Showing $U$ open in topological group $G$ $\implies$ $gU$ is open

Question

If $G$ is a topological group, and $U$ is an open set in $G$, then do we have that $gU$ is also open in $G$?

I know that since $G$ is a topological group, the mappings $\mu: G^2 \rightarrow G$ s.t. $\mu(g,h) = gh$ and $i:G\rightarrow G$ s.t. $i(g) = g^{-1}$ are continuous.

Answer 1

The map $h\mapsto gh$ is continuous, being the restriction of $\mu$ to $\{g\}\times G$. Similarly, $h\mapsto g^{-1}h$ is continuous, and these mappings are clearly inverse to each other. Hence $h\mapsto gh$ is a homeomorphism $G\to G$. Now it follows that $gU$ is open whenever $U$ is open.

Answer 2

Let $a\in gU$, say $a=gu$. Then $(g^{-1},a)\in G^2$ is a point that maps to $u\in U$ under $\mu$, hence is in the open set $\mu^{-1}(U)$. By definition of product toppology, there exists a neighborhood $U_1\times U_2\subseteq\mu^{-1}(U)$ where $U_1$ is an open set containing $g^{-1}$ and $U_2$ is an open set containing $a$. Then $g^{-1}U_2\subseteq U$, i.e. $a\in U_2\subseteq gU$. ${{{}}}$