Zero-sum subsets of decomposable sets in Abelian groups

TL;DR

This paper investigates the structure of decomposable subsets in Abelian groups, providing partial answers to an open problem about zero-sum subsets, with constructions and bounds in various groups.

Contribution

It offers new constructions of decomposable sets with specific zero-sum properties and establishes bounds for such sets in different groups.

Findings

Existence of decomposable sets with zero sum in cyclic groups of order 2^n-1.

Every decomposable subset of size ≤7 in ℝ contains a small zero-sum subset.

Construction of subsets in ℤ with specific zero-sum subset properties.

Abstract

A subset $D$ of an Abelian group is $decomposable$ if $\emptyset\ne D\subset D+D$. In the paper we give partial answer to an open problem asking whether every finite decomposable subset $D$ of an Abelian group contains a non-empty subset $Z\subset D$ with $\sum Z=0$. For every $n\in\mathbb N$ we present a decomposable subset $D$ of cardinality $|D|=n$ in the cyclic group of order $2^n-1$ such that $\sum D=0$, but $\sum T\ne 0$ for any proper non-empty subset $T\subset D$. On the other hand, we prove that every decomposable subset $D\subset\mathbb R$ of cardinality $|D|\le 7$ contains a non-empty subset $Z\subset D$ of cardinality $|Z|\le\frac12|D|$ with $\sum Z=0$. For every $n\in\mathbb N$ we present a subset $D\subset\mathbb Z$ of cardinality $|D|=2n$ such that $\sum Z=0$ for some subset $Z\subset D$ of cardinality $|Z|=n$ and $\sum T\ne 0$ for any non-empty subset $T\subset D$ of cardinality $|T|<n=\frac12|D|$. Also we prove that every finite decomposable subset $D$ of an Abelian group contains two non-empty subsets $A,B$ such that $\sum A+\sum B=0$.

Equations

$$D_n:=\{2^k:0\le k<n\}\cup \{2^k-2^n+1:0\le k<n\}$$

$$Z=\{2^k:1\le k<n\}\cup \{2-2^n\}\subset D_n$$

$$D_n:=A\cup (A-(2^n-1))$$

$$\{-3,-2,-1,1,2\}\cdot r,\{-4,-2,-1,1,2\}\cdot r,\{-3,-2,-1,2,4\}\cdot r$$

$$0=n^{\prime }+a+\bar{p}\ge n^{\prime }+p^{\prime }+p^{\prime }>p^{\prime }+n^{\prime }>p^{\prime }+n^{\prime }+n^{\prime }\ge p^{\prime }+a^{\prime }+\bar{n}=0,$$

$$d(x_{1})\in d\big{[}{\perp}_{{\downarrow}x}[{\downarrow}x\setminus\{x\}]\big{]}$$

$$A_T=\{{\perp }_{\downarrow x}(t):f(x)<t<x\}\text{mbox}and{B}_T:=\{{\perp }_{\downarrow y}(t):f(y)<t<y\}.$$

$$d(y_1)=d(g^{\prime }(y_1))=d(x_1)+d(x_2)+\sum _{t\in A_T}d(t)$$

$$d(x_1)=d(g^{\prime }(x_1))=d(y_1)+d(y_2)+\sum _{t\in B_T}d(t).$$

$$d(y_1)-d(x_1)=\sum _{t\in A_T\cup \{x_2\}}d(t)=-\sum _{t\in B_T\cup \{y_2\}}d(t)$$

Full text

1. Introduction

A subset $D$ of an Abelian group $G$ is called decomposable if $\emptyset\neq D\subset D+D:=\{x+y:x,y\in D\}$. For a finite subset $F$ of an Abelian group put $\sum F=\sum_{x\in F}x$. In this paper we discuss the following open problem, whose special case for the additive group of real numbers was posed by Gjergji Zaimi in 2010 on MathOverflow [Z].

Problem 1**.**

Let $D$ be a finite decomposable subset of an Abelian group $G$. Is $\sum Z=0$ for some non-empty set $Z\subset D$?

Observe that the answer to Problem 1 is affirmative if the group $G$ is Boolean (which means that $x+x=0$ for any $x\in G$). Indeed, take any element $a\in D$ and find elements $b,c\in D$ with $a=b+c$. If $0\in\{a,b,c\}$, then $T=\{0\}\subset\{a,b,c\}\subset D$ is a subset with $\sum T=0$. If $0\notin\{a,b,c\}$, then the elements $a,b,c$ are pairwise distinct and the set $T=\{a,b,c\}$ has $\sum T=a+b+c=2(b+c)=0$. Therefore, any decomposable set $D$ in a Boolean group contains a subset $T\subset D$ of cardinality $|T|\leq 3$ with $\sum T=0$.

