2
$\begingroup$

Is the following true for some prime $p$?

There exists some prime $\ell$ and some $n$ such that $PSL_n(\mathbb{F}_{\ell})$ contains nontrivial $p$-torsion, and moreover if $x \in PSL_n(\mathbb{F}_{\ell})$ has order $p$ and $0 < k < p$, then $x$ is conjugate to $x^k$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

Yes. Take $\ell$ to be a primitive root mod $p$ (which exists by Dirichlet's theorem), $n=p-1$, then because $p$ divides $\ell^{p-1}-1$ there is a $p$-torsion element in $PSL_n(\mathbb F_\ell$. However, because $p$ is prime to $\ell$, all such elements are semisimple, hence conjugate to their $\ell^k$th powers for any $k$. Because $\ell$ is a primitive root, these powers fill out all the conjugacy classes mod $p$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Can you explain why semisimple matrices are conjugate to their $\ell^k$th powers for any $k$? $\endgroup$
    – Will Chen
    Oct 16, 2017 at 18:18
  • $\begingroup$ @oxeimon Conjugacy of semisimple elements is determined by the characteristic polynomial, and taking the $\ell$th power of the matrix is equivalent to taking the $\ell$th power of the eigenvalues, which is the Frobenius automorphisms, so that's equivalent to taking the $\ell$th power of the coefficients of the characteristic polynomial, which does nothing as these are in $\mathbb F_\ell$. $\endgroup$
    – Will Sawin
    Oct 16, 2017 at 18:30
  • $\begingroup$ @oxeimon I guess to check the characteristic polynomial in $PSL_2(\mathbb F_\ell)$, you have to check that the centralizer contains elements of every possible determinant. This reduces to checking that finite fields contain elements of every possible norm, and crucially uses the semisimplicity. It's faster to check in this case than in general, by proving the centralizer is exactly $\mathbb F_{\ell^{p-1}}$. $\endgroup$
    – Will Sawin
    Oct 16, 2017 at 18:32
  • $\begingroup$ Why is the case $n = 2$ special? (Why do you need to check that the centralizer contains elements of every possible determinant?) $\endgroup$
    – Will Chen
    Oct 16, 2017 at 20:48
  • $\begingroup$ @oxeimon Sorry, putting the $2$ there is a typo. That's required to check in $PSL_n$ rather than in $PGL_n$, which is easier. To check that the matrix conjugating one element to another in $PGL_n$ in fact lies in $SL_n$, it suffices to find something in the center which has the same determinant as it, so you can multiply by the inverse to get something in $PSL_n$. $\endgroup$
    – Will Sawin
    Oct 16, 2017 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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