Homomorphic images of ultraproducts of groups

AbbreviationUltraproduct kernel#✓fully proved

For an ultrafilter \(U\) on \(\mathbb{N}\) and a sequence \((G_i)_{i \in \mathbb{N}}\) of groups, the normal subgroup of \(\prod_i G_i\) generated by the sequences \((g_i)_i\) with \(\{i : g_i = 1\} \in U\). These sequences already form a normal subgroup, so the kernel consists of exactly them.

abbrev ultraproductKernel (U : Ultrafilter ℕ) (G : ℕ → Type) [∀ i, Group (G i)] :
    Subgroup (∀ i, G i) :=
  Subgroup.normalClosure {f | ∀ᶠ i in (U : Filter ℕ), f i = 1}

Used by 2

Used byKourovka Problem 20.13(a) (Bergman) · Ultraproduct of a sequence of groups

AbbreviationUltraproduct of a sequence of groups#✓fully proved

For an ultrafilter \(U\) on \(\mathbb{N}\) and a sequence \((G_i)_{i \in \mathbb{N}}\) of groups, the ultraproduct \(\left(\prod_i G_i\right)/U\) is the quotient of the direct product \(\prod_i G_i\) by the ultraproduct kernel, the normal subgroup of sequences \((g_i)_i\) with \(\{i : g_i = 1\} \in U\).

abbrev Ultraproduct (U : Ultrafilter ℕ) (G : ℕ → Type) [∀ i, Group (G i)] : Type :=
  (∀ i, G i) ⧸ ultraproductKernel U G

Uses 1 · Used by 1

Statement usesUltraproduct kernel

Used byKourovka Problem 20.13(a) (Bergman)

DefinitionKourovka Problem 20.13(a) (Bergman)#✓fully proved

The proposition that for every abelian group \(A\), every sequence \((G_i)_{i \in \mathbb{N}}\) of (not necessarily abelian) groups and every nonprincipal ultrafilter \(U\) on \(\mathbb{N}\) (one containing every cofinite set), if there is a surjective homomorphism from the ultraproduct \(\left(\prod_i G_i\right)/U\) onto \(A\), then there are a sequence \((H_i)_{i \in \mathbb{N}}\) of abelian groups and a nonprincipal ultrafilter \(V\) on \(\mathbb{N}\) with a surjective homomorphism from \(\left(\prod_i H_i\right)/V\) onto \(A\).

def abelianImageOfUltraproductIsImageOfAbelianUltraproduct : Prop :=
  ∀ (A : Type) [CommGroup A] (G : ℕ → Type) [∀ i, Group (G i)] (U : Ultrafilter ℕ),
    (U : Filter ℕ) ≤ Filter.cofinite →
    (∃ φ : Ultraproduct U G →* A, Function.Surjective φ) →
    ∃ (H : ℕ → Type) (_ : ∀ i, CommGroup (H i)) (V : Ultrafilter ℕ),
      (V : Filter ℕ) ≤ Filter.cofinite ∧ ∃ ψ : Ultraproduct V H →* A, Function.Surjective ψ

Uses 2 · Used by 1

Statement usesUltraproduct of a sequence of groups · Ultraproduct kernel

Used byAbelian images of ultraproducts are images of abelian ultraproducts

DefinitionKourovka Problem 20.13(b) (Bergman)#✓fully proved

The proposition that every abelian group \(A\) that is a homomorphic image of a direct product \(\prod_{i \in I} G_i\) of an infinite family \((G_i)_{i \in I}\) of finite (not necessarily abelian) groups is also a homomorphic image of the direct product \(\prod_{j \in J} H_j\) of some family \((H_j)_{j \in J}\) of finite abelian groups.

def abelianImageOfProductOfFiniteIsImageOfAbelianProduct : Prop :=
  ∀ (A : Type) [CommGroup A] (I : Type) [Infinite I] (G : I → Type) [∀ i, Group (G i)]
    [∀ i, Finite (G i)], (∃ φ : (∀ i, G i) →* A, Function.Surjective φ) →
    ∃ (J : Type) (H : J → Type) (_ : ∀ j, CommGroup (H j)), (∀ j, Finite (H j)) ∧
      ∃ ψ : (∀ j, H j) →* A, Function.Surjective ψ

