$H$ a discrete subgroup of topological group $G$ $\implies$ there exists an open $U\supseteq\{1\}$ s.t. the $hU$ are pairwise disjoint

Question

Problem: Let $G$ be a topological group (i.e., $G$ is a topological space and $G \times G\rightarrow G$ is continuous), and let $H$ be a discrete subgroup of $G$. Prove that there is a neighborhood, $U$, of the identity, $1$, such that the sets $h \cdot U$, $h \in H$ are pairwise disjoint.

Hint: First choose a neighborhood $V \subset G$ of $1$ such that $V \cap H = \{1\}$. Now use the map $f: G \times G \rightarrow G$ defined by $f(x,y) = xy^{-1}$ to prove there is an open set $U$ containing $1$ such that $\{xy^{-1} \mid x,y \in U\} \subset V$.

Attempt:

1. Let $U$ be the set described in the hint (whose definition is unknown), and $U' = \{xy^{-1} : x,y \in U\}$.

2. Then we will have that $\{1\} \subseteq U' \subseteq V$ (for reasons yet to be determined).

3. Now let $h_1, h_2 \in H$ s.t. $h_1 \ne h_2$.

4. We aim to show that $h_1 U \ne h_2 U$ in order that the $h \cdot U$ be pairwise distinct.

How does one approach this problem? The mapping $f(x,y) = xy^{-1}$, as described in the hint, is quite similar to the mapping $(x,y) \mapsto xy$ which is continuous since $G$ is a topological group. Indeed, since the mapping $y \mapsto y^{-1}$ is also continuous (since $G$ is a topological group), we also have that if $x$ is fixed, then the composition

$$y \mapsto y^{-1} \mapsto xy^{-1}$$

is continuous (being the composition of two continuous mappings). Perhaps this fact can be used to get us closer to the result.

Answer 1

The mapping $\nu\colon G\times G\to G$ defined by $\nu(x,y)=xy^{-1}$ is continuous (easy), and $\nu(1,1)=1$; since $V$ is a neighborhood of $1$, there is a neighborhood $W$ of $(1,1)$ in $G\times G$ such that $\nu(W)\subseteq V$. Since the topology on $G\times G$ is the product topology, there exists a neighborhood $U$ of $1$ such that $U\times U\subseteq W$. As a consequence $$ x,y\in U \to (x,y)\in U\times U\to (x,y)\in W\to \nu(x,y)\in V\to xy^{-1}\in V. $$ Thus the first doubt should be solved.

We can certainly find $V$ such that $V\cap H=\{1\}$ because the topology on $G$ induces on $H$ the discrete topology, by hypothesis. So, fix $U$ found as before and suppose $h_1U\cap h_2U$, for $h_1,h_2\in H$. Let $z\in h_1U\cap h_2U$; then $z=h_1x=h_2y$, for some $x,y\in U$.

Therefore $$ xy^{-1}=h_1^{-1}h_2 $$ Now, $xy^{-1}\in V$ and $h_1^{-1}h_2\in H$. So we can conclude that …

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Now, why is $\nu$ continuous? The map $i:g\mapsto g^{-1}$ is continuous; therefore also the map $\mathit{id}\times i:G\times G\to G\times G$ defined by $(x,y)\mapsto(x,y^{-1})$ is continuous. If $\mu\colon G\times G\to G$ is the multiplication, then $\nu=\mu\circ(1\times i)$.