This simple upper bound does not hold for decomposable sets in arbitrary groups.

Example 2**.**

Let $n\geq 2$ and $G$ be a cyclic group of order $2^{n}-1$ with generator $g$. The set $D=\{2^{k}g:0\leq k<n\}$ is decomposable and has $\sum D=0$ but $\sum T\neq\emptyset$ for any subset $T\subset D$ of cardinality $0<|T|<|D|=n$.

Example 3**.**

For every $n\in\mathbb{N}$ the subset

[TABLE]

of cardinality $|D_{n}|=2n$ in the infinite cyclic group $\mathbb{Z}$ is decomposable and the subset

[TABLE]

of cardinality $|Z|=n$ has $\sum Z=0$. On the other hand, $\sum T\neq 0$ for every non-empty subset $T\subset D_{n}$ of cardinality $|T|<n$.

The properties of the set $D_{n}$ from Example 3 will be established in Section 2.

The above examples suggest to assign to every finite subset $D$ of an Abelian group $G$ the largest number $z(D)\leq|D|+1$ such that $\sum T\neq 0$ for any non-empty subset $T\subset D$ of cardinality $|T|<z(D)$. Therefore, $z(D)$ is equal to the smallest cardinality of a subset $Z\subset D$ with $\sum Z=0$ if such set $Z$ exists and $z(D)=|D|+1$ in the opposite case.

In terms of the number $z(D)$, Problem 1 can be reformulated as follows.

Problem 4**.**

Let $D$ be a finite decomposable subset of an Abelian group. Is $z(D)\leq|D|$?

The decomposable set $D_{n}$ from Example 2 has $z(D)=|D|=n$ and the decomposable set $D_{n}\subset\mathbb{Z}$ from Example 3 has $z(D_{n})=n=\frac{1}{2}|D_{n}|$.

This suggests the following refinement of Problems 1 and 4 for the infinite cyclic group $\mathbb{Z}$.

Problem 5**.**

Is $z(D)\leq\frac{1}{2}|D|$ for any finite decomposable subset $D\subset\mathbb{R}$?

A special case of this problem was posed by Aryabhata in [A].

Problem 6** (Aryabhata).**

Is $z(D)\leq 7$ for any decomposable subset $D\subset\mathbb{Z}$ of cardinality $|D|=15$?

In Section 3 we shall provide an affirmative answer to Problem 5 for decomposable subsets $D\subset\mathbb{Z}$ of cardinality $|D|\leq 7$.

Proposition 7**.**

Any decomposable subset $D\subset\mathbb{R}$ of cardinality $|D|\leq 7$ has $z(D)\leq\frac{1}{2}|D|$.

We also observe that for $n\in\{2,3\}$ the decomposable set $D_{n}=\{2^{k},2^{k}-2^{n}+1:0\leq k<n\}$ form Example 3 is unique in the following sense.

Proposition 8**.**

Every decomposable set $D\subset\mathbb{R}$ with $n=z(D)=\frac{1}{2}|D|\in\{2,3\}$ is equal to $D_{n}{\cdot}r$ for some real number $r$.

Problem 9**.**

Is every finite decomposable set $D\subset\mathbb{R}$ with $z(D)=\frac{1}{2}|D|=n\geq 2$ equal to $D_{n}{\cdot}r$ for some real number $r$?

The following proposition shows that the problems on fiinite decomposable sets in torsion-free Abelian groups can be reduced to the case of the infinite cyclic group.

Proposition 10**.**

For any finite decomposable set $D$ is a torsion-free Abelian group $G$ there exists a decomposable set $D^{\prime}\subset\mathbb{Z}$ such that $|D^{\prime}|=|D|$ and $z(D^{\prime})=z(D)$.

Proof.

