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

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The canonical map to the profinite completion has kernel equal to the finite residual and has dense image

Statement

The canonical map $ι_{G} : G \to \hat{G}$ has kernel equal to the finite residual of $G$ , and its image is dense in $\hat{G}$ .

Facts & Assumptions

Given: An abstract group $G$ with profinite completion $\hat{G}$ and canonical map $ι_{G}$ .

[L1]

The profinite completion is the inverse limit of the quotients $G / N$ , and $R_{f} (G)$ is the intersection of the finite-index normal subgroups (The profinite completion is the inverse limit of the finite quotients G over N, The canonical map sends g to its coherent system of residue classes, The finite residual is the intersection of the finite-index normal subgroups, and a group is residually finite when that intersection is trivial).

[L2]

The completion carries the inverse-limit topology from its finite discrete quotients (The profinite completion is the inverse limit of the finite quotients G over N).

Proof

technique · direct

1.1L1given

An element $g \in G$ lies in $\ker ι_{G}$ exactly when every coordinate $g N$ is the identity coset in $G / N$ . That is equivalent to $g \in N$ for every finite-index normal subgroup $N$ . By [L1], this means precisely $g \in R_{f} (G)$ .

1.2L1L2givenconstruct

By [L2], let $U$ be a nonempty basic open set of $\hat{G}$ . Then $U$ fixes finitely many coordinates, say at $N_{1}, \dots, N_{m}$ , to compatible cosets $g_{k} N_{k}$ . Let $M : = N_{1} \cap \dots \cap N_{m}$ , which is again finite-index normal. Compatibility means exactly that these finitely many coordinates come from one coset $g M$ in $G / M$ . Since $M \subseteq N_{k}$ , the image of $g M$ in $G / N_{k}$ is the prescribed coset $g_{k} N_{k}$ for each $k$ . Therefore $ι_{G} (g) \in U$ .

2.1step 1.1step 1.2∎

Step 1.2 says that every nonempty basic open set meets $ι_{G} [G]$ , so the image is dense. Together with step 1.1, this proves the theorem.

Depends on

Used by

Dependency tree · two levels

10 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)