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\hypersetup{pdftitle={A logarithmic error term for subset sums of zero-sum-free sets},pdfauthor={Chao Liu}}
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\title[Subset sums with a logarithmic error term]{A logarithmic error term for subset sums\protect\\ of zero-sum-free sets}
\author{Chao Liu}
\address{Department of Mathematics, University of Mississippi,
University, MS 38677, USA}
\email{chaoliuac@gmail.com}
\date{Revised draft, October 28, 2026}
\subjclass[2020]{11B75, 20K01}
\keywords{Subset sums, zero-sum-free sets, finite abelian groups, Olson constant}

\begin{document}
\begin{abstract}
For a positive integer $k$, let $f(k)$ be the minimum number of distinct
nonempty subset sums of a zero-sum-free $k$-element set, where the minimum
is taken over all finite abelian groups. We prove
\[
 f(k)\ge \frac{k^2}{2}-13k\log(k+1)-116k.
\]
Together with the classical upper bound, this gives
$f(k)=k^2/2+O(k\log k)$, with the optimal leading coefficient.
As a consequence, the Olson constant of every finite abelian group $G$
satisfies $\Ol(G)\le\sqrt{2|G|}+O(\log|G|)$.
The argument combines a triangular lower bound for repeated supports,
a count using blocks of two and three copies, and a greedy expansion
argument. No restriction is imposed on the order of the group or on
the stabilizer of the set of subset sums.
\end{abstract}
\maketitle

\section{Introduction}

Let $G$ be a finite abelian group, written additively. For a finite sequence
$T$ over $G$, let $\Sig(T)$ be the set of its subsequence sums, including
the empty sum. When $T$ is a set, these are its subset sums. We call $T$
\emph{zero-sum-free} if none of its nonempty subsequences has sum zero.
For such a sequence, put
\[
 \N(T)=|\Sig(T)|-1.
\]
Thus $\N(T)$ counts the distinct nonempty subsequence sums. We study
\[
 f(k)=\min\{\N(S): S\subseteq G,\ |S|=k,\ S\text{ zero-sum-free},\
                  G\text{ a finite abelian group}\}.
\]
The distinction between sets and sequences is essential in this definition:
the elements of $S$ must be distinct.

Eggleton and Erd\H{o}s introduced this invariant and proved, in particular,
that
\begin{equation}\label{eq:classical-upper}
 f(k)\le \left\lfloor\frac{k^2}{2}\right\rfloor+1
 \qquad(k\ge4);
\end{equation}
see \cite{EE}. Olson obtained a quadratic lower bound with coefficient
$1/9$ \cite{Olson}. Gao, Huang, Hui, Li, Liu, and Peng subsequently proved
the uniform estimate $f(k)\ge k^2/6$ \cite[Theorem 1.1]{GHHLLP}.
Our main result approaches the coefficient in
\eqref{eq:classical-upper} with a logarithmic loss per element.

\begin{theorem}\label{thm:main}
Let $S$ be a zero-sum-free set of $k\ge1$ elements in a finite abelian
group. Then
\begin{equation}\label{eq:main}
 \N(S)\ge \frac{k^2}{2}-13k\log(k+1)-116k,
\end{equation}
where $\log$ denotes the natural logarithm. Consequently,
\[
 \frac{k^2}{2}-O(k\log k)\le f(k)
 \le\left\lfloor\frac{k^2}{2}\right\rfloor+1.
\]
\end{theorem}

The associated zero-sum threshold is the \emph{Olson constant}
$\Ol(G)$: the smallest positive integer $t$ such that every $t$-element
subset of $G$ has a nonempty zero-sum subset.

\begin{corollary}\label{cor:olson}
For every finite abelian group of order $n$,
\[
 \Ol(G)\le\sqrt{2n}+13\log(n+1)+266.
\]
In particular, $\Ol(G)\le\sqrt{2n}+O(\log n)$, uniformly in $G$.
\end{corollary}

Hamidoune and Z\'emor proved
$\Ol(G)\le\sqrt{2|G|}+O(|G|^{1/3}\log|G|)$
\cite[Theorem 4.5]{HZ}. Thus the leading constant $\sqrt2$ in the Olson
problem is classical. Corollary~\ref{cor:olson} reduces the error in that
estimate to logarithmic order. Theorem~\ref{thm:main} controls the actual
number of subset sums; a bound on the ambient group order alone does not
give this conclusion.

The proof has three parts. First, for every integer $r\ge2$, if $U$ is
a set and the sequence containing $2r$ copies of every element of $U$
is zero-sum-free, then $r$ copies already have at least
$r|U|(|U|+1)/2$ nonempty sums. Second,
we combine blocks of two and three copies to count sums in a quotient
whose multiplicities are almost equal. This yields an alternative for
antisymmetric zero-sum-free sets: either the set already has at least
half the square of its size in nonempty subset sums, or every proper
iterated sum of its symmetrization satisfies the expansion bound in
Proposition~\ref{prop:alternative}. Finally, a greedy argument converts
this expansion into \eqref{eq:main}. Its stopping rule handles tails
that generate a proper subgroup.

AI tools developed by OpenAI were used in this work.

\section{Notation and additive tools}

For a sequence $T=t_1\cdots t_\ell$, write
$\sigma(T)=t_1+\cdots+t_\ell$, with $\sigma(\varnothing)=0$.
For a subgroup $H\le G$, let $\pi_H:G\to G/H$ be the natural quotient map.
For an integer $r\ge1$ and a set $U$, the notation $U^r$ denotes the
sequence in which each element of $U$ occurs $r$ times. The sumset $rX$,
by contrast, denotes $X+\cdots+X$, with repetitions allowed and exactly
$r$ summands. Set
\[
 \tri m=\frac{m(m+1)}2,\qquad
 C^*=C\cup(-C)\cup\{0\}.
\]
A set $C$ is \emph{antisymmetric} if $C\cap(-C)=\varnothing$.
A zero-sum-free set becomes antisymmetric after all its elements of
order two have been removed.

