Davenport's constant for groups with large exponent

TL;DR

This paper establishes new bounds on Davenport's constant for finite abelian groups based on their exponent and order, confirming a conjecture for groups with large exponents.

Contribution

It proves a conjecture relating Davenport's constant to group parameters, providing bounds depending on the group's exponent relative to its order.

Findings

D(G) xp(G)+|G|/exp(G)-1 for xp(G)|G|

D(G) sqrt{|G|}-1 for xp(G)<|G|^{1/2}

Confirmed a conjecture by Balasubramanian and the first author.

Abstract

Let $G$ be a finite abelian group. We show that its Davenport constant $D(G)$ satisfies $D(G)\leq \exp(G)+\frac{|G|}{\exp(G)}-1$, provided that $\exp(G)\geq\sqrt{|G|}$, and $D(G)\leq 2\sqrt{|G|}-1$, if $\exp(G)<\sqrt{|G|}$. This proves a conjecture by Balasubramanian and the first named author.

Equations

$$\begin{array}[]{lcl}1+N^{p}(T)+N^{2p}(T)+N^{3p}(T)&\equiv&0\pmod{p}\\ 1+N^{p}(U)+N^{2p}(U)+N^{3p}(U)+N^{4p}(U)&\equiv&0\pmod{p}\\ 1+N^{p}(S)+N^{2p}(S)+N^{3p}(S)+N^{4p}(S)+N^{5p}(S)&\equiv&0\pmod{p}\\ \end{array}$$

$$D_{k}(\mathbb{Z}_{p}^{3})\leq\max\Big{(}5p-2,\frac{3(p-1)}{2}+2p+5\Big{)},$$

$$D_{k}(\mathbb{Z}_{p}^{d})\leq\max\big{(}(6p-4)p^{d-3}+1,\frac{3(p-1)}{2}k+1+(6p-4)p^{d-3}\big{(}\frac{1}{4}+\frac{3}{2p}-\frac{3}{4p^{2}}-\frac{1}{dp}\big{)}\Big{)}$$

$$\frac{|S|-(6p-4)p^{d-3}-1}{3(p-1)/2}+\frac{(6p-4)p^{d-3}-(6p-4)p^{d-4}}{2p}+\frac{(6p-4)p^{d-4}}{d(p-1)+1}$$

$$D_{k}(\mathbb{Z}_{p}^{d})\leq(6p-4)p^{d-3}+1+\\ \max\Big{(}0,\frac{3(p-1)}{2}\big{(}k-\frac{(6p-4)p^{d-3}-(6p-4)p^{d-4}}{2p}+\frac{(6p-4)p^{d-4}}{d(p-1)+1}\big{)}\Big{)},$$

$$k=\frac{N}{(a-1)p}-\sum_{\nu=a}^{d-1}\frac{\nu+1}{\nu(\nu-1)}p^{d-a}-1\geq\frac{N}{(a-1)p}\big{(}1-\frac{1}{a(1-p^{-1})}\big{)}$$

$$\frac{N-(a+1)p^{d-a+1}}{(a-1)p}+\frac{(a+1)p^{d-a+1}-(a+2)p^{d-a+2}}{ap}+\cdots +\frac{d{p}^2-(d+1)p}{ap}+1$$

$$D(H)\le \frac{|A|-M}{p}+k,$$

$$\frac{\exp (G)}{p}+\frac{|G|}{\exp (G)p^d}-1\le \frac{|A|-M}{p}+k,$$

$$\exp (G)+\frac{|G|}{\exp (G)p^{d-1}}-p\le \exp (G)+\frac{|G|}{\exp (G)}-1-M+pk.$$

$$1-p\le p^{d-1}-1-M+pk.$$

$$\sqrt{|G|/p^d}=\sqrt{|H|}>\exp (H)=\exp (G)/p\ge \sqrt{|G|/p^2},$$

$$2p\sqrt{|H|}-p\le 2\sqrt{|G|}-1-M+pN.$$

$$(2-2p^{-(d-2)/2})\sqrt{|G|}\ge p^{d-1}-p+1.$$

$$(2-2p^{-(d-2)/2})\sqrt{|G|}\ge (2-2p^{-(d-2)/2})p^d>\frac{1}{2}{p}^d>p^{d-1}-p+1,$$

$$\big{(}\exp(H)+\frac{|H|}{\exp(H)}-1\big{)}p\leq 2\sqrt{|G|}-1-M+kp,$$

$$\exp (G)+\frac{|G|}{p^{d-2}\exp (G)}-p\le 2\sqrt{|G|}-1-p^{d-1}.$$

$$\sqrt{|G|}\ge 1+p^{d-1}+\sqrt{|G|}{p}^{2-d}-p,$$

$$D_{n_1+n_2-1}({\mathbb{Z}}_p^d)\le 2\sqrt{p^d{n}_1{n}_2}-1.$$

$$n_{1}+n_{2}-1\leq\frac{N}{(a-1)p}\big{(}1-\frac{1}{a(1-p^{-1})}\big{)}.$$

$$\frac{ap^{d-a}}{(a-1)p}\big{(}1-\frac{1}{a(1-p^{-1})}\big{)}\geq\frac{a+p^{-(d-a)}}{4a}\big{(}ap^{-a}+p^{-d}\big{)},$$

