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

How statement and proof provenance work

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.

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 $g \neq e$ 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 $g \in G$ with $g \neq e$ .

[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 $ι : N \to Z$ is injective and preserves order (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ 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).

[L2]

The cyclic subgroup $⟨ g ⟩$ is a subgroup with $∣ ⟨ g ⟩ ∣ = ord (g)$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian, If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L3]

If $A \subseteq B$ are finite and $∣ A ∣ = ∣ B ∣$ , then $A = B$ (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[F2]

If a finite set $G$ contains $e$ and $∣ G ∣ \neq 1$ , 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

1.1givenF1L1

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

2.1step 1.1L2L3

The subgroup $⟨ g ⟩ \subseteq G$ has cardinality $ord (g) = ∣ G ∣$ , so $⟨ g ⟩ = G$ .

3.1step 2.1F1F2L1∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources