A solvable group seems to be variously defined as one with a composition series where all the composition factors are Abelian, or as one with a subnormal series where all the quotients are Abelian. Unlike with the first definition, this definition does not explicitly seem to require that the quotients be simple. Are they nonetheless equivalent?
Similarly, when looking at e.g. proofs that finite $p$-groups are solvable, I have seen the inductive argument which constructs a subnormal series with all the quotients Abelian but they never seem to show that the quotients are also simple. Why? Can one give a proof which does show that they are simple, and therefore complies with the composition series definition of solvable?
Thanks in advance for assistance.
Observe that a composition series is by definition a subnormal series, hence a composition series in which all of the factor groups are abelian is a subnormal series in which all of the factor groups are abelian. Conversely, a subnormal series of maximal length is a composition series by definition. Ultimately, therefore, it suffices to show that any subnormal series whose factor groups are abelian gives rise to (or more specifically, can be extended to) a subnormal series of maximal length.
Edit: As two users have noted in the comments below, the following argument holds when $G$ is a finite group. For instance, if $\mathbb Z$ had a composition series $$0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_n = \mathbb Z,$$ then $0 \subsetneq H_1$ is a maximal strict normal subgroup of $H_1,$ hence $H_1$ must be a simple subgroup of $\mathbb Z.$ Considering that every subgroup of $\mathbb Z$ is cyclic, there exists an integer $n$ such that $H_1 = n \mathbb Z.$ But then, $2n \mathbb Z$ is a subgroup of $H_1$ --- a contradiction. Consequently, no infinite cyclic group has a composition series. That $G$ is finite is assumed in the answer linked in the comments above, and it is explicitly mentioned in the original post, hence we may assume for our purposes that $G$ is finite.
To this end, let $1 = H_1 \trianglelefteq H_2 \trianglelefteq \cdots \trianglelefteq H_n = G$ be a subnormal series such that for each integer $1 \leq i \leq n - 1,$ we have that $H_{i + 1} / H_i$ is abelian. Given that this has maximal length (i.e., all of the inclusions are strict and $H_i$ is a maximal strict normal subgroup of $H_{i + 1}$), we are done. Otherwise, we may assume that there exists an integer $i$ such that $H_i$ is contained in another normal subgroup of $H_{i + 1}.$ By hypothesis that $H_{i + 1} / H_i$ is abelian, it is solvable and therefore must have a composition series in which all of its factor groups are abelian. Explicitly, our composition series is $$1 = K_1 \triangleleft K_2 \triangleleft \cdots \triangleleft K_m = H_{i + 1} / H_i$$ with $K_{j + 1} / K_j$ is abelian for each integer $1 \leq j \leq m - 1,$ where $\triangleleft$ denotes strict inclusion as a normal subgroup. By the Fourth Isomorphism Theorem, we can identify $1$ with the subgroup $H_i$ of $G.$ Likewise, each of the subgroups $K_j$ gives rise to a strict inclusion of subgroups $N_j \triangleleft N_{j + 1}$ of $G$ that contain $H_i.$ We can therefore unravel this composition series as $H_i \subsetneq N_1 \subsetneq N_2 \subsetneq \cdots \subsetneq H_{i + 1}$ within our original subnormal series of $G$ to obtain a maximal subnormal series of $G.$
Our only remaining task is to check that $N_{j + 1} / N_j$ is an abelian group. By hypothesis, $K_{j + 1} / K_j$ is abelian, so by the Third Isomorphism Theorem, we find that $K_{j + 1} / K_j = (N_{j + 1} / H_i)/(N_j / H_i) \cong N_{j + 1} / N_j$ is likewise abelian, as desired.