I'm trying to solve this problem from my abstract algebra text book:
Being $G$ a non-abelian group of order $p^3$, with $p$ prime. Prove that $G'=Z(G)$
In my notation $G'$ is the derived subgroup of $G$. I guess I have to prove separatedly that $G'\subseteq Z(G)$ and $G'\supseteq Z(G)$. My problem is that I don't know how to use the condition $|G|=p^3$. Maybe the Sylow theorems? The first thing I noticed from $|G|$ being $p^3$ is that then $Z(G)$ isn't trivial, but I don't know how to continue. Any help will be appreciated, thanks in advance.
First of all, there are $2$ standard results which I will use here: (both very easy to prove)
$1$. If $G/Z(G)$ is cyclic then $G$ is abelian.
$2$. If $p$ is a prime then groups of order $p$ and $p^2$ are abelian.
So let $G$ be a non abelian group of order $p^3$. Then $|Z(G)|=p$. Indeed, if we had $|Z(G)|=p^2$ then $G/Z(G)$ would be cyclic and then $G$ would be abelian, a contradiction. So $|Z(G)|=p$, and so $|G/Z(G)|=p^2$. Thus $G/Z(G)$ is abelian, and this implies $G'\leq Z(G)$. By Lagrange's theorem it follows that either $G'=\{e\}$ or $G=Z(G)$. Since $G$ is not abelian it can't be the first option. So $G'=Z(G)$.