Let $N=<x,y>$ and $|x|=|y|=p$ for some prime $p$. Assume aswell that $xy=yx$ so N is Abelian, then the fundamental theorem of finitely generated abelian groups tells us that either $N \simeq \mathbb{Z}_{p^2}$ or $N \simeq \mathbb{Z}_{p} \times \mathbb{Z}_{p} $. Both options have order $p^2=|N|$ but since $gcd(p,p)\neq 1$ the Chinese remainder theorem tells us that $\mathbb{Z}_{p^2} \not\simeq \mathbb{Z}_{p} \times \mathbb{Z}_{p}$ .
I am supposed to find that the second option, $N \simeq \mathbb{Z}_{p} \times \mathbb{Z}_{p} $, is the case, but not sure what singles that one out?