For $B\subseteq G$, define its translation boundary by
\[
 \lambda_B(x)=|(B+x)\setminus B|.
\]
We use
\begin{equation}\label{eq:boundary-properties}
 \lambda_B(x)=\lambda_B(-x)=\lambda_{G\setminus B}(x),\qquad
 \lambda_B(x+y)\le\lambda_B(x)+\lambda_B(y).
\end{equation}
For $1\le j<|X|$, write $j\wedge X$ for the sums of $j$ distinct
elements of $X$.

The following standard estimates will be used throughout:
\begin{equation}\label{eq:basic}
 \N(A)\ge\max\{|A|^2/6,\,2|A|-1\}
 \quad\text{for every nonempty zero-sum-free set }A,
\end{equation}
and
\begin{equation}\label{eq:superadditive}
 \N(PQ)\ge\N(P)+\N(Q)
 \quad\text{if the concatenation }PQ\text{ is zero-sum-free}.
\end{equation}
See \cite[Theorem 1.1 and Lemmas 2.1--2.2]{GHHLLP}.
We set $\N(\varnothing)=0$.

We also use Kneser's theorem in its multiple-summand form: if $J$ is
the stabilizer of $A_1+\cdots+A_t$, then
\begin{equation}\label{eq:kneser}
 |A_1+\cdots+A_t|\ge\sum_{i=1}^t|A_i+J|-(t-1)|J|.
\end{equation}
By \cite[Theorem~3.2]{BGGH}, including its elementary two-group exception,
\begin{equation}\label{eq:restricted}
 |j\wedge X|\ge |X|-1\qquad(1\le j<|X|).
\end{equation}

We recall the structural input needed only for the repeated-support
lemma below. A set $X\subseteq Q$ is a \emph{Vosper set} if
$|X+Y|\ge\min\{|Q|-1,|X|+|Y|\}$ whenever $|Y|\ge2$.
An AP-representation or a Vosper-representation of $A\subseteq G$
means a subgroup $J$ for which
\[
 |A+J|<\min\{|G|,|A|+|J|\},
\]
and the image of $A$ in $G/J$ is, respectively, an arithmetic progression
or a Vosper set. If $0\in A$, $\gen A=G$, and $|A|\le|G|/2$, such a
representation exists \cite[Theorem 3.1]{BGGH}.
Call $U$ \emph{valid} if $|\Sig(V)|\le|\gen V|/2$ for every nonempty
$V\subseteq U$. The following consequence of
\cite[Proposition 4.1]{BGGH} handles the AP case: if $U$ is valid,
antisymmetric, generates $G$, $|U|=m\ge4$, and $U^*$ has an
AP-representation, then $|\Sig(U)|\ge\tri m+1$.
Here $\Sig$ includes the empty sum, as does the subset-sum notation
in \cite{BGGH}. The growth and boundary estimates for the Vosper case
are proved below under the four-copy zero-sum-free hypothesis used
in this paper.

Finally, Olson's boundary estimate, in the form
\cite[Lemma 2.7]{GHHLLP}, says that if $C$ is antisymmetric,
$c=|C|$, $H=\gen C$, and
$|H|\ge2b$ for $b=\min\{|B|,|G\setminus B|\}\ge1$, then
\begin{equation}\label{eq:olson-boundary}
 \max_{x\in C}\lambda_B(x)\ge\frac12\min\{b+1,c+1\}.
\end{equation}
In this assertion $H$ may be a proper subgroup of $G$.

\section{Repeated supports}

\begin{lemma}\label{lem:four-copy-growth}
Let $U$ be an $m$-element set generating a finite abelian group $G$,
where $m\ge4$. Suppose $U^4$ is zero-sum-free and $U^*$ has a
Vosper-representation. Then, for every integer $j\ge1$,
\begin{equation}\label{eq:four-copy-growth}
 |jU^*|\ge\min\{|G|,2mj+1\}.
\end{equation}
\end{lemma}

\begin{proof}
The four-copy assumption makes $U$ antisymmetric and valid. To see
validity, suppose $Z=\Sig(V)$ has size greater than $|\gen V|/2$
for some nonempty $V\subseteq U$. Then $Z+Z=\gen V$, and a
representation of $-v$ by $V^2$, followed by a third copy of any
$v\in V$, gives a zero-sum subsequence of $U^4$.

Write $X=U^*$, and let $H$ be the subgroup in a
Vosper-representation of $X$. Put
\[
 h=|H|,\quad Q=\pi_H(X),\quad u=|Q|,\quad v=|G/H|,
 \quad \delta=hu-|X|.
\]
The representation is strict, so $0\le\delta<h$ and $Q$ is a proper
generating subset of $G/H$. It is symmetric and contains zero.
Consequently $u\ge3$: a symmetric generating set of size two containing
zero would be the entire quotient of order two.

We first prove the kernel estimate
\begin{equation}\label{eq:four-copy-kernel}
 \delta\ge\frac{h-1}{2}.
\end{equation}
Let $A=U\cap H$, $a=|A|$. Since $|X\cap H|=2a+1$,
$\delta\ge h-2a-1$. If $a=0$, this proves
\eqref{eq:four-copy-kernel}. If $a>0$, validity and
\eqref{eq:basic} give
\[
 2a\le|\Sig(A)|\le |\gen A|/2\le h/2.
\]
For odd $h$ this yields $2a\le(h-1)/2$, as required.
For even $h$, equality $|\Sig(A)|=h/2$ is impossible. Indeed, set
$Z=\Sig(A)$ and $D=\Sig(A^2)=Z+Z$. The fact that $A^4$ is
zero-sum-free gives $D\cap(-D)=\{0\}$, hence $|D|\le h/2$.
Since $Z\subseteq D$ and $|Z|=h/2$, we obtain $Z+Z=Z$.
A finite set containing zero and closed under addition is a subgroup.
Thus $Z=\gen A$, contradicting validity. Therefore
$2a\le h/2-1$ when $h$ is even, which again proves
\eqref{eq:four-copy-kernel}.