$$\begin{gathered} \frac{2^d-2}{3}+\frac{2^{d-1}+2-2^{(d+1)/2}-1}{4}+\frac{2^{(d+1)/2}-d-2}{d+1}+1= \\ \frac{2^d}{4}+\frac{2^d+2}{24}-2^{(d-3)/2}+\frac{2^{(d+1)/2}-1}{d+1} \end{gathered}$$

$$\eta ({\mathbb{Z}}_3^d)\le \mathfrak{s}({\mathbb{Z}}_3^d)\le 3^{d-6}\mathfrak{s}({\mathbb{Z}}_3^6)<3^{d-1},$$

$$\eta ({\mathbb{Z}}_3^5)\le 3\lceil \frac{\eta ({\mathbb{Z}}_3^5)-16}{12}\rceil +3^4,$$

$$\Vert A{\Vert }_u:=\frac{1}{|A|}\underset{\xi \in {\mathbb{F}}_p^d\setminus \{0\}}{\max }\sum _{\alpha \in A}e(⟨\xi ,\alpha ⟩).$$

$$\frac{|A{|}^{p-1}}{p^{(p-1)d}}>\Vert A{\Vert }_u^{p-3}\left(\Vert A{\Vert }_u+\frac{p-1}{2p^{d-1}}\right)+\left(\genfrac{}{}{0ex}{}{p}{2}\right)\frac{|A{|}^{p-2}}{p^{(p-1)d}}$$

$$N\ge \frac{|A{|}^p}{p^d}-\Vert A{\Vert }_u^{p-2}|A|p^{(p-2)d}.$$

$$M\le \frac{|A{|}^{p-1}}{p^d}+\Vert A{\Vert }_u^{p-3}|A|p^{(p-3)d}.$$

$$N-M\ge \frac{|A{|}^p}{p^d}-\Vert A{\Vert }_u^{p-2}|A|p^{(p-2)d}-\frac{|A{|}^{p-1}}{p^d}-\Vert A{\Vert }_u^{p-3}|A|p^{(p-3)d},$$

$$\Vert A{\Vert }_u\le \frac{1}{|A|}\underset{0\le n_i\le 36}{\underset{n_1+\cdots +n_5=|A|}{\max }}|n_1+n_2\zeta +\cdots +n_5{\zeta }^4|.$$

Full text

1. Introduction and results

For an abelian group $G$ we denote by $D(G)$ the least integer $k$, such that every sequence $g_{1},\ldots,g_{k}$ of elements in $G$ contains a subsequence $g_{i_{1}},\ldots,g_{i_{\ell}}$ with $g_{i_{1}}+\dots+g_{i_{\ell}}=0$.

Write $G=\mathbb{Z}_{n_{1}}\oplus\dots\oplus\mathbb{Z}_{n_{r}}$ with $n_{1}|\dots|n_{r}$, where we write $\mathbb{Z}_{n}$ for $\mathbb{Z}/n\mathbb{Z}$. Put $M(G)=\sum n_{i}-r+1$. In several cases, including 2-generated groups and $p$-groups, the value of $D(G)$ matches with the obvious lower bound $M(G)$, however, in general this is not true. In fact there are infinitely many groups of rank 4 or more where $D(G)$ is greater than $M(G)$ see, for example, [11]. As far as upper bounds are concerned we have only rather crude ones. One such example, which is appealing for its simple structure, is the estimate D(G)\leq\exp(G)\big{(}1+\log\frac{|G|}{\exp(G)}\big{)}, due to van Emde Boas and Kruyswijk[4]. This bound, for the case when $\frac{|G|}{\exp(G)}$ is small, was improved by Bhowmik and Balasubramanian [1] who proved that $D(G)\leq\frac{|G|}{k}+k-1$, where $k$ is an integer $\leq\min(\frac{|G|}{\exp(G)},7)$, and conjectured that one may replace the constant $7$ by $\sqrt{|G|}$. Here we prove this conjecture. It turns out that the hypothesis that $k$ be integral creates some technical difficulties, therefore we prove the following, slightly sharper result.

Theorem 1.1**.**

For an abelian group $G$ with $\exp(G)\geq\sqrt{|G|}$ we have $D(G)\leq\exp(G)+\frac{|G|}{\exp(G)}-1$, while for $\exp(G)<\sqrt{|G|}$ we have $D(G)\leq 2\sqrt{|G|}-1$.

We notice that the first upper bound is actually reached for groups of rank 2 where $D(G)=\exp(G)+\frac{|G|}{\exp(G)}-1$. An application of our bound to random groups and $(\mathbb{Z}_{n}^{*},\cdot)$ will be the topic of a forthcoming paper.

