ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02

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For an abelian group $G$ , $Z (G) = G$ and $[G, G] = {e}$

Example

If $G$ is an abelian group with identity $e$ , then every element is central and every commutator is the identity. Consequently

$Z (G) = G and [G, G] = {e} .$

Facts & Assumptions

Given: An abelian group $G$ .

[F1]

The group axioms provide associativity, identity, and inverses (Group and abelian group).

[F2]

The center is $Z (G) = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

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

[F4]

The subgroup generated by a set is the smallest subgroup containing it (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

Verification

technique · direct

1.1

Since $G$ is abelian, every $z \in G$ commutes with every $g \in G$ . Thus every element satisfies [F2], and $Z (G) = G$ .

F2given

1.2

For $g, h \in G$ , commutativity and the group laws give $[g, h] = g h g^{- 1} h^{- 1} = g g^{- 1} h h^{- 1} = e$ . Hence the set of all commutators is ${e}$ .

F1F3given

2.1

The set ${e}$ is itself a subgroup, so by the minimality in [F4] the subgroup it generates is ${e}$ . Therefore [F3] gives $[G, G] = {e}$ .

step 1.2F3F4∎

Depends on

Used by

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Sources

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

For an abelian group GG, Z(G)=GZ(G)=G and [G,G]={e}[G,G]={e}