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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$\begingroup$ do you mean "maximal connected nilpotent" (="maximal unipotent" here)? If you really mean "maximal nilpotent", you probably also have more subgroups, including finite subgroups. $\endgroup$– YCorCommented Feb 9, 2013 at 10:00
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$\begingroup$ @Yves I do mean maximal connected nilpotent. Thank you. $\endgroup$– DavisCommented Feb 10, 2013 at 3:14
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1 Answer
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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).
