Suppose $G$ is a topological group. Let $g\in G$, suppose $V\subseteq G$ is an open neighborhood of $1_G$. Show that $gV=\{gv \mid v\in V\}$ is open.

Question

The Problem: Suppose $G$ is a topological group. Let $g\in G$, suppose $V\subseteq G$ is an open neighborhood of $1_G$. Show that $gV=\{gv \mid v\in V\}$ is open.

My Attempt: Clearly we need to make use of the fact that $(x, y)\mapsto xy^{-1}$ is a continuous map from $G^2$ to $G$; but how? For any $v\in V$, I failed to find an open set $B\subseteq G$ such that $gv\in B$. Any hint would be greatly appreciated.

Answer 1

For each $g\in G$, the function$$\begin{array}{rccc}\pi_g\colon&G&\longrightarrow&G\\&h&\mapsto&g^{-1}h\end{array}$$is continuous. Therefore, $\pi_g^{\,-1}(V)$is an open set. But $\pi_g^{\,-1}(V)=\{gv\mid v\in V\}$.