Field/Group Homomorphisms $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ Consider the following problem:
Find all Field Homomorphisms $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ and all Group Homomorphisms $(\mathbb{Z}/3\mathbb{Z})^{\times} \to (\mathbb{Z}/5\mathbb{Z})^{\times}$
Since Field Homomorphisms are always injective and for all Field Homomorphisms $f$ holds $f([0]) = [0]$ and $f([1]) = [1]$ we only need to consider $[2]$ which either maps to $[2], [3]$ or $[4]$.
Now $f([2]) = f([1] + [1]) = f([1]) + f([1]) = [1] +[1] = [2]$ so the only Field Homomorphism is $f:\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}, [a] \mapsto [a]$.
As for the Group Homomorphisms I'm a bit stuck figuring out all possible ones. Any ideas?
 A: The only group homomorphism from $\phi:\mathbb{Z}/3\mathbb{Z}\to\mathbb{Z}/5\mathbb{Z}$ is the map that sends $x\mapsto0$ for all $x\in\mathbb{Z}/3\mathbb{Z}$.
We can prove this by using the fact that for any group homomorphism $\psi:G\to H$,
$$|G|=|\ker\psi|\cdot|\text{im }\psi|.$$
Let $\phi:\mathbb{Z}/3\mathbb{Z}\to\mathbb{Z}/3\mathbb{Z}$. We have that
$$|\ker\phi|\cdot|\text{im }\phi|=|\mathbb{Z}/3\mathbb{Z}|=3.$$
So $|\text{im }\phi|$ divides $3$. But $|\text{im }\phi|$ also divides $5$, since $\text{im }\phi$ is a subgroup of $\mathbb{Z}/5\mathbb{Z}$. Hence $\text{im }\phi$ is trivial and $\phi(x)=0$ for all $x\in\mathbb{Z}/3\mathbb{Z}$.
As for field homomorphisms, since every field homomorphism $\psi:F\to K$ satisfies $\psi(x+y)=\psi(x)+\psi(y)$ for all $x,y\in F$, the only function $\mathbb{Z}/3\mathbb{Z}\to\mathbb{Z}/5\mathbb{Z}$ that could possibly be a field homomorphism is the function $\phi:\mathbb{Z}/3\mathbb{Z}\to\mathbb{Z}/5\mathbb{Z}$ which sends each $x\mapsto0$. I think most mathematicians would not consider this to be field homomorphism, since most mathematicians' definition of a field homomorphism also requires that $1\mapsto1$.
Finally, there are two group homomorphisms $\mathbb{Z}/3\mathbb{Z}^{\times}\to\mathbb{Z}/5\mathbb{Z}^{\times}$. There is the map $\phi_1:\mathbb{Z}/3\mathbb{Z}^{\times}\to\mathbb{Z}/5\mathbb{Z}^{\times}$ which sends $x\mapsto1$ for each $x\in\mathbb{Z}/3\mathbb{Z}^{\times}$, and the map $\phi_2:\mathbb{Z}/3\mathbb{Z}^{\times}\to\mathbb{Z}/5\mathbb{Z}^{\times}$ which sends $1\mapsto1$, and $2\mapsto4$.
