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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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The derived subgroup is characteristic and the abelianization is universal

Statement

For every group G, the derived subgroup G=[G,G] is characteristic, hence normal. The quotient Gab:=G/G is abelian and has the following universal property: for every homomorphism f:GA into an abelian group, there is a unique homomorphism fˉ:GabA with f=fˉq, where q:GG/G is the quotient map.

Facts & Assumptions

Given: A group G, its commutator subgroup G, and a homomorphism f:GA to an abelian group.

[F1]

[G,G] is generated by the commutators [x,y]=xyx1y1 (Commutators [g,h]=ghg1h1 and the commutator subgroup [G,G]).

[F2]

A characteristic subgroup is preserved by every automorphism (Characteristic subgroups).

[L2]

For NG, G/N is abelian if and only if [G,G]N (G/N is abelian if and only if [G,G]N).

[L3]

If NG and Nkerf, then f factors uniquely through G/N (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

Proof

technique · direct
1.1

Every automorphism α satisfies α([x,y])=[α(x),α(y)], so it maps the generating commutators of G into G; applying the same argument to α1 gives α(G)=G.

F1algebra
1.2

For x,yG, the group A is abelian, so f([x,y])=[f(x),f(y)]=1; hence every generator of G lies in kerf, and Gkerf.

givenF1algebra
2.1

Thus G is characteristic by [F2], and therefore normal by [L1].

step 1.1F2L1
3.1

Since GG, [L2] gives that G/G is abelian.

step 2.1L2
4.1

By [L3] there is a unique fˉ:G/GA with f=fˉq. Together with step 3.1, this is the asserted universal abelian quotient.

step 2.1step 1.2L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 14 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