CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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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 homomorphism of finitely generated pro-p groups is surjective exactly when the induced map on Frattini quotients is surjective

Statement

Let $f : G \to H$ be a continuous homomorphism of finitely generated pro- $p$ groups. Then $f$ is surjective if and only if the induced linear map $G / Φ (G) \to H / Φ (H)$ is surjective.

Facts & Assumptions

Given: A continuous homomorphism $f : G \to H$ of finitely generated pro- $p$ groups.

[L1]

In a finitely generated pro- $p$ group, a subset topologically generates the group exactly when its image spans the Frattini quotient (A subset topologically generates a finitely generated pro-p group exactly when its image spans the Frattini quotient over Fp).

Proof

technique · direct

1.1

If $f$ is surjective, then every quotient map induced by $f$ , including $G / Φ (G) \to H / Φ (H)$ , is surjective.

given

1.2

Conversely, suppose $G / Φ (G) \to H / Φ (H)$ is surjective, and let $K : = f (G) \leq H$ . The image of $K$ in $H / Φ (H)$ is all of $H / Φ (H)$ by hypothesis, so [L1] says that $K$ topologically generates $H$ . But $K$ is compact as the continuous image of the profinite group $G$ , hence closed in the Hausdorff group $H$ . A closed subgroup whose closure is all of $H$ must equal $H$ , so $f$ is surjective.

L1givenalgebra

2.1

Steps 1.1 and 1.2 prove both implications. In the trivial-group boundary case, both maps are automatically surjective.

step 1.1step 1.2∎

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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)