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Let $G$ be a locally compact topological group and $H$ be a subgroup of $G$. Show $G/H$ is locally compact.

This question already has an answer here: Show that if $G$ is a locally compact topological group and $H$ is a subgroup, then $G/H$ is locally compact.

But my lecturer doesn't allow me to use the fact that $\pi:G\to G/H$ is an open map. According to him, we require the fact that $H$ is closed to use this fact. I looked at https://www.uni-muenster.de/AGKramer/content/LCManuscript.pdf (Preposition 1.17) and it seems to not use the fact $H$ is closed. Is the proof provided here wrong? Do we need $H$ is closed?

Anyways, I have attempted another proof. $G$ is a topological group, so it is regular, therefore hausdorff. We have another characterization for locally compact hausdorff, which is strongly locally compact

$X$ is strongly locally compact if for all $x \in X$ for all neighborhood $U_1$ of $x$ there exists neighborhood $U_2$ of $x$ such that $\overline{U_2}$ is compact and $x \in \overline{U_2} \subset U_1$

I will try to use this to show $G/H$ is locally compact. Let $gH \in G/H$ where $g$ is arbitrary inside the coset. Let $V_1$ be a neighborhood of $gH$. $\pi$ is a quotient map so $U_1=\pi^{-1}(V_1)$ is open. Using strongly locally compact of $G$, there exists $U_2$ neighborhood of $g$ with $\overline{U_2}$ compact and $\overline{U_2} \subset U_1 = \pi^{-1}(V_1)$. So

$$ g \subset \overline{U_2} \subset U_1 $$

Now I want to say something like $$ gH \subset \overline{V_2} \subset V_1 $$ Here the straightforward candidate for $\overline{V_2}$ is $\pi(\overline{U_2})$. We have $\pi(\overline{U_2})$ is compact because $\pi$ is continuous. We also have the second inclusion $\pi(\overline{U_2}) \subset V_1 = \pi(\pi^{-1}(V_1))$ from $\overline{U_2} \subset \pi^{-1}(V_1) = U_1$ because $\pi$ is surjective ($\pi \circ \pi^{-1} = id$).

I am having difficulty producing an open set $V_2$ such that $\overline{V_2} = \pi(\overline{U_2})$. Because I'm not sure whether $\pi$ is open map, I can't say $V_2=\pi(U_2)$ therefore it is open. Is there a theorem/trick using hausdorffness to produce this? Or did we not require $H$ to be closed for $\pi$ to be an open map in the first place?

Overall I'm not even sure if this approach is correct.. Any help would be appreciated.

user651267
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    A lot of people use the definition of a topological group that includes the $T_1$ axiom or (equivalently) the Hausdorff condition. Under this definition, if you want a quotient group, $G/H$, of a topological group, $G$, to also be a topological group, you need $H$ to be a closed normal subgroup of $G$. Also, I think that this is an iff condition, i.e. $G/H$ is $T_1$/ Hausdorff iff $H$ is closed in $G$ (I'm not sure about this though). – CardioidAss22 Apr 09 '20 at 16:12
  • @Cardioid_Ass_22 If you read his linked note, you'll see that his topological groups do not include assumptions on separation axioms. – Henno Brandsma Apr 09 '20 at 21:54
  • @HennoBrandsma You are correct. I just thought it may be the case that Vladimir Putin's professor specifically may be using this definition. – CardioidAss22 Apr 09 '20 at 22:32

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You're completely right that $\pi: G \to G{/}H$ is an open continuous map (regardless of whether $H$ is closed or a normal subgroup), so the standard theorem that a locally compact space under such a map has a locally compact image applies.

Look no further.

Henno Brandsma
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