TITLE: About Lie group $G$ has this escape property？
QUESTION [6 upvotes]: Every Lie group $G$ has the following escape property: For every $x \ne e$ in a sufficiently small neighborhood $U$ of the identity $e$ in $G$, there is a integer $n$ such that $x^n$ is not in $U$.
The $\textbf{Question one }:$  is if we can find a sufficiently small neighborhood $V$ of $e$ in $G$, for any two different points $a,b \in V$ , there is a integer $m$ such that $a^m(b^{-1})^m$ is not in $V$.

Thanks for Scholar's  answer about the above question
If $G$ is a commutative group the question actually has a positive answer, since $G$ has the escape property. I guess it is true for some larger class of Lie groups. The problem is to state that the power mapping can enlarge the distance between two different points.
But I didn't find a good property of power mapping by looking up data.
meanwhile，Thanks  Michael Albanese for editing of the question.

REPLY [3 votes]: This is false in the affine group of matrices $\begin{pmatrix} \theta & \eta\\0 & 1\end{pmatrix}$, $\theta,\eta\in\mathbf{R}$, $\theta>0$.
Indeed $V$ being fixed, choose $a=\begin{pmatrix} \theta & \eta\\0 & 1\end{pmatrix}$, $b=\begin{pmatrix} \theta & 0\\0 & 1\end{pmatrix}$, so $$a^mb^{-m}=\begin{pmatrix} 1 & (\theta^{m-1}+\theta^{m-2}+\dots +\theta+1)\eta\\0 & 1\end{pmatrix}.$$
Indeed choose $\theta<1$, close enough to $1$, and then $\eta$ small enough, so that $a,b\in V$ and so that $S=\begin{pmatrix} 1 & [0,\eta/(1-\theta)]\\0 & 1\end{pmatrix}$ is contained in $V$. Then $a^mb^{-m}\in S$, hence remains in $V$.

On the other hand the property is true when $G$ is connected nilpotent. Indeed if $G$ is simply connected nilpotent, $a^mb^{-m}$ tends to infinity for all $a\neq b$, by a simple application of the BCH formula (details upon request). In the case of $G$ connected nilpotent Lie group, the same property follows if $ab^{-1}$ is not central, and if $ab^{-1}$ is central, one argues as in the abelian case.