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.


1 Answer 1


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.

  • 3
    $\begingroup$ 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
    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
    Apr 21 at 15:32
  • 1
    $\begingroup$ Sure, since the map $G\rightarrow \widehat{k[G]}$ is. $\endgroup$
    – Adrien
    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
    Apr 22 at 0:45
  • 1
    $\begingroup$ You're right of course, but group-like elements in a Hopf algebra are linearly independant so that's fine. $\endgroup$
    – Adrien
    Apr 22 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.