Subfields of a splitting field with Galois group of order 14 This is an old comp problem:
If $E/F$ is a Galois extension with $|Gal(E/F)|=14$, then

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*Show there is a unique intermediate subfield $F \subset K \subset E$ with $[K:F]=2$


*Assume there are 2 distinct intermediate subfields $F\subset L_i \subset E$ for $i=1,2$, such that $[L_i:F]=7$, show that the Galois group $Gal(E/F)$ is not abelian
My Solution

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*As the Galois group is of order 14 by Sylow's theorem there is a unique Sylow 7 subgroup $H$. By the Fundamental Theorem of Galois Theory this corresponds to a subfield $K$ under the bijection, whose index is $[K:F]=[Gal(E/F):H] = 2$


*Assume there are two distinct intermediate subfields of index 7. Then again via Sylow's theorem we know that the number of Sylow 2-subgroups of the Galois group is either 1 or 7. If there are 2 distinct intermediate subfields by the bijection in the Fundamental Theorem of Galois theory there number of Sylow 2 subgroups must be 7, and thus they are not unique and hence not normal. As not every subgroup of the Galois group is not normal it cannot be abelian.
Does this look alright? I worry that I'm missing some details.
 A: This is fine, but you don't need the full strength of the Sylow theorems; it suffices to use Cauchy's theorem. So we can get away with a bit less technology:

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*By Cauchy's theorem there's an element of order $7$, generating a subgroup of order $7$, whose fixed field is an intermediate quadratic subfield. The Galois group $G$ can only contain one such subgroup (because it only has $14$ elements, $14 - 7 = 7$ of which are not in a particular subgroup of order $7$, and of the remaining $7$ elements some must have order $2$), so the quadratic subfield is unique.


*If the Galois group $G$ is abelian then (because there's an element of order $2$ and there's an element of order $7$) $G$ must be isomorphic to $C_2 \times C_7 \cong C_{14}$, which has a unique subgroup of order $2$, corresponding to a unique subfield of degree $7$. If there are at least two such subfields then $G$ can't be isomorphic to $C_{14}$ so must be nonabelian.
A: The problem amounts to determine the structure of a group $G$ of order $2p$, where $p$ is an odd prime. By Cauchy, $G$ contains an element $s$ of order $p$, and the cyclic subgroup  $H=<s>$ has index 2, hence is normal in $G$. Then take $t \in G$ of order 2 and let it act on $H$ by conjugation. Necessarily $tst^{-1}= s^i$ with $i^2 \equiv 1$ mod $p$. Since $p$ is odd, $i\equiv 1$ or $-1$ mod $p$. The first case corresponds to $G=C_p \times C_2$ (bicyclic), the second to $G=D_{2p}$ (dihedral). This shows what you want.
A: Your solution is fine. A variant of the arguments proffered by the others.
If there are two intermediate fields of extension degree $7$, they are the fixed fields of two subgroups $P,P'$ of $G=Gal(E/F)$ of order two. Necessarily $P$ and $P'$ are both Sylow $2$-subgroups. But all Sylow $2$-subgroups are conjugate to each other in $G$, so if there is more than one conjugate, $G$ cannot be abelian.
