Questions about finitely generated nilpotent groups Suppose $G$ is a nilpotent group of class $c$, where $c > 1$. This means that if we define the lower central series by $l_1(G) = G$ and $l_k(G) = [G,l_{k-1}(G)]$ for $k \geqslant 2$, then $l_{c+1} (G) = {1}$, while $l_c(G)$ is nontrivial. Suppose also that $G$ is generated by the elements $x_1,...,x_n$. Now I have a couple of questions about this group.
1) Consider an element $h \in l_c(G)$. How to show that there are elements $g_i$ of $l_{c-1}(G)$ for $i=1,...,n$ such that 
$$
h = \prod_1^n [x_i, g_i].
$$
2) Can we invent something like induction to show that actually an arbitrary element $h\in [G,G]$ can be expressed as a such product where $g_i \in G$?
Also, a good reference on this subject would be really nice. I have never done such things before, so I do not have any good ideas on what to do actually, so sorry if this is a really dumb question.
 A: By definition:
$$l_c(G):=[G,l_{c-1}(G)]:=\langle [x,y]\;;\;x\in G\,,\,y\in l_{c-1}(G)\rangle$$
so question (1) is just definition...and so it's question 2, of course.
A: I am using the convention $x^y = y^{-1}xy$, $[x,y]=x^{-1}y^{-1}xy$. Then using the standard commutator identities:
If $g_1,g_2 \in G$ and $h \in L_{c-1}(G)$, then $[g_1g_2,h] = [g_1,h]^{g_2}[g_2,h] = [g_1,h][g_2,h]$, since $[g_1,h] \in Z(G)$.
Similarly, for $g \in G$, $h_1,h_2 \in H$, $[g,h_1h_2]= [g,h_2][g,h_1]^{h_2} = [g,h_1][g,h_2]$.
(This shows that the commutator map $G \times L_{c-1}(G) \to L_c(G)$ is bilinear.)
It follows from this that if $G = \langle x_1,x_2,\ldots,x_n \rangle$ and $L_{c-1}(G) = \langle y_1,\ldots,y_m \rangle$, then $L_c(G)  = \langle [x_i,y_j] \mid 1 \le i \le n,\, 1 \le j \le m \rangle$.
So each element of $L_c(G)$ can be written as product of elements of the form $[x_i,y_j]^{\pm 1} = [x_i,y_j^{\pm 1}]$. They all commute and, for each fixed $i$, you can group all of the $[x_i,y_j]$ terms together and replace them by a single term $[x_i,g_i]$, which answers Question (1).
This argument shows that the images of  $[x_i,x_j]$ generate $L_2(G)/L_3(G)$, and then that the images of $[x_i,[x_j,x_k]]$ generate $L_3(G)/L_4(G)$, etc, and so you can conclude that each $L_i(G)$ is finitely generated.
We can prove Qn (2) by induction on $c$. It is trivial for $c=1$.
Assume that $c>1$, and let $h \in G'$.
Applying (2) inductively to $G/L_c(G)$, there exist $g_1,g_2,\ldots,g_n \in G$ and $h' \in L_c(G)$ with $g = [x_1,g_1][x_2,g_2] \cdots [x_n,g_n]h'$.
Them applying Qn (1) to $h'$, there exist $h_1,\ldots,h_n \in L_{c-1}(G)$ with $g = [x_1,h_1][x_2,h_2] \cdots [x_n,h_n]$. Since the commutators $[x_i,h_i]$ all lie in $Z(G)$, we have
$$g = \prod_{i=1}^n [x_i,g_i][x_i,h_i] =\prod_{i=1}^n [x_i,g_i][x_i,h_i]^{g_i}= \prod_{i=1}^n [x_i,h_ig_i],$$ 
which proves (2).
