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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. [Use the preceding exercise and exercise 15 of section 4]

The previous exercise was to prove that for an odd prime $p$ and $P$ a $p$-group that is not cyclic then $P$ contains a normal subgroup $U$ such that $U \cong \mathbb{Z}/p \times \mathbb{Z}/p$.

Exercise 15 says that if $A$ and $B$ are two normal subgroups of a group $G$ such that $G/A$ and $G/B$ are abelian then $G/(A \cap B)$ is abelian.

Clearly if $P$ is cyclic then $P$ is abelian so we may assume that it is non-cyclic and hence by the previous exercise must have a subgroup $U \cong \mathbb{Z}/p \times \mathbb{Z}/p$.

My idea is to try to get two subgroups that are disjoint to each other and such that their respective quotients are abelian as then $G/\{e\} \cong G$ would be abelian. I am unsure how to produce two groups like that, any suggestions are appreciated, thanks!

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Proceed by induction on $n$ in the statement "if $P$ has order $p^n$ and every subgroup of $P$ is normal, then $P$ is abelian."

The base case is trivial since groups of order $p$ are cyclic.

Suppose the statement holds for groups of order $p^i$ where $i<n$, and let $P$ be a group of order $p^n$. If $P$ is cyclic, we are done. Otherwise $P$ has a normal subgroup $U\cong \Bbb Z/p\times \Bbb Z/p$. Within $U$ we have two subgroups $H_1$ and $H_2$ corresponding to $\Bbb Z/p\times\{0\}$ and $\{0\}\times \Bbb Z/p$. By assumption, $H_1$ and $H_2$ are normal, so we can consider the groups $P/H_1$ and $P/H_2$. Now any subgroup $K$ of $P/H_1$ corresponds to a subgroup $K'$ of $P$ containing $H_1$. By assumption $K'$ is normal in $P$, so the correspondence theorem says $K$ is normal in $P/H_1$. This shows all subgroups of $P/H_1$ are normal, so $P/H_1$, and similarly $P/H_2$ is abelian by the induction hypothesis.

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