Group $G$ is nilpotent if and only if $G^n = 1$ for some $n \geq 0$

Question

This is from the book Abstract Algebra, $3$rd edition, by Dummit & Foote; theorem $8$ on page $194$.

Definition (upper central series): For any group $G$ define the following subgroups inductively: $$Z_0(G) = 1, \qquad Z_1(G) = Z(G)$$ and $Z_{i+1}(G)$ is the subgroup of $G$ containing $Z_i(G)$ such that $$Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G)).$$ The chain of subgroups $$Z_0(G) \leq Z_1(G) \leq Z_2(G) \leq \cdots$$ is called the upper central series of $G$.

Definition (nilpotent): A group $G$ is called nilpotent if $Z_c(G) = G$ for some $c \in \Bbb Z$. The smallest such $c$ is called the nilpotence class of $G$.

Definition ($G^n$ and lower central series): For any (finite or infinite) group $G$ define the following subgroups inductively: $$G^0 = G, \qquad G^1 = [G, G], \qquad \text{ and } G^{i+1} = [G, G^i].$$ The chain of groups $$G^0 \geq G^1 \geq G^2 \geq \cdots$$ is called the lower central series of $G$.

The next theorem shows the relation between the upper and lower central series of a group.

Theorem $8$: A group $G$ is nilpotent if and only if $G^n = 1$ for some $n \geq 0$. More precisely, $G$ is nilpotent of class $c$ if and only if $c$ is the smallest non-negative integer such that $G^c = 1$. If $G$ is nilpotent of class $c$ then $$Z_i(G) \leq G^{c - i - 1} \leq Z_{i+1}(G) \qquad \text{ for all } i \in \{0, 1, \ldots , c - 1\}.$$

Proof: This is proved by a straightforward induction on the length of either the upper or lower central series. $\square$

I don't see the straightforward proof here, and would like the complete details. Is there another book or reference that includes the complete proof in detail?

Answer 1

One problem here is that the indexing is so unhelpful and confusing. Let's prove by induction on $i$ that $G^i \le Z_{c-i}(G)$, which is equivalent to the second containment you have to prove. It's true for $i=0$, since $G^0=Z_c(G)=G$. Assuming it's true for $i$, we get $G^{i+1} = [G,G^i] \le [G,Z_{c-i}(G)]$.

But $Z_{c-i}(G)/Z_{c-i-1}(G) = Z(G/Z_{c-i-1}(G))$ implies that $[G,Z_{c-i}(G)] \le Z_{c-i-1}(G)$, which completes the inductive step.

I am afraid that the left hand inequality is not true in general! Let $G = D_{16} \times C_2$ be the direct product of a dihedral group of order $16$ and a cyclic group of order $2$. This is nilpotent of class $3$, but the direct factor $C_2$ of $G$ is contained in $Z_1(G) = Z(G)$, but not in $G^1 = [G,G]$.