Wiki
Wiki

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

Updated


Statement

For a finite abelian group GG and S⊂GS\subset G, Σ∗(S)\Sigma^*(S) is the set of sums of nonempty subsets of SS (p. 143). Theorem 3.3 (p. 148). Let GG be a group of prime order pp and let S⊂GS\subset G. Then

∣S∣≥2p+5ln⁡pimplies0∈Σ∗(S).|S|\ge\sqrt{2p}+5\ln p\quad\text{implies}\quad0\in\Sigma^*(S).

Source. Y. O. Hamidoune and G. Zémor, On zero-free subset sums, Acta Arith. 78 (1996), no. 2, 143--152, DOI 10.4064/aa-78-2-143-152 (received 18 January 1996); Theorem 3.3 on printed p. 148 (PDF p. 6 of the publisher's ten-page file), read on the page image and in the text layer.

Read depth. Claims checked: the statement and the introduction's account (p. 143) of the constants c=2c=2 for prime order (via Olson) and c=3c=3 in general were read clause by clause. The half-page proof was read for structure only.

Proof pointer

If 0∉Σ∗(S)0\notin\Sigma^*(S) then S∩(−S)=∅S\cap(-S)=\emptyset; with s=∣S∣s=|S| and k=s−⌈log⁡3/2s⌉k=s-\lceil\log_{3/2}s\rceil, Corollary 3.2 (a lower bound for ∣Σ∗(S)∣|\Sigma^*(S)| from the Cauchy--Davenport theorem and the isoperimetric connectivity κ\kappa) gives a contradiction once 12k(k+1)−94k(1+ln⁡k)≥p\tfrac12k(k+1)-\tfrac94k(1+\ln k)\ge p (display (10)), which the stated hypothesis guarantees for p≥1000p\ge1000; smaller pp are covered by Theorem 2.6 (p. 144, Olson's bound: ∣S∣>4p−3|S|>\sqrt{4p-3} implies 0∈Σ∗(S)0\in\Sigma^*(S)).

Dependencies

The Cauchy--Davenport theorem (Theorem 2.1), three theorems of Olson (Section 2) and the paper's Lemma 3.1 and Corollary 3.2.

Bears on

  • Problem 540: for prime NN the threshold is 2N\sqrt{2N} up to 5ln⁡N5\ln N, close to the constant 2\sqrt2 Erdős speculated; superseded for primes by Balandraud's exact result and extended to all finite abelian groups by Theorem 4.5.