Next we show that $2X$ is $H$-periodic. For $R\in Q$, let
$X_R=X\cap\pi_H^{-1}(R)$. The total number of missing points
in these fibers is $\delta<h$. If $R\ne S$, then
$|X_R|+|X_S|>h$, so $X_R+X_S$ fills the whole coset indexed by
$R+S$. If $R\ne-R$, symmetry gives $|X_R|=|X_{-R}|>h/2$;
thus $X_R+X_R$ also fills the coset indexed by $2R$.
The remaining repeated-fiber sums have $R=-R$ and target the
zero coset. This coset is filled by $X_R+X_{-R}$ if a non-self-inverse
direction exists. Otherwise $Q\setminus\{0\}$ contains at least
two self-inverse directions. At most one of their fibers has size at
most $h/2$, since their total deficit is less than $h$. A fiber of
size greater than $h/2$ then fills $H$ by addition to itself.
Every coset of $2Q$ is therefore full in $2X$. It follows that
\begin{equation}\label{eq:periodic-iterate}
 |jX|=h|jQ|\qquad(j\ge2).
\end{equation}

Let $t$ be the last integer for which $tQ\ne G/H$. Such an integer
exists because $Q$ is proper, contains zero, and generates the
quotient. If $t=1$, \eqref{eq:four-copy-growth} follows at once from
$|X|=2m+1$ and \eqref{eq:periodic-iterate}.
Suppose $t\ge2$. The Vosper property gives
\begin{equation}\label{eq:quotient-last-layer}
 |jQ|\ge ju\quad(1\le j<t),\qquad
 |tQ|\ge tu-1,\qquad tu\le v.
\end{equation}
For completeness, induct on $j<t$. The only alternative to
$|jQ|\ge ju$ in the Vosper bound is $|jQ|=v-1$; then
$jQ+Q=G/H$, contradicting $j+1\le t$. At the last layer,
the same bound gives $|tQ|\ge\min\{v-1,tu\}$. If $tu>v$,
then $|(t-1)Q|+|Q|>v$ would force $tQ=G/H$, also a
contradiction. This proves \eqref{eq:quotient-last-layer}.

For $2\le j<t$, equations \eqref{eq:periodic-iterate} and
\eqref{eq:quotient-last-layer} give $|jX|\ge j|X|$.
The same holds for $j=t$ if $tu\le v-1$. In the only other case,
$tu=v$, equations \eqref{eq:four-copy-kernel} and
\eqref{eq:periodic-iterate} yield
\[
 |tX|\ge h(tu-1)=t|X|+t\delta-h\ge t|X|-1
 \ge2mt+1.
\]
Here $t\ge2$ and $t(h-1)/2-h\ge-1$.
The first layer has size $2m+1$, and the layers after $t$ are full.
This proves \eqref{eq:four-copy-growth}.
\end{proof}

\begin{lemma}\label{lem:four-copy-boundary}
Under the hypotheses of Lemma~\ref{lem:four-copy-growth}, let
$B\subseteq G$ satisfy
\[
 m(m-1)<b=|B|\le |G|/2.
\]
Then
\begin{equation}\label{eq:usable-vosper-boundary}
 \max_{x\in U}\lambda_B(x)>
 m-\frac{m(m-1)}b>m-1.
\end{equation}
In particular, some $x\in U$ has $\lambda_B(x)\ge m$.
\end{lemma}

\begin{proof}
Put $\alpha=\max_{x\in U}\lambda_B(x)$ and $\ell=2b-3$.
We have $1\le\ell\le |G|-1$. Write
$\ell=2mR+q$, where $0\le q<2m$. The growth estimate
\eqref{eq:four-copy-growth} permits us to choose pairwise disjoint
sets $E_j\subseteq jU^*\setminus\{0\}$ with
\[
 |E_j|=2m\ (1\le j\le R),\qquad |E_{R+1}|=q.
\]
Indeed, each ball contains enough nonzero points to add the next
required set. Let $E$ be their union. Counting ordered differences
of distinct elements of $B$ gives
\[
 \sum_{x\in E}\lambda_B(x)
 \ge\ell b-b(b-1)=b(\ell-b+1).
\]
On the other hand, subadditivity of the boundary gives
\begin{align*}
 \sum_{x\in E}\lambda_B(x)
 &\le\alpha\left(2m\sum_{j=1}^Rj+q(R+1)\right)\\
 &=\alpha(R+1)(mR+q)
 \le\frac{\alpha(\ell+m)^2}{4m}.
\end{align*}
The last inequality follows by completing the square in $q$.
Consequently
\[
 \alpha\ge\frac{4mb(b-2)}{(2b+m-3)^2}
 =\frac{m(1-2/b)}{(1+(m-3)/(2b))^2}.
\]
Since $m\ge4$ and $b>m(m-1)$, we may use
$(1+x)^{-2}>1-2x$ for $x>0$ to obtain the following strict bound.
Integrality of $\alpha$ then gives the final assertion.
\begin{align*}
 \alpha
 &>m\left(1-\frac2b\right)\left(1-\frac{m-3}b\right)\\
 &=m-\frac{m(m-1)}b+\frac{2m(m-3)}{b^2}
 >m-\frac{m(m-1)}b>m-1.\qedhere
\end{align*}
\end{proof}

\smallskip
\noindent\textit{Remark.}
Our argument follows \cite[Section~4.2]{BGGH}. In
\href{https://arxiv.org/abs/1112.1929v1}{arXiv:1112.1929v1},
the final comparison in the proof of Lemma~4.6 (p.~18) is incorrect:
in our notation, the product's $1/b$ term is $-m(m-1)/b$, as expanded
above. Lemmas~\ref{lem:four-copy-growth}--\ref{lem:four-copy-boundary}
establish the estimates needed here under the four-copy hypothesis
without invoking \cite[Lemma~4.2]{BGGH}.

