This is one part of me trying to solve exercise 3.4.8 in D&F Abstract Algebra. In particular I am proving (a) implies (b), and am frustrated with the method I found because it involves nested induction, which gets messy and long. Can you help me catch mistakes or hint at (leave some thinking to be done!) a simpler approach to the problem?
8. Let $G$ be a finite group. The following are equivalent:
(a) $G$ is solvable
(b) $G$ has a chain of subgroups $1 = H_0 \trianglelefteq H_1 \trianglelefteq H_2 \trianglelefteq \ldots \trianglelefteq H_s = G$ such that $H_{i+1}/H_i$ is cyclic.
$\vdots$
The definition for "solvable" so far in the book means that there is a chain of normal subgroups of $G$ such that the adjacent quotients are abelian.
Proof. First we prove the following lemma: If $G$ is solvable, and $G_0 \trianglelefteq G_1$ is a chain of subgroups of $G$ satisfying that $G_1 / G_0$ is abelian, then there exists a chain of subgroups $$G_0 = H_0\trianglelefteq H_1 \trianglelefteq \ldots \trianglelefteq H_{s-1} \trianglelefteq H_s = G_1$$ such that $H_{i+1}/H_i$ is cyclic. Let $\left|G_1\right| = m$, $\left|H_0\right| = n$, so that $\left|G_1/G_0\right| = \frac{m}{n}$. If $\left|G_1\right|$ is prime then we are done because $n = 1$ and groups of prime order are cyclic. If $m = n$ then we are done because the quotient group is trivial and therefore cyclic ($m=1$ is similarly easy). Otherwise $\left|G_1/G_0\right|$ has at least one prime factor. Induct on $z$, the number of such prime factors, counting repetitions. When $z = 1$, let $p$ be the only prime dividing $\frac{m}{n}$ (with no repetitions), so that $\frac{m}{n} = p$. The quotient group $G_1/G_0$ must be cyclic because its order is prime, and so the chain $G_0 \trianglelefteq G_1$ proves our lemma for the base case $z = 1$.
Now assume the lemma is true when $z = k$, i.e., for chains $G_0 \trianglelefteq G_1$ such that $\frac{m}{n}$ has $k$ prime factors, counting repetitions, where $m = \left|G_1\right|$, $n = \left|G_0\right|$. Let $G_0 \trianglelefteq G_1$ satisfy instead that $\frac{m}{n}$ (as defined before) has $k + 1$ prime factors, counting repetitions. Let $p$ be one such factor. By Cauchy's Theorem, $G_1/G_0$ has a subgroup of order $p$, which by the Fourth Isomorphism Theorem is of the form $A/G_0$ where $A$ is a subgroup of $G_1$ containing $G_0$. Since $\left|A/G_0\right| = p$, we have $\left|A\right| = pn$. Consider the chain of subgroups $A \trianglelefteq G_1$. Since $p$ is one of the $k + 1$ prime factors of $\frac{m}{n}$ (counting repetitions), $\left|G_1/A\right| = \frac{m}{pn}$ has $k$ prime factors counting repetitions. By the inductive hypothesis, there exists some chain of subgroups $A = H_1 \trianglelefteq H_2 \trianglelefteq \ldots \trianglelefteq H_s = G_1$ satisfying $H_{i+1}/H_i$ is cyclic. Since $A/G_0$ has prime order, it is cyclic, and so the chain $$G_0 = H_0 \trianglelefteq A = H_1 \trianglelefteq H_2 \trianglelefteq \ldots \trianglelefteq H_s = G_1$$ proves our lemma.
Now we prove (a) implies (b). If (a) is true, then by definition there exists a chain of subgroups $1 = G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq \ldots \trianglelefteq G_t = G$ such that $G_{i+1}/G_i$ is abelian. Induct on $t$. The base case $t = 1$ follows immediately from the lemma because the chain of subgroups has length 2. Now assume (a) implies (b) when $t$ is some positive integer $k$. Assume $t = k+1$, so that there exists a chain of subgroups of $G$: $$1 = G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq \ldots \trianglelefteq G_{k+1} = G$$ such that $G_{i+1}/G_i$ is abelian. By the inductive hypothesis, there exists a chain of subgroups $1 = H_0 \trianglelefteq H_1 \trianglelefteq \ldots \trianglelefteq H_s = G_k$ such that $H_{i+1}/H_i$ is cyclic, for some $s \in \mathbb{Z}^+$. By the lemma applied to $G_k \trianglelefteq G_{k+1}$, there exists another chain of subgroups $G_k = H_s \trianglelefteq H_{s+1} \trianglelefteq \ldots \trianglelefteq H_{s + r} = G_{k+1}$ such that $H_{i + 1}/H_i$ is cyclic, for some $r \in \mathbb{Z}^+$. So the chain $$1 = H_0 \trianglelefteq H_1 \trianglelefteq \ldots \trianglelefteq H_s = G_k \trianglelefteq H_{s+1} \trianglelefteq \ldots \trianglelefteq H_{s+r} = G_{k+1}$$ completes the inductive step.
Consider a subnormal sequence with Abelian nontrivial factors of $G$. WLOG we can assume that this sequence has the maximal possible number of terms. Suppose $M/N$ is one of the non-cyclic Abelian sections of that sequence. Then $M/N$ contains a non-trivial cyclic subgroup $H/N$. Since $M/N$ is Abelian, $H$ is a normal subgroup in $M$ and is subnormal in $G$. Thus we can add $H$ to the sequence between $M$ and $N$ and increase the length of the sequence by 1, a contradiction with maximality of the subnormal sequence.