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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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    $\begingroup$ Not an answer, but: the commutator subgroup of $F_2$ is infinite and has infinite index. So I would start looking at something similar inside SL(2,Z) viewed as a lattice inside SL(2,R) $\endgroup$ – Yemon Choi Feb 14 '17 at 16:46
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    $\begingroup$ @YemonChoi: Actually, what you wrote is an answer! $\endgroup$ – Moishe Kohan Feb 14 '17 at 18:20
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    $\begingroup$ For a hyperbolic manifold of positive first Betti number (e.g., the complement of any hyperbolic link) the commutator subgroup is a normal subgroup of infinite index. I guess it is often (always?) also an infinite group. $\endgroup$ – ThiKu Feb 14 '17 at 21:56
  • $\begingroup$ @ThiKu, yes, always. Otherwise the fundamental group of your manifold would be virtually abelian! Of course, we now know that all hyperbolic 3-manifolds have a finite-index subgroup with $b_1>0$. $\endgroup$ – HJRW Feb 16 '17 at 8:25
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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.

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  • $\begingroup$ Same for nonuniform rank 1 lattices (Arzhantseva-Minasyan-Osin). $\endgroup$ – Moishe Kohan Feb 14 '17 at 18:23
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    $\begingroup$ In "Filling Riemaniann Manifold", Gromov "remarked" that in a compact manifold with strictly negative curvature the normal subgroup of the fundamental group generated by the high power of an element is a free group, hence of infinite index ; SQ universality for hyperbolic groups is stated in his (Gromov's) paper on hyperbolic groups. These statements generated a huge literature, not only in Olshanskii's work.. $\endgroup$ – Thomas Feb 14 '17 at 18:51
  • $\begingroup$ $\Gamma_2$ contains $-Id$. $\endgroup$ – few_reps Feb 16 '17 at 12:31
  • $\begingroup$ @few_reps: Sorry, I meant in $PSL_2(\mathbb{Z})$, of course. $\endgroup$ – HJRW Feb 16 '17 at 16:15
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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.

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  • $\begingroup$ Am I right in thinking that it's open whether or not every lattice in $PU(2,1)$ is large, i.e. virtually surjects a free group? $\endgroup$ – HJRW Feb 16 '17 at 11:09
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    $\begingroup$ @HJRW: Yes, I think that's right. For the arithmetic congruence lattices of "second type", Rogawski showed that the first betti number is always zero. I think it's not known whether one of these lattices has a non-congruence subgroup with positive betti number, much less large. (see: worldscientific.com/doi/abs/10.1142/S0218196707004165 for a generalization, they don't have FA). $\endgroup$ – Ian Agol Feb 17 '17 at 4:14
  • $\begingroup$ @IanAgol: thanks! They do have FA, btw. :) $\endgroup$ – HJRW Feb 17 '17 at 8:43

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