\begin{lemma}\label{lem:repetition}
Let $r\ge2$, and let $U$ be an $m$-element set in a finite abelian group.
If $U^{2r}$ is zero-sum-free, then
\begin{equation}\label{eq:repetition}
 \N(U^r)\ge r\tri m.
\end{equation}
\end{lemma}

\begin{proof}
Fix $r$ and induct on $m$, replacing the ambient group by $\gen U$.
For $m\le2$, $\N(U)\ge\tri m$ is immediate. For $m=3$, the three
one-element sums are distinct, as are the three two-element sums.
There is at most one equality between the two families: two such
equalities would force an element of $U$ to have order two. The total
sum belongs to neither family. Hence $\N(U)\ge6$.
Equation~\eqref{eq:superadditive} proves \eqref{eq:repetition} in these
base cases.

Assume $m\ge4$. The set $U$ is antisymmetric. It is also valid.
Indeed, if $A=\Sig(V)$ had size greater than $|\gen V|/2$ for some
nonempty $V\subseteq U$, then $A+A=\gen V$. A representation of
$-v$ by $V^2$, followed by a third copy of any $v\in V$, would give a
nonempty zero-sum subsequence of $U^{2r}$.

Put $B=\Sig(U^r)$, $b=|B|$, and $n=|G|$. First suppose $b>n/2$.
Since $U^{2r}$ is zero-sum-free,
$B\cap(-B)=\{0\}$. Therefore $n=2b-1$ is odd. Let
$c=\sigma(U^r)$ and $M=b-1$, so $n=2M+1$. Taking complementary
subsequences gives $B=c-B$, and consequently
\begin{equation}\label{eq:dense-alternation}
 B\cup(B-c)=G,\qquad B\cap(B-c)=\{0\}.
\end{equation}
On every orbit of translation by $c$ not containing zero, membership
in $B$ would alternate. Such an orbit has odd length, which is
impossible. Thus $c$ generates $G$. On its one orbit,
\eqref{eq:dense-alternation} determines $B$: with $g=-2c$,
\[
 B=\{0,g,\ldots,Mg\},\qquad c=Mg.
\]
Write the distinct elements of $U$ as $w_1g,\ldots,w_mg$, where
$1\le w_i\le M$. If the integer sum of the weights of $U^r$ first
exceeded $M$ at some prefix, that prefix weight would lie in
$[M+1,2M]$, contradicting the description of $B$. Thus
$r\sum_iw_i\le M$. Its residue is $M$ modulo $2M+1$, so equality
holds. It follows that $\N(U^r)=M\ge r\tri m$.

Now suppose $b\le n/2$. Every zero-sum-free sequence has at least
its length in nonempty sums, so $b\ge rm+1\ge2m+1=|U^*|$.
The representation theorem applies to $U^*$. In the AP case, validity
and the input from Section~2 give $\N(U)\ge\tri m$, and
\eqref{eq:superadditive} finishes the proof.

In the Vosper case, the induction hypothesis gives
\[
 b\ge r\tri{m-1}+1\ge m(m-1)+1.
\]
Also $U^4$ is zero-sum-free, since $r\ge2$.
Thus Lemma~\ref{lem:four-copy-boundary} applies and supplies an
$x\in U$ with $\lambda_B(x)\ge m$. For $0\le j\le r+1$, put
\[
 B_j=\Sig((U\setminus\{x\})^r)+\{0,x,\ldots,jx\}.
\]
The increments $d_j=|B_j\setminus B_{j-1}|$ are nonincreasing:
if $y\in B_{j+1}\setminus B_j$, then $y-x\in B_j\setminus B_{j-1}$.
Indeed, $y$ must come from the new translate, and membership of
$y-x$ in $B_{j-1}$ would imply $y\in B_j$.
Since $B_r=B$ and $d_{r+1}=\lambda_B(x)$, induction yields
\[
 \N(U^r)\ge\N((U\setminus\{x\})^r)+r\lambda_B(x)
 \ge r\tri{m-1}+rm=r\tri m.\qedhere
\]
\end{proof}

\Needspace{14\baselineskip}
\section{Quotient sums and almost full fibers}

The next lemma permits zero relations at the corners of a box of
multiplicities. It excludes them only when some coordinate is strictly
between its two endpoints.

\begin{lemma}\label{lem:capacity}
Let $q_1,\ldots,q_m$ be distinct elements of a finite abelian group,
and let $7\le L_i\le h$ be integers. Suppose
\begin{equation}\label{eq:internal-nonzero}
 \sum_i j_iq_i\ne0
 \quad\text{whenever }0\le j_i\le L_i
 \text{ and some }0<j_i<L_i.
\end{equation}
Let $\mathcal I$ be the set of sums specified in
\eqref{eq:internal-nonzero}, and put $e=\sum_i(h-L_i)$. Then
\begin{equation}\label{eq:capacity}
 |\mathcal I|\ge h\tri m-4me-1.
\end{equation}
\end{lemma}

\begin{proof}
The case $m=0$ is immediate, so assume $m\ge1$.
Let $O=\{i:L_i\text{ is odd}\}$, $o=|O|$, and set
\[
 b_0=\begin{cases}h,&h\text{ even},\\h-3,&h\text{ odd},\end{cases}
 \qquad E_i=\min\{b_0,L_i-3\boldsymbol1_{i\in O}\}.
\]
Each $E_i$ is even and at least four. Form an ordered list of sequences:
first three copies of $\{q_i:i\in O\}$, if $O\ne\varnothing$,
and then, for $t=1,\ldots,\max_i E_i/2$, two copies of
\[
 V_t=\{q_i:E_i\ge2t\}.
\]
The sets $V_t$ are nested, and $V_1=V_2=\{q_1,\ldots,q_m\}$.
The total multiplicity used in direction $i$ is
$E_i+3\boldsymbol1_{i\in O}\le L_i$.

