Let pp be a prime and let GG be a finite of order pnp^n, where n≥1n\ge 1. Then the Z(G)Z(G) is nontrivial. More precisely,

p∣∣Z(G)∣.p\mid |Z(G)|.
Examples
  • If GG is abelian, then Z(G)=GZ(G)=G.
  • For D8=⟨r,s∣r4=s2=e, srs=r−1⟩D_8=\langle r,s\mid r^4=s^2=e,\ srs=r^{-1}\rangle, one has Z(D8)={e,r2}Z(D_8)=\{e,r^2\}.
Remarks

This follows from the : every noncentral has cardinality divisible by pp, so ∣Z(G)∣≡∣G∣≡0(modp)|Z(G)|\equiv |G|\equiv 0\pmod p.