Problem: If $G$ is a nonabelian group of order $p^3$ ($p$ prime), then the center of $G$ is the subgroup generated by all elements of the form $aba^{-1} b^{-1}$ ($a,b\in G$).
I know that $Z(G)$ must have order $p$ for $G$ to be nonabelian. I'm working with the equation for finite subgroups trying to reach a contradiction but I don't see it yet.
\begin{equation} |[G,G]Z(G)| = \frac{|[G,G]|\cdot|Z(G)|}{|[G,G]\cap Z(G)|} \end{equation}
Any group of order $p^2$ is abelian. Also, the center of a $p $-group is nontrivial. So the center has order $p $ (otherwise it has order $p^2$, the quotient is cyclic of order $p $, and this implies that the group is abelian).
Then $G/Z (G) $ is abelian. Thus $[G,G]\le Z (G) $. Thus $[G,G]=Z (G) $.
For the sake of future readers, I'd like to add some clarification on the solution which has been posted before.
First, note that the center of a $p$-group is nontrivial. Hence, the center is of order either $p$ or $p^2$. $|Z(G)|=p^2$ is impossible because $|Z(G)|=p^2$ implies that $|\frac{G}{Z(G)}|=p$, and hence $\frac{G}{Z(G)}$ is cyclic. Therefore, $G$ is abelian, which is against our assumption.
Therefore, $|Z(G)|=p$, and $|\frac{G}{Z(G)}|=p^2$. Since any group of order $p^2$ is abelian , we conclude that $\frac{G}{Z(G)}$ is abelian. Now, for all $x,y \in G$, we have: $(xZ(G))(yZ(G))=(yZ(G))(xZ(G))$. So, $$1Z(G)=(xZ(G))^{-1}(yZ(G))^{-1}(xZ(G))(yZ(G)) \\ =x^{-1}y^{-1}xyZ(G)=[x,y]Z(G).$$
Thus, $[x,y]\in Z(G)$ for all $x,y\in G$, so $[G,G] \leq Z(G)$. However, note that $Z(G)$ is a group of order $p$ whose only subgroups are $e$ and $Z(G)$. Since $[G,G]$ is nontrivial (if it were trivial, $G$ would have been abelian), we must have $[G,G]=Z(G).$
