Showing $(G^{i}Z(G))/Z(G) = (G/Z(G))^{i}$ If $G$ is any group, $G^{i}$ denotes the lower central series for $G$. In this question, Nicky Hekster suggested me to prove that for each $i$
$$(G^{i}Z(G))/Z(G) = (G/Z(G))^{i}$$

Here's my incomplete attempt:
Claim 1. Let $G,A$ be groups and let $\varphi : G \to A$ be a homomorphism and $H,K \le G$. Then $\phi ([H,K]) = [\varphi (H) , \varphi (K)]$.
Proof. Follows immediately from the definition of commutators.
Claim 2. Let $G$ be a group. Let $N\trianglelefteq G$ and suppose that $N \le H$ and $N\le K$. Then $[H,K]/N = [H/N, K/N]$.
Proof. Using the canonical homomorphism $\pi : G \to G/N$ and the result in Claim 1, this immediately follows.
Claim 3. Let $G$ be a group and $\overline G$ denote the quotient group $G/Z(G)$. Then $(G^{i}Z(G)) /Z(G) = (\overline G)^{i}$ for each $i$.
Proof. Note that $G^{i}$ and $(\overline{G})^{i}$ denote the lower central series for $G$ and $\overline{G}$ respectively. We prove this by induction. Clearly $G^{0}/Z(G)=G/Z(G)= (\overline G)^{0}$. Assume that $(G^{i-1}Z(G)) /Z(G) = (\overline G)^{i-1}$ for some $i>0$. Then we have that $(\overline G )^{i} = [\overline G , (\overline{G})^{i-1}] =[G/Z(G) , (G^{i-1}Z(G))/Z(G)] = [G, G^{i-1}Z(G)]/ Z(G) =(G^{i}Z(G))/Z(G)$.
Note that the first equality is simply definition, second is due to induction hypothesis and third is a consequence of Claim 2. I leave it incomplete here because I cannot justify the fourth equality.
It seems to me that the set of commutators from $G$ and $G^{i-1}Z(G)$ and the set of commutators from $G$ and $G^{i-1}$ are the same. So we have that $[G,G^{i-1}Z(G)]= [G, G^{i-1}]=G^{i}$. So how do I bring the additional $Z(G)$ here which I want?
 A: The notation is both non-standard (the symbol already has a different meaning) and the numbering is non-standard (usually $G$ is the first term of the lower central series, not the zeroth; we have $[G_n,G_m]\subseteq G_{n+m}$ with the usual numbering, but with the one you give we have $[G^n,G^m]\subseteq G^{n+m+1}$...). Dummit and Foote are just being contrarian, and badly, here...
Terms of the lower central series are verbal (they are the subgroups generated by all values of a group word), and for any set of words $\mathfrak{W}$, if $f\colon G\to K$ is a surjective homomorphism then $\mathfrak{W}(K) = f(\mathfrak{W}(G))$. But you don't need this machinery here. You just need to note the following, which is an easy variation of your first claim.
I will use the standard notation: $G_1=G$, $G_{k+1} = [G,G_k]$ (or $[G_k,G]$).
Theorem. let $f\colon G\to K$ be a surjective group homomorphism. The for all $n\geq 1$, $f(G_n) = K_n$.
Proof. Induction on $n$.  The result follows for $n=1$ because $f$ is surjective. Assume we already know that $f(G_n) = K_n$.  Let $x\in K$, $y\in K_n$; since $f$ is surjective and $f(G_n)=K_n$, there exists $g\in G$ and $h\in G_n$ such that $f(g)=x$ and $f(h) = y$. Then
$$[x,y] = [f(g),f(h)] = f([g,h]) \in  f([G,G_n]) = f(G_{n+1}).$$
Thus, every generator of $K_{n+1}$ lies in $f(G_{n+1})$, so $K_{n+1}\subseteq f(G_{n+1})$. Conversely, if $a\in G$ and $b\in G_n$, then $f([a,b]) = [f(a),f(b)]\in [K,K_{n}]=K_{n+1}$, so every generator of $G_{n+1}$ maps into $K_{n+1}$. Thus, $f(G_{n+1})\subseteq K_{n+1}$, proving equality. $\Box$
(For general verbal subgroups, if $w(x_1,\ldots,x_n)$ is a word, then $f(w(x_1,\ldots,x_n)) = w(f(x_1),\ldots,f(x_n))$, so that's why you have the result for any verbal subgroup.)
Now consider $K=G/Z(G)$ and $f\colon G\to K$ the canonical surjection. Then $f(G_i) = K_i = (G/Z(G))_i$. Since $f(G_i) = f(G_iZ(G))$, we have that $G_iZ(G)/Z(G) = f(G_iZ(G)) = f(G_i) = K_i = (G/Z(G))_i$.
