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CounterexampleConstruction: AI-adaptedVerification: 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.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Nonisomorphic groups can share the same profinite completion

Statement refuted

If two groups have the same profinite completion, then they are already isomorphic as abstract groups.

Facts & Assumptions

Given: The groups G=Z and H=Z⊕Q.

[L1]

The profinite completion depends only on the system of finite quotients (The profinite completion is the inverse limit of the finite quotients G over N, A homomorphism induces a continuous homomorphism of profinite completions).

Counterexample

technique · direct
1.1

As in FALSE: isomorphic profinite completions force the original groups to be isomorphic, every finite quotient of H factors through the projection onto Z. Hence G and H have the same finite quotients, namely the finite cyclic groups.

givenalgebra
2.1

Therefore [L1] gives isomorphic profinite completions, both equal to Z^. But G and H are not isomorphic because H contains the divisible subgroup Q and G does not.

L1step 1.1
3.1

This is a concrete counterexample to the statement.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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