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I'm having trouble on this assignment question and was hoping someone can point me in the right direction.

Let $G$ be a group equipped with a Hausdorff topology in which the group operation and inversion are continuous. Let $H$ be a subgroup. Let $g_1 \sim g_2$ if $g_1h = g_2$ for some $h \in H$ and denote by $G/H$ the set of equivalence classes. Let $G/H$ have the quotient topology.

Prove that if $H$ is dense in $G$, then $G/H$ has the trivial topology.

My attempt:

The goal is to show that if $A \subset G/H$ is nonempty and open, then $A = G/H$. Let $\rho$ denote the standard quotient map from $G$ to $G/H$ that maps each $g \in G$ to the equivalance class containing $g$. Suppose $A$ contains the element $C = \{g_1, g_2, ...\} \in G/H$. ($C$ is an equivalence class.) Then, $C \subset \rho^{-1}(A)$. But $C = gH$ for some $g \in G$, and $H$ is dense, so $C$ is also dense in $G$. So, $G = \overline{C} \subset \overline{\rho^{-1}(A)}$, and so $\overline{\rho^{-1}(A)} = G$.

I think we need to show that $\overline{\rho^{-1}(A)} = \rho^{-1}(\overline{A})$. Because that would give us $\overline{A} = \rho(G) = G/H$ (which doesn't quite show the desired result $A = G/H$, so maybe I'm going in the wrong direction). But we only have $\rho^{-1}(\overline{A}) \subset \overline{\rho^{-1}(A)}$ from the continuity of $\rho^{-1}$. This is where I'm stuck. Any help would be greatly appreciated!

Community
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Kevin Hsu
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  • You should be able to show that $\rho^{-1}(A) = G$, not just $\overline{\rho^{-1}(A)}$ – D_S Apr 09 '17 at 19:19

1 Answers1

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Since $H$ is dense in $G$, so is $xH$ for every $x \in G$.

A subset $E$ of $G$ is saturated with respect to $H$ if whenever $g \in E$, so is $gh$ for every $h \in H$. In other words, $E$ is a union of left cosets.

Let $\pi: G \rightarrow G/H$ be the projection map. It follows from the definition of the quotient topology that the open sets of $G/H$ are exactly the images $\pi(U)$, where $U$ runs through the saturated open sets of $G$.

If $U$ is a nonempty saturated open set, then $U = G$. This is because for each $x \in G$, $U$ contains a point of $xH$, hence contains every point of $xH$.

Let $\pi(U)$ be a nonempty open set in $G/H$, for $U$ a saturated open set in $G$. Since $\pi(U)$ is nonempty, neither is $U$, hence $U = G$ and $\pi(U) = G/H$.

D_S
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