Problem 5.4.7 (If $|P| = p^3$, then $P' = Z(P)$)

Proof. We know that the center of $Z(P)$ is not trivial, thus $|Z(P)|=p$ or $|Z(P)|=p^2$ ( $|Z(P)|=p^3$ is impossible since $P$ is nonabelian).

If $|Z(P)|=p^2$, then $|P∕Z(P)|=p$, so $P∕Z(P)$ is cyclic. Then by Exercise 3.1.36, $P$ is abelian, in contradition with the hypothesis. Therefore

Since $Z(P)⊴P$ and $|P∕Z(P)|=p^2$ then $P∕Z(P)$ is abelian. By Proposition 7 (4),

But $Z(P)$ is cyclic of prime order, thus $P^{\prime }=\{1\}$ or $P^{\prime }=Z(P)$, and $P$ is not abelian, thus $P^{\prime }\ne \{1\}$. Therefore

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