Let $\Gamma$ be a lattice in a (real or p-adic) Lie group.
Is it true that for a given natural number $n$ there exists a finite index subgroup $\Sigma\subset\Gamma$ such that each $\sigma\in\Sigma$ is an $n$-th power of some element of $\Gamma$?


In other words, is it true that for given $\sigma\in\Sigma$ there exists $\gamma\in\Gamma$ such that $\sigma=\gamma^n$?


If not true for general lattices, are there some restrictions under which this holds (cocompactness, arithmeticity,...)?



Property of lattices in Lie groups
The answer is no, even in higher rank groups. For example, take $\Gamma = SL\_3({\mathbb Z})$. If such a $\Sigma $ existed for any $n$, then its completion in the profinite (same as congruence) completion of $\Gamma$ would have this property that every element in $\Sigma$ would be an $n$-th power. But the completion of $\Sigma$, by strong approximation, is open. So for almost all primes, $SL\_3({\mathbb Z}\_p)$ would have this property. But you can easily cook up a prime $p$ and elements in $SL\_3({\mathbb F}\_p)$ which are not $n$-th powers. 


To see this, get large primes such that $p^3 \equiv 1\quad (mod\quad n)$ but $p-1$ is not divisible by $n$. Choose a cubic extension of ${\mathbb F}\_p$; the group of norm one elements in the cubic extension is cyclic of order $\frac{p^3-1}{p-1}$. Hence the $n$-th power map on this subgroup of $SL\_3({\mathbb F}\_p)$ is not surjective; but any $n$-th root of a generator for this subgroup already lies in this subgroup. 


[Edit] To summarise: for any lattice in a semi-simple group, and for any integer $n\geq 2$, the image of the $n$-th power map never contains a finite index subgroup. The argument using strong approximation works if the lattice has number field entries (this holds for all lattices except those in $SL(2,{\mathbb R})$) . I think that it is true for any Zariski dense finitely generated subgroup of a semi-simple group, using strong approximation due to Nori (and Weisfeiler). 


