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This is not a duplicate of this post, neither of this: they don't give an explicit description of these groups, but only some of their properties.

Using GAP to find all the non-abelian groups of order $3^3,5^3,7^3,11^3$ I found that there were in each case only two groups, namely those of the form $(C_p \times C_p) \rtimes C_p$ and $C_{p^2} \rtimes C_p$ (I hope I have the symbol $\rtimes$ right). It is my intention to prove that this is always the case for all odd primes $p$. My (incomplete) proof for a group $G$ of order $p^3$ is based on the following points:

  1. The center $Z$ of $G$ is $C_p$: Let $g \in G$ be such that $\langle g \rangle\cap Z =\{1\}$ then the group generated by $Z$ and $g$ is the direct product of $Z$ and $\langle g \rangle$, which is abelian. If the order of $Z$ were $p^2$ then $Z$ and $g$ would generate the whole group $G$ and thus $G$ would be abelian, so $Z$ has order $p$.
  2. There are two cases. G has an element of order $p^2$ or all the elements of $G \setminus \{1\}$ have order $p$. We will discuss the former case first. Let $g$ be an element of order $p^2$. If $\langle g \rangle \cap Z = \{1\}$ then $\langle g,Z \rangle = G$, which makes it abelian so $Z \subset \langle g \rangle$. Now $\langle g \rangle/Z$ is normal in $G/Z$ so $\langle g \rangle$ is normal in $G$. The automorphism group $A$ of $\langle g \rangle \cong C_{p^2}$ is cyclic of order $(p-1)p$ and contains $p-1$ elements of order $p$ so one can construct at least one non trivial map $G/\langle g \rangle \cong C_p: \rightarrow A$ to be used to define the semidirect product $C_{p^2} \rtimes C_p$. I still have to prove that all $p-1$ possible maps give rise to ismorphic groups.
  3. The case where all non identity elements have order $p$. Let $H_i$ be a group generated by $Z$ and an element $g_i$ of order $p$ not in $Z$ then $H_i$ and $H_j$ either coincide or their intersection is $Z$. Indeed let $g \in H_i \cap H_j$ and $g \notin Z$ then $g$ generates $H_i$ but also $H_j$ so they coincide.
  4. The set of groups $H_i$ has $p+1$ elements: $\#(G \setminus Z) = p^3-p$ and $\#(H_i \setminus Z) = p^2-p$; so there are $\frac{p^3-p}{p^2-p} = p+1$ groups $H_i$.
  5. At least one group $H = H_i$ is normal. $G$ acts on this set by conjugacy. Orbits have a size that is a divisor of $p^3$ so by $4.$ at least one of them has size $1$. Since $H \cong C_p \times C_p$ we have to find a non trivial map $G/H \cong C_p \rightarrow \operatorname{Aut}(C_p \times C_p)$ , but I have no idea how to obtain it.
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    Everything looks good so far. Small typo in (2.), you wrote $C_3$ instead of $C_p$. Also, the first sentence in (4.) is a bit confusing. Do you know Aut$(C_p\times C_p)$? As a hint, note that $C_p\times C_p$ can be realized as a vector space over the prime field. – 2017-02-03
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    @JMag: Indeed the automorphism group is $GL(2,p)$. I'm working on it. – 2017-02-06
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    If you want a reference, these two groups are known as extraspecial groups. – 2017-02-08
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    Yes ok please . Extraspecial, does that mean special among the spectial :-? – 2017-02-08
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    Actually, yes it does. There are special $p$-groups. The ones you are considering are extraspecial $p$-groups. Google searches can probably find something. Gorenstein's book Finite Groups has a whole subsection devoted to extraspecial groups, section 5.5. – 2017-02-08

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