How to show that $Z_{p^2}\ncong Z_p\rtimes Z_p$, where $p$ is a prime? My thoughts:
1.First show that $Z_{p^2}\ncong Z_p\times Z_p$.
2.Recall that group of order $p^2$ is isomorphic to either $Z_{p^2}$ or $Z_p\times Z_p$, so show that $Z_p\rtimes Z_p\cong Z_p\times Z_p$. 
But I am stuck on the second step, how should I continue? And is there a better way to solve this?

There is an element of order $p^2$ in the group $Z_{p^2}$. To get an element of order (divisible by) $p^2$  in the semi-direct product, you need an automorphism of order a multiple of $p$ on the group $Z_p$. But all automorphisms fix the identity, so you can't get the second $p$.
