Size threshold for sum-free multiplicative subgroups of finite fields: closing the Alon–Bourgain gap between order $p^{1/3}$ and order $q^{3/4}$

A multiplicative subgroup $H$ of $\mathbb{F}_q^{*}$ (the nonzero elements of the finite field with $q$ elements) is sum-free if it contains no $x,y,z\in H$ with $x+y=z$; dividing by $z$, this is equivalent to containing no $x,y\in H$ with $x+y=1$. Alon and Bourgain (Geom. Funct. Anal. 24 (2014), 721–739) proved that infinitely many prime fields $\mathbb{F}_p$ contain sum-free subgroups of order $\Omega(p^{1/3})$ — via a cyclotomic obstruction: for each order $d$ not divisible by 6, an explicit integer built from cyclotomic units has few prime divisors $p\equiv 1\pmod d$, and for every other such prime the order-$d$ subgroup is sum-free — while every subgroup of order at least $q^{3/4}$ must contain a solution, by the Weil bound for the Gauss periods (the exponential sums that form the nontrivial eigenvalues of the associated Cayley sum graph, the graph on $\mathbb{F}_q$ joining two distinct elements whose sum lies in $H$). In extension fields, constructions via cyclic caps — point sets in projective space with no three collinear points, carried by a cyclic group of projective transformations — give sum-free subgroups of order $\Omega(q^{2/3})$. A random model predicts the threshold $\Theta(q^{1/2})$, the constructions suggest $q^{2/3+o(1)}$, and the proven range in the prime regime is $[1/3,3/4]$. The problem is open: neither side of the gap has been improved, and neither of the two named gateways — the $1/\phi(d)$ distribution question for the cyclotomic obstruction, and square-root cancellation for Gauss periods, which would force solutions already at order $cq^{2/3}$ — has been settled. The logarithmic factors in the companion Hamiltonicity threshold were removed by Glock, Munhá Correia and Sudakov (2024), bringing that threshold to the same order $q^{3/4}$, whose tightness is equally open.

Background

Additive structure in multiplicative subgroups. The interplay between addition and multiplication in finite fields has been a central theme of additive combinatorics since the sum-product theorem of Erdős and Szemerédi for integers and its finite-field analogue by Bourgain, Katz and Tao: a subset of $\mathbb{F}_q$ (the finite field with $q$ elements, $q$ a prime power) cannot be simultaneously close to closed under both operations. The extremal case of multiplicative closure is a multiplicative subgroup: a subgroup $H$ of the cyclic group $\mathbb{F}_q^{*}=\mathbb{F}_q\setminus\{0\}$ under multiplication, necessarily of order $d$ dividing $q-1$. Alon and Bourgain asked how far such an $H$ can be from being additively closed, in particular whether it can be sum-free, meaning there are no $x,y,z\in H$ with $x+y=z$; dividing by $z\in H$ shows this is equivalent to the absence of $x,y\in H$ with $x+y=1$, i.e. $H\cap(1-H)=\varnothing$.

The sufficient condition and the Gauss-period barrier. Consider the Cayley sum graph of $\mathbb{F}_q$ with respect to $H$: the graph on vertex set $\mathbb{F}_q$ in which distinct $x,y$ are adjacent whenever $x+y\in H$. It is $d$-regular, and its nontrivial eigenvalues are the Fourier coefficients $\hat{1_H}(a)=\sum_{x\in H}e^{2\pi i\,\mathrm{Tr}(ax)/p}$ for $a\ne 0$, where $\mathrm{Tr}$ is the trace to the prime subfield; for subgroups these are classical quantities called Gauss periods. An $(n,d,\lambda)$-graph is an $n$-vertex $d$-regular graph whose nontrivial eigenvalues are at most $\lambda$ in absolute value; the Weil bound (the Riemann hypothesis for curves over finite fields) gives $\lambda\le q^{1/2}$ here, and the expander mixing lemma — the standard principle that in a graph with small nontrivial eigenvalues, the number of edges between any two vertex sets is close to what a random graph of the same density would contain — then implies that every subgroup with $|H|\ge q^{3/4}$ contains a solution of $x+y=z$. Alon and Bourgain record this as easy and well known, citing for it a paper of Sárközy on products and shifted products of residues modulo $p$. Improving the exponent $3/4$ requires beating the Weil bound for Gauss periods of large subgroups: the conjectured square-root cancellation $\max_{a\ne 0}|\hat{1_H}(a)|\ll |H|^{1/2+o(1)}$, attributed by Alon and Bourgain to Montgomery, Vaughan and Wooley, would force every subgroup of order at least $cq^{2/3}$ to contain a solution. No such improvement is known in the range $|H|\ge q^{1/2}$: the strongest published power-saving estimates for these exponential sums apply only to subgroups of order below $p^{1/2}$ in prime fields.

