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

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The commutator subgroup is normal

Statement

For every group $G$ , its commutator subgroup $[G,G]$ is normal in $G$ .

Facts & Assumptions

Given: A group $G$ , its commutator subgroup $D=[G,G]$ , and an element $x\in G$ .

[F1]

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

[L1]

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

[F2]

Conjugating a subgroup by a fixed group element produces a subgroup (Subgroup).

[L2]

A subgroup $D\le G$ is normal if $xDx^{-1}\subseteq D$ for every $x\in G$ (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

Proof

technique · direct

1.1

Direct multiplication gives $x[g,h]x^{-1}=[xgx^{-1},xhx^{-1}]$ for all $g,h\in G$ .

F1algebra

1.2

The conjugate $x^{-1}Dx=\{x^{-1}dx:d\in D\}$ is a subgroup of $G$ .

F2algebra

2.1

For every commutator $c=[g,h]$ , step 1.1 gives $xcx^{-1}\in D$ , so $c\in x^{-1}Dx$ . Thus the subgroup $x^{-1}Dx$ contains every generator of $D$ , and [L1] gives $D\subseteq x^{-1}Dx$ .

step 1.1step 1.2F1L1

3.1

Conjugating the containment in step 2.1 by $x$ gives $xDx^{-1}\subseteq D$ . Since $x$ was arbitrary, [L2] gives $D\mathrel{\trianglelefteq}G$ .

step 2.1L2algebra∎

Depends on

Used by

The abelianisation Gᵃᵇ:=G/[G,G] and its canonical map Definition

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 17 results over 13 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)