# Is Malcev completion an embedding?

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.

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.
• 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. Apr 21 at 14:26
• 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? Apr 21 at 15:32
• Sure, since the map $G\rightarrow \widehat{k[G]}$ is. Apr 21 at 20:29
• 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? Apr 22 at 0:45