Let $\mathfrak{s}_{\leq n}(G)$ be the least integer $k$, such that every sequence of length $k$ contains a subsequence of length $\leq n$ adding up to 0 and let $\mathfrak{s}_{=n}(G)$ be the least integer $k$ such that any sequence of length $k$ in $G$ contains a zero-sum of sequence of length exactly equal to $n$. In the special case where $n=\exp(G)$ we use the more standard notation of $\eta(G)$ and $\mathfrak{s}(G)$ respectively. We need the following bounds on $\eta$ and $\mathfrak{s}$.

Theorem 1.2**.**

1. (1)

We have $\mathfrak{s}(\mathbb{Z}_{3}^{3})=19$, $\mathfrak{s}(\mathbb{Z}_{3}^{4})=41$, $\mathfrak{s}(\mathbb{Z}_{3}^{5})=91$, and $\mathfrak{s}(\mathbb{Z}_{3}^{6})=225$. 2. (2)

We have $\mathfrak{s}(\mathbb{Z}_{5}^{3})=37$, $\mathfrak{s}(\mathbb{Z}_{5}^{4})\leq 157$, $\mathfrak{s}(\mathbb{Z}_{5}^{5})\leq 690$, and $\mathfrak{s}(\mathbb{Z}_{5}^{6})\leq 3091$. 3. (3)

If $p\geq 7$ is prime and $d\geq 3$, then $\eta(\mathbb{Z}_{p}^{d})\leq\frac{p^{d}-p}{p^{2}-p}(3p-7)+4$.

The above results for $\mathbb{Z}_{3}$ are due to Bose[5], Pelegrino[13], Edel, Ferret, Landjev and Storme[6], and Potechin[14], respectively. The value of $\mathfrak{s}(\mathbb{Z}_{5}^{3})$ was determined by Gao, Hou, Schmid and Tangadurai[10], the bounds for higher rank will be proven in section 4 using the density increment method. The last statement will be proven by combinatorial means in section 5.

We further need some information on the existence of zero-sums not much larger then $\exp(G)$.

Theorem 1.3**.**

Let $p$ be a prime, $d\geq 3$ an integer. Then a sequence of length $(6p-4)p^{d-3}+1$ in $\mathbb{Z}_{p}^{d}$ contains a zero-sum of length $\leq\frac{3p-1}{2}$. If $d\geq 4$, then a sequence of length $(6p-4)p^{d-4}+1$ in $\mathbb{Z}_{p}^{d}$ contains a zero-sum of length $\leq 2p$.

The proof of Theorem 1.1 uses the inductive method. To deal with the inductive step we require the following.

Theorem 1.4**.**

Let $p$ be a prime, $d\geq 2$ an integer. Then there exist integers $k,M$, such that $M\geq\eta(\mathbb{Z}_{p}^{d})$, every sequence of length $M$ contains at least $k$ disjoint zero-sums, and $M\leq p^{d-1}+pk$.

Note that the statement is trivial if $\eta(\mathbb{Z}_{p}^{d})\leq p^{d-1}$. However, this bound is false for $p=2$ and all $d$, as well as for the pairs $(3,3),(3,4),(3,5)$ and $(5,3)$. We believe that this is the complete list of exceptions. From the Alon-Dubiner-theorem and Roth-type estimates one can already deduce that the above bound for $\eta$ holds for all but finitely many pairs. However, dealing with the exceptional pairs by direct computation is way beyond current computational means.

2. Systems of disjoint zero-sums

Let $D_{k}(G)$ be the least integer $t$ such that every sequence of length $t$ in $G$ contains $k$ disjoint zero-sum sequences. The most direct way to prove the existence of many disjoint zero-sums is by proving the existence of rather short zero-sums, therefore we are interested in zero sums of length not much beyond $p$.

Lemma 2.1**.**

Every sequence of length $6p-3$ in $\mathbb{Z}_{p}^{3}$ contains a zero-sum of length $\leq\frac{3p-1}{2}$, every sequence of length $6p-3$ in $\mathbb{Z}_{p}^{4}$ contains a zero-sum of length $\leq 2p$, and every sequence of length $(d+1)p-d$ in $\mathbb{Z}_{p}^{d}$ contains a zero sum of length $\leq(d-1)p$.

Proof.

We claim that a sequence of length $6p-3$ in $\mathbb{Z}_{p}^{3}$ contains a zero sum of length $p$ or $3p$. To see this we adapt Reiher’s proof of Kemnitz’ conjecture [16]. For a sequence $S$ denote by $N^{\ell}(S)$ the number of zeo-sum subsequences of $S$ of length $\ell$. Let $S$ be a seauence of length $6p-3$ without a zero sum of length $p$ or $3p$, $T$ a subsequence of length $4p-3$, and $U$ a subsequence of length $5p-3$. Then the Chevalley-Warning theorem gives the following equations.

