Let $p$ be prime with $p > 2$.
(a) Determine the number of group homomorphisms between $\mathbb{Z}_p$ and $\mathbb{Z}_{2p}$
(b) Determine the number of ring homomorphisms between $\mathbb{Z}_p$ and $\mathbb{Z}_{2p}$
This is what i understand so far thank to everyone here. So for part (a), if $$\varphi: \mathbb{Z}_p \longrightarrow \mathbb{Z}_{2p}$$ is a homomorphism. I first think that $\mathbb{Z}_p$ is cyclic, so $\mathbb{Z}_p = \langle1\rangle$, but as a relation in $\mathbb{Z}_p$ we must have that $p\cdot1=0$. So to know all the homomorphism from $\mathbb{Z}_p \rightarrow \mathbb{Z}_{2p}$, I send the element $1 \in \mathbb{Z}_p $ to an element $x \in \mathbb{Z}_{2p}$, but the element $x$ must satisfy $p\cdot x=0$ in $\mathbb{Z}_{2p}$. In other words, we have that $p\cdot x=0$ mod $(2p)$. Now this condition implies that $2p \mid px$, but since $p$ is prime then $2 \nmid p \Rightarrow 2 \mid x$, therefore i can send the element $1 \in \mathbb{Z}_p $ to all the even numbers in $\mathbb{Z}_{2p}$, that is $0,2,4, \ldots , 2p-2$ and we thus have exactly $p$ homomorphism from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{2p}$.
Hint for (a): Let $\phi\colon\mathbb Z_p \to \mathbb Z_{2p}$ be a homomorphism. As $p = 0$ in $\mathbb Z_p$ we have $p\cdot\phi(1) = \phi(p) = \phi(0) = 0$. What elements $x \in \mathbb Z_{2p}$ satisfy $p\cdot x = 0$? If $\phi(1) = x$ then show that $\phi$ is defined by $\phi(a) = a\cdot x$ (use induction).
Hint for (b): A ring homomorphism is, in particular, a group homomorphism of the additive structure, so look at the homomorphisms you got in part (a) and check which of them satisfy the additional axioms for ring homomorphisms.