In a nonabelian p-group of order p³, the commutator subgroup and center are equal

Prove that if p is a prime and P a nonabelian group of order p^3, then P^\prime = Z(P).


Since P is nonabelian, we have Z(P) \neq P. Moreover, if |Z(P)| = p^2, then P/Z(P) \cong Z_p is cyclic, so that P is abelian; thus |Z(P)| \neq p^2. Since every p-group has a nontrivial center, we have |Z(P)| = p by Lagrange.

Now |P/Z(P)| = p^2; we know that every group of order p^2 is abelian, thus we have P^\prime \leq Z(P). Since P is nonabelian, P^\prime \neq 1; thus |P^\prime| \geq p, and we have P^\prime = Z(P).

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