Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
A partition of a group into cosets of subgroups of finite index admits multiplicity if for some distinct (p. 3).
Lemma 2.3 (p. 3). Let be a coset partition of a group without multiplicity, and put for . Then
- a) for distinct ;
- b) ;
- c) "for any ";
- d) if is a counterexample to the Herzog-Schönheim conjecture of minimal order, then for every .
Part c) is printed for any ; the paper's restatement just below (p. 3) reads it for distinct pairs, and with it holds only when , so it fails for the one-coset partition , which is not a partition the conjecture concerns.
Source. L. Margolis and O. Schnabel, The Herzog-Schönheim conjecture for small groups and harmonic subgroups, Beitr. Algebra Geom. 60 (2019), no. 3, 399--418, doi:10.1007/s13366-018-0419-1. Labels and pages are those of arXiv:1803.03569v1, the edition the source card names: the lemma is on p. 3.
Read depth. Claims checked: the statement was read clause by clause against the print. The paper gives no proof beyond its sources; nothing here is independently reviewed.
Proof pointer
p. 3. The paper calls a) and b) clear: a) restates the absence of multiplicity, and b) records that each coset takes the share of the group. Part c) follows from Lemma 2.2 (p. 3, after Ginosar and Schnabel [GS11, Corollary 2.1]): if then no coset of is disjoint from a coset of , and subgroups of coprime indices satisfy . Part d) is [GS11, Lemma 2.3].
Dependencies
Y. Ginosar and O. Schnabel, Prime factorization conditions providing multiplicities in coset partitions of groups, J. Comb. Number Theory 3 (2011), no. 2, 75--86 (the paper's [GS11]), Corollary 2.1 and Lemma 2.3.
Bears on
- Problem 274: for a finite group, an exact covering by two or more cosets of pairwise different sizes is a partition without multiplicity, so its indices satisfy a)--c), and in a counterexample of least order also d). These are the conditions from which the proof of Theorem A starts; they do not decide the problem.