I've been asked to show that if $G$ is a non-abelian group of order $p^3$, then the derived or commutator subgroup is equal to the centre, $G'$=$Z(G)$. I know that since $G'$ is a subgroup of $G$, it must have order $1$, $p$, $p^2$ or $p^3$. If I can show that |$G'$| = $p^2$, then it is abelian and a maximal proper subgroup, so it must be the centre of $G$. However I'm unsure of how to go about proving this. Thanks for your help.
Let $p$ be a prime number and $G$ a non abelian group or order $p^3$. Prove that $Z(G) = [G,G]$.   I have already figured out that $|Z(G)| = p$, and that $G'=[G,G] \lhd G$.  Also, $|G'| = p$ or $p²$.  I suppose I'd have to prove that $G' \subset Z(G)$, but I've been trying and I have no idea how. Any hints? Thanks!
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