$H$ is a normal subgroup which is not closed. How to prove the quotient group $G/H$ is not Hausdorff. $G$ is a Hausdorff topology group, $H$ is a normal subgroup which is not closed. How to prove the quotient group $G/H$ is not Hausdorff.
For example, what are the quotient group and quotient topology of $\mathbb{R}/\mathbb{Q}$?
 A: Let $X=\mathbb R$ and $G=\mathbb Q$.  Let $G$ act on $X$ by translations.  Let $\pi$ be the identification map.  For $A\subseteq X$ let $\overline{A}=\pi(A)$.   Then $\mathbb Q$ represents exactly one subset of the partition, so $\overline{\mathbb Q}$ is a single point in $X/G$.  Now suppose $U\subseteq X/G$ is an open set.  Then $\pi^{-1}(U)$ is open in $X$.  So there is an open interval $(a,b)\subseteq X$ such that $(a,b)\subseteq\pi^{-1}(U)$.  Now $\mathbb Q\cap (a,b)\not=\emptyset$.  Thus $\overline{\mathbb Q}\in U$.  Thus every open set in $X/G$ contains the point $\overline{\mathbb Q}$.  It follows that every pair of open sets in $X/G$ have non-empty intersection.  Thus $X/G$ cannot be Hausdorff.
A: the cosets of $\mathbb{Q}$ in $\mathbb{R}$ are all translations of $\mathbb{Q}$ hence dense in $\mathbb{R}$.
thus every nonempty set in the quotient topology contains the whole of $\mathbb{R/Q}$ - the quotient topology is trivial.
A: Suppose that $H\subset G$ is normal and not closed, let $G/H$ be endowed with the quotient topology, we denote by $p:G\rightarrow G/H$  the canonical projection.  Since $H$ is not closed, there exists $a$ in the adherence of $H$ which is not in $H$.  Let $e$ be the neutral element of $G$ and $U$ be any neighborhood of $a$, $U\cap H$ is not empty, this implies that $p(U)\cap p(H)$ is not empty, we know that $p(H)=e$, thus $e\in p(U)$ since $a$ is not in $H$, $p(a)\neq p(e)$ and every neighborhood of $p(a)$ for the quotient topology contains $p(e)$ so $G/H$ is not separated. done.
A: Let $f\colon G\to G'$ be a continuous homomorphism of topological groups. If $G'$ is Hausdorff, then $\{1\}$ is closed in $G'$, so $\ker f$ is closed in $G$.
In the particular case when $f\colon G\to G/H$ is the projection map, we see that $G/H$ Hausdorff implies $H$ closed.
The quotient topology on $\mathbb{R}/\mathbb{Q}$ is the largest topology such that $\pi\colon\mathbb{R}\to\mathbb{R}/\mathbb{Q}$ (canonical projection) is continuous. That is, a subset $V$ of $\mathbb{R}/\mathbb{Q}$ is open if and only if $\pi^{-1}(V)$ is open. Since $\pi^{-1}(V)\supseteq\mathbb{Q}$, saying $\pi^{-1}(V)$ is open implies $\pi^{-1}(V)=\mathbb{R}$, because $\mathbb{Q}$ is dense. Thus the quotient topology on $\mathbb{R}/\mathbb{Q}$ is indiscrete.
