For odd primes p, a p-group whose every subgroup is normal is abelian

Let p be an odd prime and let P be a p-group. Prove that if every subgroup of P is normal then P is abelian. (Note that Q_8 is a nonabelian 2-group with this property, so the result fails for p=2.)

[Disclaimer: I looked at Alfonso Gracia-Saz’s notes when solving this problem.]

We proceed by induction on k such that |P| = p^k.

For the base case, if k = 1, then P is cyclic and thus abelian.

For the inductive step, suppose the conclusion holds for all P such that |P| = p^k, where k \geq 1. Let P be a group of order p^{k+1}. If P is cyclic, then P is abelian. If P is not cyclic, then by the previous exercise there exists a normal subgroup U \leq P with U \cong Z_p \times Z_p. Choose x,y \in U such that U = \langle x \rangle \times \langle y \rangle. By the hypothesis, \langle x \rangle, \langle y \rangle \leq P are normal. Now P/\langle x \rangle and P/\langle y \rangle are groups of order p^k, and by the Lattice Isomorphism Theorem and the induction hypothesis, every subgroup of P/\langle x \rangle and P/\langle y \rangle is normal; thus P/\langle x \rangle and P/\langle y \rangle are abelian. By this previous exercise, P/(\langle x \rangle \cap \langle y \rangle) \cong P is abelian.

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