How to show that for $p$ prime, every group of order $p^5$ is metabelian? A group is metabelian if and only if its commutator subgroup $G'$ is abelian.
If $G$ is a group of order $p^5$, then $G$ has a normal subgroup $N$ of order $p^3$ (in $p$-groups, there is a normal subgroup of every possible order). Then $G/N$ is abelian and thus $G' \leq N$. Therefore the commutator subgroup has order $1$, $p$, $p^2$ or $p^3$. The only case where $G'$ could be nonabelian is when it has order $p^3$. This actually happens when $G = D_{32}$, the dihedral group of order $32$, so proving that $|G'| = p^3$ is impossible is not the way to go.
Any ideas where to go from here?