[TABLE]

By assumption $S$, and a forteriori $U$ and $T$ do not contain zero sums of length $p$ or $3p$, thus all occurrences of $N^{p}$ and $N^{3p}$ vanish. If $N^{5p}(S)\neq 0$, and $Z$ is a zero sum in $S$, then choosing for $T$ a subsequence of $Z$ of length $4p-3$ we find from the first equation that $T$ contains a zero sum $Y$ of length $2p$. But then $Z\setminus Y$ is a zero sum of length $3p$, a contradiction. We now add up the first equation over all subsequences $T$ of length $4p-3$, and the second over all subsequences of length $5p-3$, and obtain a system of three euations in the two variables $N^{2p}(S)$, $N^{4p}(S)$, which is unsolvable.

Now let $S$ be a sequence of length $6p-3$, and let $Z$ be a zero sum of length $p$ or $3p$. If $|Z|=p$, then we found a zeo sum of length $\leq\frac{3(p-1)}{2}$. Otherwise $Z$ contains a zero sum $Y$, and then either $Y$ or $Z\setminus Y$ is the desired zero-sum of length $\leq\frac{3(p-1)}{2}$.

The second claim follows similarly starting from the fact that every sequence of length $6p-3$ in $\mathbb{Z}_{p}^{4}$ contains a zero-sum subsequence of length $p,2p$ or $4p$, while the last one follows from the fact proven by Gao and Geroldinger[8, Theorem 6.7], that a sequence of length $(d+1)p-d$ contains a zero sum of length divisible by $p$. ∎

The next result is used to lift results for special groups $\mathbb{Z}_{p}^{d}$ to groups of arbitrary rank. The argument is rather wasteful, still the resulting bounds are surprisingly useful.

Lemma 2.2**.**

If $a\leq d$, then $\mathfrak{s}_{\leq k}(\mathbb{Z}_{p}^{d})\leq\frac{p^{d}-1}{p^{a}-1}(\mathfrak{s}_{\leq k}(\mathbb{Z}_{p}^{a})-1)+1$

Proof.

Let $A$ be a sequence of length $\frac{p^{d}-1}{p^{a}-1}(\mathfrak{s}_{\leq k}(\mathbb{Z}_{p}^{a})-1)+1$ in $\mathbb{Z}_{p}^{d}$. If $A$ contains 0, then we found a short zero sum. Otherwise let $U$ be a subgroup of $\mathbb{Z}_{p}^{d}$ with $U\cong\mathbb{Z}_{p}^{a}$ chosen at random. The expected number of elements of $A$, which are in $U$ is sightly bigger than $\mathfrak{s}_{\leq k}(\mathbb{Z}_{p}^{a})-1$, hence there exists a subgroup which contains at least $\mathfrak{s}_{\leq k}(\mathbb{Z}_{p}^{a})$ elements of the sequence. Restricting our attention to this subgroup we obtain the desired zero sum. ∎

Lemma 2.3**.**

We have

[TABLE]

and, for $d\geq 4$,

[TABLE]

Proof.

We only give the proof for the second inequality, the first one being significantly easier.

Let $S$ be a sequence of length at least $(6p-4)p^{d-3}+1$. Then we can find a zero sum of length $\leq\frac{3(p-1)}{2}$. We continue doing so until there are less zero-sums left. Then we remove zero sums of length $\leq 2p$, until there are less than $(6p-4)p^{d-4}+1$ points left. Among the remaining points we still find zero sums of length at most $D(\mathbb{Z}_{p}^{d})=d(p-1)+1$, hence, in total we obtain a system of at least

[TABLE]

disjoint zero sums. Hence,

[TABLE]

and our claim follows. ∎

The reader should compare our result with a similar bound given by Freeze and Schmid[7, Proposition 3.5]. In our result the coefficient of $k$ is smaller, while the constant term is much bigger. The following result is an interpolation between these results.

Lemma 2.4**.**

Let $N,d\geq 3$ be integers, $p$ a prime number, and define $a$ to be the largest integer such that $N>(a+1)p^{d-a+1}$. If $a\geq 2$, then $D_{k}(\mathbb{Z}_{p}^{d})\leq N$, where

[TABLE]

Proof.

Let $S$ be a sequence of length $N$ in $\mathbb{Z}_{p}^{d}$. We have to show that $S$ contains a systm of $k$ disjoint zero sums. Since $N>(a+1)p^{d-a+1}$, $S$ contains a zero sum of length $\leq(a-1)p$. We remove zero sums of this length, until the remaining sequence has length $<(a+1)p^{d-a+1}$. From this point onward we remove zero sums of length $\leq ap$, until the remainder has length $<(a+2)p^{d-a+2}$, and so on. In this way we obtain a disjoint system consisting of

[TABLE]

zero sums. This sum almost telescopes, yielding the first expression for $k$. For the inequality note that the sequence $\frac{\nu+1}{\nu(\nu-1)}$ is decreasing, hence the summands in the series are decreasing faster than the geometrical series $\sum p^{-\nu}$, and we conclude that the whole sum is bounded by the first summand multiplied by $(1-p^{-1})^{-1}$. Our claim now follows. ∎

The following result is a special case of a result of Lindström[12] (see also [7, Theorem 7.2, Lemma 7.4]).

