On complete subsets of the cyclic group

TL;DR

This paper proves a conjecture by Vu that a subset of the cyclic group Z_n with size greater than 1+2√(n-4) is complete, meaning every element of the subgroup can be expressed as a sum of distinct elements from the subset.

Contribution

The paper establishes the exact threshold size for completeness of subsets in cyclic groups, confirming Vu's conjecture and extending previous results.

Findings

Proves that subsets larger than 1+2√(n-4) are complete in Z_n.

Confirms Vu's conjecture on the size threshold for completeness.

Extends Olson's and Erdős-Heilbronn's results to a more general setting.

Abstract

A subset $X$ of an abelian $G$ is said to be {\em complete} if every element of the subgroup generated by $X$ can be expressed as a nonempty sum of distinct elements from $X$. Let $A\subset \Z_n$ be such that all the elements of $A$ are coprime with $n$. Solving a conjecture of Erd\H{o}s and Heilbronn, Olson proved that $A$ is complete if $n$ is a prime and if $|A|>2\sqrt{n}.$ Recently Vu proved that there is an absolute constant $c$, such that for an arbitrary large $n$, $A$ is complete if $|A|\ge c\sqrt{n},$ and conjectured that 2 is essentially the right value of $c$. We show that $A$ is complete if $|A|> 1+2\sqrt{n-4}$, thus proving the last conjecture.

Equations

$$k\wedge X=\left\{\sum_{x\in A}x\ \ \ \Big{|}\ \ \ A\subset X\mbox{ and }|A|=k\right\}.$$

$$S_X=\bigcup _{k\ge 1}k\wedge X.$$

$$S_X^0=S_X\cup \{0\}.$$

$$|X+Y|\ge \min (n,|X|+|Y|-1).$$

$${\lambda }_B(x)=|(B+x)\setminus B|.$$

$$|B|\ge |S_{Y\setminus \{y\}}^0|+{\lambda }_B(y),$$

$$|S_{Y\cup \{z\}}^0|=|S_Y^0|+{\lambda }_B(z).$$

$${\lambda }_B(x)$$

$${\lambda }_B(x+y)$$

$$\sum _{x\in C}{\lambda }_B(x)$$

$$|(B+x)\cap \overline{B}|$$

$${\lambda }_B(x+y)$$

$$\sum _{x\in C}{\lambda }_B(x)$$

$$\underset{x\in A}{\max }{\lambda }_B(x)>a-\frac{a(a-3)}{b}.$$

$$\underset{x\in A}{\max }{\lambda }_B(x)\ge a-1.$$

$$t=2ma+r,m\ge 0,0\le r\le 2a-1.$$

$$\sum _{x\in E}{\lambda }_B(x)$$

$$\alpha \ge \frac{4a{\sum }_{x\in E}{\lambda }_B(x)}{(t+a{)}^2}\ge \frac{4ab(t-b+1)}{(t+a{)}^2}.$$

$$\alpha$$

$$|S_A^0|\ge \min \{\frac{n+2}{2},3+\frac{|A|(|A|-1)}{2}\}.$$

$$|B|\ge |S_{A\setminus \{x\}}^0|+{\lambda }_B(x)\ge 3+(a-2)(a-1)/2+a-1=3+\frac{a(a-1)}{2},$$

$$|S_X^0|+|S_Y^0|\le n+1,$$

$$3+\lfloor \frac{k}{2}\rfloor (\lfloor \frac{k}{2}\rfloor -1)/2<(n+2)/2,$$

$$n/2>2+\frac{k}{2}(\frac{k}{2}-1)/2=2+k(k-2)/8,$$

$$|S_{A_2}^0|\ge 3+a(a-1)/2.$$

$${\lambda }_B(y)\ge a-1.$$

$$|S_{C_2}^0|\ge |S_{A_2}^0|+{\lambda }_B(y)\ge 3+a(a-1)/2+a-1=2+\frac{a(a+1)}{2}.$$

