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The Malcev completion of an abstract group $G$ over a field $k$ of characteristic zero is defined by $$\hat G = \mathbb{G}(\widehat{k[G]}) ,$$ the group-like part of the completed (by the augmentation ideal $I$) group ring.

Question: Suppose that $G$ is residually nilpotent and torsion-free. Then, is the natural map $G\to \hat G$ an injection?

I've heard that this is true for free groups, but without any proofs. If $G$ is further nilpotent, I could prove it using the fact that $G_n = (1+I^n)\cap G$ (for the notation and the proof, see this MO post and Theorem 12.1.6 of CDBooK by Chmutov-Duzhin-Mostovoy, respectively).

Any comments or references are appreciated.

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This map is an embedding if and only if $G$ is residually torsion-free-nilpotent, which is much stronger that being both torsion free and residually nilpotent.

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    $\begingroup$ For a specific example, fix a prime $p$ and an integer $n\geq 3$ and let $G$ be the corresponding congruence subgroup $\ker(\text{SL}_n(\mathbb{Z})\to \text{SL}_n(\mathbb{Z}/(p))$. Then $G$ is torsion free and, by conisdering deeper conruence subgroups, it is residually (finite-)nilpotent. However $G$ has no non-trivial torsion free nilpotent qoutient (e.g by proerty T), so $\hat{G}$ is trivial. $\endgroup$
    – Uri Bader
    Commented Apr 21 at 14:26
  • $\begingroup$ Thank y'all for the answer and a counterexample. Under the residual torsion-free-nilpotency, is it also true that the map $k[G]\to\widehat{k[G]}$ is also injective? $\endgroup$
    – Qwert Otto
    Commented Apr 21 at 15:32
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    $\begingroup$ Sure, since the map $G\rightarrow \widehat{k[G]}$ is. $\endgroup$
    – Adrien
    Commented Apr 21 at 20:29
  • $\begingroup$ I'm sorry; I think that's not necessarily the case (at least in general). Even if $X\to V$ is an injection for a set $X$ and a vector space $V$, this may not induce an injective linear map. Are there other reason that I'm missing? $\endgroup$
    – Qwert Otto
    Commented Apr 22 at 0:45
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    $\begingroup$ You're right of course, but group-like elements in a Hopf algebra are linearly independant so that's fine. $\endgroup$
    – Adrien
    Commented Apr 22 at 7:27

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