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!
Best Answer
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
$$\gamma_n:=[\gamma_{n-1},G]=1\iff \forall\,x\in\gamma_{n-1}\;\;and\;\;\forall\,g\in G\;,\;\;x^{-1}g^{-1}xg=1\iff xg=gx\implies$$
$$\implies \gamma_{n-1}\le Z(G)\implies Z(G)\neq 1$$