A finite group is solvable if and only if all its composition factors are cyclic of prime order

Suppose $G$ is solvable. Each composition factor is a quotient of a subgroup of $G$, hence is solvable by [L3].

Conversely, take a composition series $G=G_0▹\cdots ▹G_n=1$ whose factors have prime order. The trivial group $G_n$ is solvable, and if $G_{i+1}$ is solvable then $G_i\mathrm{/}{G}_{i+1}$ is cyclic, hence abelian and solvable, so [L4] makes $G_i$ solvable. Finite upward induction gives $G=G_0$ solvable.

Let $S$ be a simple solvable composition factor. Its derived subgroup $S^{\mathrm{\prime }}$ is normal by [L7], so simplicity gives $S^{\mathrm{\prime }}=1$ or $S^{\mathrm{\prime }}=S$; solvability excludes $S^{\mathrm{\prime }}=S$, and therefore $S$ is abelian.

Choose $1\mathrm{\ne }x\in S$. Since $S$ is abelian, $⟨x⟩⊴S$, so simplicity gives $S=⟨x⟩$. By [L6] choose a prime $p$ dividing $\mathrm{\mid }S\mathrm{\mid }$; [L5] gives a subgroup of order $p$, which is nontrivial and normal in the abelian group $S$, hence equals $S$. Thus $S$ is cyclic of prime order.

Steps 1.1, 2.1, and 3.1 prove that solvability forces prime-order composition factors, while step 1.2 proves the converse; [L2] makes the condition independent of the chosen composition series.