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Recall that a group is virtually torsion-free if it admits a finite index subgroup which is torsion-free.

Question. Is $\mathrm{SL}_n(\mathbb{Q}_p)$ virtually torsion-free for $n > 1$?

Comments.

  1. Note that $\mathrm{GL}_1(\mathbb{Q}_p) = \mathbb{Q}_p^*$ is virtually torsion-free.
  2. We know by a theorem of Selberg that for a field $K$ of characteristic 0, any finitely generated subgroup of $\mathrm{GL}_n(K)$ is virtually torsion-free. However, this does not apply to $\mathrm{SL}_n(\mathbb{Q}_p)$ as it is not finitely generated; the diagonal matrices give a copy of $\mathbb{Q}_p^*$, which is uncountably infinite.
  3. A related question can be found here where it is shown that $\mathrm{SL}_n(\mathbb{Z}_p)$ is virtually torsion-free as it is a compact $p$-adic analytic group.

Thanks in advance for the help!

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    $\begingroup$ $SL_n(\mathbb Q)$ is simple modulo centre and hence does not have subgroups of finite index. It also does have torsion elements. Therefore, it is not virtually torsion free. $\endgroup$
    – Venkataramana
    May 22, 2019 at 22:44
  • $\begingroup$ $SL_1(Q_p)$ is not $Q_p^*$, it is just the trivial group. $\endgroup$
    – YCor
    May 23, 2019 at 12:16
  • $\begingroup$ @YCor That was rather silly of me! Thank you for the comment. I have removed that comment from the question. $\endgroup$
    – Jackson Morrow
    May 23, 2019 at 13:13
  • $\begingroup$ @Venkataramana Thank you for the comment! I believe this answers the question since virtual torsion freeness is inherited by subgroups. $\endgroup$
    – Jackson Morrow
    May 23, 2019 at 13:22
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    $\begingroup$ By the way, I think that the proof that ${\rm SL}(n, \mathbb{Z}_{p})$ is virtually torsion free is not that complicated: the matrices congruent (elementwise) to the identity (mod $p$) form a torsion-free normal subgroup of finite index, and the quotient group by that normal subgroup is finite. $\endgroup$
    – Geoff Robinson
    May 23, 2019 at 13:24

1 Answer 1

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No (this was already answered in comments).

$\mathrm{SL}_n(\mathbf{Q}_p)$ is generated by its unipotent 1-parameter subgroups isomorphic to $\mathbf{Q}_p$, hence it has no proper subgroup of finite index. Hence, if it were virtually torsion-free, it would be torsion-free, which is not the case (for $n\ge 2$) as it has an element of order $2$.

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