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CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)
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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.

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A finite group of prime order is cyclic and every nonidentity element generates it

Statement

Let G be a finite group such that the positive integer ι(∣G∣) is prime. Then every g≠e has order ∣G∣, satisfies ⟨g⟩=G, and hence generates G. In particular, G is cyclic.

Facts & Assumptions

Given: A finite group G with identity e, with ι(∣G∣) prime, and an element g∈G with g≠e.

[F1]

A prime integer p satisfies p>1, and every positive divisor of p is 1 or p (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L1]

The natural ord⁡(g) is positive, equals 1 exactly when g=e, and its image in Z divides ι(∣G∣); the embedding ι:N→Z is injective and preserves order (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, The order of every element of a finite group divides the order of the group, The naturals embed in the integers).

[L3]
[F2]

If a finite set G contains e and ∣G∣≠1, then some element of G differs from e: otherwise G={e}, whose cardinality is 1 (The cardinality ∣A∣ of a finite set).

Proof

technique · direct
1.1

The positive integer ι(ord⁡(g)) divides the prime ι(∣G∣), so it is 1 or ι(∣G∣). It is not 1 because g≠e, hence ord⁡(g)=∣G∣ by injectivity of ι.

givenF1L1
2.1

The subgroup ⟨g⟩⊆G has cardinality ord⁡(g)=∣G∣, so ⟨g⟩=G.

step 1.1L2L3
3.1

Thus every nonidentity element generates G. Since ι(∣G∣)>1=ι(1) by [F1], these two integers differ; injectivity in [L1] gives ∣G∣≠1, and [F2] supplies a nonidentity element. Consequently G is cyclic.

step 2.1F1F2L1∎

Depends on

Used by

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Sources