Let $f(x)$ be an irreducible polynomial of degree $4$ with rational coefficients, let $\alpha$ be a root of $f$ and set $L=\mathbb{Q}(\alpha)$ (say $\alpha \in \mathbb{C}$). Let $K$ be the splitting field of $f(x)$ over $L$ and assume that $L$ contains a quadratic subfield $\mathbb{Q} \subset M \subset L$ for $M$ a quadratic extension. Suppose $K$ properly contains $L$. Then, show that $K/\mathbb{Q}$ has as Galois group $D_{8}$.
The options for $\mathrm{Gal}(K/\mathbb{Q})$ (the extension is Galois) would be the transitive subgroups of $S_{4}$, namely the Klein 4-group, the cyclic group generated by $(1234)$, $D_{8}$, $A_{4}$ and $S_{4}$ (up to isomorphism).
Now, if $M$ is a quadratic extension of $\mathbb{Q}$, I have shown that $\alpha$ will satisfy a polynomial $g(x)$ of degree $2$ with coefficients in $M$ (in particular, this will be the minimal polynomial of $\alpha$ in $M$). Thus, it follows that $g|f$ in $M[x]$ since $\alpha$ is as well a root of $f$. One has:
$[K:\mathbb{Q}]=[K:L][L:Q]=4[K:L]$
However, I'm not sure on how to show that $[K:L]=2$ for the result to hold, and as well on how to use the fact that $g|f$ to get the result.
What would be the best way to proceed in this case? Thanks.