DefinitionDefinition: Literature-sourcedProof: Not applicableSession-authored (Fable 5 assisted)audited 2026-08-11

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The abelianisation $G^{\mathrm{ab}}:=G/[G,G]$ and its canonical map

Definition

Let $G$ be a group. Its abelianisation is the quotient

$G^{\mathrm{ab}}:=G/[G,G],$

where $[G,G]$ is the commutator subgroup of Commutators $[g,h]=ghg^{-1}h^{-1}$ and the commutator subgroup $[G,G]$ . This subgroup is normal by The commutator subgroup is normal, so the quotient is defined. The abelianisation map is the canonical surjective homomorphism

$q_G:G\longrightarrow G^{\mathrm{ab}},\qquad q_G(g)=g[G,G],$

of The canonical projection $\pi:G\to G/N$ , $\pi(g)=gN$ , is a surjective group homomorphism.

Depends on

Used by

The abelianisation of a free group on X is a free abelian group on X Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 15 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

John McKernan, Presentations and Groups of Small Order, Lecture 12 (standard reference, not scraped)
Encyclopedia of Mathematics, Commutator subgroup (standard reference, not scraped)

The abelianisation Gab:=G/[G,G]G^{\mathrm{ab}}:=G/[G,G]Gab:=G/[G,G] and its canonical map