We lose no generality assuming that $G$ is finitely-generated and hence is isomorphic to $\mathbb{Z}^{n}$ for some $n\in\mathbb{N}$. Let $e_{1},\dots,e_{n}$ be the standard generators of the group $\mathbb{Z}^{n}$. Find $m\in\mathbb{N}$ such that $\{\sum F:F\subset D\}\subset\{\sum_{i=1}^{n}x_{i}e_{i}:(x_{i})_{i=1}^{n}\in[-m,m]^{n}\}=[-m,m]^{n}$. Consider the homomorphism $h:\mathbb{Z}^{n}\to\mathbb{Z}$ such that $h(e_{i})=(2m+1)^{i}$ for all $1\leq i\leq n$. It is easy to see that the restriction $h{\restriction}[-m,m]^{n}$ is injective. Consequently, a subset $F\subset D$ has $\sum F=0$ if and only if $\sum h(F)=0$. This implies that for the set $D^{\prime}=h(D)$ we have the equalities $|D^{\prime}|=|D|$ and $z(D^{\prime})=z(D)$. ∎

Our final result provides an affirmative answer to a weak version of Problem 1. The following theorem will be proved in Section 4.

Theorem 11**.**

For any finite decomposable subset $D$ of an Abelian group there are two non-empty sets $A,B\subset D$ such that $\sum A+\sum B=0$.

Corollary 12**.**

For any finite decomposable subset $D$ of an Abelian group, there exists a non-empty subset $T\subset D$ and a function $f:T\to\{1,2\}$ such that $\sum_{x\in T}f(x)\cdot x=0$.

A decomposable subset $D$ of an Abelian group is called minimal decomposed if no proper subset of $D$ is decomposed. It is clear that every finite decomposed set contains a minimal decomposed set.

Corollary 12 can be compared with the following result that was essentially proved by Hsien-Chih Chang [C] in his answer to the problem of Zaimi [Z].

Proposition 13**.**

For every finite minimal decomposable subset $D$ of an Abelian group there exists a function $f:D\to\omega$ such that $\sum_{x\in D}f(x)=|D|$ and $\sum_{x\in D}f(x)\cdot x=0$.

Proof.

By the decomposability of $D$, there exist functions $\alpha,\beta:D\to D$ such that $x=\alpha(x)+\beta(x)$ for every $x\in X$. For every $x\in D$ let $g(x)=|\alpha^{-1}(x)|+|\beta^{-1}(x)|$ and observe that $\sum_{x\in D}g(x)=|D|+|D|$. The minimal decomposability of $D$ ensures that $D=\alpha(D)\cup\beta(D)$ and hence $f(x):=g(x)-1\geq 0$ for every $x\in D$. Then $\sum_{x\in D}f(x)=(|D|+|D|)-|D|=|D|$. It follows that $\sum D=\sum_{x\in D}(\alpha(x)+\beta(x))=\sum_{y\in D}g(y)\cdot y$ and hence $\sum_{x\in D}f(x)\cdot x=\sum_{x\in D}(g(x)-1)\cdot x=0$. ∎

2. Properties of the decomposable set in Example 3

Given a natural number $n$ consider the subsets $A=\{1,2,4,\dots,2^{n-1}\}$ and

[TABLE]

in the group $\mathbb{Z}$ of integers. Then $|D_{n}|=2n$.

The set $D_{n}$ is decomposable, because, clearly, each element of the set $A\setminus\{1\}$ is decomposable, $1=2^{n-1}+(2^{n-1}-(2^{n}-1))$, $2^{k}-(2^{n}-1)=2^{k-1}+(2^{k-1}-(2^{n}-1))$ for every positive $k<n$, and $1-(2^{n}-1)=(2^{n-1}-(2^{n}-1))+(2^{n-1}-(2^{n}-1))$.

It is clear that the set $Z=\{2^{k}:1\leq k<n\}\cup\{2^{n}-2\}$ has cardinality $|Z|=n=\frac{1}{2}|D_{n}|$ and $\sum Z=0$.

Next, we prove that every subset $T\subset D_{n}$ of cardinality $|T|<n$ has $\sum T\neq 0$. Assuming that $\sum T=0$, we conclude that $n\geq 3$ and $T$ contains at most $n-2$ positive elements. Since all of them are distinct elements of $A$, their sum is at most $2^{n}-4$. On the other hand, the largest negative element of the set $A-(2^{n}-1)$ is $-2^{n-1}+1=-(2^{n}-2)/2$. Thus if $T$ contains at least two negative elements then $\sum T<0$. If $T$ contains exactly one negative element $2^{k}-2^{n}+1$ then $\sum T=0$ implies that we have a representation of $2^{n}-1$ as a sum of at most $n-1$ powers of $2$ with at most one power used twice. This representation collapses to a sum of at most $n-1$ distinct powers of $2$. If the representation contains a power $2^{l}$ with $l\geq n$ then it is bigger than $2^{n}-1$. Otherwise the sum is at most $2^{1}+2^{2}+\dots+2^{n-1}=2^{n}-2<2^{k}-1$. Thus $z(D_{n})\geq n$.

