CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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A nontrivial finite $p$ -group is cyclic exactly when $d (P) = 1$

Statement

A nontrivial finite $p$ -group $P$ is cyclic if and only if $d (P) = 1$ , while $d (1) = 0$ .

Facts & Assumptions

Given: A finite $p$ -group $P$ .

[L1]

A subset $X \subseteq P$ is minimally generating exactly when the quotient map restricts to a bijection from $X$ onto a basis of $P / Φ (P)$ (Burnside Basis Theorem).

[F1]

The generator rank $d (P)$ is the common size of a basis of $P / Φ (P)$ (The generator rank $d (P)$ of a finite $p$ -group).

[F2]

A group $G$ is cyclic when $G = ⟨ g ⟩$ for some $g \in G$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

Proof

technique · direct

1.1

For the reverse direction, if $d (P) = 1$ , choose a one-element quotient basis and lift it. By [L1] the lift is a one-element generating set, so $P$ is cyclic by [F2].

givenL1F1F2

1.2

For the forward direction, if nontrivial $P$ is cyclic with generator $g$ , then ${g}$ is minimally generating because the empty set generates only the trivial subgroup. By [L1], its quotient image is a one-element basis, so $d (P) = 1$ by [F1].

givenL1F1algebra

2.1

If $P = 1$ , its Frattini quotient has the empty basis, hence $d (P) = 0$ by [F1]; this is why nontriviality is needed in the biconditional.

step 1.1step 1.2F1∎

Depends on

Used by

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Sources

K. Conrad, Generating Sets, §6 (standard reference, not scraped)

A nontrivial finite pp-group is cyclic exactly when d(P)=1d(P)=1