Every point closed $\stackrel{?}{\Rightarrow}$ space is Hausdorff

Solution 1:

A space $X$ has the property that all singletons are closed if and only if it is $T_1$, meaning that whenever $x,y\in X$ and $x\ne y$, there is an open set $U$ such that $x\in U$ and $y\notin U$. The definition is symmetric, so there is also an open set $V$ such that $y\in V$ and $x\notin V$, but there is no guarantee that $U$ and $V$ can be chosen to be disjoint. For example, if $X=\Bbb N$, and the open sets are $\varnothing$ and the sets whose complements in $\Bbb N$ are finite, then $X$ is $T_1$ but not Hausdorff: in this space $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open sets, but for any distinct $m,n\in X$, $X\setminus\{n\}$ is an open nbhd of $m$ that does not contain $n$.

Solution 2:

Notice that $H$ is a closed normal subgroup of $G$, for that see e.g. this for proof that it is a closed subgroup (equal to $\operatorname{cl} \{e\}$), and for normality just notice that conjugation preserves the neighbourhoods of identity (as a set), so it does preserve intersection as well.

From that we see that $G/H$ is a topological group.

It is a known fact that for topological groups, $T_0$ implies completely regular Hausdorff. Every point being closed is equivalent to $T_1$, from which $T_{3\frac {1}{2}}$, so in particular $T_2$, follows.

A proof can be found in many places, e.g. Engelking's General Topology iirc.

A short one for closed $\{e\}\implies T_2$: notice that Hausdorffness is equivalent to the diagonal being closed. But the diagonal is the preimage of identity by the map $(x,y)\mapsto xy^{-1}$.

In general, we do not have the implication, as shown by e.g. the cofinite topology on an infinite space.