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Let $G$ be a locally compact topological group, $K$ a compact subgroup and $\Gamma$ a discrete subgroup. I try to find a neighbourhood $U$ of the identity such that $\Gamma \cap UK = \Gamma \cap K$. How can I construct such a neighbourhood? If $K$ is trivial, the existence of $U$ follows from the discreteness of $\Gamma$.

Do you have some good references on topological groups?

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Hewitt and Ross, "Abstract Harmonic Analysis" Volume I, Chapter 2 is a good source for this. They may not answer this exact question, but I believe they present enough for you to answer it yourself. –  William DeMeo Feb 22 '12 at 6:47
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It seems that you do not need the local compactness of $G$, $K$ may be an arbitrary nonempty compact subset of $G$ and you can construct the neighborhood $U$ by the following way. Since $\Gamma$ is discrete then for each element $x\in G$ there are neighborhoods $V_x$ and $U_x$ of the identity such that $V_xx\cap\Gamma=\{x\}\cap\Gamma$ and $U_x^2\subset V_x$. Let $F$ be a finite subset of $K$ such that $K\supset\bigcup \{U_xx:x\in F\}$. Put $U=\bigcap \{U_x:x\in F\}$. Suppose now that $y\in\Gamma\cap UK$. Then there is an element $z\in K$ such that $y\in Uz$. There is an element $x\in F$ such that $z\in U_xx$. Then $y\in Uz\subset UU_xx\subset U^2_xx\subset V_xx$. Since $y\in\Gamma$, we see that $y\in V_xx\cap \Gamma=\{x\}\cap\Gamma\subset K\cap\Gamma$.

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