About connected Lie Groups

Question

How can I prove that a connected Lie Group is generated by any neighborhood of the identity?

The result is almost trivial for $R^n$ but I tried using the open subgroup generated by this neighborhood.

Answer 1

An open subgroup $H$ of a topological group $G$ is closed because $$ G \smallsetminus H = \bigcup_{g \notin H} gH $$ is open as union of the open sets $gH$.

Now take your neighborhood $U$ of the identity, let $H = \bigcup_{n \in \mathbb{Z}} U^{n}$ and check that $H$ is an open (hence closed) subgroup of $G$. By connectedness $G = H$.

Answer 2

Is the question is similar to: Prove that any open subgroup of a connected lie group is the whole group then here is my answer, if not then please ignore it.

Take any $g\in G$ define $l_g:G\rightarrow G, l_g(k)=gk$, Let $H$ be any open subgroup of $G$,clearly as $l_g$ is a homeomorphism, and $H$ is open so $l_g(H)=gH$ is also open, but $G$ is connected so $H=G$