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I encountered the following problem:

Let $f\in\mathbb{Z}[x]$ be an irreducible quartic whose splitting field has Galois group $S_{4}$over $\mathbb{Q}$. Let $\theta$ be a root of $f$ and denote $K=\mathbb{Q}(\theta)$ prove that $[K:\mathbb{Q}]=4$ which has no proper subfield.

which has been solved here.

I am trying to understand what DonAntonio is talking about in this answer. Quoting:

An idea: $\,S_4\,$ has a unique subgroup of order $\,4\,$ which is then normal and contained in $\,A_4\,$ , as can be swiftly checked.

If there was a subextension of order two this would be mean $\,A_4\, $ has a subgroup of order $\,6\,$ (of or index $\,2\,$, however you want to attack this), which is does not, as we know.

I am okay with the group-theoretic claims made, but I don't understand the connection with the problem at hand (even though I sketched (part of) the Galois correspondence for the splitting field of $f(x)$ over $\mathbb Q$; I am not seeing what corresponds to what in DonAntonio's answer).

Can someone please explain this answer?

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  • $\begingroup$ DonAntonio's answer is not the clearest. IMHO the best argument here is that $K$ corresponds to a point stabilizer, that is a subgroup like $S_3\le S_4$. And there cannot be any subgroups properly between $S_{n-1}$ and $S_n$ for any $n$, where $S_{n-1}$ consists of the permuations that keep $n$ fixed. For if $H$ were such a subgroup, then there would be a permutation $\sigma\in H$ such that $\sigma(n)\neq n$. But, then $H$ is automatically transitive. But, a transitive subgroup with point stabilizer $Stab_H(n)=S_{n-1}$ is clearly all of $S_n$. $\endgroup$ Aug 16, 2017 at 20:47

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The quartic subfield is the fixed field of a subgroup of $S_4$ of order 6. If the quartic field had a quadratic subfield, then the quadratic subfield would be fixed by a subgroup of $S_4$ of order $12$ containing the subgroup of order $6$. $S_4$ has a unique subgroup of order $12$, namely $A_4$, and since $A_4$ contains no subgroup of order $6$ this is impossible.

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  • $\begingroup$ Thank you! However, where should one use the fact mentioned by DonAntonio in the first sentence? (that $S_4$ has a unique subgroup of order $4$, which is normal and contained in $A_4$) $\endgroup$
    – Cauchy
    Aug 16, 2017 at 17:42
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    $\begingroup$ @Cauchy Good question, but I don't see a purpose for it. In fact it's not true. $S_4$ does have the Klein $4$-group as a normal subgroup, but it isn't "the unique subgroup of order $4$," since $S_4$ also has non-normal cyclic subgroups of order $4$. $\endgroup$
    – sharding4
    Aug 16, 2017 at 17:48

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