Determining Galois Group of Polynomial of Degree 4 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
 A: 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.
