# Non-trivial nilpotent group has non-trivial center

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!

 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!

• Here is the proof of a stronger result math.stackexchange.com/questions/127001/… – benh Dec 23 '13 at 2:56
• @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 3:57
• 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
• @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

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.

• 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
• @MaryStar - no! The primes $p_i$/Sylow $p_i$-subgroups are different. – Nicky Hekster Mar 28 '16 at 18:44
• 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