ would lead to a contradiction.
I am not sure how to proceed further. I somehow must be looking at the group order/index of various subgroups of $S_4$ and $A_4$...