3. Proof of Propositions 7 and 8

We divide the proof of Propositions 7 and 8 into five lemmas. In fact, Proposition 8 follows from Lemmas 15 and 18 proved below.

Lemma 14**.**

Every decomposable subset $D\subset\mathbb{R}$ of cardinality $|D|\leq 3$ contains zero and hence has $z(D)=1\leq\frac{1}{2}|D|$.

Proof.

To derive a contradiction, assume that $0\notin D$. Replacing $D$ by $-D$, if necessary, we can assume that $D$ contains a unique positive element $p$. Since $p=a+b$ for some elements $a,b\in D$, one of the numbers $a$ or $b$ is positive, so it equals $p$ and the other summand is zero. ∎

The following lemma was proved by Ingdas [Z]. We present a short proof for convenience of the reader.

Lemma 15** (Ingdas).**

Any decomposable subset $D\subset\mathbb{R}$ with $|D|=4$ and $z(D)\geq 2$ is equal to the set $\{-2,-1,1,2\}{\cdot}r=D_{2}{\cdot}r$ for some positive real number $r$.

Proof.

Since $z(D)\geq 2$, the set $D$ does not contain zero. Then $D$ should contains at least two positive numbers and at least two negative numbers (otherwise $D$ will be not decomposable). Since $|D|=4$, the set $D$ contains exactly two positive and two negatice numbers. Let $c$ be the largest positive element of $D$. Since $D$ is decomposable, $c=a+b$ for some $a,b\in D$. Since $D$ does not contain zero, the maximality of $c$ ensures that the elements $a,b$ are strictly positive and hence coincide with the unique positive element $r$ of the set $D\setminus\{c\}$. Therefore, $c=a+b=2r$. By the same reason, the subset $\{x\in D:x<0\}$ of negative numbers is equal to $\{2n,n\}$ for some negative real number $n$. Write the element $r$ as $r=x+y$ for some $x,y\in D$ with $x\leq y$. Taking into account that $D$ contains no zero and $r,2r$ are unique positive elements of $D$, we conclude that $y=2r$ and then $x=r-y=r-2r=-r\in\{2n,n\}$. If $-r=2n$, then $D=\{-r,-\frac{1}{2}r,r,2r\}$ is not decomposable as $-\frac{1}{2}r\notin D+D$. So, $r=-n$ and hence $D=\{-2r,-r,r,2r\}=\{-2,-1,1,2\}{\cdot}r$. ∎

Lemma 16**.**

Any decomposable subset $D\subset\mathbb{R}$ of cardinality $|D|\in\{4,5\}$ contains a subset $T\subset D$ of cardinality $|Z|\in\{1,2\}$ with $\sum Z=0$. Consequently, $z(D)\leq 2\leq\frac{1}{2}|D|$.

Proof.

If $0\in D$, then $Z=\{0\}$ has $\sum Z=0$ and witnesses that $z(D)=0\leq\frac{1}{2}|D|$.

So, assume that $0\notin D$. Replacing $D$ by $-D$ we can assume that $D$ has at most two positive elements. Let $c$ be the largest positive element of $D$. Since $D$ is decomposable, $c=a+b$ for some $a,b\in D$. Since $D$ does not contain zero, the maximality of $c$ ensures that the elements $a,b$ are strictly positive and hence coincide with the unique positive element of the set $D\setminus\{c\}$. Therefore, $c=a+b=2a$. Write the element $a$ as $a=x+y$ for some $x,y\in D$ with $x\leq y$. Taking into account that $D$ contains no zero and $a,c$ are unique positive elements of $D$, we conclude that $y=c$ and then $x=a-y=a-c=a-2a=-a$. Then the set $Z=\{-a,a\}$ has $\sum Z=0$ and witnesses that $z(D)\leq 2\leq\frac{1}{2}|D|$.∎

Remark 17**.**

It can be shown that each decomposable subset $D\subset\mathbb{R}$ of cardinality $|D|=5$ with $0\notin D$ is equal to one of the sets

[TABLE]

for some non-zero number $r$.

Lemma 18**.**

Any decomposable subset $D\subset\mathbb{R}$ of cardinality $|D|=6$ with $z(D)\geq 3$ coincides with $\{1,2,4,-3,-5,-6\}\cdot r$ for some real number $r$ and hence has $z(D)=3=\frac{1}{2}|D|$.

