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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The answer to Problem 1179 is yes: for every fixed 0<ϵ<10<\epsilon<1,

gϵ(N)=(1+oϵ(1))log⁡2N.g_\epsilon(N)=(1+o_\epsilon(1))\log_2N.

The lower bound is trivial, since the 2k2^k subset sums of a kk-element set can meet every element of a group of order NN only when 2k≥N2^k\ge N. The upper bound is the Theorem of P. Erdős and R. R. Hall, Probabilistic methods in group theory, II: in a finite abelian group GG of order nn, with R(g)R(g) the number of representations g=ϵ1g1+⋯+ϵkgkg=\epsilon_1g_1+\cdots+\epsilon_kg_k, ϵi∈{0,1}\epsilon_i\in\{0,1\}, and η>0\eta>0 fixed, almost all choices of g1,…,gkg_1,\ldots,g_k (all but o(nk)o(n^k) of the nkn^k ordered choices) satisfy (1−η)2k/n<R(g)<(1+η)2k/n(1-\eta)2^k/n<R(g)<(1+\eta)2^k/n for every g∈Gg\in G, provided

k≥log⁡nlog⁡2(1+O(log⁡log⁡log⁡nlog⁡log⁡n)),k\ge\frac{\log n}{\log2} \Bigl(1+O\Bigl(\frac{\log\log\log n}{\log\log n}\Bigr)\Bigr),

the implied constant depending only on η\eta; the result also holds with η→0\eta\to0 as long as log⁡(1/η)=O(log⁡n/log⁡log⁡n)\log(1/\eta)=O(\log n/\log\log n). The paper chooses the kk elements independently with repetition allowed, where the problem takes a uniformly random kk-element subset AA and counts FA(g)F_A(g) over subsets of AA. With k=O(log⁡n)k=O(\log n) the probability of a repeated element is O(k2/n)→0O(k^2/n)\to0, the ordered choice conditioned on distinct entries is a uniformly random ordered kk-subset, and R(g)=FA(g)R(g)=F_A(g) for distinct entries, so the paper's "almost all" statement is the problem's probability-tending-to-one statement; this bridge is this page's, not the paper's. The proof is a second-moment argument combined with Watson's Lemma 1, which bounds the number of choices satisfying a system of 00-11 linear equations, and conditional-probability estimates for coinciding subset sums; the authors call the theorem sharp except for the OO-terms. The source card erdos_1976_probabilistic_methods_group_theory records the paper. The earlier Theorem 1 of Erdős and Rényi (1965), an accepted partial claim on [[problems/additive_combinatorics/E1179/claims/1965_12_01_erdos_renyi|its claim page]], gives gϵ(N)≤(2+o(1))log⁡2N+Oϵ(1)g_\epsilon(N)\le(2+o(1))\log_2N+O_\epsilon(1), and its authors conjectured that the factor 22 could not be removed without structural hypotheses on the group; the 1976 theorem removes it.

Formalization. Boris Alexeev's lean-proofs repository holds, since 2026-08-17, a Lean 4 development that declares itself a formalization of the Erdős–Hall solution, with Erdős and Hall as its informal authors and Codex and GPT-5.6 Sol as its formal authors (src/latest/ErdosProblems/Erdos1179.lean, linked above). Its theorem erdos_1179 proves the trivial lower bound, that an explicit Erdős–Hall size erdos1179Size N, divided by log⁡2N\log_2N, tends to 11, and that with that size the success probability tends to 11 along every sequence of finite abelian groups whose orders grow; it also formalizes the transfer from independent ordered samples to uniformly random kk-subsets, the bridge stated above. The formal-conjectures statement file for the problem, added 2026-09-20 and linked above, points its formal_proof attributes at this file. This corpus has not built the development, so no formalized evidence is listed.

Depends on. Nothing in this wiki.

Acceptance. Refereed publication: Houston J. Math. 2 (1976), no. 2, 173--180, received 1 December 1975 as the paper prints; the paper has no DOI, the issue month is not printed, and the year is filled to its first day for this page's name. Reviewed: the site's curator, Thomas Bloom, labels the problem proved and records the Erdős–Hall bound [ErHa76] as the answer beside the trivial lower bound in the problem page's commentary; he is independent of the authors. The proof is not compiled in this corpus.