Given a countable group $\Gamma$ and a finite index subgroup $\Gamma_0$ and an infinite subset $S \subset \Gamma$ such that $S \cap \Gamma_0 = \emptyset$ can one conclude that $\Gamma_0 \cup S \Gamma_0 = \Gamma$?
2026-03-27 21:19:18.1774646358
Translates by infinite subset of finite index subgroup
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Let $\Gamma = \mathbb{Z}$, $\Gamma_0 = 3\mathbb{Z}$, and $S = 3\mathbb{Z} + 1$.