Lemma 2.5**.**

Every sequence of length $2^{d-1}+1$ in $\mathbb{Z}_{2}^{d}$ contains a zero-sum of length $\leq 3$, and this bound is best possible. Every sequence of length $2^{(d+1)/2}+1$ in $\mathbb{Z}_{2}^{d}$ contains a zero-sum of length $\leq 4$.

3. Proof of Theorem 1.1

In this section we show that Theorem 1.4 implies Theorem 1.1.

Lemma 3.1**.**

Let $G$ be an abelian group of rank $r\geq 3$. Assume that Theorem 1.1 holds true for all proper subgroups of $G$. Then it holds true for $G$ itself.

Proof.

Let $p$ be a prime divisor of $|G|$. Choose an elementary abelian subgroup $U\cong\mathbb{Z}_{p}^{d}$ of $G$, such that $d\geq 3$, $\exp(G)=p\exp(G/U)$, and $|U|$ is minimal under these assumptions. Put $H=G/U$. Let $A$ be a sequence consisting of $\exp(G)+\frac{|G|}{\exp(G)}-1$ or $2\lfloor\sqrt{|G|}\rfloor-1$ elements, depending on whether $\exp(G)>\sqrt{|G|}$ or not. Denote by $\overline{A}$ the image of $A$ in $H$. Then we obtain a zero-sum, by choosing a large system of disjoint zero-sums in $\mathbb{Z}_{p}^{d}$, and then choosing a zero-sum among the elements in $H$ defined by these sums, provided that

[TABLE]

where $M\geq\eta(\mathbb{Z}_{p}^{d})$ and $k=k(p,d,M)$ is defined as in Theorem 1.4. The left hand side can be estimated using the inductive hypothesis. We have $\exp(H)=\frac{\exp(G)}{p}$, $|H|=\frac{|G|}{p^{d}}$. Assume first that $\exp(G)\geq\sqrt{|G|}$ and $\exp(H)\geq\sqrt{|H|}$. Then our claim follows, provided that

[TABLE]

inserting the choice of $A$ and rearranging terms this becomes

[TABLE]

The quotient of $G$ by its largest cyclic subgroup contains at least $\mathbb{Z}_{p}^{d-1}$, hence, $\frac{|G|}{\exp(G)}\geq p^{d-1}$. Clearly, by replacing $\frac{|G|}{\exp(|G|}$ with a lower bound we lose something, hence, it suffices to establish the relation

[TABLE]

However, this relation is implied by Theorem 1.4.

Next suppose that $\exp(G)\geq\sqrt{|G|}$ and $\exp(H)<\sqrt{|H|}$. Then

[TABLE]

thus $d<2$, but this case was excluded from the outset.

If $\exp(G)<\sqrt{|G|}$ and $\exp(H)<\sqrt{|H|}$, the same argument as in the first case yields $D(G)\leq 2\sqrt{G}-1$, provided that

[TABLE]

Since $|H|=\frac{|G|}{p^{d}}$ and $M-pN\leq p^{d-1}$ this becomes

[TABLE]

As $\exp(H)<\sqrt{H}$ we have that $H$ is of rank at least 3, which by our assumption on the size of $H$ implies that $|G|\geq p^{2d}$. This implies

[TABLE]

and our claim is proven.

If $\exp(G)<\sqrt{|G|}$ and $\exp(H)\geq\sqrt{H}$, the theorem follows provided that

[TABLE]

that is

[TABLE]

The bounds for $\exp(G)$ and $\exp(H)$ imply $\sqrt{|G|}p^{d/2-1}\leq\exp(G)<\sqrt{|G|}$, and in this range the left hand side is increasing as a function of $\exp(G)$, hence, this inequality is certainly true if

[TABLE]

which follows from $\sqrt{|G|}\geq p^{d}$. If this is not the case, then $|H|<p^{d}$, and by the choice of $p$ we have that $H$ has rank at most 2, that is, $H=\mathbb{Z}_{n_{1}}\oplus\mathbb{Z}_{n_{2}}$ and $G=\mathbb{Z}_{p}^{d-2}\oplus\mathbb{Z}_{pn_{1}}\oplus\mathbb{Z}_{pn_{2}}$, say. Then $D(H)=n_{1}+n_{2}-1$, thus it suffices to prove

[TABLE]

Denote the right hand side by $N$. Then Lemma 2.4 shows that our claim holds true, provided that

[TABLE]

Using the trivial bound $n_{1}+n_{2}-1\leq n_{1}n_{2}$ we find that this inequality follows from

[TABLE]

and by direct inspection we see that our claim follows for all $a\geq 2$, with exception only the case $(p,a)=(2,2)$. In this case our claim follows from Lemma 2.5, provided that $d>3$. Finally, if $p=2$ and $d=3$, then $D(G)=M(G)$ was shown by van Emde Boas[3] under the assumption that Lemma 5.1 holds true for all prime divisors of $|H|$, which we today know to hold for all primes. Hence the proof is complete. ∎

We know that $D(\mathbb{Z}_{n_{1}}\oplus\mathbb{Z}_{n_{2}})=n_{1}+n_{2}-1$, hence Theorem 1.1 holds true for all groups of rank $\leq 2$. Hence Theorem 1.1 follows by induction over the group order.

