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This question is a follow-up to this post, from which I quote:

Let $\mathfrak{e}$ be the 3-dimensional Lie algebra with basis $(H,X,Y)$ and bracket $[H,X]=Y$, $[H,Y]=-X$, $[X,Y]=0$. It is isomorphic to the Lie algebra of the group of isometries of the plane. Its central extension $\tilde{\mathfrak{e}}$ is defined as the 4-dimensional Lie algebra defined by adding a central generator $Z$ and the additional nonzero bracket $[X,Y]=Z$.

It seems to me that since $\mathrm{exp}$ is a diffeomorphism on a Carnot group then the latter's Lie algebra $\mathfrak{g}$, should not contain either $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$. However, where can I find a proof of the fact that no Carnot group contains either $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$ (without going through the Sato-Dixmier result of the quoted post).

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    $\begingroup$ It's immediate that $\mathfrak{e}$ (and hence $\tilde{\mathfrak{e}}$) is not nilpotent. $\endgroup$
    – YCor
    Commented May 28, 2019 at 3:23

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The Lie algebra of a Carnot group is nilpotent. But a nilpotent Lie algebra cannot contain nonzero elements $H,X,Y$ satisfying $[H,X]=Y$ and $[H,Y]=-X$, for then arbitrarily long brackets of the form $$[H, [H, [H, \dots,[H,X]\dots]]]$$ would equal something nonzero (either $\pm X$ or $\pm Y$). This rules out the possibility of containing either $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$.

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  • $\begingroup$ Oh, I feel silly now. Thank Nate. $\endgroup$
    – ABIM
    Commented May 28, 2019 at 7:55

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