Let $G$ be a topological group, $S\subseteq G$ is a syndetic set if there is a compact set $K\subseteq G$ with $G=KS$.
In my research I need to find non-empty set $R\subseteq G$, such that for every syndetic set $S$ there is $g\in R$ such that $S\cap g^{-1}S\neq \emptyset$.
$R\subseteq G$ is called a thick set, if for every finite set $F\subseteq G$, we have $R\cap_{g\in F}gR\neq \emptyset$.
Is it true that if $R$ is a thick set, then for every syndetic set $S$, there is $g\in R$ such that $S\cap g^{-1}S \neq\emptyset$
If you do not have any restriction on $G$, this may not be true. Let $G$ be a compact topological group, for example $S^1$, then $S=\{e\}$ where $e$ is the identity of the group, is syndetic since $\{e\}G=G$ and $G$ is compact. Meanwhile, the set $R=G\backslash\{e\}$ is thick since for any finite set $F$ pick some $h\notin F^{-1}\cup\{e\}$, then $h\in R\cap_{g\in F}gR$ which implies that $R$ is thick. Note that $S\cap g^{-1}S \neq\emptyset$ if and only if $g$ is the identity, and it is not in $R$, which makes this as a counterexample.
If $G$ is discrete then it is true since in this case a set in $G$ is compact if and only if it it is finite. In details, if $S$ is syndetic, then there exists finite set $K$ such that $G=KS$. Since $R$ is thick there exist some $X$ such that $x\in R\cap_{g\in K}gR$. So $g^{-1}x\in R$ for any $g\in K$. Note that $x^{-1}KS=x^{-1}G=G$ and so there exists some $g\in K$ such that $x^{-1}gS\cap S\neq \emptyset$. And $x^{-1}g=(g^{-1}x)^{-1}$ is that we want.