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?

If the (already known to be abelian) group $H$ in question is not cyclic, pick any nonzero element $a$. Its order must be $p$, so $\langle a\rangle$ is a subgroup of order $p$. Finally, consider $H/\langle a\rangle$, or, alternatively, pick another arbitrary element $b$ from $H\setminus\langle a\rangle$, and aim to prove that $\langle a\rangle \cap \langle b\rangle=\{ e\}$ and $\langle a,b\rangle=H$.
