Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sorry if this question is below the level of this site: I've read that the quotient of a Hausdorff topological group by a closed subgroup is again Hausdorff. I've thought about it but can't seem to figure out why. Is it obvious? A simple yes or no (with reference is possible) is all I need.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Edit: Below I expand my crude original answer "Yes" as requested by the community.


Yes. Let $G$ be the group and $H$ be the closed subgroup. The kernel of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. Hence the kernel is closed. According to this $G/H$ is Hausdorff.

share|improve this answer
    
Given that the result is false without group structure, and given that your link is to the wiki article rather than a particular subsection, -1. Will upvote if you expand on this –  Yemon Choi Dec 15 '09 at 19:12
    
I am not sure a question like this needs much more explanation. See the discussion at mathoverflow.net/questions/9014/field-structure-for-rn –  jvp Dec 15 '09 at 19:21
    
But then don't answer because this link is useless :) –  user717 Dec 15 '09 at 19:32
    
Yes, the last line certainly implies G/H Hausd iff H closed. However, why is the last line true. (My apologies is this glaringly obvious.) –  Dyke Acland Dec 15 '09 at 19:46
    
+1 (i.e. revert now we have expanded on this). As people have said, the key is that in a top group a weak separation axiom boosts up to an a priori stronger one. –  Yemon Choi Dec 15 '09 at 20:32

In fact, an even stronger statement holds: If $G$ is a topological group and $H$ is an (abstract) subgroup, then $G/H$ is Hausdorff if and only if $H$ is closed (cf Bourbaki, General Topology, III.2.5, prop 13). It's not hard to prove.

share|improve this answer
    
Yes, before I posted the question I saw that H closed implies G/H T_0. What I couldn't see was that T_0 implies T_2 for a topological group. Is it obvious? –  Dyke Acland Dec 15 '09 at 19:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.