Proof.

Let $D$ be a decomposable set consisting of six real numbers. If $0\in D$ then $z(D)=1$. If $D$ contains exactly one (resp. two) positive (or negative) elements then similarly to the case $n\leq 3$ (resp. $4\leq n\leq 5$) we can show that $z(D)=1$ (resp. $z(D)\leq 2$).

So it remains to consider the case when $D$ consists of three positive and three negative numbers. Let $p_{\max}$ (resp. $n_{\min}$) be the largest positive (resp. the smallest negative) element of $D$. Write $p_{\max}$ as $p_{\max}=p+p^{\prime}$ for some numbers $p,p^{\prime}\in D$ with $p\leq p^{\prime}$. By the maximality of $p_{\max}$, both numbers $p$ and $p^{\prime}$ are positive. Assume that $p<p^{\prime}$ and write $p=\bar{p}+a$, $p^{\prime}=\bar{p}^{\prime}+a^{\prime}$ for some elements of $D$ with positive $\bar{p}$ and $\bar{p}^{\prime}$. If $\bar{p}=p_{\max}$ (resp. $\bar{p}^{\prime}=p_{\max}$) then $p^{\prime}+a=0$ (resp. $p+a^{\prime}=0$), so $z(D)=2$. Therefore, we can assume that $\bar{p}\neq p_{\max}\neq\bar{p}^{\prime}$. In this case $\bar{p}=p^{\prime}$ and $\bar{p}^{\prime}=p$. Consequently, $a=p-\bar{p}=p-p^{\prime}=\bar{p}^{\prime}-p^{\prime}=-a^{\prime}$ and $Z=\{a,a^{\prime}\}$ is a set with $\sum Z=0$, witnessing that $z(D)\leq 2$.

Thus it remains to consider the case when $p_{\max}=2p^{\prime}$ for some $p^{\prime}\in D$ and, by the symmetry, $n_{\min}=2n^{\prime}$ for some $n^{\prime}\in D$. Let $p^{\prime}$,$2p^{\prime}=p_{\max},p^{\prime\prime}$ be positive elements of the set $D$ and $n^{\prime}$, $2n^{\prime}=n_{\min},n^{\prime\prime}$ be its negative elements. Then $n^{\prime}$ is a sum of two elements of $D$ at least one of which is negative. If $n^{\prime}=2n^{\prime}+x$ for some $x\in D$ then $n^{\prime}+x=0$ and $z(D)=2$. Thus $n^{\prime}=n^{\prime\prime}+a$ for some $a\in D$. If $n^{\prime\prime}=n^{\prime}+b$ for some $b\in D$ then $a+b=0$ and $z(D)=2$. Thus $n^{\prime\prime}=2n^{\prime}+\bar{p}$ for some positive $\bar{p}\in D$. Similarly, $p^{\prime}=p^{\prime\prime}+a^{\prime}$ for some $a^{\prime}\in D$ and $p^{\prime\prime}=2p^{\prime}+\bar{n}$ for some negative $\bar{n}\in D$. If $a>0$ and $a^{\prime}<0$ then $\max\{\bar{n},a^{\prime}\}\leq n^{\prime}$ and $p^{\prime}\leq\min\{\bar{p},a\}$. So

[TABLE]

a contradiction. Thus $a<0$ or $a^{\prime}>0$. If $a<0$, then since $n^{\prime}=n^{\prime\prime}+a$ we have $n^{\prime}=2n^{\prime\prime}$. Similarly, if $a^{\prime}>0$, then $p^{\prime}=2p^{\prime\prime}$. Reverting the signs of elements of $D$, if needed, we can suppose that $p^{\prime}=2p^{\prime\prime}$. Then $\{p^{\prime\prime},2p^{\prime\prime},4p^{\prime\prime}\}\in D$, so $\bar{n}=-3p^{\prime\prime}$ and $n^{\prime\prime}=\bar{p}+2n^{\prime}$. The following cases are possible:

$n^{\prime}=\bar{n}=-3p^{\prime\prime}$, so $2n^{\prime}=-6p^{\prime\prime}$.

• 1.1.

If $\bar{p}=p^{\prime\prime}$ then $n^{\prime\prime}=-5p^{\prime\prime}$ so the proposition claim holds.

• 1.2.

If $\bar{p}=2p^{\prime\prime}$ then $n^{\prime\prime}=-4p^{\prime\prime}$, so $n^{\prime\prime}+4p^{\prime\prime}=0$ and $z(D)=2$.

