TITLE: a neighbourhood of identity $U$ generates $G$ where $G$ is a connected lie group
QUESTION [19 upvotes]: Let $G$ be a connected Lie group and  $U$ any neighbourhood of the identity element. How to prove that $U$ generates $G$.

REPLY [29 votes]: By replacing $U$ with $U \cap U^{-1}$ if necessary, assume that $U = U^{-1}$.
Consider the set generated by $U$:
$$S = \{g_1 \cdots g_n : g_1,\cdots, g_n \in U \text{ for some $n$} \}$$
We want to show that $S = G$ by showing that $S$ is nonempty, open, and closed. Connectedness of $G$ would then imply $S = G$. 
Non-emptiness is evident.
For openness, note that for any $g \in S$, $gU \subset S$.
For closedness, note that if $g \notin S$, then $gU$ is disjoint with $S$. Otherwise if $gu \in S$, we have $g = guu^{-1} \in S$ as well.