Margulis' normal subgroup theorem states that any normal subgroup of a higher rank irreducible lattice is either finite or of finite index. What are the known counter-examples in rank $1$ ?
I am especially interested by $PU(2,1)$.
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Sign up to join this communityMargulis' normal subgroup theorem states that any normal subgroup of a higher rank irreducible lattice is either finite or of finite index. What are the known counter-examples in rank $1$ ?
I am especially interested by $PU(2,1)$.
Any cocompact lattice in a rank-one Lie group is word-hyperbolic. Olshanskii proved that such groups are SQ-universal, meaning in particular that they have uncountably many normal subgroups. Similar results are known for non-uniform lattices, which are relatively hyperbolic.
As Yemon Choi suggests in comments, the simplest example comes from $PSL(2,\mathbb{Z})$. The 2-congruence subgroup $\Gamma(2)$ is free of rank two, and hence surjects every two-generator group.
Also interesting in this context is the existence of compact complex surfaces whose universal cover is the ball and admitting an holomorphic map onto a Riemann surface with connected fibers. The kernel of the induced map is a finitely generated group of infinite index. This is similar to the existence (Rips counterexample) of finitely generated normal subgroup in certain small cancelation groups with arbitrary complicated f.p. quotient.