Erdős (Erdos) Problem #274: status in 2026

Erdős Problem #274

If G is a group, can there exist an exact covering of G by more than one coset of different sizes? (i.e. each element is contained in exactly one of the cosets.) The conjectured answer is no: in every such exact covering, two of the subgroups have the same cardinality.

Status (as of 2026-11-11): Open according to erdosproblems.com (status last updated 2025-08-31).

Current state (tracked automatically)

erdosproblems.com lists Erdős problem #274 as open (status updated 31 August 2025).

All tracked facts and sources

Fact	Value	Source
Status of Erdős problem #274 on erdosproblems.com	open 2025-08-31	erdosproblems.com
Formal Conjectures: research statements in the file	2 open, 1 solved	Formal Conjectures

Read twice a day from public sources by deterministic collectors (no AI; the sentences above are filled-in templates): erdosproblems.com (checked 2026-11-12), Formal Conjectures (checked 2026-11-12). How it works.

From the catalogue. Imported from The Formal Conjectures Authors (Google DeepMind and contributors) (Apache-2.0) — original. Nobody has started on it here yet: tasks are created as soon as someone asks for one or submits a claim.

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Cite

@misc{cairn-erdos-274,
  title        = {Erdős Problem #274},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/erdos-274}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY 4.0. Accessed 2026-11-12}
}

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Claims

Verified

Disputed

Refuted

On the literature board

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What a result looks like: C-16, a newcomer working with an AI model found C(85,24,3) ≤ 76 (previous best 77), and a computer verified it within minutes. A documented dead end counts too.

The problem

The question

erdos_274. If $G$ is a group, can there exist an exact covering of $G$ by more than one coset of different sizes? (i.e. each element is contained in exactly one of the cosets.)

The conjectured answer is no: in every such exact covering, two of the subgroups have the same cardinality.

herzog_schonheim. Let $G$ be a group, and let $A = {a_{1} G_{1}, \dots, a_{k} G_{k}}$ be a finite system of left cosets of subgroups $G_{1}, \dots, G_{k}$ of $G$ .

Herzog and Schönheim conjectured that if $A$ forms a partition of $G$ with $k > 1$ , then the indices $[G : G_{1}], \dots, [G : G_{k}]$ cannot be distinct.

Formal statement (Lean 4)

From Formal Conjectures, module FormalConjectures.ErdosProblems.«274» (2 statements).

theorem erdos_274 : ∃ (G : Type*) (_ : Group G),
    1 < ENat.card G ∧ ∃ (ι : Type*) (_ : Fintype ι) (P : Group.ExactCovering G ι),
    1 < Fintype.card ι ∧ ∀ i j, i ≠ j → #(P.parts i) ≠ #(P.parts j)

theorem herzog_schonheim {G : Type*} [Group G] (hG : 1 < ENat.card G) {ι : Type*} [Fintype ι]
    (hι : 1 < Fintype.card ι) (P : Group.ExactCovering G ι) :
    ∃ i j, i ≠ j ∧ (P.parts i).index = (P.parts j).index

What counts as progress

A Lean proof of one of the statements above, pinned as the claim's formal statement.
Partial results: special cases, weaker bounds, reductions — as verified claims.
Computations and numerical evidence with published code (reproducible).
Literature: the problem may have been solved or partly solved already — check erdosproblems.com/274. Report it as a literature claim.
A precise flaw in the formal statement (a misformalisation) — report it upstream too.

References

Source and licence

Imported from Formal Conjectures (Erdős problems), commit 2f91d251c07f. Statements and descriptions © The Formal Conjectures Authors, Apache License 2.0; reformatted for this page.