Increasing unions of isomorphic subgroups

DefinitionClaim of Scottish Book Problem 62#✓fully proved

The claim that for every abelian group \(G\) and every sequence of subgroups \(H_0 \le H_1 \le H_2 \le \cdots\) of \(G\) whose union is \(G\) and each of which is isomorphic to \(H_1\), the group \(G\) is itself isomorphic to \(H_1\).

The problem is posed for groups; the claim here is restricted to abelian groups, which loses nothing for its refutation since Baer's counterexample is abelian.

def ScottishBookProblem62Claim : Prop :=
  ∀ (G : Type) [AddCommGroup G] (H : ℕ → AddSubgroup G),
    (∀ n, H n ≤ H (n + 1)) →
    (∀ n, Nonempty (↥(H n) ≃+ ↥(H 1))) →
    (⨆ n, H n) = ⊤ →
    Nonempty (G ≃+ ↥(H 1))

Used by 1

Used byRefutation of Scottish Book Problem 62

TheoremRefutation of Scottish Book Problem 62#●ready to prove

The claim of Scottish Book Problem 62 is false: there are an abelian group \(G\) and subgroups \(H_0 \le H_1 \le H_2 \le \cdots\) of \(G\) whose union is \(G\), each isomorphic to \(H_1\), such that \(G\) is not isomorphic to \(H_1\).

R. Baer's counterexample is \(G = \mathbb{Q}\) with \(H_n = \frac{1}{n!}\mathbb{Z}\): each \(H_n\) is isomorphic to \(\mathbb{Z}\) and their union is \(\mathbb{Q}\), but \(\mathbb{Q}\) is not isomorphic to \(\mathbb{Z}\).

theorem_wanted mazurUlam62_refuted : ¬ ScottishBookProblem62Claim

Sources

• S. Mazur and S. Ulam, Problem 62 of the Scottish Book; the typescript Ad 62 records R. Baer's remark giving the counterexample, with G the group of rational numbers and G 1 the integers: the rationals are the union of an increasing chain of subgroups each isomorphic to the integers (for example the subgroups (1 / n !) ℤ), but the rationals are not isomorphic to the integers.

Uses 1

Statement usesClaim of Scottish Book Problem 62

1 fully proved1 ready to prove

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