# p-groups such that all subgroups are normal must be 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. [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!

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 $$i, and let $$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.