$G$ a connected topological group, U open, then $\mathcal{U} = \cup_{i \in \mathbb{N}}\; U^i$ is a subgroup of $G$. Let $G$ a connected topological group, and $U$ an open subset of $G$ that contains the identity element of $G$.  I want to solve the following problem:

If $U^n$ is the set of the products $u_1 u_2 ...u_n$, with $u_i$ in $U$, show that $\mathcal{U} = \bigcup_{i \in \mathbb{N}}\; U^i$ is an open subgroup of $G$, and hence $\mathcal{U}=G$ .

The only part that is giving me problem is to prove that $\mathcal{U}$ is a subgroup. A hint given to the problem is to consider first an open set $V$, such that if $g$ in $V$ then $g^{-1}$ is in $V$. Obviously with $V$ in place of $U$, the property of subgroup is of direct comprobation. Any help?

Choose an open set $V\subseteq U$ containing the identity such that if $g\in V$ then $g^{-1}\in V$ (for instance, $V=U\cap U^{-1}$).  As you say, you can then easily show that $\mathcal{V}=\bigcup_{i\in\mathbb{N}} V^i$ is an open subgroup of $G$.  Thus $\mathcal{V}=G$.  But $\mathcal{V}\subseteq\mathcal{U}$ since $V\subseteq U$, and so you must also have $\mathcal{U}=G$.
(Note that without the assumption that $G$ is connected, it is not necessarily true that $\mathcal{U}$ is a subgroup.  For instance, take $G=\mathbb{Z}$ and $U=\{0,1\}$.)
