How can I show that these two topologies are the same?

Question

$G$ is a topological group and $H$ an open subgroup of $G$ I want to show that the quotient topology of $G/H$, $\tau_{G/H}$ is equal to the discrete topology $\tau_{\text{discrete}}$.

I mean clearly we know that $\tau_{G/H}\subseteq \tau_{\text{discrete}}$ since $\tau_{\text{discrete}}$ is the finest topology. But now to show the other implication, I'm a bit lost. Since in $\tau_{\text{discrete}}$ every subset of $G$ is open I know that if $U\in \tau_{\text{discrete}}$, then $U\subset G$. But to show that $U\in \tau_{G/H}$, I need to show that $\pi^{-1}(U)\subset G$ is open. Where $\pi:G\rightarrow G/H$ is the projection. But now I'm confused what topology we have on $G$ because since it is a topological group I only know that the product map and the inverse map are continuous.

Can someone help me?

Answer 1

You can prove it this way, without resorting to showing equality of sets.

Lemma: A topological group is discrete iff the set $\{1_G\}$ of the neutral element is open.

Proof: It is found in Dikran, Example 3.1.2.

Theorem: Let G be a topological group and H be a subgroup of G. Then the quotient group is discrete iff H is open.

$\Rightarrow$ We say that the quotient map $\pi : G \longrightarrow \dfrac{G}{H}$ is an open map. If $H$ is open in $G$, then

$$\pi(H) = \{H\},$$

is open in $\dfrac{G}{H}$. By the Example 3.1.2, the quotient group is discret. Therefore, the quotient topology is discrete.

This happens because:

1. How $\dfrac{G}{H}$ is a a topological group, the translations maps are homeomorphism. Let $X \subset \dfrac{G}{H}$, $xH \in X$ and the homeomorphism T : $\dfrac{G}{H} \longrightarrow \dfrac{G}{H}$, given by T(gH) = gxH. How T is an open map, T({H}) = {xH} is open. Then X is an union of open sets, therefore is open.

$\Leftarrow$ If $\dfrac{G}{H}$ is discrete, then every subset of $\dfrac{G}{H}$ is open. Then {H} is open. By definition of the quotient topology, we have that

$$\pi^{-1}(\{H\}) = H$$ is open in G. Then the quotient topology, induced by $G$, is discrete.

Answer 2

A subset of $U \subseteq G/H$ is open iff the pre-image $\pi^{-1}(U)$ is an open set in $G$.

In particular, since $\pi^{-1}(\{H\}) = H$, we know that $\{H\}$ is open in $G/H$. What about the other sets in $G/H$? Well they all have the form

$$ \bigcup_{x\in S} \{xH\} $$

Where $S\subseteq G$ is some subset of $G$. So if we can show that $\{xH\}$ is always open, then any set in $G/H$ is open. We now prove that $\{xH\}$ is always open.

Let $x\in G$. Then since the map

\begin{align} \mathbf{x^{-1}} : G&\to G \\ g&\mapsto x^{-1}\cdot g \end{align} Is continuous, the pre-image of $H$ under $\mathbf{x^{-1}}$ must be open, however, this pre-image is

\begin{align} (\mathbf{x^{-1}})^{-1}(H) = xH \end{align}

So $xH$ must be open in $G$. Then since

$$\pi^{-1}(\{xH\}) = xH $$

We conclude that $\{xH\}$ is open in $G/H$, which is what we wanted to show. So every subset of $G/H$ is open, completing the proof that the topology on $G/H$ is the discrete topology

Edit:This is essentially the same as https://math.stackexchange.com/users/1088667/gleberson-antunes answer, with the difference being that only the continuity of $\pi$ is used, and not its openness, which as far as I am aware is not guaranteed in general, (but is true for a quotient by an open subgroup, though I don't see how to prove it without first proving that this topology is discrete)