For each sequence in this list take its set of sums, including zero,
and translate it by the total sum of all preceding sequences. Consecutive
translated sets have a common endpoint. We check all other intersections.
If a point from a later block equals a point from an earlier block,
subtracting their multiplicity vectors gives $w$ with
$0\le w_i\le L_i$ and $\sum_iw_iq_i=0$. By
\eqref{eq:internal-nonzero}, every $w_i$ is either $0$ or $L_i$.

For consecutive blocks, $w_i\le5<L_i$; hence $w=0$, and the
intersection is precisely their common endpoint. For nonconsecutive
two-copy blocks, if the earlier is not the first two-copy block,
choose a direction active in the later block. It has already occurred
before the earlier block and also occurs in a block strictly between
the two. Thus $0<w_i<L_i$, a contradiction.
If the earlier block is the first two-copy block, the intervening
second two-copy block makes every $w_i$ positive. An odd direction,
if present, has already used three copies and gives $w_i<L_i$.
If no odd direction is present, all $w_i=L_i$, which is possible
only at the global initial and final endpoints. Finally, between the
initial three-copy block and a nonadjacent block, the first two-copy
block again makes all coordinates positive; only the same global
endpoint collision can remain. No intermediate endpoint participates
in this collision, because its multiplicity vector has an internal
coordinate.

The six-copy sequence on the whole support is zero-sum-free, since
$6<L_i$ for all $i$. Lemma~\ref{lem:repetition}, with $r=3$ for
the initial block and $r=2$ for the other blocks, therefore applies.
Let $n_j$ count the nonempty sums of block $j$. Adjacent overlaps
leave $1+\sum_j n_j$ points if the global endpoints are distinct;
deleting both leaves $\sum_j n_j-1$. If they coincide, the union has
one fewer point but only one endpoint is deleted, giving the same
bound. All remaining multiplicity vectors have an internal coordinate.
The repeated-support bounds therefore give
\begin{equation}\label{eq:block-count}
 |\mathcal I|\ge3\tri o+2\sum_t\tri{|V_t|}-1.
\end{equation}

To estimate the right side, note that removing one occurrence of a
direction from a layer decreases its triangular count by at most $m$.
If $h$ is even, $o\le e$ and $\sum_i(h-E_i)=e+3o$, whence
\[
 3\tri o+2\sum_t\tri{|V_t|}
 \ge h\tri m-me-3mo+3\tri o\ge h\tri m-4me.
\]
If $h$ is odd, $m-o\le e$ and
$\sum_i((h-3)-E_i)\le e$. Hence
\[
 3\tri o+2\sum_t\tri{|V_t|}
 \ge(h-3)\tri m-me+3\tri o
 \ge h\tri m-4me,
\]
using $\tri m-\tri o\le m(m-o)$. Combining these inequalities with
\eqref{eq:block-count} proves the lemma.
\end{proof}

\begin{lemma}\label{lem:opposite}
Let $C$ be antisymmetric and zero-sum-free, let $K$ be a subgroup of
order $h$, and put $A=C\cap K$, $a=|A|\ge2$. Write
\[
 C\cap(g+K)=g+X,\qquad C\cap(-g+K)=-g+Y
\]
for two distinct opposite cosets. If $X,Y$ are both nonempty, then
\begin{equation}\label{eq:opposite}
 h-|X|-|Y|\ge a-1.
\end{equation}
\end{lemma}

\begin{proof}
Antisymmetry gives $X\cap(-Y)=\varnothing$. Suppose
$e=h-|X|-|Y|<a-1$. Set $Z=\Sig(A)$, so $|Z|\ge2a$ by
\eqref{eq:basic}. The set $D=Z+X+Y$ avoids zero. For
$J=\stab_K(D)$ and $j=|J|$, Kneser's theorem gives
\[
 |Z+J|+|X+J|+|Y+J|-2j\le h-j,
 \qquad |Z+J|\le e+j.
\]
If $j\le e$, then $|Z|\le2e<2a$, a contradiction. Thus $j>e$,
and divisibility by $j$ gives $Z\subseteq J$ and
$|X+J|+|Y+J|=h$.
Since $D+J=D$ avoids zero and $0\in Z$, we have
$(X+J)\cap(-Y+J)=\varnothing$. These sets therefore partition $K$.
In $L=K/J$ this gives $\bar Y=-(L\setminus\bar X)$. For $u\in L$,
\[
 u\notin\bar X+\bar Y\quad\Longleftrightarrow\quad
 \bar X=u+\bar X.
\]
Thus the complement of $\bar X+\bar Y$ is the subgroup
$P=\stab_L(\bar X)$. This subgroup also stabilizes
$L\setminus P=\bar X+\bar Y$. Since $J$ was the full stabilizer
of $D$, the quotient sumset is aperiodic, so $P$ is trivial.
In particular, $\bar X+\bar Y=L\setminus\{0\}$.

