p-groups such that all subgroups are normal must be abelian

Question

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!

Answer 1

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.

Answer 2

Here's an alternative solution that isn't the most efficient, but which you might nevertheless enjoy reading:

Theorem: Let $p$ be an odd prime, and let $G$ be a $p$-group, whose every subgroup is normal. Then $G$ is abelian.

Proof:

The proof is by induction on $|G|$. If $G=1$, we have nothing to prove, so suppose $G>1$.

Let $x\in G$ have order $p$. Let $Z=\langle x\rangle$. Since $Z\triangleleft G$ by assumption, by a standard result on $p$-groups, $Z\cap\mathbf{Z}(G)>1$. But $Z$ has order $p$, so this forces $Z\subseteq\mathbf{Z}(G)$.

By the inductive hypothesis, $G/Z$ is abelian. This means that $G'\subseteq Z$. Clearly we may assume that $G'>1$. Since $G'\subseteq Z$ and $Z$ has order $p$, this must mean that $G'=Z$.

Since we chose $Z$ to be an arbitrary subgroup of order $p$, $G'$ must be the unique subgroup of order $p$ in $G$. (Edit: In particular, every element of order $p$ in $G$ must be contained in $G'$).

Since $G'$ is central, $G$ has nilpotence class 2. In particular, the following two facts are true for any $x,y,z\in G$:

1. $[xy,z]=[x,z][y,z]$
2. $(xy)^n=x^ny^n[x,y]^{n(n-1)/2}$

Since $p$ is odd, $p(p-1)/2$ is divisible by $p$. Since $[x,y]\in G'$ has order at most $p$, we must have that for any pair of elements, $[x,y]^{p(p-1)/2}=1$.

Hence, for any pair of elements $x,y\in G$, $(xy)^p=x^py^p$.

The map $x\mapsto x^p$ is thus a homomorphism. The kernel consists of elements of order at most $p$, which we have shown earlier must all be contained in $G'$. Since Cauchy's theorem tells us that the kernel cannot be trivial, the kernel must be precisely $G'$.

Let $G^p=\{x^p | x\in G\}$. The map $x\mapsto x^p$ is a surjective homomorphism from $G$ onto $G^p$, and the kernel of this map is $G'$. We have shown that $G/G'\cong G^p$, and thus $|G/G^p|=|G'|=p$.

Since $G/G^p$ has prime order, it must be cyclic. To conclude that $G$ is abelian, it suffices to show that $G^p$ is central.

Let $x,y\in G$ be any pair of elements. Since the map $[\cdot, y]$ is a homomorphism (Fact 1 listed above),

$[x^p,y]=[x,y]^p=1$

where the final equality holds because $|G'|=p$.

We have now shown that $G^p$ is central in $G$. Since $G/G^p$ is cyclic, we may conclude that $G$ is abelian, which contradicts the assumption that $G'>1$. $\blacksquare$