Let $G$ be an abstract group. The Malcev completion $\widehat{G}$ of $G$ (over $\mathbb{Q}$) is the set of group-like elements in the complete Hopf algebra $\widehat{\mathbb{Q}[G]} = \lim\_n \mathbb{Q}[G]/I^n, I = \langle g-1 \,|\, g\in G\rangle.$ The nilpotent completion of $G$ is also defined by $G^{\mathrm{nil}}=\lim\_n G/G^{(n)}, G^{(1)}=G, G^{(n+1)}= (G,G^{(n)}).$ These limit definitions induce two natural filtrations (both denoted by $F$) on Malcev and nilpotent completion. Also in both cases $\forall m,n, \,(F\_n,F\_m)\subset F\_{n+m}$ and so there exists a natural structure of Lie algebras on the direct sum of graded quotients: $E\widehat{G} = \oplus F\_n \widehat{G}/F\_{n+1}\widehat{G}, EG^{\mathrm{nil}}= \oplus F\_n G^{\mathrm{nil}}/F\_{n+1}G^{\mathrm{nil}}$. 


I have three questions on these concepts:


1- What is the relation between these two completion processes? I know (By theorems in Chapter 8 of Fresse's book: Homotopy of operads and Grothendieck-Teichmuller groups) in the case of finitely generated free groups, $E\widehat{G} \simeq EG^{\mathrm{nil}}\otimes \mathbb{Q}.$ Does this hold in general? If not, does there exist simple sufficient conditions for this to be true? 


2-It's evident that $\widehat{G}^{(n)}\subset F\_n(\widehat{G})$. Does there exist examples of $G$ for which this inclusion is strict for some $n$?


3-Is $\widehat{G}=\widehat{\widehat{G}}$ in general?


Thanks!



Some questions about the Malcev completion
A nice exposition of all of this can be found in section 12,1 of the book of Chmutov-Duzhin-Mostovoy (<http://www.pdmi.ras.ru/~duzhin/papers/cdbook/>).


Assume for the sake of safety that $G$ is finitely generated. The series $G^{(n)}$ is characterized by the property that $G/G^{(n)}$ is the largest quotient of $G$ which is nilpotent of class $n$. There is another series $G\_n$ characterized by the fact that the successive quotient are the largest **torsion free** quotient of $G$ which are nilpotent of class $n$. More or less by definition:
$$G\_n=\{x \in G\ |\ \exists k, x^k \in G^{(n)}\}$$


Now one can show that $G\_n$ coincides with the filtration induced on $G$ by the augmentation ideal of $\mathbb{Q}[G]$, i.e. $G\_n=(1+I^n)\cap G$.


The moral is that the difference between $G^{nil}$ and $\hat{G}$ is about torsion, hence disapears after tensoring with $\mathbb{Q}$. So:


1) is always true, it's a theorem of Quillen


I'm pretty confident that 2) is always an equality since you already completed everything


3) is definitely true: $\hat{G}$ is characterized by the property that the map $G\rightarrow \hat{G}$ is universal among maps from $G$ to pro-unipotent groups. Hence the pair $(\hat{G},id:\hat{G}\rightarrow \hat{G})$ realizes $\hat{G}$ as its own Malcev completion.