Choose a nonzero $u\in L$ and representations
$\bar x_1+\bar y_1=u$, $\bar x_2+\bar y_2=-u$.
Fix actual $x_2,y_1$, and vary $x_1,y_2$ in their indicated
$J$-fibers. These two fibers together miss at most $e$ points,
so at least $j-e$ pairs solve $x_1+x_2+y_1+y_2=0$ in $K$.
Since $j\ge|Z|\ge2a$ and $e<a-1$, more than two solutions exist.
Excluding $x_1=x_2$ and $y_1=y_2$ eliminates at most two. A
remaining solution gives four distinct elements of $C$ summing to
zero, a contradiction.
\end{proof}

\begin{lemma}\label{lem:fibers}
Let $C$ be antisymmetric and zero-sum-free, $K$ a subgroup of order
$h$, and $A=C\cap K$ with $a=|A|\ge2$. Suppose its occupied
nonzero $K$-cosets are $q_1,\ldots,q_m$, with no self-inverse
coset and no opposite pair. If their fiber sizes $L_i$ satisfy
$L_i\ge7$ and $e=\sum_i(h-L_i)<a-1$, then
\begin{equation}\label{eq:fibers}
 \N(C)\ge h\bigl(h\tri m-4me-1\bigr)+\N(A).
\end{equation}
\end{lemma}

\begin{proof}
For $Z=\Sig(A)$, \eqref{eq:basic} gives $|Z|\ge2a>e+1$.
If $X_i$ is a fiber translated into $K$, then
\eqref{eq:restricted} and $h-L_i\le e$ imply
\[
 Z+(j\wedge X_i)=K\qquad(0<j<L_i).
\]
Consequently every multiplicity vector having an internal coordinate
fills the entire coset of its quotient sum. That quotient sum cannot
be zero, since all these sums use an external element. Thus
\eqref{eq:internal-nonzero} holds. Each point of $\mathcal I$ contributes
$h$ nonempty sums outside $K$, and $A$ contributes $\N(A)$ more.
Apply Lemma~\ref{lem:capacity}.
\end{proof}

\section{An expansion alternative}

\begin{proposition}\label{prop:alternative}
Let $C$ be an antisymmetric zero-sum-free set, $c=|C|$, and
$H=\gen C$. Then at least one of the following holds:
\begin{enumerate}
\item $\N(C)\ge c^2/2$;
\item for every integer $r\ge1$ for which $rC^*\ne H$,
\begin{equation}\label{eq:expansion}
 |rC^*|\ge2r(c-100).
\end{equation}
\end{enumerate}
\end{proposition}

\begin{proof}
There is nothing to prove if $c\le100$. Suppose
$rC^*\ne H$ and $|rC^*|<2r(c-100)$. Put
$K=\stab_H(rC^*)$, $h=|K|$, $A=C\cap K$, and $a=|A|$.
Let $p$ count the nonzero $K$-cosets met by $C^*$.
Kneser's theorem gives
\begin{equation}\label{eq:expansion-kn}
 |rC^*|\ge(rp+1)h.
\end{equation}
Here $p\ge2$. Indeed, $p=0$ would imply $K=H$; and if $p=1$,
the sole nonzero quotient element is self-inverse and generates
$H/K$, forcing $rC^*=H$.
Antisymmetry also gives $2(c-a)\le ph$. Thus, writing
\[
 e=\frac{ph}{2}-(c-a)\ge0,
\]
the assumed failure and \eqref{eq:expansion-kn} imply
\begin{equation}\label{eq:kernel-budget}
 a>100,\qquad e<a-100,\qquad a^2<6h,
 \qquad h>1600,\qquad a<h/16.
\end{equation}
The third inequality follows from $\N(A)+1\le h$ and
\eqref{eq:basic}; the last two follow from $a>100$ and $a^2<6h$.

No distinct opposite pair of external cosets can both be occupied.
Otherwise its contribution to $e$ is at least $a-1$ by
Lemma~\ref{lem:opposite}, contradicting \eqref{eq:kernel-budget}.
There is no occupied self-inverse nonzero coset either. If one has
size $L$, its elements lie in a subgroup of order $2h$, so
\eqref{eq:basic} gives $L<\sqrt{12h}$. Its contribution to $e$ is
$h/2-L$, whereas $e<a$ gives
\[
 L>h/2-a>h/2-\sqrt{6h}>\sqrt{12h}
 \quad(h>1600),
\]
a contradiction.

We may therefore write $p=2m$, with one occupied coset per opposite
pair. If their sizes are $L_i$, then
\[
 e=\sum_i(h-L_i),\qquad c=mh+a-e,\qquad
 L_i\ge h-e>h-a+100>7.
\]
Lemma~\ref{lem:fibers} applies. After dropping its nonnegative
$\N(A)$ term, we obtain
\begin{align*}
 \N(C)-\frac{c^2}{2}
 &\ge \frac{mh^2}{2}-mha-3mhe-\frac{(a-e)^2}{2}-h\\
 &>\frac{mh^2}{2}-4mha-4h
   =\frac{mh}{2}\left(h-8a-\frac8m\right)>0.
\end{align*}
The last inequality uses $a<h/16$ and $h>1600$.
Thus every failure of \eqref{eq:expansion} gives the first alternative.
\end{proof}

\section{From expansion to subset sums}

\begin{lemma}\label{lem:exp-boundary}
Let $C\subseteq G$, $H=\gen C$, and $d>0$. Suppose, for every
$r\ge1$, either $rC^*=H$ or $|rC^*|\ge2rd$.
If $b=\min\{|B|,|G\setminus B|\}>d$ and $|H|\ge2b$, then
\begin{equation}\label{eq:exp-boundary}
 \max_{x\in C}\lambda_B(x)>d-d^2/b.
\end{equation}
\end{lemma}

\begin{proof}
By complementation assume $|B|=b$. Suppose all boundaries are at
most $M=d-d^2/b>0$. Put $\rho(x)=|B\cap(B+x)|$.
For $x\in rC^*$, subadditivity gives $\rho(x)\ge b-rM$, whereas
\begin{equation}\label{eq:correlation}
 \sum_{x\in H}\rho(x)=\sum_{Q\in G/H}|B\cap Q|^2\le b^2.
\end{equation}
Write $x_0=b/d=R+\theta$, with $R=\lfloor x_0\rfloor$ and
$0\le\theta<1$. The hypotheses imply
$|rC^*|\ge2rd$ for $1\le r\le R$ and
$|(R+1)C^*|\ge2b$. Also $t_0=b-(R+1)M>0$.
Integrating the level sets of $\rho$ over the consecutive intervals
of length $M$, and then over $[0,t_0]$, yields
\[
 \sum_{x\in H}\rho(x)\ge dMR(R+1)+2bt_0.
\]
The right side, divided by $b^2$, equals
\[
 1+\frac{1-\theta+\theta^2+\theta(1-\theta)/x_0}{x_0^2}>1,
\]
contradicting \eqref{eq:correlation}.
\end{proof}

