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If $A$ has positive density in a finite abelian group, then $A+A-A-A$ contains a structured neighbourhood of zero. In a finite-dimensional vector space this neighbourhood can be taken to be a large <vector subspace>; in a cyclic group it can be taken to be a <Bohr set>.

Bogolyubov lemma

A <Bohr set> of rank $d$ and width $\rho$ satisfies
$$
|B(\Gamma,\rho)|\geq\left(\frac{\rho}{8}\right)^d|G|.
$$
The proof partitions each circle coordinate into arcs and applies translation averaging and the pigeonhole principle.

Lower bound for the size of a Bohr set

For $\Gamma\subseteq\widehat G$ and $\rho>0$, the <Bohr set> is
$$
B(\Gamma,\rho)
=\{x\in G:|\gamma(x)-1|\leq\rho\text{ for every }\gamma\in\Gamma\}.
$$
Writing $d=|\Gamma|$, the <lower bound for the size of a Bohr set> is
$$
|B(\Gamma,\rho)|\geq
\left(\frac{\rho}{8}\right)^d|G|.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 129 / 2 / a / Solution

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For a finite abelian group $G$, frequencies $\Gamma\subseteq\widehat G$, and $\rho>0$, the Bohr set is
$$
B(\Gamma,\rho)=\{x\in G:|\gamma(x)-1|\leq\rho\text{ for every }\gamma\in\Gamma\}.
$$
Its rank is $|\Gamma|$ and its width is $\rho$.


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 Incoming links: Bohr set