Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01
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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 order of every element of a finite group divides the order of the group

Statement

If G is finite and g∈G, then g has finite order and

ι(ord⁡(g))∣ι(∣G∣)

in Z, where ι:N→Z is the canonical embedding.

Facts & Assumptions

Given: A finite group G and an element g∈G.

[F1]

The generated set ⟨g⟩ is a subgroup of G and equals the set of integer powers of g (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups, ⟨g⟩={ gn:n∈Z }, and every cyclic group is abelian).

[L1]

If g has finite order, then ⟨g⟩ is finite and ∣⟨g⟩∣=ord⁡(g); every element of a finite group has finite order (If ord⁡(g)=n then gk=e iff k is an integer multiple of n, the powers g0,…,gn−1 are distinct, and ⟨g⟩ has exactly n elements; if g has infinite order then gj=gk only for j=k, The order ∣G∣ of a finite group and the order ord⁡(g) of an element, with ord⁡(g)=∞ when no positive power of g is the identity).

[L2]

Lagrange's theorem gives ∣G∣=[G:H]∣H∣ for every subgroup H of a finite group and consequently ι(∣H∣)∣ι(∣G∣) (Lagrange's theorem: ∣G∣=[G:H]∣H∣ for every subgroup H of a finite group G, Divisibility in Z: d∣a when a=dq for some integer q, The naturals embed in the integers).

Proof

technique · direct
1.1F1L1

The element g has finite order, and ⟨g⟩≤G has order ∣⟨g⟩∣=ord⁡(g).

2.1step 1.1L2∎

Apply [L2] to H=⟨g⟩ to obtain ι(ord⁡(g))∣ι(∣G∣).

Depends on

Dependency tree · 2 levels

Sources