Is $SL_n(\mathbb{R})$ actually simple?

Question

It's probably not hard to prove that $\frak{sl}_n\mathbb{R}$ is simple, so that $SL_n\mathbb{R}$ has no nontrivial connected normal subgroups. But do there exist discrete normal subgroups of $SL_n(\mathbb{R})$ aside from $\{I, -I\}$?

Answer 1

Any normal subgroup $N\subset SL_n(\mathbb{R})$ has $\pi(N)$ normal in $PSL_2(\mathbb{R})$, where $\pi$ is the projection map. It's a standard classical result that $PSL_n(k)$ is simple except for $PSL_2(\mathbb{F}_2) = S_3$ and $PSL_2(\mathbb{F}_3) = A_4$. Given that, if $N$ is discrete, then $\pi(N)$ must therefore be $1$; that is, $N$ is trivial or $\{\pm 1\}$.