We now prove the global estimate for antisymmetric sets. For $s\ge1$
write
\[
 L(s)=\lceil\log_{3/2}s\rceil,\qquad
 \calH_s=\sum_{j=1}^s\frac1j.
\]

\begin{lemma}\label{lem:quarter}
Every antisymmetric zero-sum-free set $T$ of size $r\ge1$ satisfies
\begin{equation}\label{eq:quarter}
 \N(T)\ge\frac{r^2}{4}-\frac{r+1}{2}L(r)-1.
\end{equation}
\end{lemma}

\begin{proof}
Arrange $T=\{a_1,\ldots,a_r\}$ greedily, always choosing the largest
boundary relative to the current prefix sumset. Let
$B_j=\Sig(\{a_1,\ldots,a_j\})$ and $\sigma_j=|B_j|$, with
$\sigma_0=1$. Before step $i$, set
$C_i=\{a_i,\ldots,a_r\}$, $c_i=r-i+1$, and $H_i=\gen{C_i}$.
Stop before the first step for which
\begin{equation}\label{eq:capacity-stop}
 |H_i|<2\min\{\sigma_{i-1},|G\setminus B_{i-1}|\},
 \qquad G=\gen T.
\end{equation}
Before stopping, zero-sum-freeness and \eqref{eq:basic} imply
$|G\setminus B_{i-1}|\ge\N(C_i)\ge2c_i-1$. Hence
\eqref{eq:olson-boundary} gives
\begin{equation}\label{eq:weak-growth}
 \sigma_i-\sigma_{i-1}\ge\frac12\min\{\sigma_{i-1}+1,c_i+1\}.
\end{equation}
While $\sigma_{i-1}<c_i$, the quantity $\sigma_i+1$ increases by
a factor at least $3/2$. This regime has ended by $L(r)$ steps if
the process has not already stopped. Once it ends, it does not
restart. Summing the lower bounds $(c_i+1)/2$ thereafter, and
discarding the first $L(r)$ ideal increments, gives, for each
available prefix,
\begin{equation}\label{eq:quarter-prefix}
 \sigma_j\ge\frac{r^2-(r-j)^2}{4}-\frac{r+1}{2}L(r).
\end{equation}
For $j\le L(r)$ the right side is nonpositive, so it also covers
an early stop.

If there is no stop, subtract one from \eqref{eq:quarter-prefix}
at $j=r$. Otherwise the stopping subgroup $H$ is proper in $G$.
Let $v$ be the last position with $a_v\notin H$, and put
$T'=\{a_{v+1},\ldots,a_r\}$, $t=r-v$. Both $v$ and $v-1$
precede the stopping position. Since $T'\subseteq H$ and $a_v\notin H$,
the two parts of
\[
 \Sig(a_vT')=\Sig(T')\sqcup\bigl(a_v+\Sig(T')\bigr)
\]
lie in distinct $H$-cosets. Hence $\N(a_vT')=2\N(T')+1$, and
\eqref{eq:superadditive} gives
\[
 \N(T)\ge\sigma_{v-1}+2\N(T')
 \ge\frac{r^2-(t+1)^2}{4}-\frac{r+1}{2}L(r)+\frac{t^2}{3}.
\]
Finally, $t^2/3-(t+1)^2/4=(t-3)^2/12-1\ge-1$.
\end{proof}

\begin{proposition}\label{prop:antisymmetric}
Every antisymmetric zero-sum-free set $S$ of size $s\ge1$ satisfies
\begin{equation}\label{eq:antisymmetric}
 \N(S)\ge\frac{s^2}{2}-10s\log(s+1)-116s.
\end{equation}
\end{proposition}

\begin{proof}
Use the same greedy notation, now in $G=\gen S$. Stop before the
first step $i$ for which either
\begin{equation}\label{eq:two-stops}
 \N(C_i)\ge c_i^2/2
 \quad\text{or}\quad
 |H_i|<2\min\{\sigma_{i-1},|G\setminus B_{i-1}|\}.
\end{equation}
Give the first condition priority if both hold. Before stopping,
\begin{equation}\label{eq:complement}
 |G\setminus B_{i-1}|\ge\N(C_i)
 \ge\max\{2c_i-1,c_i^2/6\}.
\end{equation}
The first inequality follows because the prefix sumset avoids the
negative nonempty tail sums. Thus \eqref{eq:weak-growth} remains valid.

Let $c=c_i$, $\sigma=\sigma_{i-1}$, and
$b=\min\{\sigma,|G\setminus B_{i-1}|\}$. By
Proposition~\ref{prop:alternative} and Lemma~\ref{lem:exp-boundary},
with $d=c-100$, the actual increment satisfies
\begin{equation}\label{eq:strong-growth}
 \sigma_i-\sigma_{i-1}\ge c-106-\frac{c^2}{\sigma}.
\end{equation}
For clarity, if $d\le0$ this is automatic. If $d>0$ and $b\le d$,
\eqref{eq:complement} forces $b=\sigma$, and the right side is
negative. Otherwise Lemma~\ref{lem:exp-boundary} applies. When
$b=\sigma$, it gives the stronger $c-100-c^2/\sigma$; when
$b=|G\setminus B_{i-1}|$, its error $d^2/b$ is at most six by
\eqref{eq:complement}. This proves \eqref{eq:strong-growth} in
every case.

