If $G$ is a finite $p$-group, and $|G':G''| \le p^2$, then $G'$ is an abelian group.
I'm reading its proof but I cannot understand a part: Suppose $G''\neq1$, then by a theorem, there exists a normal subgroup $K$ of $G$ contained as a subgroup of index $p$ in $G''$. Then $(G/K)'=G'/K$ and $(G/K)''=G''/K \neq 1$. So by another theorem, the center $Z/K$ of $G'/K$ is not cyclic, and so $|Z:K| \ge p^2$. "" Hence $|G':Z| \le p$ "", and so $G'/K/Z/K$ is cyclic. Thus $G'/K$ is abelian, contradiction.
Why is it true that $|G':Z| \le p$?
cf. The above proof looks simple but uses other theorems a lot. Does anyone know another good proof?