# Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent?

I was told that a polynomial group law on (all of) $\mathbb{R}^n$ gives automatically a nilpotent (Lie, of course) group.

Is it true? Where can I find a proof?

A counterexample for open subsets of $\mathbb{R}^n$ is furnished by the halfplane with the $ax+b$ law.

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Torsten, make it an answer (not just a comment). –  Sammy Black Apr 17 '10 at 19:13
Thank you, Torsten! –  Gian Maria Dall'Ara Apr 17 '10 at 19:49
Moved a comment to an answer as per instructions. –  Torsten Ekedahl Apr 17 '10 at 20:36

This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"

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This short paper is apparently not available online, but a version of Lazard's theorem is also written down in the 1970 book by Demazure-Gabriel, Groupes algebriques, I: see IV, section 4, 4.1. –  Jim Humphreys Apr 17 '10 at 20:53