There is the following problem in the book "Fourier Analysis on Number Fields" by D. Ramakrishnan and R. Valenza:
Let $G$ be a locally compact abelian subgroup with Haar measure $\mu$, and $\Gamma$ a subgroup of $G$ (not necessarily closed) such that $G=\Gamma+\Phi$ for some compact subset $\Phi\subset G$. If $X$ is a compact subset of $G$ such that $\mu(X)>\mu(\Phi)$ then there exist two different elements $x_1,x_2\in X$ such that $x_1-x_2\in \Gamma$.
I was only able to show this for $\Gamma$ discrete (or countable). I put a sketch of my partial solution below (I wanted to make it hidden, to not suggest an approach that may not possibly work in general, but I am doing something wrong with the command). Any hints or solutions are welcome.
Assuming the contrary, all the translations $\gamma+X$, for $\gamma\in\Gamma$, are pairwise disjoint. Let $$S=\{\gamma\in\Gamma:(\gamma+X)\cap \Phi\neq \emptyset\}=\Gamma\cap (X-\Phi).$$ The set $S$ is finite since $\Gamma$ is discrete and $X-\Phi$ is compact. Setting $$Y_\gamma=(\gamma+X)\cap\Phi,\ \ \ Z_\gamma=(Y_\gamma-\gamma)\subset X,$$ we obtain $$\bigsqcup_S Y_\gamma \subset \Phi\implies \sum_S\mu(Y_\gamma)\leq\mu(\Phi)$$ $$\bigcup_S Z_\gamma=X\implies \sum_S\mu(Z_\gamma)\geq \mu(X)$$ which gives the contradiction since $\mu(Y_\gamma)=\mu(Z_\gamma)$.