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 (lying in some $SL(l,\mathbb{R})$), and $\Gamma$ is a lattice in $G$, then $\Gamma$ is not solvable, and $[\Gamma, \Gamma]$ is infinite.

What I've managed to show so far is that $\Gamma$ can't be abelian. I did this by modding out by the compact factors $K$ of $G$, after which $\Gamma/(K \cap \Gamma)$ projects densely on the
maximal compact factor $\{e\}$, allowing me to apply a version
of the Borel density theorem that states $\mathcal{C}_G(\Gamma) = Z(G)$. Since $Z(G)$ is finite, $\Gamma$ can't be contained in it (because $G$ is non-compact), and this shows $\Gamma$ is not abelian.
What I'd like to know is how do I upgrade this method to prove the stronger claims that $\Gamma$ is not solvable and $[\Gamma, \Gamma]$ is infinite? I'd appreciate some help at this part. Thanks!
 A: For this problem, it's much easier to prove the stronger claim that solvable discrete subgroups cannot be Zariski dense in semisimple linear algebraic groups. We prove this by induction on the solvable length $n$ of derived series of solvable discrete subgroups. When $n = 1$, the solvable discrete group $\Gamma$ is abelian, and Zariski density tells us it's contained in the center $Z(G)$, which is finite, and hence $\Gamma$ can't be Zariski dense. Now suppose we know that all solvable groups of solvable length at most $n-1$ are not Zariski dense. Consider the derived series for $\Gamma$.
$$
\{e\} = \Gamma_0 \vartriangleleft \Gamma_1 \vartriangleleft \cdots \vartriangleleft \Gamma_{n-1} \vartriangleleft \Gamma_n = \Gamma
$$
For contradiction's sake, assume $\Gamma_n$ is Zariski dense in $G$. Let $H$ be the Zariski closure of $\Gamma_{n-1}$, and $H^{\circ}$ be its identity component in the analytic topology. It's easy to see that $\Gamma_n$ normalizes $H$, and in particular, $H^{\circ}$, which means $\mathfrak{h}$ is an invariant subspace for the $\Gamma$ conjugation action on $\mathfrak{g}$. By Zariski density, $\mathfrak{h}$ is also invariant under the $G$ action, which means $\mathfrak{h}$ is a proper ideal in $\mathfrak{g}$, and hence semisimple since $\mathfrak{g}$ is semisimple. If $\mathfrak{h} = 0$, that means $H$ is a discrete algebraic subgroup, which means it's finite. By Proposition 4.5.4 in Morris, it's contained in $Z(G)$, which means we can quotient it out to get an abelian lattice in $G/H$, which leads to a contradiction. If $\mathfrak{h} \neq 0$, then $H$ is a semisimple linear algebraic group in which $\Gamma_{n-1}$ is Zariski dense. But this can't happen because the induction hypothesis says that all solvable groups of length at most $n-1$ are not Zariski dense. We also end up getting a contradiction in this case, which proves the result.
