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

$$G$$ 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?

• Maybe think of this. If every coset of $H$ in $G$ is an open set (for the topology of $G$), then every point of $G/H$ is an open set (for the quotient topology). Sep 24 at 18:23

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.

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)