TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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

Statement

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

$[G, G] \subseteq N .$

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 $(g N) (h N) = g h N$ and $(g N)^{- 1} = g^{- 1} N$ , with identity $N$ (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

[F1]

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

[L2]

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

[L3]

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

1.1givenL1L3F2algebra

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

1.2givenL1L3F2algebra

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 $g N$ and $h N$ is $N$ . Multiplying the equality $(g N) (h N) (g N)^{- 1} (h N)^{- 1} = N$ on the right by $(h N) (g N)$ gives $(g N) (h N) = (h N) (g N)$ . Thus $G / N$ is abelian.

2.1step 1.1F1L2

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

3.1step 1.1step 2.1step 1.2∎

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

Depends on

Dependency tree · 2 levels

Sources

Encyclopedia of Mathematics, Commutator subgroup