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.
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. 


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. 

