Corollary

Let $G$ be a finite group and

a Jordan-Hölder series of $G$ For $G$ to be solvable, it is necessary and sufficient that the quotients $G^k/G^{k+1}$ be cyclic of prime order.

If the quotients of a composition series of $G$ are cyclic and hence commutative, $G$ is solvable by Proposition 10. Conversely, if $G$ is solvable, the group $G^k/G^{k+1}$ is, for all $k$, solvable and simple ($§ 4$, no. 7, Proposition 9). Now, every solvable simple group $H$ is cyclic of prime order. For $D(H)$ is a normal subgroup of $H$; $D(H) = H$ is impossible for in that case $D^k(H) = H$ for all $k$; then $D(H) = \{e\}$ and $H$ is commutative. The corollary then follows from $§ 4$, no. 10, Corollary to Proposition 20.