I have some difficulty proving the following fundamental statement on the topic of topological groups; a topic that I have just started to study independently.
Let $G$ be a topological group. Consider an arbitrary element $g$ of $G$. It follows that a set $U$ is a neighborhood of $g$ if and only if $g^{-1}U$ is a neighborhood of the identity, $e$ of $G$.
I believe that the right approach will be along the line of, first defining specific continuous function $G\times G\to G$, exploiting that $G$ is a topological group. Then perhaps viewing the inverse of the function defined in order to deduce something on open sets. I know it is vogue, but this is the only thing that I can work with.
Best Answer
Consider the map from $G$ into itself defined by $h\mapsto g^{-1}h$. It is a homeomorphism which maps $g$ into $e$ (its inverse is $h\mapsto gh$), and therefore it maps $U$ into a neighborhood of $e$.