Alphabeta Math
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 ιG:GG^ has kernel equal to the finite residual of G, and its image is dense in G^.

Facts & Assumptions

Given: An abstract group G with profinite completion G^ and canonical map ιG.

[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
1.1

An element gG lies in kerιG exactly when every coordinate gN is the identity coset in G/N. That is equivalent to gN for every finite-index normal subgroup N. By [L1], this means precisely gRf(G).

L1given
1.2

By [L2], let U be a nonempty basic open set of G^. Then U fixes finitely many coordinates, say at N1,,Nm, to compatible cosets gkNk. Let M:=N1Nm, which is again finite-index normal. Compatibility means exactly that these finitely many coordinates come from one coset gM in G/M. Since MNk, the image of gM in G/Nk is the prescribed coset gkNk for each k. Therefore ιG(g)U.

L1L2givenconstruct
2.1

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

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources