Constructing all Semidirect Products $C_p \rtimes C_p$ for $p$ Prime

Question

I'd just like to check my understanding and to see if my thinking is right.

I'm tasked with constructing all semidirect products of $C_p$ by $C_p$ where $C_p$ is the cyclic group of prime order $p$.

Firstly, $C_p \cong \mathbb{Z} /p\mathbb{Z}$, and $Aut(\mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$. To construct all semidirect products, I then simply need to consider all homomorphisms $\tau: \mathbb{Z}_p \to Aut(\mathbb{Z}_p)$. However, these are cyclic groups of coprime order, and so the only homomorphism is the trivial one, $x \mapsto 0$, and so $\mathbb{Z} /p\mathbb{Z} \rtimes \mathbb{Z} /p\mathbb{Z} \cong \mathbb{Z} /p\mathbb{Z} \times \mathbb{Z} /p\mathbb{Z}$, where the latter is the direct product.

Thanks.

Answer 1

Yes, this is correct. ${}{}{}$

Answer 2

To fill in on Jyrki's comment, let $A$ be a group of order $p^2$, and let us show it is abelian. To this end, recall that any $p$-group has nontrivial center, and also recall that the center-quotient $G/Z$ of any non-abelian group cannot be cyclic. Since any group of order $p$ is cyclic, any group of order $p^2$ is thus abelian. Remark that every group of order $p$ is abelian, but the Heisenberg group (of parameter $p$) is an example of a non-abelian group of order $p^3$.