Cocompact lattice has infinite commutator subgroup Let $ G $ be a noncompact semisimple Lie group and $ \Gamma $ a cocompact lattice in $ G $.
Is the commutator subgroup of $ \Gamma $ always infinite?
EDIT (I deleted two sub-questions, also I'm adding this example of lots of lattices in simple Lie groups that must have infinite commutator subgroup):
Let $ M $ be a closed Hyperbolic $ n $ manifold. Then
$$
SO_{n,1}(\mathbb{R})/\pi_1(M)
$$
is the total space of an $ SO_n(\mathbb{R}) $ principal bundles over $ M $. So $ \Gamma:= \pi_1(M) $ is a cocompact lattice in $ SO_{n,1}(\mathbb{R}) $.
But $ M $ is closed, aspherical and not a torus (since it is hyperbolic) so by the lemma below $ \Gamma=\pi_1(M) $ must have infinite commutator subgroup.
For example when $ n=2 $ we can take surface of genus $ g $ and then $ \pi_1(\Sigma_g) $ is a lattice in $ PSL_2(\mathbb{R})=SO_{2,1}(\mathbb{R}) $. Indeed $ SO_{2,1}(\mathbb{R})/\pi_1(\Sigma_g) $ is the unit tangent bundle of $ \Sigma_g $.
In fact, using ideas along these lines one can prove that every cocompact lattice in $ \widetilde{SL_2(\mathbb{R}) }$ (which is torsion free since its maximal compact subgroup is trivial) must has infinite commutator subgroup.
Lemma: A closed aspherical manifold whose fundamental group has finite commutator subgroup must be a torus.
Proof: Let $ M $ be closed and aspherical. Suppose that the fundamental group of $ M $ has finite commutator subgroup. Since $ M $ is aspherical the fundamental group is torsion free. Since we just assumed the commutator subgroup is finite that implies that the commutator subgroup vanishes and the fundamental group is abelian. But $ M $ is closed so the fundamental group is finitely generated. And as previously noted the fundamental group of an aspherical space is torsion free. So we can conclude that $ \pi_1(M) $ is a finitely generated torsion free abelian group $ \mathbb{Z}^k $. But the Borel conjecture is true for closed aspherical manifolds with abelian fundamental group (even virtually solvable fundamental group) thus $ X $ must be the a torus $ T^k $.
 A: If $G$ is a noncompact semisimple Lie group, then every lattice $\Gamma< G$ has infinite commutator subgroup. This is an application of the results (primarily due to Borel and known in the literature as "Borel density theorem") in Chapter 5 of
Raghunathan, M. S., Discrete subgroups of Lie groups, Ergebnisse der Mathematik und ihrer Grenzgebiete. Band 68. Berlin-Heidelberg-New York: Springer-Verlag. VIII,227 p. (1972). ZBL0254.22005.
For instance, if $[\Gamma,\Gamma]$ were to be finite, then a finite index subgroup in $\Gamma$ would be abelian. But a finite index subgroup in a lattice is again a lattice. Now, read the statement of Corollary 5.18 in the book: If $G$ has no compact factors, then the centralizer of a lattice in $G$ equals the center of $G$. (Raghunathan proves the result for all subgroups satisfying Property (S), but all lattices do, as proven earlier in the chapter.) The general case easily reduces to the case when $G$ has no compact factors by taking the quotient of $G$ by its maximal normal compact subgroup.
All in all, if you are interested in lattices in Lie groups, you should at least browse through Raghunathan's book.
