Let $G$ be a topological group and $g\in G$ , $U$ is a neighborhood of $g$ . Prove that there exists a symmetric neighborhood $V$ of $e$ such that $VgV^{-1}\subset U$.
If $g=e$, l have proved it. But if $g\neq e$ , l have no idea. So how to prove ?
Every element $g$ of $G$ has a symmetric neighborhood $V$ of $e$ such that $VgV^{-1}\subset U$
topological-groups
Best Answer
Let $f: G \to G$ be $f(x) = xgx^{-1}$. Then $f(e) = g$. So by continuity, we can find some open neighborhood $V$ of $e$ for which $f(V) \subset U$, i.e. for which $VgV^{-1} \subset U$. To make $V$ symmetric, we can just replace $V$ with $V\cap V^{-1}$, which is a subset of $V$ (so that $VgV^{-1} \subset U$ still holds).