## 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.