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Q : Let $G$ be a nontrivial finite group. Then $G$ is solvable if and only if every nontrivial quotient of $G$ has a nontrivial abelian normal subgroup.\\

A : $(\Leftarrow)$ Suppose $G$ is solvable and $H = G/N$ be a nontrivial quotient of $G$. Then we have a derived series $1\triangleleft H^{(n)}\triangleleft\cdots\triangleleft H^{(1)}\triangleleft H$. Note that if $\varphi:H^{(i-1)}\to H^{(i-1)}$ is any endomorphism then for $g,h\in H^{(i-1)}$, $\varphi([g,h]) = [\varphi(g),\varphi(h)]$ so that $H^{(i)}$ is fully invariant subgroup of $H^{(i-1)}$. In particular, if we consider the conjugation map, then the fact that $H^{(n)}$ is a fully invariant subgroup of $H^{(n-1)}$ and $H^{(n-1)}\triangleleft H^{(n-2)}$ implies $H^{(n)}$ is a normal subgroup of $H^{(n-2)}$. Inductively we conclude that $H^{(n)}$ is a normal subgroup of $H$. Since $H^{(n)}$ is abelian, done.\\

$(\Rightarrow)$ We induction on $|G|$. Let $N$ be a normal abelian subgroup of $G$. Since any quotient of $G/N$ is a quotient of $G$, the assumption holds for $G$ also holds for $G/N$. Then by induction, $G/N$ is solvable and as $N$ is abelian, $G$ is solvable.