TITLE: Is $\mathrm{SL}_n(\mathbb{Q}_p)$ virtually torsion-free?
QUESTION [5 upvotes]: 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. 

Note that $\mathrm{GL}_1(\mathbb{Q}_p) = \mathbb{Q}_p^*$ is virtually torsion-free.
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.
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!

REPLY [2 votes]: 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 $2$.