# Showing $(G^{i}Z(G))/Z(G) = (G/Z(G))^{i}$

group-theory

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?

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 $$G_iZ(G)/Z(G) = f(G_iZ(G)) = f(G_i) = K_i = (G/Z(G))_i$$.