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Can a lattice in $PSL_2(\mathbb{R})$ have a normal abelian subgroup? It looks to me that it doesn't, but where can I read a proof?

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    In any semisimple connected Lie group with trivial center, any lattice is Zariski-dense. Hence any normal abelian subgroup $N$ should have its Zariski closure normal in the whole group, which forces $N=1$. – YCor Jan 13 '15 at 20:54
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    Migrating this from MO seems to show a certain trigger happiness. – Igor Rivin Jan 13 '15 at 23:56

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A somewhat different argument from that of Yves comes from Delzant's theorem SOUS-GROUPES DISTINGUÉS ET QUOTIENTS DES GROUPES HYPERBOLIQUES, which states that the normal closure of a pair of noncommuting hyperbolic elements contains a free group. Showing that any normal subgroup contains hyperbolic elements is not hard.

Igor Rivin
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In any semisimple connected Lie group with trivial center, any lattice is Zariski-dense. Hence any normal abelian subgroup N should have its Zariski closure normal in the whole group, which forces $N=1$.

YCor
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