$$|S_{C_1}^0|\ge 3+\frac{a(a-1)}{2}.$$

$$n+1\ge |S_{C_1}^0|+|S_{C_1}^0|\ge 3+a(a-1)/2+2+a(a+1)/2=5+a^2.$$

Full text

Let $A\subset\mathbb{Z}_{n}$ be such that all the elements of $A$ are coprime with $n$. Solving a conjecture of Erdős and Heilbronn, Olson proved that $A$ is complete if $n$ is a prime and if $|A|>2\sqrt{n}.$ Recently Vu proved that there is an absolute constant $c$, such that for an arbitrary large $n$, $A$ is complete if $|A|\geq c\sqrt{n},$ and conjectured that $2$ is essentially the right value of $c$.

We show that $A$ is complete if $|A|>1+2\sqrt{n-4}$, thus proving the last conjecture.

1 Introduction

The additive group of integers modulo $n$ will be denoted by $\mathbb{Z}_{n}$.

Let $G$ be a finite Abelian group and let $X\subset G$. The subgroup generated by a subset $X$ of $G$ will be denoted $\langle{X}\rangle$. For a positive integer $k$, we shall write

[TABLE]

Following the terminology of [12] we write

[TABLE]

The set $X$ is said to be complete if $S_{X}=\langle{X}\rangle.$ The reader may find the connection between this notion and the corresponding notion for integers in [12]. We shall also write

[TABLE]

Note that $S_{X}^{0}=\sum_{x\in X}\{0,x\}$.

Let $p$ denote a prime number and let $A\subset\mathbb{Z}_{p}\setminus\{0\}$. Erdős and Heilbronn [4] showed that $A$ is complete if $|A|\geq\sqrt{18}\sqrt{p}$, and conjectured that $\sqrt{18}$ can be replaced by $2$. This conjecture was proved by Olson[8]. More precisely, Olson’s Theorem states that $A$ is complete if $|A|\geq\sqrt{4p-4}.$ This result was sharpened by Dias da Silva and one of the authors [1] by showing that $|k\wedge A|=p$, if $|A|\geq\sqrt{4p-4},$ where $k=\lceil\sqrt{p-1}\ \rceil$. They also showed that $|(j\wedge A)\cup((j+1)\wedge A)|=p$, if $|A|\geq\sqrt{4p-8},$ where $j=\lceil\sqrt{p-2}\ \rceil$.

Let $G$ be a finite abelian group and let $A\subset G\setminus\{0\}.$ Complete sets for general abelian group were investigated by Diderrich and Mann [3]. Diderrich [2] proved that, if $|G|=pq$ is the product of two primes, then $A$ is complete if $|A|\geq p+q-1.$

Let $p$ be the smallest prime dividing $|G|.$ Diderrich conjectured [2] that $A$ is complete, if $|G|/p$ is composite and $|A|=p+|G|/p-2.$ This conjecture was finally proved by Gao and one of the authors [5]. More precise results were later proved by Gao and the present authors [6]. Note that the bound of Diderrich is best possible, since one may construct non complete sets of size $p+|G|/p-3$.

However the result of Olson was extended recently by Vu [13] to general cyclic groups. Let $A\subset\mathbb{Z}_{n}$ be such that all the elements of $A$ are coprime with $n$. Vu proved that there is an absolute constant $c$ such that, for an arbitrary large $n$, $A$ is complete if $|A|\geq c\sqrt{n}.$ The proof of Vu is rather short and depends on a recent result of Szemerédi and Vu [11]. In the same paper Vu conjectures that the constant is essentially $2$.

Our main result is the following:

Theorem 1.1

Let $A$ be a subset of $\mathbb{Z}_{n}$ be such that all the elements of $A$ are coprime with $n$. If $|A|>1+2\sqrt{n-4}$ then $A$ is complete.

