Math-1774 · mathematics · Combinatorics

Math-1774

Exact subgroup self-sumsets and a robust quotient-character leakage bound

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Statement

Definitions and mathematical statement

For an odd prime $p$, let $\mathbb F_p^*$ be the multiplicative group of its nonzero elements. A proper multiplicative subgroup $H<\mathbb F_p^*$ is required to differ from the whole multiplicative group. For $A\subseteq\mathbb F_p$, write $A+A=\{a+b:a,b\in A\}$, allowing equal summands. The containment $A+A\subseteq H$ includes every such sum, so in particular it excludes zero from $A+A$.

Define \[ Q(p)=\max\{|A|:A\subseteq\mathbb F_p, \exists H<\mathbb F_p^*\text{ with }A+A\subseteq H\}. \] The maximum includes the empty set and is therefore always defined. The explicit containment version of the workshop question asks whether \[ Q(p)=p^{o(1)}\qquad(p\to\infty\text{ through odd primes}), \] meaning that for every $\varepsilon>0$ there is $p_0(\varepsilon)$ such that $Q(p)\leq p^\varepsilon$ for all primes $p\geq p_0$. Equivalently, can a fixed positive power of $p$ ever be exceeded along infinitely many primes by a set whose every pairwise sum lies in one proper multiplicative subgroup? A negative answer to the proposed subpower bound requires such an infinite family for some fixed exponent. The subgroup may depend on the prime and on $A$; uniformity over all proper subgroups is part of the assertion. The source also asks about exact equality $A+A=H$. The classification in [S3] rules out that equality for $|A|>1$, while the containment target formulated here is a different remaining question.

Short English statement

Can a set of polynomial size in a prime field have every pairwise sum inside one proper multiplicative subgroup?

Sources

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