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Let $G$ be a topological group and $H$ be a normal subgroup. I want to show that $G/H$ is a topological group. I have managed to prove that $m:G/H\times G/H\to G/H, \bar a \bar b\to \bar{ab}$ and $i:G/H\to G/H: \bar a\to \bar {a^{-1}}$ is continuous, but I cannot figure out how to prove that $G/H$ is Hausdorff. I know that proving that $H$ is closed would solve the problem, but how can I prove it? Any hint is appreciated.

Runyang Wang
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    $\mathbb{Q}$ is normal in $\mathbb{R}$ but not closed. – user10354138 Oct 28 '23 at 14:10
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    I think that you may be a victim of conflicting conventions. Some people require "Hausdorff" as part of the definition of topological group, while others do not. The fact that any quotient of a topological group is a topological group with the quotient topology is true only if "topological group" doesn't include Hausdorffness. Generally the quotient is Hausdorff iff $H$ is closed, so the most you can say is that "the quotient of $G$ by the closure of $H$ is a Hausdorff topological group". – Izaak van Dongen Oct 28 '23 at 14:15

1 Answers1

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Proving via function composition is much easier.

Theorem 1. Let $(S, \cdot)$ be a group and $\tau_S$ be a topology on $S$. Then $(S,\cdot,\tau)$ is a topological group iff the function

\begin{aligned} f \ \colon \ S \times S & \longrightarrow S\\ (x,y) &\longmapsto xy^{-1}, \end{aligned}

is continuous.

Proof: Exercise. (Dikran presents this as a definition but it is possible to prove it through function composition. One tip is to use translation homeomorphisms together with the operation and group inversion.)

Let $(G, \cdot, \tau_G)$ be a topologial group and $H$ be a normal subgroup of $G$. Consider the quotient $\dfrac{G}{H}$ endowed by the quotient topology induced by $G$.

We have that:

  1. The function

\begin{aligned} f \ \colon \ G \times G & \longrightarrow G\\ (x,y) &\longmapsto xy^{-1} \end{aligned}

is continuous, because $(G, \cdot, \tau_G)$ is a topological group.

  1. The function

\begin{aligned} q \ \colon G & \longrightarrow \dfrac{G}{H}\\ g &\longmapsto gH \end{aligned}

is continuous by definition. (Note that a set $U \subset \dfrac{G}{H}$ is open iff $q^{-1}(U) \in \tau_G$).

We can easily show that the function

\begin{aligned} q \ \colon G \times G & \longrightarrow \dfrac{G}{H} \times \dfrac{G}{H} \\ (x,y) &\longmapsto (xH, yH) \end{aligned}

is continuous with respect to product topolgy. We then have all the tools to prove this statement. Then, consider the function:

\begin{aligned} F \ \colon \dfrac{G}{H} \times \dfrac{G}{H} & \longrightarrow \dfrac{G}{H} \\ (xH,yH) &\longmapsto xy^{-1}H. \end{aligned}

Note that that the diagram

$\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ composition

commutes. Therefore

$$q \circ f = F \circ (q \times q).$$

The first member of this equality is a continuous application in $(x, y)$. We then have that $F \circ (q \times q)$ must be continuous in $(x, y)$. Since the map $q \times q$ is continuous in $(x, y)$, we have that $F$ is continuous in $(xH, yH) = (q \times q)(x, y).$ By the Theorem 1, whe have that $\bigg(\dfrac{G}{H}, \cdot, \tau_{\frac{G}{H}}\bigg)$ is a topological group.

Gleberson Antunes
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