Why do we say the group of order $p^3$ is not always abelian? For any integer $n$, there is a cyclic group $\Bbb Z(n)$.
So for any integer $n$ there is always an abelian group $G$.
So why we say the group of order $p^3$ is not always abelian?
 A: Yes, we can speak about "the" cyclic group of order $p^3$ (up to isomorphism). However, this does not mean that there aren't other, non-cyclic groups of that same order.
Since there's more than one isomorphism class of groups of order $p^3$, there is no single  "the" group of order $p^3$ to talk about.
A: EDIT
You can't speak about the group of order $n$, because given $n\in\Bbb N$ there can be more than one group of that order.
If $n=p$ is prime, then (up to isomorphism) there exists exactly one group: $C_p$ (or $\Bbb Z_p$ if you prefer), which is the cyclic group of order $p$ (it's easy to prove).
In other cases it's often difficult to know how many non isomorphic groups of a given order $n$ there exist.
However the case $n=p^3$ is completely solved: there is a classification of finite groups of order $p^3$ (not only of order $8$).
Three of them are abelian: $C_{p^3}$, $C_{p^2}\times C_p$ and $C_p\times C_p\times C_p$.
The remaining two are not abelian (they are expressed as semidirect product).
If $n=8=2^3$ the non abelian groups are the quaternion group $Q_8$ and the dihedral group of the square $D_4$.
