Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The center of a group is a normal subgroup

Statement

For every group G, the center Z(G) is a normal subgroup of G.

Facts & Assumptions

Given: A group G with identity e and center Z(G).

[F1]

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

[F2]

A subset of a group is a subgroup when it contains the identity and is closed under products and inverses (Subgroup).

[L1]

A subgroup N≤G is normal if gNg−1=N for every g∈G (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

Proof

technique · direct
1.1

The identity lies in Z(G). If x,y∈Z(G) and g∈G, then (xy)g=x(yg)=x(gy)=(xg)y=(gx)y=g(xy), so xy∈Z(G). If x∈Z(G), then xg=gx implies x−1g=gx−1 after multiplying by x−1 on both sides, so x−1∈Z(G). Hence Z(G)≤G.

F1F2algebra
1.2

If z∈Z(G) and g∈G, then gzg−1=zgg−1=z. Therefore gZ(G)g−1=Z(G) for every g∈G.

F1algebra
2.1

Steps 1.1 and 1.2 show that Z(G) is a subgroup invariant under conjugation, so it is normal.

step 1.1step 1.2L1∎

Depends on

Used by

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