TITLE: How to show that topological groups are automatically Hausdorff?
QUESTION [27 upvotes]: On page 146, James Munkres' textbook Topology(2ed),

Show that $G$ (a topological group) is Hausdorff. In fact, show that if $x \neq y$, there is a neighborhood $V$ of $e$ such that $V \cdot x$ and $V \cdot y$ are disjoint.

Noticeably, the definition of topological group in Munkres's textbook differs from that in wikipedia.

A topological group $G$ is a group that is also a topological space satisfying the $T_1$ axiom, such that the map of $G \times G$ into $G$ sending $x \times y$ into $x \cdot y$ and the map of $G$ into $G$ sending $x$ into $x^{-1}$, are continuous maps.

REPLY [4 votes]: By the above argument, there are two cases: 

If $\{1\}$ is closed then the diagonal is closed (as it is the inverse image of $\{1\}$ under the continuous map $g_1 \times g_2\mapsto g_1g_2^{-1}$). So $G$ is Hausdorff. 
If $\{1\}$ is open then $G$ is discrete, so it's Hausdorff.