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Let $G$ be a connected Lie group and $U$ any neighbourhood of the identity element. How to prove that $U$ generates $G$.

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up vote 5 down vote accepted

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.

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so this holds for ant topological group too? – Koushik Jan 20 '13 at 3:17
@K.Ghosh, I believe so. – user27126 Jan 20 '13 at 3:24

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