# Lattices are not solvable in non-compact semisimple Lie groups

I'm trying to prove the following result.

If $$G$$ is a non compact semisimple Lie group with no compact factors (lying in some $$SL(l,\mathbb{R})$$), and $$\Gamma$$ is a lattice in $$G$$, then $$\Gamma$$ is not solvable, and $$[\Gamma, \Gamma]$$ is infinite.

So far, I've managed to prove that $$\Gamma$$ can't be abelian, since $$\Gamma$$ projects densely onto the maximal compact factor $$\{e\}$$, and a version of the Borel density theorem tells us that $$\mathcal{C}_G(\Gamma) = Z(G)$$, and $$Z(G)$$ is finite, but $$\Gamma$$ isn't, which rules out the possibility that $$\Gamma$$ is abelian. My question is whether I can refine this argument to prove that $$\Gamma$$ isn't solvable as well, or do I need some additional machinery to prove that? Thanks!

• Yes you can refine the argument to get solvability, and also use the Borel density for the derived subgroup being infinite. However, this seems to better fit MathSE. – YCor May 18 at 6:13
• Already crossposted at MSE here. – Dietrich Burde May 18 at 16:59
• One can prove this as a consequence of the Tits alternative, or maybe of its proof. en.wikipedia.org/wiki/Tits_alternative – Ian Agol May 22 at 3:43