The Large Davenport Constant I: Groups with a Cyclic, Index 2 Subgroup

TL;DR

This paper determines the large Davenport constant for finite groups with a cyclic, index 2 subgroup, revealing it equals the small Davenport constant plus the size of the commutator subgroup.

Contribution

It provides a precise formula for the large Davenport constant in groups with a cyclic, index 2 subgroup, extending previous results and clarifying the relationship between the constants.

Findings

For non-cyclic groups, (G)=|G|/2.

For cyclic groups, (G)=|G|-1.

The large Davenport constant is (G)+|G'|.

Abstract

Let $G$ be a finite group written multiplicatively. By a sequence over $G$, we mean a finite sequence of terms from $G$ which is unordered, repetition of terms allowed, and we say that it is a product-one sequence if its terms can be ordered so that their product is the identity element of $G$. The small Davenport constant $\mathsf d (G)$ is the maximal integer $\ell$ such that there is a sequence over $G$ of length $\ell$ which has no nontrivial, product-one subsequence. The large Davenport constant $\mathsf D (G)$ is the maximal length of a minimal product-one sequence---this is a product-one sequence which cannot be factored into two nontrivial, product-one subsequences. It is easily observed that $\mathsf d(G)+1\leq \mathsf D(G)$, and if $G$ is abelian, then equality holds. However, for non-abelian groups, these constants can differ significantly. Now suppose $G$ has a cyclic, index 2 subgroup. Then an old result of Olson and White (dating back to 1977) implies that $\mathsf d(G)=\frac12|G|$ if $G$ is non-cyclic, and $\mathsf d(G)=|G|-1$ if $G$ is cyclic. In this paper, we determine the large Davenport constant of such groups, showing that $\mathsf D(G)=\mathsf d(G)+|G'|$, where $G'=[G,G]\leq G$ is the commutator subgroup of $G$.