4. Proof of Theorem 1.4: The case $p\leq 7$

4.1. The primes 2 and 3

To prove Theorem 1.4 for $p=2$, we want to show that in a set of $2^{d}$ points we can find a system consisting of many disjoint zero-sums. We first remove one zero-sum of length $\leq 2$, then zero-sums of length $\leq 3$, until this is not possible anymore, and then we switch to zero-sums of length 4. Finally we remove zero-sums of length $\leq d+1$, which is possible in view of $D(\mathbb{Z}_{2}^{d})=d+1$. In this way we obtain at least

[TABLE]

zero-sums. Disregarding the last fraction we see that this quantity is $\geq 2^{d-2}$, provided that $d\geq 7$. For $3\leq d\leq 6$ we obtain our claim by explicitly computing this bound.

Next we consider $p=3$. For $d\geq 6$ we have

[TABLE]

hence, Theorem 1.4 holds true with $N=0$, $M=3^{d-1}$. For $d=5$ it follows from Lemma 2.1 that a sequence of length $\eta(\mathbb{Z}_{3}^{5})-3$ contains a system of $N=\lceil\frac{\eta(\mathbb{Z}_{3}^{5})-2d-6}{3d-3}\rceil$ disjoint zero-sums, hence, our claim follows provided that

[TABLE]

that is, $89\leq 21+81$. In the same way we see that for $d=4$ a sequence of length 39 in $\mathbb{Z}_{3}^{4}$ contains a system of 4 disjoint zero-sums, thus our claim follows from $39\leq 12+27$. Finally it is shown in [2, Proposition 1], that a sequence of length 15 in $\mathbb{Z}_{3}^{3}$ contains a system of 3 disjoint zero-sums. Together with $\eta(\mathbb{Z}_{3}^{3})=17$ our claim follows in this case as well.

4.2. The prime 5

We begin by proving the second statement of Theorem 1.2. We do so by using a density increment argument together with explicit calculations. Define the Fourier bias $\|A\|_{u}$ of a sequence $A$ over $\mathbb{F}_{p}^{d}$ as

[TABLE]

Then we have the following.

Lemma 4.1**.**

Let $p\geq 3$ be a prime number, $A$ be a sequence over $\mathbb{F}_{p}^{d}$. Then $A$ contains a zero-sum of length $p$, provided that

[TABLE]

Proof.

Let $N$ be the number of solutions of the equation $a_{1}+\dots+a_{p}=0$ with $a_{i}\in A$. From [17, Lemma 4.13] we have

[TABLE]

A solution $a_{1}+\dots+a_{p}=0$ corresponds to a zero-sum of $A$, if $a_{1},\ldots,a_{p}$ are distinct elements in $A$. Using Möbius inversion over the lattice of set partitions one could compute the over-count exactly, however, it turns out that the resulting terms are of negligible order, which is why we bound the error rather crudely. The number of solutions $M$ in which not all elements are different is at most $\binom{p}{2}$ times the number of solutions of the equation $2a_{1}+a_{2}+\dots+a_{p-1}=0$. Since multiplication by 2 is a linear map in $\mathbb{F}_{p}^{d}$ we have that $\|2A\|_{u}=\|A\|$, using [17, Lemma 4.13] again we obtain

[TABLE]

Hence the number of zero-sums is at least

[TABLE]

and our claim follows. ∎

We now use this lemma recursively to obtain bounds for $\mathfrak{s}(\mathbb{Z}_{5}^{d})$, starting from $\mathfrak{s}(\mathbb{Z}_{5}^{3})=37$.

Consider a 3-dimensional subgroup $U$, and let $\xi\in\mathbb{Z}_{5}^{4}$ be a vector such that $v\bot U$. Let $n_{1},\ldots,n_{5}$ be the number of elements of $A$ in each of the 5 cosets of $U$, $\zeta$ be a fifth root of unity. If $\max(n_{i})\geq 37$, we have a zero-sum of length $p$ in one of the hyperplanes. Hence

[TABLE]

Since $1+\zeta+\dots+\zeta^{4}=0$, we have

[TABLE]

that is,

[TABLE]

For $|A|\geq 144$ the right hand side equals $180-|A|$, and we obtain a zero-sum, provided that

[TABLE]

One easily finds that this is the case for $|A|=157$, and we deduce $\mathfrak{s}(\mathbb{Z}_{5}^{4})\leq 157$. The same argument yields for $d=5$ the inequality

[TABLE]

which is satisfied for $|A|\geq 690$, that is, we obtain $\mathfrak{s}(\mathbb{Z}_{5}^{5})\leq 690$. Finally for $\mathbb{Z}_{5}^{6}$ we obtain

[TABLE]

which is satisfied for $|A|\geq 3091$, thus the last inequality follows as well.

