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 ?
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Welcome to MSE. What is $U$? – José Carlos Santos Nov 27 '18 at 09:48
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$U$ is any neighborhood of $g$. – water graph Nov 27 '18 at 09:51
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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).
mathworker21
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But $f(V) = {x g x^{-1}:x\in V } $ whereas $VgV^{-1} = {x_1 g x_2^{-1}:x_1,x_2\in V } $ .Thus $f(V)\subset U $ does not imply $VgV^{-1} \subset U $. – alagris Oct 04 '22 at 16:58
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@alagris You are right, thank you. What I think should be done is to define $f : G\times G \to G$ by $f(x,y) = xgy^{-1}$. Then by continuity there's some open neighborhood $W \subseteq G\times G$ if $(e,e)$ with $f(W) \subseteq U$. But any open neighborhood $W$ of $(e,e)$ contains $V_1\times V_2$ for some open neighborhoods $V_1,V_2$ of $e$. The desired $V$ will then be $V_1\cap V_2 \cap V_1^{-1}\cap V_2^{-1}$. – mathworker21 Oct 10 '22 at 10:04