There exists only two groups of order $p^2$ up to isomorphism. I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$. 
Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.
Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on?

A proof of that could be:
The center of a group is a subgroup, so it's order must divide $p^2$, but it's a known fact that if a group has order $p^m$, with $p$ prime, then the center of the group is different from $p^{m-1}$ and different from $1$, so in our case, the center has order $p^2$, so it's abelian. 
That's what you already have, know, by the theorem of structure of finite abelian groups, the group can be either $\mathbb{Z}_p\times\mathbb{Z}_p$ or $\mathbb{Z}_{p^2}$. But from that same theorem, one can deduce that a group $\mathbb{Z}_m\times\mathbb{Z}_n$ is isomorphic to a group $\mathbb{Z}_{nm}$ iff $\gcd(m,n)=1$, so in our case, those two groups are not isomorphic, and there are only two groups of order $p^2$.
