Definition of a nilpotent group.

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 series for $$G$$ 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 be nilpotent if it has a central series.

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

The descending central series of a group $$G$$ is $$G = \gamma_1(G) \supseteq \gamma_2(G) \supseteq ...,$$

where $$\gamma_{i+1}(G) = [\gamma_i(G),G].$$ A group $$G$$ is called nilpotent if the lower central series reaches $$\{1\}$$; that is, if $$\gamma_n(G)=\{1\}$$ for some $$n$$.

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.

• The descending central series (oftens called the lower central series) is the central series that descends most rapidly. You might also like to look up the definition of the upper (or ascending) central series. This the one that ascends most rapidly, and gives rise to another equivalent definition of a nilpotent group. – Derek Holt Jul 18 '13 at 8:24
• Where the definition of nilpotent groups appears first in the literature? No English book (and some German book) gives a historical reference to the definition of nilpotent group. Zassenhaus mentions in his book (1937) that the name comes from "nilpotent Lie ring". Whereas, Hall mentions in his famous paper(1939) "classification of prime power groups" that the upper and lower series of a group (if they terminate) have same length, and he refers to the German book of A. Speiser (I have no copy of this book, so I didn't see further reference to definition of nilpotent group). – Beginner Jul 19 '13 at 7:10

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 series for 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 be nilpotent if 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.)