Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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G/N is abelian if and only if [G,G]⊆N

Statement

Let N⊴G. Then G/N is abelian if and only if

[G,G]⊆N.

Facts & Assumptions

Given: A group G, a normal subgroup N⊴G, and the quotient group G/N.

[L1]

In G/N, products and inverses satisfy (gN)(hN)=ghN and (gN)−1=g−1N, with identity N (For N⊴G, the cosets form a group with identity N and inverse (gN)−1=g−1N).

[F1]

The commutator subgroup [G,G] is generated by the elements [g,h]=ghg−1h−1 (Commutators [g,h]=ghg−1h−1 and the commutator subgroup [G,G]).

[L2]

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

[L3]

For x∈G, one has xN=N if and only if x∈N (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈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∈G, the commutator of the cosets gN and hN is the identity, so [L1] gives [g,h]N=N; hence [g,h]∈N by [L3].

givenL1L3F2algebra
1.2

Conversely, suppose [G,G]⊆N. Then for any g,h∈G, one has [g,h]∈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]⊆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

Dependency tree · two levels

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