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I saw two different definitions of a nilpotent group, but I’m not really sure how these definitions are equivalent.

The first one is from *Basic Abstract Algebra* (Robert Ash):

A

central seriesfor $G$ is a normal series $1 = G_0 \trianglelefteq G_1 \trianglelefteq … \trianglelefteq G_r = G$ such that $G_i/G_{i-1} \subseteq Z(G/G_{i-1})$ for every $i = 1, … ,r$. An arbitrary group $G$ is said to benilpotentif it has a central series.

The second one is from *Advanced Modern Algebra* (Rotman):

The

of a group $G$ is $$ G = \gamma_1(G) \supseteq \gamma_2(G) \supseteq …,$$descending central serieswhere $\gamma_{i+1}(G) = [\gamma_i(G),G].$ A group $G$ is called

if the lower central series reaches $\{1\}$; that is, if $\gamma_n(G)=\{1\}$ for some $n$.nilpotent

I can’t really see how these definitions are equivalent, because they are defining it in very different ways… so I would appreciate it if anybody could clarify this for me.

Thanks in advance

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Observe that

$$\frac{G_{i+1}}{G_i}\subseteq Z\left(\frac{G}{G_i}\right)\iff \left[\frac{G_{i+1}}{G_i},\frac{G}{G_i}\right]\subseteq\frac{G_i}{G_i}\iff [G_{i+1},G]\subseteq G_i. \tag{$\circ$}$$

Thus by induction

$$[[\cdots[[G_i,\overbrace{G],G],\cdots],G}^{\large i}]\subseteq G_0=1.$$

In particular, if $G_n=G$ then $\gamma_{n+1}(G)=1$. So existence of an ascending series implies the descending series terminates. Conversely, the descending series is also an ascending series if it actually has $1$ at its base, since its terms $G_i:=\gamma_{n+1-i}(G)$ satisfy the right-hand side of $(\circ)$.

Allow me first to change the notation in your first definition, just so as not to give us both a headache:

A

central seriesfor G is a normal series $G = G_1 \unrhd G_2 \unrhd \dots \unrhd G_n = \{1\} $ such that $G_i/G_{i+1}\subseteq Z(G/G_{i+1})$ for every $i=1,…,n-1$. An arbitrary group G is said to benilpotentif it has a central series.

Check I haven’t changed anything (modulo potential typos).

For one direction: show that, if the descending central series of $G$ terminates, then it is a central series (set $G_i = \gamma_i(G)$ and show that $G_i/G_{i+1}\subseteq Z(G/G_{i+1})$ using the definition of $\gamma$). (In spite of the name, the descending central series is only a central series when it terminates.)

For the other direction: suppose you have any central series $G_i$ (as above) for $G$. Show that the descending central series descends at least as fast as your central series, i.e. that $\gamma_i(G) \subseteq G_i$ (you may want to start with $i = 2$), and hence that the descending central series terminates. (We sometimes say that the descending central series is the *fastest descending* central series.)

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