Exercise 8.1.8

Proof. Suppose that

with $G_i◃G_{i-1},i=1,\cdots ,n$.

(a)

Suppose that $G_{i-1}∕G_i$ is Abelian for $i=1,\cdots ,n$. Then $G_{i-1}∕G_i$ is solvable (Proposition 8.1.5), so we can find a composition series ${({\tilde{G}}_{i,k})}_{0\le k\le n_i}$ of $G_{i-1}∕G_i$, with cyclic quotients of prime order. The proof of Theorem 8.1.4 (see Exercises 2,3,4) shows that the pre-images $G_{i,k}={\pi }^{-1}({\tilde{G}}_{i,k})$ of ${\tilde{G}}_{i,k}$ by the canonical projection $\pi :G_{i-1}\to G_{i-1}∕G_i$ form a composition series $$G_i=G_{i,n_i}\subset G_{i,n_i-1}\subset \cdots \subset G_{i,k}\subset G_{i,k-1}\subset \cdots \subset G_{i,0}=G_{i-1},$$

such that $G_{i,k}◃G_{i,k-1},i=1,\cdots n_i$ and such that $(G_{i,k-1}:G_{i,k})$ is prime.

If we glue together all these composition series for $i=1,\cdots ,n$, we obtain a composition series of $G$ where all quotients are of prime order, so $G$ is solvable according to Definition 8.1.1.

(b)

The proof of part (a) shows that it is sufficient that the quotients $G_{i-1}∕G_i$ are solvable to prove that $G$ is solvable.