It is known that a subextion $L/F/K$ of an abelian (Galois) field extension $L/K$ is also abelian. The converse is not true: even when assuming that $L/K$ is Galois and $L/F$ and $F/K$ are abelian, $L/K$ might not be abelian.
I am looking for an explicit counterexample of such Galois non abelian $L/K$ and abelian subextensions $L/F$ and $F/K$. I understand that these cases might arrise when the corresponding extensions of the Galois groups are something like
$$ 1\rightarrow C_3 \rightarrow S_3, $$ but I coudn't find such construction.
Let $L$ be the splitting field of the polynomial $f=x^3-2$ over $\mathbb{Q}$. This polynomial is irreducible, its discriminant is negative and in particular not a square of a rational number. So $\operatorname{Gal}(L/\mathbb{Q})\cong S_3$. Also, it is easy to see that $L=\mathbb{Q}\left(\sqrt[3]{2}, e^{\frac{2\pi i}{3}}\right)$. So $M=\mathbb{Q}\left(e^{\frac{2\pi i}{3}}\right)$ is an intermediate field, the extensions $L/M, M/\mathbb{Q}$ are Galois extensions and the Galois groups have order less than $6$, so they are Abelian.