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](https://mathoverflow.net/questions/265058/is-sl-n-mathbbz-p-virtually-torsion-free) 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!



Is $\mathrm{SL}_n(\mathbb{Q}_p)$ virtually torsion-free?
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 $2$. 


