CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The canonical map is injective exactly when the group is residually finite

Statement

The canonical map to the profinite completion is injective if and only if the group is residually finite.

Facts & Assumptions

Given: An abstract group $G$ .

[L1]

The kernel of the canonical map $ι_{G} : G \to \hat{G}$ is the finite residual $R_{f} (G)$ (The canonical map to the profinite completion has kernel equal to the finite residual and has dense image).

[F1]

Residual finiteness means $R_{f} (G) = {1}$ (The finite residual is the intersection of the finite-index normal subgroups, and a group is residually finite when that intersection is trivial).

Proof

technique · direct

1.1

If $ι_{G}$ is injective, then its kernel is trivial. By [L1], the finite residual is therefore trivial, and [F1] says that $G$ is residually finite.

L1F1given

1.2

If $G$ is residually finite, then [F1] gives $R_{f} (G) = {1}$ . By [L1], this is exactly the statement that $\ker ι_{G} = {1}$ , so $ι_{G}$ is injective.

L1F1given

2.1

The two implications prove the equivalence.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

6 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

Brian Osserman, Math 6112 notes on inverse limits and profinite groups (standard reference, not scraped)
H. W. Lenstra, Profinite groups and Galois groups (standard reference, not scraped)