• 1.3.

If $\bar{p}=4p^{\prime\prime}$ then $n^{\prime\prime}=-2p^{\prime\prime}$, so $n^{\prime\prime}+2p^{\prime\prime}=0$ and $z(D)=2$.

• 2.

$2n^{\prime}=\bar{n}=-3p^{\prime\prime}$, so $n^{\prime}=-1.5p^{\prime\prime}$.

• 2.1.

If $\bar{p}=p^{\prime\prime}$ then $n^{\prime\prime}=-2p^{\prime\prime}$, so $n^{\prime\prime}+2p^{\prime\prime}=0$ and $z(D)=2$.

• 2.2.

If $\bar{p}=2p^{\prime\prime}$ then $n^{\prime\prime}=-p^{\prime\prime}$, so $n^{\prime\prime}+p^{\prime\prime}=0$ and $z(D)=2$.

• 2.3.

If $\bar{p}=4p^{\prime\prime}$ then $n^{\prime\prime}=p^{\prime\prime}>0$, a contradiction.

• 3.

$n^{\prime\prime}=\bar{n}=-3p^{\prime\prime}$. Then $\bar{p}+2n^{\prime}+3p^{\prime\prime}=0$.

• 3.1.

If $\bar{p}=p^{\prime\prime}$ then $n^{\prime}=-2p^{\prime\prime}$, so $n^{\prime}+2p^{\prime\prime}=0$ and $z(D)=2$.

• 3.2.

If $\bar{p}=2p^{\prime\prime}$ then $n^{\prime}=-2.5p^{\prime\prime}$, so $2n^{\prime\prime}=-5p^{\prime\prime}$ and $n^{\prime}\not\in D+D$, a contradiction.

• 3.3.

If $\bar{p}=4p^{\prime\prime}$ then $n^{\prime}=-3.5p^{\prime\prime}$, so $2n^{\prime\prime}=-7p^{\prime\prime}$ and $n^{\prime}\not\in D+D$, a contradiction.

∎

Lemma 19**.**

Every decomposable subset $D\subset\mathbb{R}$ of cardinality $|D|=7$ has $z(D)\leq 3\leq\frac{1}{2}|D|$.

Proof.

Let $D$ be a decomposable set consisting of seven real numbers. If $0\in D$ then $z(D)=1$. If $D$ contains exactly one (resp. two) positive (or negative) elements then similarly to the case $n\leq 3$ (resp. $4\leq n\leq 5$) we can show that $z(D)=1$ (resp. $z(D)\leq 2$). So, reverting the signs of elements of $D$, if needed, it remains to consider the case when $D$ consists of four positive and three negative numbers. Let $p_{\max}$ (resp. $n_{\min}$) be the largest positive (resp. the smallest negative) element of $D$. Similarly to the proof of Lemma 18 we can show that $n_{\min}=2n^{\prime}$ for some $n^{\prime}\in D$.

Let $n^{\prime}$, $2n^{\prime}=n_{\min},n^{\prime\prime}$ be negative elements of the set $D$. Similarly to the proof of Lemma 18 we can show that $n^{\prime}=n^{\prime\prime}+a$ for some $a\in D$ and $n^{\prime\prime}=2n^{\prime}+\bar{p}$ for some positive $\bar{p}\in D$. Then $n^{\prime}+a+\bar{p}=0$. If these numbers are distinct then $z(D)\leq 3$. So we assume the converse. Since $n^{\prime}=n^{\prime\prime}+a$, $a\neq n^{\prime}$, so $a=\bar{p}=n^{\prime}-n^{\prime\prime}=n^{\prime\prime}-2n^{\prime}$ and $3n^{\prime}=2n^{\prime\prime}$. Divide all elements of the set $D$ by $\frac{1}{2}|n^{\prime}|$. Then it will have elements $-2$, $-3$, $-4$, and $1$.

Depending on the representation of $1$ as a sum of elements of $D$, the following cases are possible:

$1=3-2$, and $3\in D$. Since $-3\in D$, we have $z(D)\leq 2$.

• 2.

$1=4-3$, and $4\in D$. Since $-4\in D$, we have $z(D)\leq 2$.

• 3.

$1=5-4$, and $5\in D$. Since $-2,-3\in D$, we have $z(D)\leq 3$.

• 4.

$1$ is a sum of distinct positive elements of $D$. Then $p_{\max}<2$ so a sum of each positive and negative elements of $D$ is negative. Then the smallest positive element of $D$ does not belong to $D+D$, a contradiction.

