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For $p$ an odd prime, I have seen proofs providing the classification of nonabelian groups of order $p^3$ as being isomorphic to either the Heisenberg group $Heis(\mathbb{Z}_p)$ or to the group $$\left\{ \begin{bmatrix}a & b\\ 0 & 1\end{bmatrix}:a,b\in \mathbb{Z}_{p^2},\; a\equiv 1\pmod{p} \right\}.$$

The proofs involved some clever reasoning with generators, as hinted in Hungerford's exercise of classifying order $p^3$ groups.
I attempted a different method.

Let $G$ be an order $p^3$ group and let $H$ be any subgroup of $G$ of order $p^2$, which is necessarily normal.
Take any $x\notin H$ and let $K=\langle x \rangle$.

Now, either $H\cong \mathbb{Z}_{p^2}$ or $H\cong \mathbb{Z}_{p} \times \mathbb{Z}_{p}$. Moreover, we have $K\cong \mathbb{Z}_{p}$ or $K\cong \mathbb{Z}_{p^2}$.

Let's first consider $H\cong \mathbb{Z}_{p^2}$.
If $K\cong \mathbb{Z}_{p}$, we can define two semidirect products through homomorphisms $\phi:K \rightarrow Aut(\mathbb{Z}_{p^2}) \cong \mathbb{Z}_{p(p-1)} $.
If $\phi$ were trivial, $G$ would be the abelian group $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p}$.
If $\phi$ were nontrivial, then $x$ conjugates the elements of $H$ by acting as an order $p$ automorphism of $H$.

Let's now consider $K\cong \mathbb{Z}_{p^2}$ so that $H\cap K \cong \mathbb{Z}_p$ and $H$ is normal in $G$.
Though $G=HK= H \vee K$ is no longer a semidirect product of $H$ and $K$, I thought that I could still reason out as I did prior.
Indeed, if $x$ acts trivially on the elements of $K$ under conjugation, $G=HK$ would be abelian.
Otherwise, $x$ conjugates the elements of $H$ by acting as an order $p$ automorphism $\phi$ of $H$.
In this nonabelian case, $\ker\phi = H\cap K$, being the unique order $p$ subgroup of $K$.

I thought that since each element of $HK$ can be written $|H\cap K|$ times and since $H\cap K$ is also $\ker \phi$, that $G=HK$ is isomorphic to the nonabelian case when $K\cong\mathbb{Z}_p$. Loosely, I thought that the action of $x$ on the elements of $H$ matches that in the case where $K\cong \mathbb{Z}_p$, modded out by the elements of $H\cap K$. However, I got stuck. Is this a feasible approach? How would I show that the two cases are (or are not) isomorphic?

F.Tomas
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    Instead of taking "any" $x \not\in H$ where $|H| = p^2$ and then worrying about what to do if this $x$ has order $p^2$, prove there's an element order $p$ outside $H$ and take $x$ to be such an element. Then $|H| = p^2$, $|K| = p$, and $G = HK$ with $H \lhd G$, so you have a semidirect product. And at that point you need to figure out how to show there are just two possible semidirect product structures, up to isomorphism. At some point you need to really use the condition $p \not= 2$, because $Q_8$ is not a semidirect product of nontrivial subgroups. – KCd Apr 14 '23 at 18:33
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    See also this post. – Dietrich Burde Apr 15 '23 at 13:49

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Your requirement on $H$ is too weak: it is allowed to be any subgroup of $G$ with order $p^2$. You need some element of order $p$ outside $H$ in order to build a semidirect product, but some $G$ have a subgroup of order $p^2$ containing all the elements of $G$ with order $p$: for your second group, consisting of mod $p^2$ matrices $$ \begin{pmatrix} a&b\\0&1 \end{pmatrix} $$ where $a \equiv 1 \bmod p$ and $b$ is anything mod $p^2$, the subgroup $H$ consisting of such matrices where $b \equiv 0 \bmod p$ (so $a$ and $b$ each have $p$ values) contains all the elements of $G$ with order $p$. So you do not want to use this $H$ as your subgroup of order $p^2$.

In Section 8 here, nonabelian groups of order $p^3$ are shown to be isomorphic to two possible semidirect products of nontrivial groups after first showing how the two concrete models of nonabelian groups of order $p^3$ both occur as semidirect products.

KCd
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