I'm trying to prove that if $N$ is a normal subgroup of $G$, with $N$ and $G/N$ solvable, then $G$ is solvable.
Proving that $G/N$ is abelian would of course suffice, but I'm not sure if that's a necessary condition or not. I suppose it's possible that I could find some normal subgroup $K$ in $G$ such that $N\subseteq K\subseteq G$ with both $G/K$ and $K/N$ abelian, but I can't see how to go about constructing it.
I've tried a couple things with the isomorphism theorems, and looking at properties preserved under homomorphisms and such, but nothing's panned out.
Can anyone help me here, I'm a bit stuck, thanks.