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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!

Boyu Zhang
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  • Here is the proof of a stronger result http://math.stackexchange.com/questions/127001/infinite-nilpotent-group-any-normal-subgroup-intersects-the-center-nontrivially – benh Dec 23 '13 at 02:56
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    @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 $;k;$ – DonAntonio Dec 23 '13 at 03:57
  • I've removed the tag "geometric group theory" as this is just plain-vanilla group theory, not geometric. – user1729 Dec 23 '13 at 10:49
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    @DonAntonio: Oh, I misunderstood the definition. Thank you for correcting me! I have already revised my problem. Now this problem is trivial... – Boyu Zhang Dec 23 '13 at 15:34
  • @BoyuZhang , any time. – DonAntonio Dec 23 '13 at 15:35

2 Answers2

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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

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

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

DonAntonio
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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.

Nicky Hekster
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    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. – Derek Holt Dec 23 '13 at 14:52
  • Derek, I totally agree, but wanted to provide another angle at this. – Nicky Hekster Dec 23 '13 at 17:51
  • Hello @NickyHekster !! When we have that $G \cong P_1 \times \dots \times P_n$ does it mean that $G$ is a $p$-group? – Mary Star Mar 28 '16 at 12:55
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    @MaryStar - no! The primes $p_i$/Sylow $p_i$-subgroups are different. – Nicky Hekster Mar 28 '16 at 18:44
  • Ah ok... Since $Z(P_i)\neq 1$, we have that $Z(G)\neq 1$. $$$$ For a $p$-group, say $P$, we have that $Z(P)\neq 1$ and if $N$ is a normal subgroup of $P$ then $N\cap Z(P)\neq 1$. Do we also have that $N\cap Z(G)\neq 1$ since $Z(G) \cong Z(P_1) \times \dots \times Z(P_n)$ ? @NickyHekster – Mary Star Mar 28 '16 at 19:54
  • Or can we not say anything about $N\cap Z(G)$ knowing that $N\cap Z(P_i)\neq 1$ ? @NickyHekster – Mary Star Mar 28 '16 at 21:13
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    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 $N \cap Z(G) \neq 1$. – Nicky Hekster Mar 28 '16 at 21:33
  • How exactly do we conlcude that $N\cap Z(G)\neq 1$ ? We have that $N\cap Z(G)=N\cap (Z(P_1)\times \cdots \times Z(P_n))$. Is this equal to $(N\cap Z(P_1))\times \cdots (N\cap Z(P_n))$ ? – Mary Star Mar 29 '16 at 10:13
  • Also, why is $N\cap P_i$ normal in $P_i$ ? @NickyHekster – Mary Star Mar 29 '16 at 10:22
  • @Mary, let us take the discussion off line in a chat room – Nicky Hekster Mar 29 '16 at 11:31