TITLE: Every neighborhood of identity in a topological group contains the product of a symmetric neighborhood of identity.
QUESTION [7 upvotes]: Let $(G,\cdot)$ be a topological group and $U$ be a neighborhood of $1$.  Then there exists a symmetric neighborhood of $1$, $V^{-1} = V$, such that $V\cdot V \subset U$.  Having a hard time proving this.  $V^{-1} = \{v^{-1}: v \in V\}$.  I know that an open set times any set is also open.  And a hint is that $VV^{-1}$ is symmetric and an open neighborhood of $1$ when $V$ is an open neighborhood of $1$ contained in $U$, but showing $VV^{-1}$ or an expression involving it is a subset of $U$ or $V$ requires something else.

REPLY [4 votes]: Take the continuous function $m:G\times G\to G$ where $(x,y)\mapsto xy$. Then $m^{-1}(U)$ is open and maps to $U$. We can then take a neighborhood of $1$, let's say $V$, such that $V\times V\subseteq m^{-1}(U)$ (since $m^{-1}(U)$ is open using the product topology). By replacing $V$ with $V\cap V^{-1}$, we can assume that $V$ is symmetric. Therefore $m(V,V)=VV\subseteq U$ and $V$ is symmetric.