I don't understand why the following two definitions of nilpotence are equivalent:

Definition 1. $G$ is $0$-step nilpotent if $G=\{e\}$. G is $k+1$th-step nilpotent if G is not $k$-step nilpotent, but $G/Z(G)$ is.

Definition 2. Let $\gamma_{i+1}(G)$ such that $\gamma_0(G) = \{e\}, \gamma_1(G) = Z(G)$ and $\gamma_{i}(G) \subseteq \gamma_{i+1}(G)$ and $\gamma_{i+1}(G)/\gamma_i(G) = Z(G/\gamma_i(G))$. For any G, we can define $\gamma_0 \subseteq \gamma_1 \subseteq \gamma_2...$. G is nilpotent if $\gamma_k(G)=G$ for some $k$.

The first definition is clear, but for the second definition, I cannot figure out the implications of having such $\gamma_{i+1}(G)$ which satisfies $\gamma_{i+1}(G)/\gamma_i(G) = Z(G/\gamma_i(G))$ which would make it equivalent to the first definition.

  • $\begingroup$ Possible duplicate of Definition of a nilpotent group. $\endgroup$
    – Ted
    Dec 6, 2015 at 0:59
  • $\begingroup$ @Ted I've read that post before adding my question, and I couldn't relate it to the first definition that I posted (which isn't on there). $\endgroup$ Dec 6, 2015 at 1:00
  • $\begingroup$ Sorry, I read your question too fast and thought you were asking a different one. $\endgroup$
    – Ted
    Dec 6, 2015 at 1:05
  • $\begingroup$ You seem to be missing $\gamma_0(G)=0$? Otherwise you could take $\gamma_i(G)=G$ "proving every group is nilpotent". $\endgroup$
    – Nex
    Dec 6, 2015 at 2:20
  • $\begingroup$ @Nex Good catch. I've updated the definition. $\endgroup$ Dec 6, 2015 at 2:54

1 Answer 1


The two definitions are related as follows:

$G$ is $k$-step nilpotent iff $k$ is the minimum value such that $\gamma_k(G) = G$.

This follows from the easily-checked fact $\gamma_{i+1}(G)/Z(G) = \gamma_i(G/Z(G))$, i.e., the $\gamma$ sequence for $G/Z(G)$ is obtained from the $\gamma$ sequence for $G$ by dropping the $\gamma_0(G)$ (trivial) term, then taking the quotient of the remaining groups by $Z(G)$.

To check the above fact, we must prove that $\overline{\gamma_{i+1}(G)} = \gamma_i(\overline{G})$ where the bar denotes projection from $G$ to $G/Z(G)$. The case $i=0$ is easy. For $i>0$, we proceed by induction.

By definition $$\gamma_{i+1}(G) / \gamma_i(G) = Z(G/\gamma_i(G)).$$

Since $\gamma_i(G) \supseteq \gamma_1(G) = Z(G)$ for $i>0$, we have the natural isomorphism $$G/\gamma_i(G) \stackrel{\cong}{\longrightarrow} \overline{G} / \overline{\gamma_i(G)}.$$ The centers of the two sides correspond under the isomorphism: $$\gamma_{i+1}(G) / \gamma_i(G) \stackrel{\cong}{\longrightarrow} Z(\overline{G} / \overline{\gamma_i(G)}).$$

Using the induction hypothesis (*) as well as the definition of the $\gamma$ groups for $\overline{G}$ (**), $$Z(\overline{G} / \overline{\gamma_i(G)}) \stackrel{(*)}{=} Z(\overline{G} / \gamma_{i-1}(\overline{G})) \stackrel{(**)}{=} \gamma_i(\overline{G}) / \gamma_{i-1}(\overline{G}) \stackrel{(*)}{=} \gamma_i(\overline{G}) /\overline{\gamma_i(G)}.$$ Thus we have $$\gamma_{i+1}(G) / \gamma_i(G) \stackrel{\cong}{\longrightarrow} \gamma_i(\overline{G}) /\overline{\gamma_i(G)}.$$ But by the definition of the isomorphism $\cong$, we have $$\gamma_{i+1}(G) / \gamma_i(G) \stackrel{\cong}{\longrightarrow} \overline{\gamma_{i+1}(G)}/\overline{\gamma_i(G)}.$$ Comparing the last 2 displayed equations, we get $\overline{\gamma_{i+1}(G)} = \gamma_i(\overline{G})$, as desired.

I should add that all this looks more complicated than it really is. If you think through the construction of the $\gamma$'s, you see the construction for $\overline{G}$ is really just "one step ahead" of the corresponding construction for $G$ at all times (up to the identification by $\cong$), thus the sequence for $\overline{G}$ terminates one step earlier.

  • 1
    $\begingroup$ How do we have $\gamma_{i+1}(G)/Z(G) = \gamma_i(G/Z(G))$ from the original $\gamma_{i+1}(G)/\gamma_i(G) = Z(G/\gamma_i(G))$? $\endgroup$ Dec 6, 2015 at 20:16
  • $\begingroup$ @user2193268 See edits. $\endgroup$
    – Ted
    Dec 7, 2015 at 0:50

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