Used by 1

Used byAbelian images of products of finite groups

TheoremAbelian images of ultraproducts are images of abelian ultraproducts#●ready to prove

Let \(A\) be an abelian group, let \((G_i)_{i \in \mathbb{N}}\) be a sequence of (not necessarily abelian) groups, and let \(U\) be a nonprincipal ultrafilter on \(\mathbb{N}\), that is, one containing every cofinite set. If there is a surjective homomorphism from the ultraproduct \(\left(\prod_i G_i\right)/U\) onto \(A\), then there are a sequence \((H_i)_{i \in \mathbb{N}}\) of abelian groups and a nonprincipal ultrafilter \(V\) on \(\mathbb{N}\) with a surjective homomorphism from \(\left(\prod_i H_i\right)/V\) onto \(A\).

public theorem_wanted abelianImageOfUltraproductIsImageOfAbelianUltraproduct_holds :
    abelianImageOfUltraproductIsImageOfAbelianUltraproduct

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 20.13 (G. M. Bergman); resolution: S. M. Corson, On homomorphic images of ultraproducts, preprint, arXiv:2503.09228 (2025), Theorem 17 with Proposition 10: an abelian group is cotorsion iff it is an image of (∏ G_i)/(⊕ G_i) for some sequence of groups, iff an image of a product of finite cyclic groups, iff an image of a product of finite groups, and cotorsion groups are images of nonprincipal ultraproducts of abelian groups; Questions 1 and 2 (Kourovka 20.13(a), (b)) are answered affirmatively, https://arxiv.org/abs/2503.09228.

Uses 1

Statement usesKourovka Problem 20.13(a) (Bergman)

TheoremAbelian images of products of finite groups#●ready to prove

Let \(A\) be an abelian group. If \(A\) is a homomorphic image of a direct product \(\prod_{i \in I} G_i\) of an infinite family \((G_i)_{i \in I}\) of finite (not necessarily abelian) groups, then \(A\) is a homomorphic image of the direct product \(\prod_{j \in J} H_j\) of some family \((H_j)_{j \in J}\) of finite abelian groups.

Corson proves more: such an \(A\) is a homomorphic image of a direct product of finite cyclic groups.

public theorem_wanted abelianImageOfProductOfFiniteIsImageOfAbelianProduct_holds :
    abelianImageOfProductOfFiniteIsImageOfAbelianProduct

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 20.13 (G. M. Bergman); resolution: S. M. Corson, On homomorphic images of ultraproducts, preprint, arXiv:2503.09228 (2025), Theorem 17 with Proposition 10: an abelian group is cotorsion iff it is an image of (∏ G_i)/(⊕ G_i) for some sequence of groups, iff an image of a product of finite cyclic groups, iff an image of a product of finite groups, and cotorsion groups are images of nonprincipal ultraproducts of abelian groups; Questions 1 and 2 (Kourovka 20.13(a), (b)) are answered affirmatively, https://arxiv.org/abs/2503.09228.

Uses 1

Statement usesKourovka Problem 20.13(b) (Bergman)

DefinitionGroups locally an image of the ultrapower \(\mathbb{Z}^{\mathbb{N}}/U\)#✓fully proved

Let \(U\) be an ultrafilter on \(\mathbb{N}\) and \(B\) a group. Then \(B\) is locally an image of the ultrapower \(\mathbb{Z}^{\mathbb{N}}/U\) when every element \(b \in B\) lies in a subgroup \(H\) of \(B\) that is a homomorphic image of \(\mathbb{Z}^{\mathbb{N}}/U\): there is a surjective homomorphism \(\psi\) from the additive group \(\mathbb{Z}^{\mathbb{N}}\) of integer sequences onto \(H\) such that \(\psi(f) = 1\) for every sequence \(f\) with \(f_i = 0\) for \(U\)-almost every \(i\).

