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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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For an abelian group GG, Z(G)=GZ(G)=G and [G,G]={e}[G,G]=\{e\}

Example

If GG is an abelian group with identity ee, then every element is central and every commutator is the identity. Consequently

Z(G)=Gand[G,G]={e}.Z(G)=G\quad\text{and}\quad[G,G]=\{e\}.

Facts & Assumptions

Given: An abelian group GG.

[F1]

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

[F2]

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

[F3]

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

[F4]

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

Verification

technique · direct
1.1

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

F2given
1.2

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

F1F3given
2.1

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

step 1.2F3F4∎

Depends on

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