This result implies the validity of the last conjecture of Vu. We conjecture the following:

Conjecture 1.2

Let $A\subset\mathbb{Z}_{n}$ be such that all the elements of $A$ are coprime with $n$ and $|A|\geq\sqrt{4n-4}$. Then $|k\wedge A|=n$, where $k=\lceil\sqrt{n-1}\ \rceil$.

2 Some tools

In this section we present known material and some easy applications of it. We give short proofs in order to make the paper self-contained.

Recall the following well-known and easy lemma.

Lemma 2.1

Let $G$ be a finite group. Let $X$ and $Y$ be subsets of $G$ such that $X+Y\neq G.$ Then $|X|+|Y|\leq|G|$.

*Proof.  *  Take $a\in G\setminus(X+Y)$. We have $(a-Y)\cap X=\emptyset$.  $\Box$

We use also the Chowla’s Theorem [7, 10] :

Theorem 2.2** (Chowla [7, 10])**

Let $n$ be a positive integer and let $X$ and $Y$ be non-empty subsets of $\mathbb{Z}_{n}$. Assume that $0\in Y$ and that the elements of $Y\setminus\{0\}$ are coprime with $n$. Then

[TABLE]

*Proof.  *  The proof is by induction on $|Y|$, the result being obvious for $|Y|=1.$ Assume first that $Y\subset X-x,$ for all $x\in X.$ Then $X+Y\subset X$, and hence $X+Y=X$. It follows that $X+Y=X+nY=\mathbb{Z}_{n}.$

Assume now that $Y\not\subset X-x,$ for some $x\in X.$ Then $0\in Y\cap(X-x)$ and $|Y\cap(X-x)|<|Y|.$ By the induction hypothesis, $|X|+|Y|-1\leq|((X-x)\cup Y)+((X-x)\cap Y)|\leq|(X-x)+Y|.$  $\Box$

Let $B\subset G$ and $x\in G$. Following Olson, we write

[TABLE]

The following result is implicit in [8]:

Lemma 2.3** (Olson, [8])**

Let $Y$ be a nonempty subset of $G\setminus\{0\}$, $z\notin Y$ and $y\in Y$. Put $B=S_{Y}^{0}$. Then

[TABLE]

and

[TABLE]

*Proof.  *  Clearly we have $B\cap(\overline{B}-y)\subset B\setminus S_{Y\setminus\{y\}}^{0}$ and hence $\lambda_{B}(y)=|B\cap(\overline{B}-y)|\leq|B|-|S_{Y\setminus\{y\}}^{0}|$.

¿From $S_{Y\cup\{z\}}^{0}=B+\{0,z\}$ we have $|S_{Y\cup\{z\}}^{0}|=|B|+|(B+z)\setminus B|\}=|B|+\lambda_{B}(z).$  $\Box$

We need the following helpful result also due to Olson:

Lemma 2.4** (Olson [8])**

Let $B$ and $C$ be nonempty subsets of an abelian group $G$ such that $0\not\in C$. Then,

[TABLE]

*Proof.  *  For each $x\in G$ we have

[TABLE]

proving (3). Let $x,y\in G$. Then,

[TABLE]

proving (4). Finally,

[TABLE]

proving (5).  $\Box$

3 The main result

The next Lemma is the key tool for our main result.

Lemma 3.1

Let $A$ and $B$ be nonempty subsets of $\mathbb{Z}_{n}$. Assume that $A\cap(-A)=\emptyset$ and that each element in $A$ is coprime with $n$. Put $a=|A|$ and $b=|B|$. Assume also that $a\geq 3$ and $2b\leq n+2$. Then

[TABLE]

In particular, if $2b\geq a(a-3)$, then

[TABLE]

*Proof.  *  Put $A^{*}=A\cup(-A)\cup\{0\}$. Let $t<n$ be a positive integer and set

[TABLE]