• 5.

$1=0.5+0.5$ and $0.5\in D$. If the smallest positive element $p_{\min}$ of $D$ is less than $0.5$ then $p_{\max}\leq 1+1=2$. Then a sum of each positive and negative elements of $D$ is non-positive and $p_{\min}\not\in D+D$, a contradiction. Thus $p_{\min}=0.5$ and one of numbers $2.5$, $3.5$, and $4.5$ belongs to $D$. But $2.5+0.5+(-3)=0$ and $3.5+0.5+(-4)=0$, so either $z(D)\leq 3$ or $4.5$ belongs to $D$. We assume the last case. Then $D^{\prime}=\{-4,-3,-2,0.5,1,4.5\}\subset D$. Let $p$ be a unique element of $D\setminus D^{\prime}$. Since $D^{\prime}+D^{\prime}\not\ni 4.5$, $4.5=p+s$ for some $s\in D$. Since $p\geq 4.5/2=2.25$, $p\not\in D^{\prime}+D^{\prime}$ and hence $p=4.5+s^{\prime}$ for some $s^{\prime}\in D$. Then $s+s^{\prime}=0$ and so $z(D)\leq 2$.

∎

4. Proof of Theorem 11

First we introduce some notation. For a function $f:X\to Y$ and subset $A\subset X$ put $f[A]=\{f(a):a\in A\}$.

By a tree we understand any non-empty finite partially ordered set $(T,\leq)$ such that for every $x\in T$ the set ${\downarrow}x=\{t\in T:t\leq x\}$ is linearly ordered.

Let $T$ be a tree. By $\min T$ we denote the smallest element of $T$ and by $\max T$ the set of all maximal elements of $T$. A branch in a tree is a maximal linearly ordered subset $B\subset T$, which can be identified with the largest element of $B$.

For an element $x$ of a tree $T$ let ${\uparrow}x:=\{y\in T:x\leq y\}$ and $\mathrm{succ}_{T}(x):=\min({\uparrow}x\setminus\{x\})$ be the set of immediate successors of $x$ in the tree $T$. For any $x\in\max T$ we have $\mathrm{succ}_{T}(x)=\emptyset$. For any element $x\neq\min T$ let $\mathrm{prec}_{T}(x)$ be the unique element $y\in T$ such that $x\in\mathrm{succ}_{T}(y)$.

A tree $T$ is called binary if for each $x\in T\setminus\max T$ the set $\mathrm{succ}_{T}(x)$ has cardinality 2.

For a branch $B$ in a binary tree, let $\perp_{B}:B\setminus\max B\to T$ be the function assigning to each element $x\in B\setminus\max B$ the unique element of the set $\mathrm{succ}_{T}(x)\setminus B$.

A function $f:\max T\to T$ is called regressive if $f(x)<x$ for each $x\in\max T$.

Lemma 20**.**

For any regressive function $f:\max T\to T$ on a binary tree $T$ there are distinct elements $x,y\in\max T$ such that $f(x)=f(y)=\max({\downarrow}x\cap{\downarrow}y)$.

Proof.

The proof if by induction on the height $\hbar(T):=\max\{|{\downarrow}x|:x\in T\}$ of the binary tree $T$. If $\hbar(T)=1$, then no regressive function $f:\max T\to T$ exists, so the statement of the lemma holds.

Assume that the lemma has been proved for all binary trees of height $<n$. Take a binary tree $T$ of height $n>1$. Let $\min T$ be the root of $T$ and $x_{1},x_{2}$ be two immediate successors of $\min T$ in $T$. Then $T_{1}:={\uparrow}x_{1}\setminus\{\min T\}$ and $T_{2}:={\uparrow}x_{2}\setminus\{\min T\}$ are trees of height $<n$. Two cases are possible.

1. $f(\max T_{i})\subset T_{i}$ for some $i\in\{1,2\}$. In this case we can apply the inductive assumption and find two elements $x,y\in\max T_{i}$ such that $f(x)=f(y)=\max({\downarrow}x\cap{\downarrow}y)$.

2. For every $i\in\{1,2\}$ there exists an element $t_{i}\in\max T_{i}$ such that $f(t_{i})=\min T$. Then $x=t_{1}$ and $y=t_{2}$ are two elements with $f(x)=f(y)=\max({\downarrow}x\cap{\downarrow}y)$. ∎

Now we can present the proof of Theorem 11. Given a finite decomposable subset $D$ of an Abelian group, we should find two non-empty subsets $A,B\subset D$ with $\sum A+\sum B=0$.

