TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$G/N$ is abelian if and only if $[G,G]\subseteq N$

Statement

Let $N\mathrel{\trianglelefteq}G$ . Then $G/N$ is abelian if and only if

$[G,G]\subseteq N.$

Facts & Assumptions

Given: A group $G$ , a normal subgroup $N\mathrel{\trianglelefteq}G$ , and the quotient group $G/N$ .

[L1]

In $G/N$ , products and inverses satisfy $(gN)(hN)=ghN$ and $(gN)^{-1}=g^{-1}N$ , with identity $N$ (For $N\mathrel{\trianglelefteq}G$ , the cosets form a group with identity $N$ and inverse $(gN)^{-1}=g^{-1}N$ ).

[F1]

The commutator subgroup $[G,G]$ is generated by the elements $[g,h]=ghg^{-1}h^{-1}$ (Commutators $[g,h]=ghg^{-1}h^{-1}$ and the commutator subgroup $[G,G]$ ).

[L2]

A subgroup generated by a set is contained in every subgroup containing that set (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups).

[L3]

For $x\in G$ , one has $xN=N$ if and only if $x\in N$ ( $x\in aH$ iff $a^{-1}x\in H$ , and $aH=bH$ iff $a^{-1}b\in H$ ).

[F2]

A group is abelian when every two of its elements commute (Group and abelian group).

Proof

technique · direct

1.1

Suppose $G/N$ is abelian. For $g,h\in G$ , the commutator of the cosets $gN$ and $hN$ is the identity, so [L1] gives $[g,h]N=N$ ; hence $[g,h]\in N$ by [L3].

givenL1L3F2algebra

1.2

Conversely, suppose $[G,G]\subseteq N$ . Then for any $g,h\in G$ , one has $[g,h]\in N$ , so [L3] and [L1] show that the commutator of $gN$ and $hN$ is $N$ . Multiplying the equality $(gN)(hN)(gN)^{-1}(hN)^{-1}=N$ on the right by $(hN)(gN)$ gives $(gN)(hN)=(hN)(gN)$ . Thus $G/N$ is abelian.

givenL1L3F2algebra

2.1

The subgroup $N$ contains every commutator, so it contains the subgroup they generate: $[G,G]\subseteq N$ .

step 1.1F1L2

3.1

Steps 1.1 and 2.1 prove the forward implication, and step 1.2 proves the reverse implication.

step 1.1step 2.1step 1.2∎

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: 25 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

Encyclopedia of Mathematics, Commutator subgroup (standard reference, not scraped)

G/NG/NG/N is abelian if and only if [G,G]⊆N[G,G]\subseteq N[G,G]⊆N