MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,...)?

share|cite|improve this question
It seems that your question concerns semisimple (real or $p$-adic) Lie groups. – YCor Feb 24 '13 at 21:46
up vote 9 down vote accepted

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).

share|cite|improve this answer
In general, suppose $\Gamma = G({\mathbb Z})$ is a "higher rank" arithmetic group. You need only work with the congruence completion. If for some $n$, such a $\Sigma$ exists, then for the same reasons as above, the $n$-th power map on $G({\mathbb F}_p)$ would be surjective, for almost all $p$ by strong approximation for $\Sigma$. But you can get anisotropic tori in $G({\mathbb F}_p)$ to replace the cubic extension in the above. – Venkataramana Feb 24 '13 at 9:36
If the $n$-th power map is surjective on the finite group $G_p=SL_3({\mathbb F}_p)$, then it is also injective; by Dirichlet's theorem on infinitude of primes in arith progressions, there are infinitely many primes such that $G_p$ has order divisible by $p$; then the n-th power map cannot be injective. – Venkataramana Feb 24 '13 at 12:18
in the last comment, "divisible by $p$" is to be replaced by "divisible by $n$". – Venkataramana Feb 24 '13 at 12:37
I know nothing about strong approximation or profinite completion; what does the hypothesis on $\Sigma$ being finite index amount to in your example? – J. Martel Feb 24 '13 at 22:38
@J.Martel, let us consider (for the sake of simplicity) the case $\Sigma \subset SL_3({\mathbb Z})$ of finite index. Strong approximation says essentially, that for all but a finite set of primes, the composite map $\Sigma \subset SL_3({\mathbb Z})\rightarrow SL_3({\mathbb F}_p)$ is a $surjection$. This is a non-trivial statement in general (but easily proved for finite index subgroups $\Sigma $ of $SL_3({\mathbb Z})$. If $\Sigma$ contains all $n$-th powers, this says then that the $n$-th power map on the finite group is surjective. By choosing suitable primes, we can see that this can't b – Venkataramana Feb 25 '13 at 7:12

This is false for uniform lattices in rank one semi simple Lie groups and large $n$ by a result of Ivanov and Olshanskii, which implies that the normal subgroup generated by $n$th powers is infinite index for certain large $n$.

share|cite|improve this answer
The rank one case tends to be a bit exceptional. Is anything known about arithmetic subgroups? – anon Feb 24 '13 at 7:26
@Algol: can you give a reference for this result? I am particularly interested in the p-adic case of higher rank, anything known there? – Anton Feb 24 '13 at 8:12
@anon: I wouldn't say that rank one is exceptional, at least in terms of importance in mathematics. – YCor Feb 24 '13 at 13:44
The rank one case is exceptional in that there are lots of nonarithmetic irreducible lattices. No one was saying that it isn't important. – anon Feb 24 '13 at 18:17
@anon: you mean arithmetic groups in rank 1? In higher rank (semisimple), Margulis' superrigidity theorem implies that all lattices are arithmetic. Also, Margulis' normal subgroup theorem implies that all normal subgroups are finite index, so for example the subgroup generated by nth powers is finite-index, as opposed to the rank 1 case. – Ian Agol Feb 24 '13 at 18:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.