Constructions of large sum-free subgroups. For an integer $d$ not divisible by 6, let $\xi=e^{2\pi i/d}$ and let $A(\xi)$ be the product, over $0<j<d$ and $0<j_1\le j_2<d$, of the algebraic integers $\xi^j-1$ and $\xi^{j_1}+\xi^{j_2}-1$. The divisibility restriction is forced: the only way two complex roots of unity can sum to $1$ is as a primitive sixth root of unity together with its inverse, and if $6\mid d$ then $H$ contains an element $z$ of order 6 with $z+z^5=1$, so every subgroup whose order is divisible by 6 automatically contains a solution. For $6\nmid d$ none of the factors vanish, their product $A(\xi)$ is a nonzero rational integer (invariant under Galois conjugation), and reducing all the non-vanishing relations modulo a prime $p\equiv 1\pmod d$ shows that the order-$d$ subgroup of $\mathbb{F}_p^{*}$ is sum-free whenever $p$ does not divide $A(\xi)$. Alon and Bourgain proved that the set $E(d)$ of exceptional primes (divisors of $A(\xi)$ lying in the class $1\bmod d$) has size at most $d^2/\log d$, and, combining this with Linnik's theorem on the least prime in an arithmetic progression and the Bombieri–Vinogradov theorem on the average distribution of such primes, that admissible $p$ exist with $p=O(d^3)$ for most orders $d$ (and $p=O(d^5)$ for every admissible $d$). This yields infinitely many prime fields with sum-free subgroups of order $\Omega(p^{1/3})$. The authors suspect that in fact only about a $1/\phi(d)$ fraction of the primes dividing $A(\xi)$ lie in the class $1\bmod d$ ($\phi$ the Euler function), which would improve the lower bound to $p^{1/2-o(1)}$, but they state they cannot prove this. In extension fields much larger examples are known: subgroups with no solution of $x+y=1$ correspond to cyclic caps — point sets in projective geometry containing no three collinear points and permuted cyclically by a Singer cycle (a cyclic group of projective linear transformations acting transitively on the points of the space) — and modifying a construction of Yekhanin gives, for infinitely many prime powers $q$ of the form $2^{4^{k+1}}$, sum-free subgroups of order $\Omega(q^{2/3})$.

Companion patterns and competing heuristics. Alon and Bourgain also proved that for $|H|$ of order $q^{3/4}$ up to polylogarithmic factors, the elements of $H$ admit a cyclic ordering in which every consecutive sum again lies in $H$ — equivalently, the induced subgraph of the Cayley sum graph on $H$ has a Hamilton cycle (a cyclic tour through all its vertices visiting each once); the polylogarithmic factors were later removed by Glock, Munhá Correia and Sudakov, so $|H|\ge cq^{3/4}$ suffices. The same cyclotomic method extends to forbidding several equations simultaneously (such as $x+y=z$ together with $x+y+z=w$), at the cost of weaker size guarantees. A random model in which a subgroup of order $d$ behaves like a random $d$-subset predicts about $d^2/q$ solutions of $x+y=1$ and hence a threshold $\Theta(q^{1/2})$, while the extension-field constructions suggest the true threshold may be as large as $q^{2/3+o(1)}$; which, if either, is correct is unknown.

Problem Statement

Determine, up to lower-order factors, the size threshold at which multiplicative subgroups of finite fields are forced to contain additive relations, in both the prime-field and the general prime-power regime. Call a subgroup $H\le\mathbb{F}_q^{*}$ sum-free if it contains no $x,y,z\in H$ with $x+y=z$, equivalently (dividing by $z\in H$) no $x,y\in H$ with $x+y=1$.

(a) Upper direction: prove that every subgroup $H\le\mathbb{F}_q^{*}$ with $|H|\ge q^{3/4-\delta}$ contains a solution for some absolute $\delta>0$ (or identify the obstructions and show $3/4$ is best possible); in particular, decide whether the square-root cancellation $\max_{a\ne 0}|\sum_{x\in H}e^{2\pi i\,\mathrm{Tr}(ax)/p}|\ll |H|^{1/2+o(1)}$ holds for subgroups with $|H|\ge q^{1/2}$, which would force solutions already at order $cq^{2/3}$.

(b) Lower direction: determine how large sum-free subgroups can be. In prime fields, improve upon $\Omega(p^{1/3})$ — ideally to $p^{1/2-o(1)}$ — or show it is essentially tight; any construction must use orders $d$ not divisible by 6, since subgroups of order divisible by 6 always contain a solution. In general prime powers, improve upon $\Omega(q^{2/3})$ or show it is essentially tight.

(c) Threshold determination: exhibit the infimum exponent $\theta$ (separately for $q$ prime and $q$ a proper prime power, if the regimes differ) such that for every $\varepsilon>0$ and all large $q$, every subgroup of order at least $q^{\theta+\varepsilon}$ contains a solution, matching it with constructions of sum-free subgroups of order at least $q^{\theta-\varepsilon}$ for infinitely many $q$; in the prime-field regime this decides between the random-model prediction $\theta=1/2$ and the algebraically suggested value $\theta=2/3$, while in the prime-power regime the known $\Omega(q^{2/3})$ constructions already exclude $\theta=1/2$, so there the question is whether $\theta=2/3$ is tight.

The source flags two quantitative gateways: (i) the prime-distribution statement that only about a $1/\phi(d)$ fraction of the primes dividing the cyclotomic obstruction $A(\xi)$ lie in the class $1\bmod d$ (with the least such non-exceptional prime $p=O(d^{2+o(1)})$), giving sum-free subgroups of order $p^{1/2-o(1)}$ in prime fields; (ii) the Gauss-period cancellation statement in (a). A companion question of the same flavor: determine whether the threshold for the cyclic-ordering pattern (consecutive sums in $H$; now known by Glock et al. to be at most $cq^{3/4}$) coincides with the threshold for the single relation, as the random-graph heuristic suggests. The one-sided directions (a) and (b) and the matched threshold (c) all concern the same exponent $\theta$, and the two gateways and the companion question above also bear on that threshold.

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