Put $L=L(s)$. After the first $L+1$ steps, weak growth gives, for
each available $i\ge L+2$,
\begin{equation}\label{eq:harmonic}
 \sigma_{i-1}\ge\frac{(i-1-L)(c_i+2)}2,
 \qquad \frac{c_i^2}{\sigma_{i-1}}
       \le\frac{2s}{i-1-L}.
\end{equation}
Indeed, by step $L$ the initial multiplicative regime has ended.
For $L+1\le j\le i-1$, the weak increment at step $j$ is at
least $(c_j+1)/2\ge(c_i+2)/2$. Summing these $i-1-L$
increments gives the first inequality. If $L+2>s$, there are no
such steps.
Discard the first $L+1$ ideal increments at cost at most $s(L+1)$,
and sum \eqref{eq:strong-growth} and \eqref{eq:harmonic}. Since
\[
 \sum_{i=1}^j c_i\ge\frac{s^2-(s-j)^2}{2},
\]
every prefix formed before stopping obeys
\begin{equation}\label{eq:half-prefix}
 \sigma_j\ge\frac{s^2-(s-j)^2}{2}
       -s\bigl(L+107+2\calH_s\bigr).
\end{equation}
Its right side is nonpositive during the discarded initial steps,
so an early stop is included.

If the first condition in \eqref{eq:two-stops} occurs, apply
\eqref{eq:superadditive} to prefix and tail. The tail contribution
$c_i^2/2$ cancels the missing square in \eqref{eq:half-prefix}.
If no stop occurs, use $j=s$ and subtract the empty sum.

For a stop caused only by subgroup capacity, let $H$ be the stopping
subgroup, $v$ the last position outside $H$, and $T=S_{>v}$,
$r=s-v$. As in Lemma~\ref{lem:quarter},
\[
 \N(S)\ge\sigma_{v-1}+2\N(T).
\]
Apply \eqref{eq:half-prefix} at $v-1$ and
\eqref{eq:quarter} to $T$. For $r=0$ use $\N(T)=0$ directly.
Since $r+1\le s$ and $L(r)\le L(s)$ when $r\ge1$, all cases give
the convenient common bound
\begin{equation}\label{eq:explicit-transfer}
 \N(S)\ge\frac{s^2}{2}-108s-(2s+1)L(s)
                        -2s\calH_s-\frac52.
\end{equation}
To simplify it, use
$L(s)\le\tfrac52\log(s+1)+1$ and
$\calH_s\le\log(s+1)+1$. The error in
\eqref{eq:explicit-transfer} is at most
\[
 (7s+\tfrac52)\log(s+1)+112s+\tfrac72
 \le10s\log(s+1)+116s,
\]
because $(3s-\tfrac52)\log(s+1)+4s-\tfrac72\ge0$ for $s\ge1$.
This proves \eqref{eq:antisymmetric}.
\end{proof}

\section{Proof of the main theorem and consequences}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Let $I$ be the set of all order-two elements in $S$, let $q=|I|$,
and set $X=S\setminus I$, $s=k-q$. Zero-sum-freeness makes $I$
linearly independent over $\mathbb F_2$. Therefore
\begin{equation}\label{eq:involutions}
 2^q=|\Sig(I)|\le\N(S)+1.
\end{equation}
If $\N(S)\ge k^2/2$ there is nothing to prove. Otherwise
\[
 q\le\log_2(\N(S)+1)<3\log(k+1),
\]
since $\N(S)+1<k^2/2+1\le(k+1)^2$ and $2/\log2<3$.
The set $X$ is antisymmetric. Proposition~\ref{prop:antisymmetric}
and $s^2/2\ge k^2/2-kq$ yield
\begin{align*}
 \N(S)&\ge\N(X)\\
 &\ge\frac{k^2}{2}-kq-10s\log(s+1)-116s\\
 &\ge\frac{k^2}{2}-13k\log(k+1)-116k.
\end{align*}
If $s=0$, use $\N(X)=0$ in the same estimate; its right side is
then nonpositive. This also covers all small values of $k$.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{cor:olson}]
Let $S$ be a zero-sum-free set of maximum size $k=\Ol(G)-1$.
The case $k=0$ is immediate. Since $\N(S)\le n-1$,
\eqref{eq:basic} gives $k\le\sqrt{6n}$. Put
$a=13\log(k+1)+116$. Theorem~\ref{thm:main} implies
$k^2/2-ak\le n-1$, whence
\[
 k\le a+\sqrt{a^2+2(n-1)}\le\sqrt{2n}+2a.
\]
Also $k+1\le(\sqrt6+1)\sqrt{n+1}$. It follows that
\[
 \Ol(G)\le\sqrt{2n}+13\log(n+1)
                   +26\log(\sqrt6+1)+233
 <\sqrt{2n}+13\log(n+1)+266.
\]
\end{proof}

The coefficient $\sqrt2$ is necessary uniformly over finite abelian
groups. For $n>\tri t$, the residues $1,2,\ldots,t$ form a
zero-sum-free set in $\mathbb Z/n\mathbb Z$. Taking
$n=\tri t+1$ shows that $\Ol(\mathbb Z/n\mathbb Z)\ge t+1$
along a sequence with $t\sim\sqrt{2n}$.

Theorem~\ref{thm:main} leaves a gap of order $k\log k$ from the
classical upper bound. It does not establish the following sharper
possibility.

\begin{conjecture}\label{conj:exact}
For every $k\ge8$,
\[
 f(k)=\left\lfloor\frac{k^2}{2}\right\rfloor+1.
\]
\end{conjecture}

The small values must be treated separately. In particular,
$f(4)=8$ and $f(7)=24$; see \cite{EE,GHHLLP}. Even an improvement
of the error in Theorem~\ref{thm:main} to $O(k)$ would be a further
step. The present proof loses a logarithmic number of initial greedy
steps and also incurs a harmonic error from reciprocal prefix sizes.
Improving the error term to $O(k)$ would require a sharper treatment
of these steps, or additional structural information.

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\end{document}