Hence, Theorem 1.2(2) is proven.

We have $\eta(\mathbb{Z}_{5}^{3})=33$, and among $33$ elements we can find one zero-sum of length $\leq 5$, one of length $\leq 10$, and one more among the remaining $18\geq D(\mathbb{Z}_{5}^{3})=13$ points. Hence we can take $M=33,N=3$, and Theorem 1.4 follows. Moreover we have $\eta(\mathbb{Z}_{5}^{4})\leq\mathfrak{s}(\mathbb{Z}_{5}^{4})-4\leq 153$, and among 153 elements we can find one zero-sum of length $\leq 5$, 13 zero-sums of length $\leq 10$, and one more zero-sum, that is, we can take $N=15$, and Theorem 1.4 follows for $d=4$ as well.

For $d=5$ we have $\eta(\mathbb{Z}_{5}^{5})\leq\mathfrak{s}(\mathbb{Z}_{5}^{5})-4\leq 686$, and among 686 points in $\mathbb{Z}_{5}$ we find 24 disjoint zero-sums of length $\leq 20$, thus taking $M=686$, $N=24$, our claim follows since $M\leq 625+120$. For $d\geq 6$ we have

[TABLE]

and our claim becomes trivial.

5. Proof of Theorem 1.4: The case $p\geq 7$

We begin by proving the last statement of Theorem 1.2.

Lemma 5.1**.**

Let $A$ be a sequence of length $3p-3$ in $\mathbb{Z}_{p}^{2}$ without a zero-sum of length $\leq p$. Then $A=\{a^{p-1},b^{p-1},c^{p-1}\}$ for suitable elements $a,b,c\in\mathbb{Z}_{p}^{2}$.

Proof.

A prime $p$ is said to satisfy property B if in every maximal zero-sum free subset of $\mathbb{Z}_{p}^{2}$ some element occurs with multiplicity at least $p-2$. Gao and Geroldinger[9] have shown that the condition of the above lemma holds true if $p$ has property B, and Reiher[15] has shown that every prime has property B. ∎

For $p=7$ we need a little more specific information.

Lemma 5.2**.**

Let $A$ be a sequence of length 15 over $\mathbb{Z}_{7}^{2}$, which does not contain a zero-sum of length $\leq 7$. Then there exist a cyclic subgroup which contains 3 elements of $A$.

Proof.

The proof can be done either by a mindless computer calculation or by a slightly more sophisticated human readable argument, however, as the latter also boils down to a sequence of case distinction we shall be a little brief. Let $A$ be a counterexample, that is, a zero-sum free sequence of length 15, such that every cyclic subgroup contains at most 15 points. We shall deduce properties of $A$ in a bootstrap manner.

Without loss we may assume that $A$ contains no two elements $x,y$ with $y=2x$. Suppose that $A$ contains two such elements. Then replacing $y$ by $x$ gives a new sequence $A^{\prime}$, such that for an element in $\mathbb{Z}_{p}^{2}$ the shortest representation as a subsum of $A^{\prime}$ is at least as long as the shortest representation as a subsum of $A$. In particular, $A^{\prime}$ contains no short zero sum.

There is at most one subgroup which contains two different elements. Without loss we may assume that $(1,0),(3,0),(0,1),(0,3)$ are in $A$. The subgroup generated by $(1,1)$ can contain either $(5,5)$ with multiplicity 2, or one of $(1,1),(2,2),(5,5)$ with multiplicity 1. If $(5,5)$ occurs twice, the remaining elements of the sequence must be among $\{(2,3),(2,4),(3,2),(3,5),(4,2),(4,5),(5,3),(5,4)\}$, which can easily be ruled out. If $(5,5)$ does not occur twice, then all subgroups different from $\langle(1,0)\rangle,\langle(0,1)\rangle,\langle(1,1)\rangle$ contain one element with multiplicity 2. The only possible elements in $\langle(1,-1)\rangle$ are $(1,6),(6,1)$, and by symmetry we may assume that $(6,1)$ occurs twice. Now $\langle(1,2)\rangle$ must contain $(6,5)$, and we conclude that the remaining points are $(2,6)$, $(3,1)$, $(5,4)$, and we obtain the zero-sum $(5,4)+(6,1)+(3,1)+(0,1)$.

There exist 3 different elements $x,y,z$, each of multiplicity 2 in $A$, such that $x+y\in\langle z\rangle$. Otherwise there are 6 elements of $\mathbb{Z}_{7}^{2}$, such that no two generate the same subgroup, and the sum of two different of them is contained in two fixed cyclic subgroups, which easily gives a contradiction.

