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Let $G$ be a topological group, $U$ is a neighborhood of $e$ which is the unit element of $G$.

My question is does there exist a neighborhood $H \subseteq U$ of $e$ s.t.

  1. $H^{-1}=H$
  2. $H\cdot H\subseteq U$?

I have tried to found such a $H$ is a subgroup, but in some cases it doesn't work.

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4  
Yes. Use the continuity of multiplication. Hint: if $V$ contains $e$ then $S := V \cap V^{-1}$ does also and it's symmetric: $S = S^{-1}$. –  KCd Nov 5 '12 at 2:12

1 Answer 1

up vote 0 down vote accepted

Thank to KCd's hint. Let me have a trying.

Proof.

It's harmless to suppose $U$ is open.

Since $G$ is topological group, $\cdot^{-1}[U]$ is open and contains $(e,e)$, hence there is a $A \times B$, in which $A$ and $B$ are both open, contains $(e,e)$ and $\subseteq \cdot^{-1}[U]$. Consequently $V \times V$ is an open neighborhood of $(e,e)$ too, in which $V=(A \cap B) \cap (A \cap B)^{-1}$. Note that $\cdot$ is an open mapping, hence $V \cdot V$ is open. Besides $e \in V \cdot V \subseteq U$ since $(e,e) \in V \times V \subseteq \cdot^{-1}[U]$. Moreover $V^{-1}=V$.

Therefore $V$ satisfies all the conditions required.

$\Box$

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