A book I'm reading quotes the following result without any explanation:

Any non-trivial nilpotent group has a non-trivial center.

(The definition of "nilpotent group" is as follows: Suppose $G$ is a group, define $G^{(1)}=[G,G]$ to the commutator subgroup, and recurrsively define $G^{(m)}=[G^{(m-1)},G^{(m-1)}]$. A group $G$ is said to be nilpotent if $G^{(m)}=0$ for sufficiently large $m$.)

The group in the claim does not have to be finite. I have thought about this claim for a while and it doesn't seem easy. Could you please help me? Thank you very much!

[Edit] As pointed out by DonAntonio, the definition of "nilpotent group" given here is not correct. The correct definition is that if we define $\gamma^n=[\gamma^{n-1},G]$ then $G$ is nilpotent if and only if $\gamma^n=0$ for sufficiently large $n$. Now the conclusion follows easily. Thank you for your help!

  • $\begingroup$ Here is the proof of a stronger result math.stackexchange.com/questions/127001/… $\endgroup$ – benh Dec 23 '13 at 2:56
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    $\begingroup$ @Boyu Zhang, Your definition of Nilpotent Group fits, in fact, to "solvable (or soluble) group": what you've defined there is the commutator or derived series. For nilpotent you need a central series, and whatis closest to what you wrote is the lower central series, defined : $$\gamma_1:=G, \gamma_2:=[\gamma_1,G]=G'\;...\;\gamma_n:[\gamma_{n-1},G]$$ and now yes: a group is nilpotent iff $\;\gamma_k=1\;$ for some finite $\;k\;$ $\endgroup$ – DonAntonio Dec 23 '13 at 3:57
  • $\begingroup$ I've removed the tag "geometric group theory" as this is just plain-vanilla group theory, not geometric. $\endgroup$ – user1729 Dec 23 '13 at 10:49
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    $\begingroup$ @DonAntonio: Oh, I misunderstood the definition. Thank you for correcting me! I have already revised my problem. Now this problem is trivial... $\endgroup$ – Boyu Zhang Dec 23 '13 at 15:34
  • $\begingroup$ @BoyuZhang , any time. $\endgroup$ – DonAntonio Dec 23 '13 at 15:35

Suppose $\;\gamma_n=1\;$ but $\;\gamma_{n-1}\neq 1\;$ (according to my definition, the correct one, and thus $\;G\;$ is of class $\;n\;$), then

$$\gamma_n:=[\gamma_{n-1},G]=1\iff \forall\,x\in\gamma_{n-1}\;\;and\;\;\forall\,g\in G\;,\;\;x^{-1}g^{-1}xg=1\iff xg=gx\implies$$

$$\implies \gamma_{n-1}\le Z(G)\implies Z(G)\neq 1$$


In case of finite groups: one can prove that finite nilpotent groups are precisely those groups that are a (internal) direct product of their Sylow subgroups: $G \cong P_1 \times \dots \times P_n$. Hence looking at the centers: $Z(G) \cong Z(P_1) \times \dots \times Z(P_n)$ and it is well-known that centers of $p$-groups are non-trivial.

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    $\begingroup$ But he specifically said that the group need not be finite. With the standard definition of a nilpotent group using central series, the nontriviality of the centre follows immediately from the definition, so it seems strange to bring in Sylow subgroups. $\endgroup$ – Derek Holt Dec 23 '13 at 14:52
  • $\begingroup$ Derek, I totally agree, but wanted to provide another angle at this. $\endgroup$ – Nicky Hekster Dec 23 '13 at 17:51
  • $\begingroup$ Hello @NickyHekster !! When we have that $G \cong P_1 \times \dots \times P_n$ does it mean that $G$ is a $p$-group? $\endgroup$ – Mary Star Mar 28 '16 at 12:55
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    $\begingroup$ @MaryStar - no! The primes $p_i$/Sylow $p_i$-subgroups are different. $\endgroup$ – Nicky Hekster Mar 28 '16 at 18:44
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    $\begingroup$ If $N$ is non-trivial then $N \cap P_i \neq 1$ for some $i$.But $N \cap P_i$ is normal in $P_i$, so $N \cap P_i \cap Z(P_i) = N \cap Z(P_i) \neq 1$. Hence $N \cap Z(G) \neq 1$. $\endgroup$ – Nicky Hekster Mar 28 '16 at 21:33

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