Topological group with trivial identity subgroup closed implies Hausdorff I am trying to prove that a topological group with trivial identity subgroup closed implies Hausdorff. This implies that every singleton is closed within the group. I know that in an arbitrary topological space Hausdorff implies every singleton set is closed, but I am not sure how to show that for a topological group, the converse should be true. 
Any hints would be appreciated!
 A: Let $G$ denote the topological group. Given any two distinct points $g,h\in G$ we want to find disjoint open neighborhoods for them. Consider the function $f(x,y)=xy^{-1}$ from $G\times G$ (with the product topology) to $G$. It is a continuous function, by the definition of a topological group. Since $g\not=h$, we have $f(g,h)\not= e$ (the identity element). Since $f$ is continuous and $\{e\}$ is closed in $G$, the set
$$
\{(x,y)\in G\times G\colon f(x,y)\not=e\}
$$
is open in $G\times G$. Thus, by definition of the product topology there exists a neighborhood $U_g\times U_h$ with $U_g,U_h$ open in $G$ such that
$$
(g,h)\in U_g\times U_h\subseteq \{(x,y)\in G\times G\colon f(x,y)\not=e\}.
$$
Now $U_g$ and $U_h$ must be disjoint, and it gives the result.
A: If $\;G\;$ is the topological group, define $\;f:G\times G\to G\,,\;\;f(x,y):=xy^{-1}\;$ . By the requirements of top. groups, the map $\;f\;$ is continuous (in the cartesian product we take, of course, the product topology). Observe then that
$$f^{-1}(\{1\})=\Delta_G:=\left\{\,(x,x)\in G\times G\,\right\}=\text{ the diagonal in the product}$$
and since it is given $\;\{1\}\;$  is closed, we get by continuity that also $\;\Delta_G\;$ is closed...and we're done, since the diagonal being closed in the product is equivalent with the space being Hausdorff (simple proof)
