Question

Let G be a nontrivial nilpotent group. Prove that G has nontrivial center.

Answer 1

a group G is called nilpotent if it admits a normal series

{e} = G_{​​​​​​0} ⊲ G_{​​​​​​1} ⊲ G_{​​​​​​2} ⊲ · · ·⊲ G_{​​​​​​r} = G
in which G_{​​​​​​i} ⊲ G and G_{​​​​​​i+1}/G_i ⊂ Z(G/G_{​​​​​​i}) for all i.

A minimal normal subgroup is a nontrivial normal subgroup that contains no other nontrivial normal subgroup. A maximal subgroup is a proper subgroup not contained in any other proper subgroup. Nontrivial finite groups obviously have minimal normal subgroups and maximal subgroups. But an infinite abelian (hence nilpotent and solvable) group need not contain minimal normal subgroups (try Z)
or maximal subgroups (for example Q).

Proof of the main theorem:

Let N be a minimal normal subgroup of nilpotent group G and Z be the center. Since N is nontrivial and normal, N ∩ Z is nontrivial by nilpotency of G. (Since if N  ∩ Z is trivial then then N must be trivial witch is contradiction as N is nontrivial).

Now since N ∩ Z is a normal subgroup of G, by minimality of N we have N = N ∩ Z , so N ⊂ Z.

Hence center of the nontrivial nilpotent group G has nontrivial center.