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?

  • 5
    $\begingroup$ You should say what you mean by $G^n$. The standard meaning is $\langle g^n \mid g \in G \rangle$. I am guessing that you mean the lower central series, but that seems very confusing and non-standing. You shouid also say what definition of nilpotence you are using, because lower central series reaching $1$ is sometimes used as the definition! $\endgroup$ – Derek Holt Nov 5 '15 at 12:48
  • $\begingroup$ @DerekHolt I apologize. I included all the definitons I think are necessary to clarify the statement of the theorem, but I was really just looking for an external source of the proof since I didn't think anyone would actually prove this as an answer on here $\endgroup$ – mr eyeglasses Nov 5 '15 at 13:07
  • $\begingroup$ What a horrible notation. If you really wanted to do that, it would be more intuitive to let $G^1=G$, $G^2=[G,G]$, etc. $\endgroup$ – Derek Holt Nov 5 '15 at 16:41

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 $G^1 = [G,G]$.

  • $\begingroup$ What about the converse? That is, suppose I know that $G^n=1$ and I want to show that $G$ is nilpotent. I can not use the inequality $G^i\leq Z_{c-i}(G)$, since to prove the inequality we have already used the fact that $G$ is nilpotent of nilpotency class $c$. $\endgroup$ – Babai Mar 10 at 20:30
  • $\begingroup$ It's a similar argument. If $G^n=1$ then prove by induction on $i$ that $G^{n-1} \le Z_i(G)$. $\endgroup$ – Derek Holt Mar 10 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.