About Lie group $G$ has this escape property？ 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.
 A: Let me rewrite a comment I made that has a typo, since it's too late to edit it. The property fails for any compact connected noncommutative Lie group. Note that if $G$ is a compact group then it has an equivariant metric, and therefore a "distance to $e$ function" $|x|: = d(x,e)$ such that the function $$d(x,y): = |xy^{-1}|$$ is a metric, and such that the topology defined by this metric is the standard topology on your group $G$. Thus any open set $U$ containing the origin must also contain some $\epsilon-$ball $B_\epsilon: = \{x\in G\mid |x|<\epsilon\}.$ Now let $a\in G$ be any element, $g\in G$ be any "small" element with $|g|<\epsilon/2$ which doesn't commute with $a$ and let $b: = gag^{-1}.$ Then $d(a,ga)<\epsilon/2, d(ga, gag^{-1})<\epsilon/2,$ so by the triangle inequality $d(a,b) = d(a,gag^{-1}) < \epsilon.$ But the same argument also shows that $d(a^n, b^n) = d(a^n, ga^ng^{-1}) <\epsilon,$ so $|a^nb^{-n}|<\epsilon$ and $|a^nb^{-n}|\in U$. Thus your conjectured property fails so long as your group is compact and contains noncommuting matrices arbitrarily close to $1$, something that is automatically true if your group is a compact connected noncommutative group. More generally, if your group contains a compact connected noncommutative group then this property fails as well (by simply taking $x,y$ elements of the compact subgroup as above).
A: 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.
