For syndetic set $S\subseteq G$, is there $e\neq g\in G$ such that $gS\cap S$ is a syndetic set? Let $G$ be a topological group, $S\subseteq G$ is a syndetic set if there is a compact set $K\subseteq G$ with $G=KS$.
Let $\mathcal{P}$ be family of all syndetic sets in $G$,
Question. For every $S\in \mathcal{P}$, Is there $e\neq g\in G$ with $gS\cap S\in \mathcal{P}$?
 A: In general, no. Indeed, let $G$ be the group of reals with the standard topology. Let $S$ be any dense subset of $\Bbb R$ consisting of real numbers linearly independent over $\Bbb Q$. Such a set $S$ can be constructed inductively as follows. Fix a countable base $\{U_n\}$ of the space $G$ (consisting of non-empty sets) and for each $n$ pick any element $s_n\in U_n\setminus\langle s_1,\dots, s_{n-1}\rangle\Bbb Q$ (we recall that $\langle A\rangle$ denotes the subgroup generated by a subset $A$ of $G$). Since the group $G$ is locally compact and $S$ is a dense subset of $G$, $S$ is syndetic in $G$. On the other hand, suppose that there exists a non-zero element $g$ of $G$ such that $|(g+S)\cup S|\ge 2$. Then there exist elements $x_1, x_2, y_1, y_2\in G$ such that $x_1\ne x_2$, $x_1+g=y_1$, and $x_2+g=y_2$. Thus $g=y_1-x_1=y_2-x_2$. If $x_1$ is different from each of $x_2$, 
$y_1$, and $y_2$ then we obtain a contradiction with the linear independence of the set $S$. Since $x_1\ne x_2$ and $x_1\ne x_1+g=y_1$, the only remaining possibility is $x_1=y_2$. But then $2x_1=y_1+x_2$ and we again obtain a contradiction with the linear independence of the set $S$.
