An association between finite nilpotent groups and Lie rings The following line turns up in a paper without much of a reference:
Let $G$ be a nilpotent groups of nilpotency class $c$, and suppose $V$ be its associated Lie ring, where $V=V_1\oplus V_2\cdots \oplus V_c$ and,
$$V_i=\gamma_i(G)/\gamma_{i+1}(G),$$
where $\gamma_1(G)=G$ and, $\gamma_k(G)=[G,\gamma_{k-1}(G)]$.
Now the problem is that the author doesn't clearly define the Lie bracket. I suppose that is something naturally defined. If we take the simplest example of groups of nilpotency class 2, then
$$V=G/[G,G]\oplus [G,G].$$
I am trying to define the Lie bracket as follows:
$$[g_1\gamma_2(G)+g_2,h_1\gamma_2(G)+h_2]=[g_1,h_1].$$
But with this definition, I don't see why this should be well defined at all. To prove well definedness, I need to show if $g_1\gamma_2(G)=g_1'\gamma_2(G)$ and $h_1\gamma_2(G)=h_1'\gamma_2(G)$, then $[g_1,h_1]=[g_1',h_1']$. But I don't see why this should hold at all! I am looking for some help in this direction. Maybe the Lie bracket is defined in some different way! I don't understand what way it is. Additionally, if it is well defined (which I don't see), how would one prove the Jacobi identity in this case? Thanks in advance for any kind of help.
 A: As pointed out in the linked slides in Dietrich Burde's comment (by the way, a classical reference for the theory is Lazard [1]), the Lie bracket is first defined on homogenous elements in the graded abelian group $V$, and then extended to the whole object via bilinearity. What is maybe not pointed out explicitly enough in the slides is that for homogeneous elements $x,y$ of degrees $i$ and $j$, respectively, the bracket $[x,y]$ is by definition an element of degree $i+j$.
So in your example, we first break down, via bilinearity,
$$[g_1\gamma_2(G)+g_2,h_1\gamma_2(G)+h_2]= [g_1\gamma_2(G),h_1\gamma_2(G)]+[g_2,h_1\gamma_2(G)]+[g_1\gamma_2(G),h_2]+[g_2,h_2].$$
All four brackets on the RHS are between homogeneous elements, of respective degrees (1 and 1), (2 and 1), (1 and 2), and (2 and 2).
That means that the last three of the brackets will just vanish, because they would live in degree 3 or higher (indeed e.g. the group commutator $[g_2, h_1 \tilde \gamma] := g_2^{-1} (h_1 \tilde \gamma)^{-1}g_2 (h_1 \tilde \gamma)$ is $=1$ i.e. trivial for any $\tilde \gamma \in \gamma_2(G)$ here, etc.).
So all that is left is
$$ = [g_1\gamma_2(G),h_1\gamma_2(G)]$$
which is defined as an element of $V_2 = \gamma_2(G) = [G,G]$, namely, the group commutator $$[g_1, h_1] := g_1^{-1}h_1^{-1}g_1 h_1$$ which, as you can check for yourself, is independent of the choice of representatives $g_1, h_1$. And in general, i.e. even for a group where $\gamma_3(G)$ would not be trivial, that commutator would still be well-defined "modulo $\gamma_3(G)$" i.e. as an element of $V_2 := \gamma_2(G)/\gamma_3(G)$; it's probably good to show this right away, and then just notice that in your example, $\gamma_3(G)$ is trivial.

[1] Lazard, Michel. Sur les groupes nilpotents et les anneaux de Lie. Annales scientifiques de l'École Normale Supérieure, Série 3, Tome 71 (1954) no. 2, pp. 101-190. doi : 10.24033/asens.1021. http://www.numdam.org/articles/10.24033/asens.1021/
