TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Extensions and finite direct products of solvable groups are solvable

Statement

Let $N ⊴ G$ . If $N$ and $G / N$ are solvable, then $G$ is solvable. Every finite direct product of solvable groups is solvable; the empty product is the trivial group.

Facts & Assumptions

Given: A normal subgroup $N ⊴ G$ with $N$ and $G / N$ solvable, and solvable groups $H_{1}, \dots, H_{t}$ .

[F1]

A group is solvable when some term of its derived series is trivial (The derived series, solvable groups, and derived length).

[L1]

A quotient map satisfies $q (G^{(r)}) = (G / N)^{(r)}$ , and inclusions give $K^{(s)} \leq H^{(s)}$ for $K \leq H$ (Homomorphisms respect commutator subgroups and derived series).

[L2]

The external direct product has coordinatewise multiplication and inverses ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

Proof

technique · direct

1.1

Choose $r, s$ with $(G / N)^{(r)} = 1$ and $N^{(s)} = 1$ .

givenF1choose

1.2

Coordinatewise calculation using [L2] gives $[(g, h), (g^{'}, h^{'})] = ([g, g^{'}], [h, h^{'}])$ , so $(G_{1} \times G_{2})^{'} = G_{1}^{'} \times G_{2}^{'}$ and therefore $(G_{1} \times G_{2})^{(k)} = G_{1}^{(k)} \times G_{2}^{(k)}$ for every $k$ .

L2algebra

2.1

By [L1], $q (G^{(r)}) = 1$ , so $G^{(r)} \leq N$ . Repeatedly applying the subgroup inclusion in [L1] gives $G^{(r + s)} = (G^{(r)})^{(s)} \leq N^{(s)} = 1$ .

step 1.1L1

3.1

Hence $G$ is solvable by [F1].

step 2.1F1

4.1

Choosing a common bound for the derived lengths in a nonempty finite family and using step 1.2 inductively proves its product solvable; for the empty family the product is the trivial group, which has derived length zero.

step 1.2F1algebra∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)