Neighborhood in topological groups Let $G$ be a topological group, $e$ the neutral element and $U$ a neighborhood of $e$.
Claim: Then there exists a neighborhood $V$ of $e$, such that $V^2 \subseteq U$.
This should follow easily from the continuity of the group multiplication. But how?

The map $m\colon (x,y)\in G^2\mapsto x\cdot y$ is continuous, where $G^2$ is endowed with the product topology. Therefore, there is $(e,e)\in O\subset G^2$ open such that $m(O)\subset U$. By definition of the product topology, there is $V_1,V_2$ open set such that $e\in V_1\cap V_2$ and $V_1\times V_2\subset O$. Now take $V:=V_1\cap V_2$. 
