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Suppose I have the Heisenberg group H say over the $p$-adic integers $\mathbb{Z}_p$, which is the set of $3\times 3$ uni-upper-trianglar matrices over $\mathbb{Z}_p$ . Its Lie algebra $h$ is the set of all $3\times 3$ strictly-upper-trianglar matrices over $\mathbb{Z}_p$.

The commutator relations of the presentation of $H$ carry over exactly to give the Lie bracket for $h$. My question is: does this hold for any arbitrary nilpotent Lie group G in place of the Heisenberg group?

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$\newcommand\Lie{\mathrm{Lie}}$If we endow the Lie algebra $\Lie(G)$ of a unipotent algebraic Lie group $G$ over a field of characteristic zero with the product given by the Baker-Campbell-Hausdorff formula —it thereby becomes a Lie group— the exponential map $\Lie(G)\to G$ becomes an isomorphism.

This is what you are seeing in the case of the Heisenberg group. Since its nilpotency index is $2$, the B-H-C series collapses to a very short Lie polynomial.

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    $\begingroup$ A useful keyword to search for in this context is Malcev completion. $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '12 at 3:02

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