TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A group is solvable if and only if it has a subnormal series with abelian factors

Statement

A group $G$ is solvable if and only if it has a finite subnormal series $G = G_{0} ⊵ G_{1} ⊵ \dots ⊵ G_{n} = 1$ whose factors $G_{i} / G_{i + 1}$ are abelian. Moreover, for every such series, $G^{(i)} \leq G_{i}$ for $0 \leq i \leq n$ .

Facts & Assumptions

Given: A group $G$ .

[F1]

A subnormal series has $G_{i + 1} ⊴ G_{i}$ at every adjacent pair (Subnormal and normal series, factors, refinements, and equivalence).

[F2]

$G$ is solvable exactly when $G^{(n)} = 1$ for some $n$ (The derived series, solvable groups, and derived length).

[L1]

Derived series terms are functorial under inclusions and quotient maps (Homomorphisms respect commutator subgroups and derived series).

[L2]

If $N ⊴ H$ , then $H / N$ is abelian if and only if $H^{'} \leq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

Proof

technique · direct

1.1

Suppose $G$ is solvable, and choose $n$ with $G^{(n)} = 1$ . The derived chain $G = G^{(0)} ⊵ \dots ⊵ G^{(n)} = 1$ is subnormal, and [L2] makes every factor $G^{(i)} / G^{(i + 1)}$ abelian.

assume-hypF1F2L2

1.2

Conversely, suppose $G = G_{0} ⊵ \dots ⊵ G_{n} = 1$ is subnormal with abelian factors. By [L2], $G_{i}^{'} \leq G_{i + 1}$ for every $i < n$ .

assume-hypF1L2

2.1

Starting with $G^{(0)} = G_{0}$ , if $G^{(i)} \leq G_{i}$ , then [L1] gives $G^{(i + 1)} = (G^{(i)})^{'} \leq G_{i}^{'} \leq G_{i + 1}$ ; hence $G^{(i)} \leq G_{i}$ for every $i \leq n$ .

step 1.2L1F2

3.1

Step 2.1 gives $G^{(n)} \leq G_{n} = 1$ , so $G$ is solvable by [F2]. Steps 1.1 and 3.1 prove both directions.

step 1.1step 2.1F2∎

Depends on

Used by

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Sources

J. S. Milne, Group Theory, Chapter 6 (standard reference, not scraped)
K. Conrad, Subgroup Series I (standard reference, not scraped)
K. Igusa, Notes on Jordan-Hölder, section 5 (standard reference, not scraped)