TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04

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 subset topologically generates a finitely generated pro-p group exactly when its image spans the Frattini quotient over Fp

Statement

Let $G$ be a finitely generated pro- $p$ group and let $S \subseteq G$ . Then $S$ topologically generates $G$ if and only if its image spans the elementary abelian quotient $G / Φ (G)$ over $F_{p}$ .

Facts & Assumptions

Given: A finitely generated pro- $p$ group $G$ and a subset $S \subseteq G$ .

[L1]

For such a group, $Φ (G) = \overline{[G, G] G^{p}}$ (For a finitely generated pro-p group, the Frattini subgroup is the closure of [G,G]G^p).

[L2]

In a finite $p$ -group, a subset generates exactly when its image in the Frattini quotient spans that quotient (Burnside Basis Theorem).

[L3]

For a finite group, generation is detected modulo the Frattini subgroup (Generation of a finite group is detected modulo its Frattini subgroup).

Proof

technique · direct

1.1

Suppose $S$ topologically generates $G$ . Then for every open normal subgroup $N$ , the image of $S$ generates the finite $p$ -group $G / N$ . By [L3], its image therefore generates $(G / N) / Φ (G / N)$ , and by [L2] that is the same as spanning the Frattini quotient over $F_{p}$ . Passing over all finite quotients shows that the image of $S$ spans $G / Φ (G)$ .

L1L2L3givenalgebra

2.1

Conversely, suppose the image of $S$ spans $G / Φ (G)$ . Let $N$ be any open normal subgroup. The image of $S$ in $(G / N) / Φ (G / N)$ then spans by functoriality of the Frattini quotient and [L1], so [L2] says that the image of $S$ generates $G / N$ . Hence the closure of the subgroup generated by $S$ surjects onto every finite quotient $G / N$ , and therefore equals $G$ . So $S$ topologically generates $G$ .

L1L2step 1.1algebra

3.1

Steps 1.1 and 2.1 prove the equivalence. The empty set fits the statement when $G = 1$ , because then $G / Φ (G) = 0$ .

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gareth Wilkes, Profinite Groups and Group Cohomology lecture notes (standard reference, not scraped)
Alexander Lubotzky, Combinatorial group theory for pro-p groups (standard reference, not scraped)