Showing order $p^3$ groups are regular for odd $p$

Question

As the title indicates, I'm trying to prove that, when $o(G) = p^3$ for $p$ an odd prime, $G$ must be regular, i.e. for any $a, b \in G$, $(ab)^p = a^pb^pc^p$ for some $c \in \langle a, b \rangle^1$. I am aware there is a classification theorem for order $p^3$ groups, but I would rather not route through it if possible since I don't think it's the pedagogical function of the exercise as it appears in my text (Scott's Group Theory).

Here is where I am so far:

We want to show for a given $a, b \in G$ that $\exists c \in \langle a, b \rangle^1 : (ab)^p = a^pb^pc^p$. If $a$ and $b$ commute, the problem is obviously easy, so let us assume not. We can have $o(\langle a, b \rangle) = 1$, $= p$, $= p^2$, or $= p^3$.

• If $o(\langle a, b \rangle) = 1$, it is trivial.
• If $o(\langle a, b \rangle) = p$, then $\langle a, b \rangle$ is cyclical, and thus Abelian, and thus $a$ and $b$ commute and we are done.
• If $o(\langle a, b \rangle) = p^2$, then we distinguish further by cases:
  • If $o(ab) = p$, then $(ab)^p = e$ and so it is trivial.
  • If $o(ab) = p^2$, then $o(\langle a, b \rangle) = o(ab)$, so $\langle a, b \rangle = \langle ab \rangle$. Since $\langle ab \rangle$ is Abelian, so is $\langle a, b \rangle$, and we are done.
• If $o(\langle a, b \rangle) = p^3$, then we distinguish again by cases:
  • If $o(ab) = p$, then we are done.
  • If $o(ab) = p^3$, then by similar reasoning as before $\langle a, b \rangle$ is Abelian and we are done.

Thus, the only case we need consider is where $o(\langle a, b \rangle) = p^3$ (thus $\langle a, b \rangle = G$) and $o(\langle ab \rangle) = p^2$. If $o(G^1) = 1$ then $G$ Abelian and we are done. As $G$ is nilpotent, we have $G^1 \subset Fr(G) < G$, where $Fr(G)$ is the Frattini subgroup of $G$. Thus $1 < o(G^1) < p^3$, i.e. $o(G^1) = p$ or $= p^2$.

Suppose $o(G^1) = p^2$. Then $G^1 = Fr(G)$, and $[G : Fr(G)]$ is prime so that $G/Fr(G)$ is cyclic, meaning G is cyclic and so Abelian and we are done.

This leaves only the case of $o(G^1) = p$. If $Fr(G) > G^1$, then again $G/Fr(G)$ is cyclic and thus G is cyclic, contrary to supposition. Thus $Fr(G) = G^1 \subset \langle ab \rangle$. I can see that, for any $c \in G^1$, $c^p = e$ and thus $(ab)^p = a^pb^pc^p$ iff $(ab)^p = a^pb^p$. But beyond that I am lost.

Answer 1

Your approach may well be completed, but there is a very simple and transparent route that avoids all of the exhausting case analysis – and it actually yields a stronger statement as a by-product.

Theorem. Let $G$ be a group of order $p^{3}$ with $p$ an odd prime. Then for all $a,b\in G$, $$ (ab)^{p}=a^{p}b^{p}. $$ In particular, $G$ is a regular $p$-group.

Here are the main steps of the proof. Given their simplicity, I think they may be left to the reader as exercises. Since there is nothing to prove in the abelian case, we may restrict our attention to non-abelian groups.

1. If $G$ is a non-abelian group of order $p^{3}$ with $p$ an prime, then $ |Z(G)| = p. $

2. If $G$ is a non-abelian group of order $p^{3}$, then $$ G' \subseteq Z(G)\quad\text{and}\quad |G'| = p. $$

3. If $G$ has class $2$ (that is, the quotient group (G/Z(G)) is abelian), then for all $a,b\in G$ and every integer $n\ge 1$, $$ (ab)^{n} = a^{n} b^{n} [b,a]^{\binom{n}{2}}. $$ This formula is most easily proved by induction on $n$

4. For $n=p$ odd, $$ (ab)^{p} = a^{p} b^{p} [b,a]^{p(p-1)/2}=a^{p} b^{p} ([b,a]^p)^{(p-1)/2}=a^{p} b^{p}. $$

Thus, $$ (ab)^{p} = a^{p} b^{p}. $$

Edit.
This gives the inductive step in the proof of the third point. By repeatedly using the inductive hypothesis, the fact that $[b,a]=b^{-1}a^{-1}ba$ lies in the center of the group, and the identity $b^{n} a = a b^{n} [b,a]^{n}$ (which can itself be proved by induction on $n$), we obtain \begin{align*} (ab)^{n+1} &= (ab)^n (ab) \\ &= a^n b^n [b,a]^{\binom{n}{2}}\, a b \\ &= a^n b^n a b\, [b,a]^{\binom{n}{2}} \\ &= a^{n+1} b^{n+1} [b,a]^n [b,a]^{\binom{n}{2}} \\ &= a^{n+1} b^{n+1} [b,a]^{\binom{n}{2} + n} \\ &= a^{n+1} b^{n+1} [b,a]^{\binom{n+1}{2}}. \end{align*}