Every neighborhood of identity in a topological group contains the product of a symmetric neighborhood of identity.

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.

Solution 1:

Use that the multiplication is continuous and $1\cdot1=1$. Let $U$ be a neighborhood of $1$. By continuity there are neighborhoods $A,B\ni1$ such that $AB\subseteq U$. Let $W=A\cap B$ and then $V:=W\cap W^{-1}$ is a symmetric neighborhood of $1$ such that $VV\subseteq U$.

Solution 2:

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.