Local Lie Groups

Question

Hello i am trying to prove the following proposition :

Let $G$ be a connected Lie group, and $U\subset G$ a neighborhood of the identity element. Also, let $U^k = \{g_1 . g_2 . \dots g_k : g_i \in U\}$ be the set of k - fold products of elements of U.

Then, $G=\cup_{k=1} ^ \inf U^k$.

I read somewhere that, this result follows immediately from connectedness of $G$ but it is not so obvious to me.

Thank you for your time!

Answer 1

This is true for any connected topological group $G$.

Let $V = \bigcup_{k=1}^\infty U^k$. It is clear that $V$ is open. On the other hand, let $g \in \bar{V}$. Since $gU^{-1}$ is a neighborhood of $g$, it must intersect $V$. Let $h \in V \cap gU^{-1}$. Then

• $h = gu^{-1}$ for some $u \in U$, and
• $h = u_1 \ldots u_k$ for some $k \in \mathbb{N}$ and $u_1, \ldots, u_k \in U$.

Hence $g = u_1 \ldots u_k u \in U^{k+1} \subset V$, proving $V$ is closed. Since $G$ is connected, and $V$ is both open and closed, we have $V = G$.