Why does $K/k$ finite, solvable, Galois and $F/k$ abelian mean that $FK/F$ solvable?

Question

This is a step from the proof of Lang 6.7.2 (solvable iff solvable by radicals). The full details are as follows:

We assume that $E/k$ is solvable and that $k\subseteq E \subseteq K$ with $K$ finite, solvable, and Galois over $k$. We let $m$ be the product of all primes that divide $[K:k]$ and are not equal to the characteristic and define $F=k(\zeta)$ where $\zeta$ a primitive $m$-th root of unity. $F/k$ is abelian which Lang then leverages to conclude that the lifting of $K$ over $F$, $KF/F$, is solvable.

Any hints as to how this last step works?

Answer 1

I (at least at first) don't want to spoiler too much, but there is a relation between the Galois group of the compositum and the individual Galois groups of the involved extensions which gives you the solvability.