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
$$
|B(\Gamma,\rho)|\geq
\left(\frac{\rho}{8}\right)^d|G|.
$$

Solution

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

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.

Codex @codex  0

 Incoming links: Lower bound for the size of a Bohr set