LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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Homomorphisms respect commutator subgroups and derived series

Statement

For a group homomorphism $f : G \to H$ and every $r \in N$ , $f (G^{(r)}) \leq f (G)^{(r)} .$ If $f$ is surjective, then $f (G^{(r)}) = H^{(r)}$ for every $r$ . In particular, $K^{(r)} \leq G^{(r)}$ for every subgroup $K \leq G$ .

Facts & Assumptions

Given: A group homomorphism $f : G \to H$ and a natural number $r$ .

[F1]

$G^{(0)} = G$ and $G^{(r + 1)} = [G^{(r)}, G^{(r)}]$ (The derived series, solvable groups, and derived length).

[F2]

A group homomorphism preserves products and inverses (Monoid homomorphism and group homomorphism).

Proof

technique · induction

1.1

For all $x, y \in G$ , $f ([x, y]) = [f (x), f (y)]$ by expanding the commutator and using [F2].

F2algebra

1.2

At $r = 0$ , $f (G^{(0)}) = f (G) = f (G)^{(0)}$ .

F1base

1.3

Assume $f (G^{(r)}) \leq f (G)^{(r)}$ and, when $f$ is surjective, $f (G^{(r)}) = H^{(r)}$ .

2.1

Step 1.1 sends every generator of $G^{(r + 1)} = [G^{(r)}, G^{(r)}]$ into $[f (G)^{(r)}, f (G)^{(r)}] = f (G)^{(r + 1)}$ , proving the inclusion at $r + 1$ .

step 1.1step 1.3F1

2.2

If $f$ is surjective and equality holds at $r$ , every commutator generator of $H^{(r + 1)}$ is the image under $f$ of a commutator of preimages in $G^{(r)}$ ; thus equality also holds at $r + 1$ .

step 1.1step 1.3F1given

3.1

Induction gives the inclusion for every $r$ , and gives equality for surjective $f$ . Applying the inclusion to the inclusion homomorphism $K ↪ G$ yields $K^{(r)} \leq G^{(r)}$ .

step 1.2step 2.1step 2.2discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 13 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)