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This is related to the question $G=\langle a\rangle H$ for subgroup $H$ raised a few days ago. Suppose $\Gamma $ is a higher rank lattice (for example, $SL_3({\mathbb Z})$). As Misha says in his comments, by this, I mean an irreducible lattice in a linear semi-simple Lie group $G$ of real rank at least two such that $G$ has no compact factors. Can one find a subgroup $H$ of infinite index in $\Gamma$ and an infinite cyclic subgroup $A$ of $\Gamma$ such that $\Gamma$ is the set theoretic product $AH$?

When I was a graduate student, this question was "floating around"; I believe the question is due to Robert Zimmer.

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  • $\begingroup$ You should add the assumption that the lattice is irreducible and in a semisimple Lie group. $\endgroup$
    – Misha
    Commented Feb 17, 2013 at 5:26
  • $\begingroup$ Aakumadula: Do you know the answer even for $SL(2,Z)$? At least, I do not see an obvious way to find an AH-decomposition as the group has finite abelianization. $\endgroup$
    – Misha
    Commented Feb 17, 2013 at 5:56
  • $\begingroup$ Sorry; I don't know if this is true for $SL_2({\mathbb Z})$. Since this is asked for higher rank lattices, I suspect that perhaps for real rank one lattices, it may be false. $\endgroup$
    – Venkataramana
    Commented Feb 17, 2013 at 6:07
  • $\begingroup$ Such decompositions of course exist for some rank 1 lattices: The ones with infinite abelianization. Maybe this is the reason the question was asked in the higher rank case. $\endgroup$
    – Misha
    Commented Feb 17, 2013 at 6:27
  • $\begingroup$ Yes, for some (many) lattices in $SU(n,1)$ and $SO(n,1)$ one has infinite abelianisation. Don't know what happens for $Sp(n,1)$ lattices (or lattices in the real rank one form of $F_4$). $\endgroup$
    – Venkataramana
    Commented Feb 17, 2013 at 7:18

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