Let $N/K$ be a finite Galois extension such that $G = Gal(N/K)$ is an abelian group, and let $M$ be an intermediate field of $N/K$. Show that $M/K$ is normal and that $Gal(M/K)$ is abelian.
Here is what I have: Let $\alpha \in N$ have the minimal polynomial $f(x)$ over $K$, and the minimal polynomial $g(x)$ over $M$. Then $g(x)$ divides $f(x)$. Now as $N/K$ is normal and has one root $\alpha \in N$ we know that $f(x)$ splits into linear factors over $N$ and that these factors are all distinct. Therefore $g(x)$ splits into distinct factors in $N$, meaning $N/M$ is normal and separable.
Also, if $\alpha \in M$ then $f(x)$ can't have a repeated root in $M$ since it doesn't in $N$, meaning $M/K$ is separable.
To prove that $M/K$ is normal we need to show that $M$ contains all of the roots of $f(x)$, but I'm not sure where to go from here.
Hint: it is always the case that $N/M$ is Galois, as you observe. The fundamental theorem of Galois theory says that the extension $M/K$ is Galois if and only if $\mathrm{Gal}(N/M)$ is a normal subgroup of $\mathrm{Gal}(N/K)$. Furthermore, it tells you that $\mathrm{Gal}(N/K)/\mathrm{Gal}(N/M) \cong \mathrm{Gal}(M/K)$. Can you see how to use the information you're given now?