Let $D$ be a finite subset of an abelian group with $D\subset D+D$. By a binary $D$-tree we understand a pair $(T,d)$ consisting of a binary tree $T$ and a function $d:T\to D$ such that each non-maximal element $x\in T$ we have $d(x)=d(y)+d(z)$, where $\{y,z\}=\mathrm{succ}_{T}(x)$. A binary $D$-tree is called $\perp$-injective if for each branch $L\subset T$ the restriction $d{\restriction}{\perp}_{L}[L\setminus\max L]$ is injective. This implies that \big{|}{\perp}_{L}[L\setminus\max L]\big{|}\leq|D| and hence $|L|\leq|D|+1$. Consequently, each $\perp$-injective binary $D$-tree is finite, so we can choose a maximal $\perp$-injective binary $D$-tree $T$. Since $D\subset D+D$, the tree $T$ is a subtree of a binary $D$-tree $\tilde{T}=T\cup\max\tilde{T}$ such that $T\cap\max\tilde{T}=\emptyset$. The maximality of the tree $T$ ensures that for any $x\in\max T$, there exists an element $x^{\prime}\in\mathrm{succ}_{\tilde{T}}(x)$ such that $d(x^{\prime})\in d[{\perp}_{{\downarrow}x}[{\downarrow}x\setminus\{x\}]$. Let $M_{2}=\{x\in\max T:d[\mathrm{succ}_{\tilde{T}}(x)]\subset d[{\perp}_{{\downarrow}x}[{\downarrow}x\setminus\{x\}]\}$ and $M_{1}=\max T\setminus M_{2}$.

For every $x\in M_{1}$ let $x_{1}$ be the unique immediate successor of $x$ such that

[TABLE]

and $g(x)\in{\downarrow}x$ be a (unique) point such that $d(x_{1})=d({\perp}_{{\downarrow}x}(g(x)))$. Let $f(x):=g(x)$.

For every $x\in M_{2}$ and every point $x^{\prime}\in\mathrm{succ}_{\tilde{T}}(x)$ there is a point $g(x^{\prime})\in{\downarrow}x$ such that $d(x^{\prime})=d({\perp_{{\downarrow}x}}(g(x^{\prime})))$. Let $f(x):=\max\{g(x^{\prime}):x^{\prime}\in\mathrm{succ}_{\tilde{T}}(x)\}$.

Now observe that we have defined a regressive function $f:\max T\to T$. By Lemma 20, there are two maximal elements $x,y\in\max T$ such that $f(x)=f(y)=\max({\downarrow}x\cap{\downarrow}y)$.

By the definition of $f(x)$ the set $\mathrm{succ}_{\tilde{T}}(x)$ contains a point $x_{1}$ such that $f(x)=g(x_{1})$. Let $x_{2}$ be the unique point of $\mathrm{succ}_{\tilde{T}}(x)\setminus\{x_{1}\}$. The definition of the function $f$ guarantees that $d(x_{2})\notin\{d(\perp_{{\downarrow}x}(t)):f(x)<t<x\}$. By analogy the set $\mathrm{succ}_{\tilde{T}}(y)$ can be written as $\{y_{1},y_{2}\}$ such that and $f(y)=g(y_{1})$ and $d(y_{2})\notin\{d(\perp_{{\downarrow}y}(t)):f(y)<t<y\}$.

Let $g^{\prime}(x_{1})={\perp}_{{\downarrow}x}(g(x_{1}))={\perp}_{{\downarrow}x}(f(x))$ and $g^{\prime}(y_{1})={\perp}_{{\downarrow}y}(g(y_{1}))={\perp}_{{\downarrow}y}(f(y))$. Observe that $\{g^{\prime}(x_{1}),g^{\prime}(y_{1})\}=\mathrm{succ}_{T}(f(x))$, $g^{\prime}(x_{1})\in{\downarrow}y$ and $g^{\prime}(y_{1})\in{\downarrow}x$.

Let

[TABLE]

It follows from the definition of $f(x)=f(y)$ that $d(x_{2})\notin d[A_{T}]$ and $d(y_{2})\notin d[B_{T}]$.

By induction it can be shown that

[TABLE]

and

[TABLE]

Then

[TABLE]

and finally $\sum A=-\sum B$ for the sets $A=d[A_{T}]\cup\{d(x_{2})\}$ and $B=d[B_{T}]\cup\{d(y_{2})\}$.