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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 $A_4$...

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  • $\begingroup$ Since $u$ is a root of $f$ and $f$ is irreducible, $f$ must be the minimal polynomial of $f$. I would proceed to show that $G(F/K(u))$ isomorphic to $S_3 \subseteq S_4$ or $A_3 \subseteq A_4$ depending on whether $G$ is $S_4$ or $A_4$. $\endgroup$ – michaelhowes Jun 2 '18 at 22:38
  • $\begingroup$ For the converse, we know that $f$ is irreucible so that $G$ must be a transitive subgroup of $S_4$. By looking at the subgroups of $S_4$ we should be able to deduce that given there is no subgroup between $G(F/K(u))$ and $G$, $G$ must be $S_4$ or $A_4$. $\endgroup$ – michaelhowes Jun 2 '18 at 22:40
  • $\begingroup$ I'll write up a more detailed answer now. $\endgroup$ – michaelhowes Jun 2 '18 at 23:19
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Let $u_1,u_2,u_3,u_4 \in F$, be the four roots of $f$. As you noted above because $Char(K)=0$ and $f$ is irreducible, $f$ is separable and each $u_i$ is distinct. Without loss of generality let $u=u_4$.

Now note that for all $\sigma \in G$, $\sigma \in G(F/K(u))$ if and only if $\sigma(u)=u$. Thus, if we view $G$ as a subgroup of Perm$\{u_1,u_2,u_3,u_4\}\cong S_4$, we have that $G(F/K(u))=G \cap S_3$ where $S_3$ is the subgroup of Perm$\{u_1,u_2,u_3,u_4\}$ of all permutations that fix $u_4$.

Thus if $G=S_4$ or $G=A_4$, we have that $G(F/K(u))=S_4 \cap S_3=S_3$ or $G(F/K(u))=A_4\cap S_3=A_3$ respectively. From your lattice of subgroups, we can note that there are no proper subgroups between $S_3$ and $S_4$ and $A_3$ and $A_4$.

Now for the converse. Since $f$ is irreducible, $G$ is a transitive subgroup of $S_4$. The transitive subgroups of $S_4$ are $S_4$, $A_4$, the dihedral group $D_8$ and its conjugate, the Klien four group $V_4$ and $C_4$ and its conjugates (see here for more about transitive subgroups of $S_4$).

For each of these groups we can look at $G(F/K(u))=S_3 \cap G$. And note that when $G\neq S_4,A_4$, there exists a proper subgroup $S_3\cap G \subsetneq H \subsetneq G$. For example when $G=D_8$, $D_8 \cap S_3$ is the trivial subgroup and thus there are many such subgroups $H$.

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