def IsLocallyUltrapowerImage (U : Ultrafilter ℕ) (B : Type) [Group B] : Prop :=
  ∀ b : B, ∃ (H : Subgroup B) (ψ : Multiplicative (ℕ → ℤ) →* H), b ∈ H ∧ Function.Surjective ψ ∧
    ∀ f : ℕ → ℤ, (∀ᶠ i in (U : Filter ℕ), f i = 0) → ψ (Multiplicative.ofAdd f) = 1

Used by 1

Used byBergman's question on images of countable ultraproducts (Kourovka 21.12)

DefinitionBergman's question on images of countable ultraproducts (Kourovka 21.12)#✓fully proved

There is a nonprincipal ultrafilter \(U\) on \(\mathbb{N}\) such that every group \(B\) that is locally an image of the ultrapower \(\mathbb{Z}^{\mathbb{N}}/U\) (every element of \(B\) lies in a subgroup of \(B\) that is a homomorphic image of \(\mathbb{Z}^{\mathbb{N}}/U\)) is a homomorphic image of an ultraproduct \((\prod_{i \in \mathbb{N}} G_i)/U\): there are groups \(G_i\), \(i \in \mathbb{N}\), and a surjective homomorphism \(\varphi \colon \prod_i G_i \to B\) such that \(\varphi(f) = 1\) for every family \(f\) with \(f_i = 1\) for \(U\)-almost every \(i\).

The question is posed for a fixed nonprincipal \(U\); it is stated here for some \(U\), so that its refutation covers every \(U\).

def locallyUltrapowerImageIsUltraproductImage : Prop :=
  ∃ U : Ultrafilter ℕ, (U : Filter ℕ) ≤ Filter.cofinite ∧
    ∀ (B : Type) [Group B], IsLocallyUltrapowerImage U B →
      ∃ (G : ℕ → Type) (_ : ∀ i, Group (G i)) (φ : (∀ i, G i) →* B), Function.Surjective φ ∧
        ∀ f : ∀ i, G i, (∀ᶠ i in (U : Filter ℕ), f i = 1) → φ f = 1

Uses 1 · Used by 1

Statement usesGroups locally an image of the ultrapower \(\mathbb{Z}^{\mathbb{N}}/U\)

Used byBergman's question 21.12 has a negative answer

TheoremBergman's question 21.12 has a negative answer#●ready to prove

For every nonprincipal ultrafilter \(U\) on \(\mathbb{N}\) there is a group \(B\) that is locally an image of the ultrapower \(\mathbb{Z}^{\mathbb{N}}/U\) (every element of \(B\) lies in a subgroup of \(B\) that is a homomorphic image of \(\mathbb{Z}^{\mathbb{N}}/U\)) but is not a homomorphic image of any ultraproduct \((\prod_{i \in \mathbb{N}} G_i)/U\) of groups: for no groups \(G_i\), \(i \in \mathbb{N}\), is there a surjective homomorphism \(\varphi \colon \prod_i G_i \to B\) with \(\varphi(f) = 1\) for every family \(f\) such that \(f_i = 1\) for \(U\)-almost every \(i\).

Corson shows that the abelian group \(\bigoplus_n C_n\), with \(C_n\) finite cyclic of unbounded orders, is such a group for every nonprincipal \(U\): it is not a homomorphic image of any nonprincipal countable ultraproduct. The Kourovka Notebook marks 21.12 solved ("No, it need not").

public theorem_wanted locallyUltrapowerImageIsUltraproductImage_refuted :
    ¬ locallyUltrapowerImageIsUltraproductImage

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 21.12 (G. M. Bergman); resolution: S. M. Corson, On homomorphic images of ultraproducts, J. Algebra 681 (2025), 306-317, arXiv:2503.09228, Section 2, proof of the negative answer to Question 3 (Bergman's Question 19): for every nonprincipal ultrafilter U on ω the group ⊕ₙ Cₙ, Cₙ finite cyclic of unbounded orders, has each element in a homomorphic image of (∏ ℤ)/U but is not a homomorphic image of any nonprincipal ultraproduct (∏ Gᵢ)/V; Kourovka Notebook (21st issue), Problem 21.12: No, it need not, https://arxiv.org/abs/2503.09228.

Uses 1

Statement usesBergman's question on images of countable ultraproducts (Kourovka 21.12)

6 fully proved3 ready to prove

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