Abelian extensions $E|F$ and $F|K$ $\implies E|K$?

Question

Let $E,F,K$ fields such that $K\subset F \subset E$

If $F|K$ and $E|F$ are abelian extensions, then $E|K$ is an abelian extension?

I can't find a counter example

Answer 1

$K=\Bbb Q$, $F=\Bbb Q(\sqrt2)$, $E=\Bbb Q(\sqrt[4]2)$.

Answer 2

As the other answer shows, it is not even necessarily the case that $E/K$ is Galois, let alone abelian.

Another way to think about finding a counterexample would be to apply the fundamental theorem of Galois theory to obtain a question about groups. For simplicity suppose that $E$ is Galois, let $G = Gal(E/K)$ and $H = Gal(E/F)$. Then your question is equivalent to asking about groups if $H$ and $G/H$ abelian imply that $G$ is abelian. Some counterexamples are $G=S_3$ and $H=A_3$ or $G=D_4$ and $H$ the subgroup of rotations; these might suggest some fields and counterexamples for you.

For example, in the first case, any extension $E$ with Galois group $S_3$ will work, for instance one coming from an irreducible cubic with nonsquare discriminant. Then $F$ will be the fixed of $A_3$, which you can write explicitly as $K$ adjoin the square root of the discriminant. The splitting field of $x^3 - 2$ over $\mathbb Q$ will work for $E$, with $F=\mathbb Q(\sqrt{-108})$.