$(1,0)$, $(0,1)$ and $(2,2)$ cannot all occur with multiplicity $2$. Suppose otherwise. Then the only further elements which can occur with multiplicity 2 are $(1,6)$, $(2,4)$, $(4,2)$, $(4,6)$, $(6,1)$, and $(6,4)$. Moreover, two elements which are exchanged by the map $(x,y)\mapsto(y,x)$ cannot both occur in $A$, hence we may assume that $(6,1)$ occurs twice in $A$, while $(1,6)$ does not. Then $(2,4)$ and $(4,6)$ occur twice in $A$, and we get the zero-sum $2\cdot(6,1)+(1,6)+(1,0)$.

$(1,0)$, $(0,1)$ and $(1,1)$ cannot all occur with multiplicity $2$. Using the previous result one finds that all further elements of multiplicity 2 have one coordinate equal to 1. By symmetry we may assume that there are two further elements of the form $(1,t)$. If there is an element of the form $(x,y)$, $2\leq x\leq 5$, this immediately gives a zero-sum of length $8-x$, hence all elements in $A$ are $(1,0)$, or of the form $(1,t)$, $(6,t)$. Since there are at least 8 different elements in $A$, there are at least 6 different elements of the form $(x,0)$, which can be written as the sum of one element of the form $(1,t)$ and one of the form $(6,t)$. Hence we obtain a zero-sum of length 2 or 3.

$(1,0)$, $(0,1)$ and $(4,4)$ cannot all occur with multiplicity $2$. There are at least 6 elements occurring with multiplicity 2, thus there are at least two further elements outside the subgroup $\langle(1,-1)\rangle$. But every element different from $(2,4),(3,5),(4,2),(5,3)$ immediately gives a zero-sum, and $(2,4)$ and $(4,2)$ as well as $(5,3)$ and $(3,5)$ cannot both occur at the same time, thus we may assume that $(5,3)$. The only possible element in $\langle(3,1)\rangle$ is $(1,5)$, and this element can only occur once. Hence $(2,4)$ becomes impossible, and we conclude that $(4,2)$ occurs with multiplicity 2. But then all elements in $\langle(1,-1)\rangle$ yield zero-sums.

We can now finish the proof. We know that there exist two elements $x,y\in A$, both with multiplicity 2, such that $\langle x+y\rangle$ contains an element of multiplicity 2. We may set $x=(1,0)$, $y=(0,1)$, and let $(t,t)$ be the element in $\langle x+y\rangle$. Then $t=0,3,5,6$ immediately yields z short zero-sum, while $t=1,2,4$ was excluded above. Hence no counterexample exists. ∎

Now suppose that $p\geq 7$ is a prime number, and $A$ is a sequence in $\mathbb{Z}_{p}^{d}$ with $|A|=n=\frac{p^{d}-p}{p^{2}-p}(3p-7)+4$ without zero-sums of length $\leq p$. Let $\ell$ be a one-dimensional subgroup of $\mathbb{Z}_{p}^{d}$, such that $m=|\ell\cap A|$ is maximal. Now consider all 2-dimensional subgroups containing $\ell$. Each such subgroup contains $p^{2}-p$ points outside $\ell$. Each point of $A$ is either contained in $\ell$ or occurs in $\frac{p^{2}-p}{p^{d}-p}$ of all such subgroups. Hence among all subgroups there is one which contains $\lceil\frac{p^{2}-p}{p^{d}-p}(n-m)\rceil$ points outside $\ell$. Call this subgroup $U$. Therefore $U$ contains at least

[TABLE]

elements of $A$. Since $\eta(\mathbb{Z}_{p}^{2})=3p-2$, this quantity is $\leq 3p-3$, which implies $m\leq 4$. Hence $m\leq 3$, which implies that $\frac{m-4}{p^{d-2}+\dots+1}$ is negative, and we find that $U$ contains $3p-6+m\leq 3p-4$ points, that is, $m\leq 2$. However, this implies that each of the $p+1$ one-dimensional subgroups of $U$ contain at most 2 elements of $A$, thus $3p-6\leq|A\cap U|\leq 2p+2$, which implies $p\leq 8$, hence, by our assumption $p=7$. In the case $p=7$ we obtain that $U\cong\mathbb{Z}_{7}^{2}$ contains a sequence $A$ of 15 elements, such that no cyclic subgroup contains more than 2 of them, and $A$ contains no zero-sum of length $\leq 7$.

We can now prove Theorem 1.4 for $p\geq 7$. We take $M=\frac{p^{d}-p}{p^{2}-p}(3p-7)+4$, and let $k$ be the largest integer for which Lemma 2.3 ensures $D_{k}(\mathbb{Z}_{p}^{d})\leq M$. Then the claim of Theorem 1.4 becomes

[TABLE]

for $d=3$, and

[TABLE]

for $d\geq 4$. After some computation one reaches the inequalities $4p^{2}\geq 6p+25$ and $28p^{4}\geq 144p^{3}+p^{2}-33$, which are satisfied for $p\geq 7$. Hence the proof of Theorem 1.4 is complete.