Derived subgroup of a finite non-Abelian p-group is proper?

Question

How do I show that the derived subgroup of a finite p-group is always proper? In Abelian groups, it's trivial. In non-Abelian groups, my intuition is that there should be some way to relate $G/Z(G)$ to $G'$, but I have no idea how.

edit: finite

Answer 1

A finite $p$-group $G$ has non trivial center $Z(G)$; this may be proved with the class equation. Since the center is characteristic, $G/Z(G)$ has non trivial center and we can define the ascending center series: $$ \{1\}\subset Z(G)=Z_1(G)\subset Z_2(G)\subset\dotsb $$ For some $n$ we must have $Z_n(G)=G$; choose $n$ minimal. Then $G'\subseteq Z_{n-1}(G)$ because, by definition, $Z_n(G)/Z_{n-1}(G)$ is abelian.