TITLE: Non-trivial nilpotent group has non-trivial center
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: 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.