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For the Lie group $SO(n,1)$ I believe the maximal nilpotent subgroups are conjugate to either a diagonal group times a compact group or a unipotent group times a compact group. In either case the compact group will commute with the other group. Is this true and if so how do I prove it?

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do you mean "maximal connected nilpotent" (="maximal unipotent" here)? If you really mean "maximal nilpotent", you probably also have more subgroups, including finite subgroups. –  YCor Feb 9 '13 at 10:00
    
@Yves I do mean maximal connected nilpotent. Thank you. –  Davis Feb 10 '13 at 3:14

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up vote 2 down vote accepted

Let's look first at maximal solvable subgroups, i.e. Borel subgroups. If $G=KAN$ is an Iwasawa decomposition of $G$, Borel subgroups are conjugate to $MAN$, where $M$ is the centralizer of $A$ in $K$. In the case of $SO(n,1)$, we have $K\simeq SO(n),A\simeq\mathbb{R}$(this is the maximal diagonalizable subgroup), $N\simeq\mathbb{R}^{n-1}$and $M\simeq SO(n-1)$. Your conjecture follows from this (observe that nilpotent subgroups of compact groups are abelian-by-finite, by Lie-Kolchin).

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