Let $G \neq \{1\}$ be a finite group. Prove that $G$ is solvable if and only if every nontrivial quotient of $G$ has a nontrivial abelian normal subgroup.
Attempt: If $G$ is solvable then every quotient is solvable. Let $H = G/N$ be such a quotient and $$1 \triangleleft H^{(n)} \triangleleft H^{(n-1)} \triangleleft \dots \triangleleft H' \triangleleft H$$ be it's derived series. Then $H^{(n)}$ is a nontrivial abelian normal subgroup of $H=G/N$. My problem is with the 'only if' direction. I thought about considering a composition series of $G$ and showing that every quotient is abelian but I was not able to proceed further.
Is my solution correct so far? How should I proceed for the 'only if' direction?
Any help is appreciated!
Yes, it's correct in that direction.
In the other direction, argue by induction on $|G|$. Let $N$ be a nontrivial Abelian normal subgroup of $G = G/\{1\}$. Since $G/N$ clearly satisfies the same hypotheses as $G$, we conclude by the induction hypothesis that $G/N$ is solvable. Since $N$ is Abelian, this proves that $G$ is itself solvable.
Actually, I cheated a bit here by not considering the case $N = G$. It's cleaner if you don't make the unnecessary assumption that $G$ is nontrivial.