I'm looking for an upper bound for the number $k(G)$ of a finite group $G$, defined as follow:

Let $\mathcal{F}_k$ be the family of subsets of $G$ with size $k$, and we define $k(G)$ be the minimum $k$ such that every subset $X \in \mathcal F_k$ contains a non-empty sum-full set $S$, which is a set satisfies $$ S \subseteq S+S := \{ x+y \mid x,y \in S \}. $$

Note that the inequality $k(G) \leq |G|$ holds trivially since there is only one subset in $\mathcal F_{|G|}$ which is $G$ itself, and $G$ is a semigroup indeed.

Are there any papers or references about this number $k(G)$? Does it have a name? I'm interesting in particularly upper bounds of $k(G)$, but any related results are fine.


The restricted Davenport number $\hat{D}(G)$ of a group $G$, is defined as the smallest number $d$ such that given a subset $A \in \mathcal F_d$, there exists a zero-sum non-empty subset $S \subseteq A$, that is,

$$ \sum_{x \in S} x = 0, $$

where $0$ is the identity in $G$. In the paper "On a conjecture of Erdos and Heilbronn", Szemeredi has proved:

$$\hat{D}(G) = O(\sqrt{|G|}). $$

Hamidoune and Zemor set a precise bound $\sqrt{2}$ on the constant of the big-O notation.

I'm trying to provide a link between $\hat{D}(G)$ and the number $k(G)$; it seems to me that the size of sum-full sets in $G$ may related to the zero-sum problem. I'll provide the justification in another post, which is highly related.

  • $\begingroup$ Oops, I messed up with the definition of a semigroup. What I'm looking for is a sum-full set $A \subseteq A+A$, where a semigroup is a set with $A+A \subseteq A$. I'll modify the question. $\endgroup$ Commented Oct 31, 2010 at 7:30
  • $\begingroup$ You are considering only Abelian groups? $\endgroup$
    – user6976
    Commented Oct 31, 2010 at 12:56
  • $\begingroup$ When dealing with Davenport number, one usually considers Abelian groups. But I'm also interested in non-Abelian ones. $\endgroup$ Commented Nov 1, 2010 at 2:17

1 Answer 1


First of all the $k(G)$ cannot be smaller than the size of any proper subgroup of $G$, because if $H$ is a proper subgroup, $gH$ is a coset, $g\not\in H$, then $gHgH$ does not intersect $gH$ (if $ghgh'=gh''$, then $g\in H$). For example, if $G$ is Abelian, $|G|$ is not prime (i.e. $G$ is not cyclic of prime order), then $k(G)>\sqrt{|G|}$: look at the size of a maximal proper subgroup.

  • $\begingroup$ Mark, the question asks that each set of size $k$ should contain a sum-full set. It does not demand that any set of size $k$ should be sum-full itself. Am I completely misunderstanding your answer? $\endgroup$
    – Alex B.
    Commented Oct 31, 2010 at 15:52
  • 2
    $\begingroup$ @Alex: I corrected my answer. The coset $gH$ cannot contain a non-empty subset $X$ with $X\subseteq XX$. This is not a complete answer because I did not treat the case when $|G|$ is prime $\gt 3$, for example. But in that case $G=\langle a\rangle$, and one can take the set of all odd powers $\le |G|/2$. This subset does not contain any subset $X$ with $X\subseteq XX$. So in this case $k$ cannot be smaller than $|G|/4$. $\endgroup$
    – user6976
    Commented Oct 31, 2010 at 18:06
  • $\begingroup$ Yes, the fix was easy enough. Nice answer! $\endgroup$
    – Alex B.
    Commented Nov 1, 2010 at 0:10

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