Let $C_{j}=jA^{*}$. By Chowla’s theorem, $|C_{j}|\geq\min\{n,2ja+1\}=2ja+1$, for $j\leq m$. Therefore we can choose a set $C\supset A^{*}$ of cardinality $t+1$ which intersects $C_{j}$ in exactly $2ja$ elements $j=2,\ldots,m$, and intersects $C_{m+1}$ in exactly $r$ elements. Let $E=C\setminus\{0\}$. Let $\alpha=\max\{\lambda_{B}(x):\;x\in A\}$. By (3) we have $\lambda_{B}(x)\leq\alpha,$ for all $x\in A^{*}$. For an element $x$ in $C_{j}$ there are elements $x_{1},\cdots,x_{j}\in A^{*}$ such that $x=x_{1}+\cdots+x_{j}$. In view of (4) we have $\lambda_{B}(x)\leq\lambda(x_{1})+\cdots+\lambda(x_{j})\leq j\alpha.$ Therefore,

[TABLE]

By using (5) we have

[TABLE]

In particular, since $2b\leq n+2$, we can set $t=2b-3$ to get,

[TABLE]

where we have used $a\geq 3.$ In particular, if $2b\geq a(a-3)$, then $\alpha>a-2$ so that $\alpha\geq a-1$. This completes the proof.  $\Box$

Lemma 3.1 gives the following estimation for the cardinality of the set of subset sums.

Lemma 3.2

Let $A\subset\mathbb{Z}_{n}$ such that $A\cap(-A)=\emptyset$ and every element of $A$ is coprime with $n$. Also assume $|A|\geq 2.$ Then

[TABLE]

*Proof.  *  We shall prove the result by induction on $a=|A|$, the result being obvious for $a=2$. Suppose $a>2$. Put $B=S_{A}^{0}$. We may assume $b=|B|\leq\frac{n}{2}+1$ so that $2b\leq n+2$. By the induction hypothesis, $2b\geq 6+(a-1)(a-2)>a(a-3)$.

By (7) there is an $x\in A$ with $\lambda_{B}(x)\geq a-1$. Then, by Lemma 2.3,

[TABLE]

as claimed. $\Box$

We are now ready for the proof of Theorem 1.1.

Proof of Theorem 1.1. Suppose $A$ non complete and put $|A|=k$. Let $X,Y$ be disjoint subsets of $A$. We clearly have $S_{X}+S_{Y}^{0}\subset S_{A}\neq\mathbb{Z}_{n}$. Since $|S_{X}|\geq|S_{X}^{0}|-1,$ we have

[TABLE]

by Lemma 2.1.

Partition $A=A_{1}\cup A_{2}$ into two almost equal parts, i.e. $|A_{1}|=\lceil k/2\rceil$ and $|A_{2}|=\lfloor k/2\rfloor$, such that $A_{i}\cap(-A_{i})=\emptyset$, $i=1,2$.

We must have

[TABLE]

since otherwise, by Lemma 3.2, we have $|S_{A_{1}}^{0}|+|S_{A_{2}}^{0}|\geq n+2,$ contradicting (8).

Case 1. $k$ even.

Then we have by (9)

[TABLE]

and hence $(k-1)^{2}+16\leq 4n,$ a contradiction.

Case 2. $k$ odd.

Put $a=\frac{k-1}{2}$. In view of (9), Lemma 3.2 implies

[TABLE]

By (7) with $B=S_{A_{2}}^{0}$, there is a $y\in A_{1}$ such that

[TABLE]

Put $C_{1}=A_{1}\setminus\{y\}$ and $C_{2}=A_{2}\cup\{y\}$. Then we have, by Lemma 2.3,

[TABLE]

On the other hand, from (9) and Lemma 3.2 we get

[TABLE]

By (8),

[TABLE]

Therefore $4n\geq 16+4a^{2}=16+(k-1)^{2}$, a contradiction. This completes the proof. $\Box$