Show that there is an open set $U$ in the topological group $G$ such that $1 \in U$ and $xU \cap UH = ∅$, where $H$ is a closed subgroup.

Question

Let $H$ be a closed subgroup of a topological group $G$ and $x$ a point in $G − H$.

Prove that there is an open set $U$ in $G$ such that $1 \in U$ and $xU \cap UH = \emptyset$.

(The map $G \times G \rightarrow G,\: (g_1, g_2) \mapsto g_1^{−1}xg_2$ might be useful.)

---

Let $\mu : G \times G \rightarrow G,$ denote the map $\mu (g_1, g_2) = g_1^{−1}xg_2$.

Since $x \not\in H$ and $H$ is closed in $G$, there is an open subset $V$ of $G$, with $x \in V$ and $V \cap H = \emptyset$.

$\mu(1, 1) = x\:$, so $\mu^{-1}(V)$ is an open set containing $(1, 1)$.

Thus there are open sets $U_1$ and $U_2$ in $G$, both containing $1$, such that $U_1 \times U_2 \subset \mu^{-1}(V)$.

Let $U = U_1 \cap U_2$.

Then $\mu(U \times U) = U^{−1}xU \subset V$ and $U^{−1} \times U \cap H = \emptyset$.

---

I am not sure on how to continue from here. A hint would be helpful.

Thanks.

Answer 1

You are almost done. Note that by your definition of $U$, you have $U^{-1}xU \cap H = \emptyset$, or $$u_1^{-1}xu_2 \ne h, $$ all $u_i \in U$, $h \in H$, multiplication by $u_1$ gives $$ xu_2 \ne u_1 h,\qquad u_i \in U, h \in H$$ or $xU \cap UH \ne \emptyset$.

Answer 2

You are almost done.

Suppose there were a $y\in xU\cap UH$ then $y=xu_1=u_2h$ for some $u_1,u_2\in U$ and $h\in H$. So $h=u_2^{-1}xu_1\Rightarrow h\in U^{-1}xU\subset V$ which is a contradiction.