Quotients of solvable groups are solvable

Statement

Let $G$ be a solvable group and let $\varphi \colon G \to K$ be a surjective group homomorphism. Then $K$ is solvable. In particular, $G/N$ is solvable for each normal subgroup $N$ of $G$.

Remarks

A proof uses the derived series: $\varphi(G^{(i)}) = K^{(i)}$ for each $i$, because $\varphi$ maps commutators onto commutators. So $G^{(n)} = \{e\}$ implies $K^{(n)} = \{e\}$. The second part follows with the projection $G \to G/N$.

Explanations

L1 human

A surjective homomorphism sends the commutators of $G$ onto the commutators of $K$. So the derived series of $K$ is the image of the derived series of $G$. If one reaches the identity, the image does too.