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Let $g(x)$ of degree 4 be an irreducible polynomial in $\Bbb Q$. Let $\alpha \in \Bbb C$ be a root of $g(x)$, and let $R=\Bbb Q(\alpha)$. We also let $K$ be the splitting field of $g(x)$ over $R$. Further, assume that there exists a quadratic field $L$ such that $\Bbb Q \subset L \subset R $, and that $[L:\Bbb Q]=2$, and assume that $K$ properly contains $R$ (i.e. $R \neq K$).

Prove that $[K:\Bbb Q]=8$ and find $Gal(K/\Bbb Q)$.

I have found that $Gal(K/\Bbb Q)$ needs to be a transitive subgroup of $S_4$, so then the options are $S_4, A_4, D_4$, Klein-Four subgroup, $\Bbb Z/4\Bbb Z$.

At this point, I would want to show that $[K:L]=4$ implying the first result. The second result most likely comes from using the resolvent cubic, but I don't see how to use it.

Any hints or help is appreciated, thanks

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  • $\begingroup$ I think you mean $L$ is quadratic, not quartic and that $K$ properly contains $R$ i.e. $K\ne R$. $\endgroup$ – Adam Hughes Mar 29 '16 at 1:11
  • $\begingroup$ @AdamHughes I definitely meant that, thanks! $\endgroup$ – user218512 Mar 29 '16 at 1:15
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First note that the Galois group has order equal to the degree of the splitting field, so really your only option is $D_4$ since it has order $8$, and Sylow's theorem says that all Sylow $p$ subgroups of a group are isomorphic (even conjugate). So as long as you can prove the degree is $8$, you're good to go.

For that you see that $4||\operatorname{Gal}(K/\Bbb Q)$ since $[R:\Bbb Q]=4$.

(Groups of order 4) We can rule out any group of order $4$--i.e. $\Bbb Z/4\Bbb Z$ and the Klein-$4$ group--since $[K:\Bbb Q]>[R:\Bbb Q]=4$.

($A_4$) We can rule out $A_4$ because $A_4$ has no subgroup of order $6$ (a basic result from group theory) and since $[L:\Bbb Q]=2$ is the index of $\operatorname{Gal}(K/L)$ in $\operatorname{Gal}(K/\Bbb Q)$.

($S_4$) Finally, to rule out $S_4$ we note that since $L\subset R\subset K$ means that we would have $L=K^{A_4}$ necessarily since $A_4$ is the unique subgroup of index $2$ in $S_4$, and by the FTGT we'd have that $\operatorname{Gal}(K/R)\le \operatorname{Gal}(K/L)$ where the order of $\operatorname{Gal}(K/R)$ is $[K:R]$ and if this is $6$--i.e. if $[K:\Bbb Q]=24$ as we are assuming--again we'd have a subgroup of $A_4$ of order $6$. Hence $S_4$ is ruled out.

This only leaves $D_4$, which must be the Galois group.

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  • $\begingroup$ How does $A_4$ having no subgroup of order $6$ make it impossible for it to be the Galois group for $K$ in this scenario? I understand what your argument for ruling out $S_4$ is, but I do not understand how you ruled out $A_4$. $\endgroup$ – Noble Mushtak Mar 29 '16 at 1:39
  • $\begingroup$ @NobleMushtak Because the FTGT says that $\operatorname{Gal}(K/L)$ has index $2$ in the big group, but index $2$ is order $6$ by Lagrange's theorem. $\endgroup$ – Adam Hughes Mar 29 '16 at 1:41
  • $\begingroup$ Oh, that makes sense! Thanks for the explanation! $\endgroup$ – Noble Mushtak Mar 29 '16 at 1:42
  • $\begingroup$ I am simply someone who answers question on Math StackExchange a lot, saw your answer, and wanted to understand it. I am not the person who asked the question. $\endgroup$ – Noble Mushtak Mar 29 '16 at 19:05

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