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?
Best Answer
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 $G^1 = [G,G]$.