How to show that $Z_{p^2}\ncong Z_p\rtimes Z_p$, where $p$ is a prime?

Question

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?

Answer 1

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 $p$.

Answer 2

To talk about a semi-direct product, you have to describe how one group acts on the other. Why don't you begin by computing the possible ways that $\mathbb Z_p$ can act as automorphisms of itself? Then you might have a better shot of proving your step 2.